Analysis III:

Measure and Integration Theory

of Several Variables

Peter Philip∗

Lecture Notes

Created for the Class of Winter Semester 2016/2017 at LMU Munich

February 14, 2020

Contents

1 Measure Theory 4

1.1 Motivation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4

1.2 Measure Spaces and Generators . . . . . . . . . . . . . . . . . . . . . . . . . . . 5

1.3 Extension . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.1 Semirings, Rings, Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.3.2 Contents, Premeasures . . . . . . . . . . . . . . . . . . . . . . . . . . . . 13

1.3.3 Lebesgue Premeasure on Rn . . . . . . . . . . . . . . . . . . . . . . . . . 20

1.3.4 Outer Measures, Caratheodory Extension Theorem . . . . . . . . . . . . 24

1.3.5 Dynkin Systems, Uniqueness of Extensions . . . . . . . . . . . . . . . . 28

1.3.6 Completion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33

1.4 Inverse Image, Trace . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35

1.5 Lebesgue Measure on Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 37

1.5.1 Intervals and Countable Sets . . . . . . . . . . . . . . . . . . . . . . . . 37

1.5.2 Approximation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38

1.5.3 Cantor Set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 39

1.6 Measurable Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.6.1 Definition, Composition . . . . . . . . . . . . . . . . . . . . . . . . . . . 42

1.6.2 Pushforward Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43

1.6.3 Invariance of Lebesgue Measure under Euclidean Isometries . . . . . . . 44

∗E-Mail: philip@math.lmu.de

1

CONTENTS 2

1.6.4 Nonmeasurable Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 48

1.6.5 R-Valued Measurable Maps . . . . . . . . . . . . . . . . . . . . . . . . . 50

1.6.6 Products, Projections, and Borel Sets . . . . . . . . . . . . . . . . . . . 59

2 Integration 62

2.1 Integration of R-Valued Measurable Maps . . . . . . . . . . . . . . . . . . . . . 62

2.1.1 Simple Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 62

2.1.2 Nonnegative Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64

2.1.3 Integrable Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 67

2.2 Null Sets, Properties Holding Almost Everywhere . . . . . . . . . . . . . . . . . 72

2.3 Convergence Theorems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

2.3.1 Fatou’s Lemma and Dominated Convergence . . . . . . . . . . . . . . . 74

2.3.2 Parameter-Dependent Integrals: Continuity, Differentiation . . . . . . . 76

2.4 Riemann versus Lebesgue . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

2.5 Products . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

2.5.1 Product Measure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 84

2.5.2 Theorems of Tonelli and Fubini . . . . . . . . . . . . . . . . . . . . . . . 91

2.6 Lp-Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 97

2.6.1 Definition, Norm, Completeness . . . . . . . . . . . . . . . . . . . . . . . 97

2.6.2 Integration with Respect to Pushforward Measures . . . . . . . . . . . . 104

2.6.3 Dense Subsets, Separability . . . . . . . . . . . . . . . . . . . . . . . . . 105

2.6.4 Convolution and Fourier Transform . . . . . . . . . . . . . . . . . . . . 110

2.7 Change of Variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 125

3 Integration Over Submanifolds of Rn 130

3.1 Submanifolds of Rn . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 130

3.2 Metric Tensor, Gram Determinant . . . . . . . . . . . . . . . . . . . . . . . . . 137

3.3 Integration Over Submanifolds . . . . . . . . . . . . . . . . . . . . . . . . . . . 138

3.4 Tangent Space and Normal Space . . . . . . . . . . . . . . . . . . . . . . . . . . 147

3.5 Gauss-Green Theorem and Green’s Identities . . . . . . . . . . . . . . . . . . . 150

A Order on, Arithmetic in, and Topology of R 159

B Algebraic Structure on Rings of Subsets 161

C Initial and Final σ-Algebras 163

CONTENTS 3

D Riemann Integral on Intervals in Rn 166

E Wallis Product 173

F Improper Riemann Integral of the Sinc Function 175

G Topological and Measure-Theoretic Supplements 177

G.1 Transitivity of Dense Subsets . . . . . . . . . . . . . . . . . . . . . . . . . . . . 177

G.2 Inverse Image and Trace for Semirings . . . . . . . . . . . . . . . . . . . . . . . 177

G.3 Measurability with Respect to Completions . . . . . . . . . . . . . . . . . . . . 178

G.4 Interchanging Integrals with Uniform Limits . . . . . . . . . . . . . . . . . . . . 178

H Polar Coordinates 179

References 180

1 MEASURE THEORY 4

1 Measure Theory

1.1 Motivation

In Analysis I, we have already measured the size of intervals I := [a, b] ⊆ R by defining|I| := |a−b|. Moreover, we defined the Riemann integral of Riemann-integrable functionsf : I −→ R on such intervals. We also mentioned that, for a Riemann-integrable f :I −→ R+

0 , the Riemann integral∫If can be interpreted as the size of the area between

the graph of f and the horizontal axis. Major goals of this class include measuring thesize of subsets of Rn and integrating functions over subsets of Rn.

While for n-dimensional intervals I =∏n

j=1 Ij ⊆ Rn with one-dimensional intervalsIj ⊆ R, an obvious choice to assign a volume is |I| := ∏n

j=1 |Ij|, for general subsets ofRn, it is far from obvious what a reasonable volume should be. For certain sets thathave a simple relation with intervals, this relation can be used to assign a volume (e.g.triangles can be decomposed into halfs of rectangles, which leads to the formula for itsarea that is taught in high school). Another common sense rule is that the volume v ofthe union of N disjoint sets with known volume vj should be the sum of the volumes,

v =∑N

j=1 vj. And it makes sense to extend this rule to countable disjoint unions: Forexample, if

∀j∈N

Ij :=] 12j,

1

2j−1

],

then the Ij are clearly disjoint,

I := [0, 1] = 0 ∪∞⋃

j=1

Ij,

and, due to the formula for the sum of the geometric series,

|I| = 1 =1

1− 12

− 1 = 0 +∞∑

j=1

2−j = |[0, 0]|+∞∑

j=1

|Ij|.

For example, such a rule then allows to compute the area of a circle by decomposing itinto countably many triangles (however, the details are a bit cumbersome and we willnot pursue this here).

Another common sense property that a volume function should satisfy is translationinvariance, i.e. v(A + x) = v(A) should hold for each A ⊆ Rn and each x ∈ Rn, whereA+ x := a+ x : a ∈ A.Definition 1.1. Let n ∈ N. We call a function v : P(Rn) −→ [0,∞] an ideal n-dimensional volume if, and only if, the following conditions (i) – (iii) are satisfied:

(a) If a, b ∈ Rn, a ≤ b, I := [a, b], then

v(I) =n∏

j=1

|bj − aj|.

1 MEASURE THEORY 5

(b) v is translation invariant, i.e., for each A ∈ P(Rn) and each x ∈ Rn,

v(A+ x) = v(A), A+ x := a+ x : a ∈ A.

(c) v satisfies countable additivity, i.e. if A,Aj ∈ P(Rn), j ∈ N, and if A is the disjointunion of the Aj, then

v(A) =∞∑

j=1

v(Aj).

—

Unfortunately, using the axiom of choice, we will be able to show that no ideal volumeexists (see Sec. 1.6.4 and, in particular, Th. 1.74(b)). However, we will see that thereexist volume functions (notably so-called Lebesgue-Borel and Lebesgue measure) thatsatisfy Def. 1.1(a)–(c), except that they are not defined on all of P(Rn) (Lebesgue-Borelmeasure is defined on a strictly smaller set than Lebesgue measure and is the restrictionof Lebesgue measure to this set, cf. Ex. 1.47).

1.2 Measure Spaces and Generators

Now that we have discussed some desirable properties of volume functions, which wewill usually call measures µ from now on, we will do the mathematicians’ job andcondense these properties into an abtract definition, then proceeding to studying somegeneral consequences. As we have already indicated that it would be too restrictive torequire measures to be defined on the entire power set P(X) of a set X, we start byconsidering subsets A ⊆ P(X) that are suitable as the domain of a measure. Fromour discussion in the previous section, it makes sense to require A to be closed undercountable unions. To be useful, A should at least contain ∅ and X. A finite measure µshould also have the property µ(X \ A) = µ(X)− µ(A). We will see that this actuallyfollows from additivity as long as A is also closed under taking complements. Withtranslation invariance in mind, one might also want A to be closed under translations(i.e. one might want A+ x ∈ A for each A ∈ A and x ∈ X), but, as X might not comewith an addition defined on it, we will not include such a requirement for the domainof a measure. Altogether, this leads to the following definition:

Definition 1.2. Let X be a set. A collection of subsets A ⊆ P(X) is called a σ-algebraon X if, and only if, it satisfies the following three conditions:

(a) ∅ ∈ A.

(b) If A ∈ A, then Ac = X \ A ∈ A.

(c) If (An)n∈N is a sequence of sets in A, then⋃

n∈NAn ∈ A.

1 MEASURE THEORY 6

If A is a σ-algebra on X, then the pair (X,A) is called a measurable space. The setsA ∈ A are called A-measurable (or merely measurable if A is understood).

—

Measures will be defined on σ-algebras (see Def. 1.10 below). Note that there aresimilarities between the definition of a σ-algebra and the definition of a topology: Forexample, both topologies and σ-algebras are always subsets of the power set of somegiven set of interest; they also both contain at least the empty set and the whole space.Of course, in general, a topology is not a σ-algebra and a σ-algebra is not a topology.Still, we will notice repeatedly, that similar concepts and techniques are useful for bothmeasure spaces and topological spaces (there is a branch of mathematics called CategoryTheory that systematically investigates such aspects – one finds that the category oftopological spaces has a lot in common with the category of measurable spaces).

Lemma 1.3. Let X be a set and let A ⊆ P(X) be a σ-algebra on X.

(a) A is closed under finite unions: Let n ∈ N. Then

A1, . . . , An ∈ A ⇒n⋃

j=1

Aj ∈ A.

(b) A is closed under countable (both finite and infinite) intersections: Let I be anonempty countable index set and Aj ∈ A, j ∈ I. Then

⋂

j∈IAj ∈ A.

Proof. (a) follows from Def. 1.2(a),(c), as

n⋃

j=1

Aj = A1 ∪ · · · ∪ An ∪ ∅ ∪ ∅ . . .

(b) follows from Def. 1.2(b),(c) and (a), since

(⋂

j∈IAj

)c

=⋃

j∈IAc

j,

completing the proof of the lemma.

Example 1.4. Let X be set.

(a) Clearly, ∅, X and P(X) are σ-algebras on X, where ∅, X sometimes called thetrivial σ-algebra on X.

1 MEASURE THEORY 7

(b) A := A ⊆ X : A countable or Ac countable constitutes a σ-algebra on X(clearly, ∅ ∈ A; if A ∈ A, then Ac ∈ A; if An ∈ A, n ∈ N, then A :=

⋃n∈NAn ∈ A:

If at least one Acn0, n0 ∈ N, is countable, then Ac =

⋂n∈NA

cn ⊆ Ac

n0is count-

able; if each Acn is uncountable, then each An must be countable, implying A to be

countable).

—

The next result will allow us to generate an abundance of σ-algebras:

Proposition 1.5. Arbitrary intersections of σ-algebras yield again σ-algebras: Let Xbe a set, let I be a nonempty index set, and let (Ai)i∈I be a family of σ-algebras on X.Then

A :=⋂

i∈IAi

is a again a σ-algebra on X.

Proof. Since ∅ is in each Ai, it is also in A. If A ∈ A, then A ∈ Ai for each i ∈ I, i.e.Ac ∈ Ai for each i ∈ I, implying Ac ∈ A. Similarly, if An ∈ A for each n ∈ N, thenAn ∈ Ai for each i ∈ I and each n ∈ N. Thus, A :=

⋃n∈NAn ∈ Ai for each i ∈ I,

showing A ∈ A.

Caveat 1.6. The union of (even two) σ-algebras is, in general, not a σ-algebra: Forexample, let X := 0, 1, 2, and consider

A1 :=∅, 0, 1, 2, X

,

A2 :=∅, 1, 0, 2, X

,

B := A1 ∪ A2.

Then, clearly, A1 and A2 are σ-algebras on X, but B is not (for example 0, 1 ∈ B,but 0, 1 /∈ B).

Definition and Remark 1.7. Let X be a set. If E is a collection of subsets of X, i.e.E ⊆ P(X), then let σ(E) := σX(E) denote the intersection of all σ-algebras on X thatare supersets of E (i.e. that contain all the sets in E). According to Prop. 1.5, σ(E) is aσ-algebra on X. Obviously, it is the smallest σ-algebra on X containing E . It is, thus,called the σ-algebra generated by E ; E is called a generator of σ(E).

Example 1.8. Let X be a set.

(a) Both ∅ and X are generators of the trivial σ-algebra on X. The set E := x :x ∈ X generates the σ-algebra defined in Ex. 1.4(b) above.

(b) Let T be a topology on X. Then B := σ(T ), i.e. the σ-algebra generated by theopen sets of X, is called the Borel σ-algebra on (X, T ) (or on X if the topology Tis understood). The elements of B are called Borel sets.

1 MEASURE THEORY 8

(c) Let n ∈ N. It is often useful to know that the Borel σ-algebra Bn on Rn (withrespect to the norm topology on Rn) is also generated by each of the following sets(in each case, it is a simple consequence of the following Lem. 1.9):

Cn := A ⊆ Rn : A closed,Kn := A ⊆ Rn : A compact,Inc := [a, b] : a, b ∈ Rn, a ≤ b ∪ ∅,

Inc,Q := [a, b] : a, b ∈ Qn, a ≤ b ∪ ∅,Ino := ]a, b[: a, b ∈ Rn, a ≤ b,

Ino,Q := ]a, b[: a, b ∈ Qn, a ≤ b,In := ]a, b] : a, b ∈ Rn, a ≤ b,InQ := ]a, b] : a, b ∈ Qn, a ≤ b,

InΣ := ⋃N

k=1 Ik : N ∈ N, I1, . . . , IN ∈ In disjoint,InΣ,Q := ⋃N

k=1 Ik : N ∈ N, I1, . . . , IN ∈ InQ disjoint.

Lemma 1.9. Let (X, T ) be a topological space and let B := σ(T ) be the correspondingset of Borel sets. Moreover, let C be the set of closed subsets of X, and let K be the setof compact subsets of X. Then the following holds true:

(a) B = σ(C).

(b) If (X, T ) is Hausdorff (i.e. a T2 space) and there exists a sequence (Kn)n∈N ofcompact sets such that X =

⋃∞n=1Kn, then B = σ(K).

(c) If E ⊆ B is such that every open set O ∈ T is a countable union of sets from E ,then B = σ(E).

Proof. (a): Since C = A ⊆ X : Ac ∈ T , one has C ⊆ B and, thus, σ(C) ⊆ B (since Bis a σ-algebra). Analogously, one also has T ⊆ σ(C) and B ⊆ σ(C), proving (a).

(b): According to [Phi16b, Prop. 3.9(b)], every compact subset ofX is closed, i.e. K ⊆ C,implying σ(K) ⊆ σ(C) = B. On the other hand, if A ∈ C, then

A = A ∩X = A ∩∞⋃

n=1

Kn =∞⋃

n=1

(A ∩Kn),

i.e. A is a countable union of sets from K, showing C ⊆ σ(K) and B ⊆ σ(K), proving(b).

(c): The hypothesis of (c) implies T ⊆ σ(E) and B ⊆ σ(E). As E ⊆ B is assumed, thisalready proves (c).

Keeping in mind our considerations of the previous section, we are now ready to definewhat we mean by a measure:

Definition 1.10. Let (X,A) be a measurable space.

1 MEASURE THEORY 9

(a) A map µ : A −→ [0,∞] is called a measure on (X,A) (or on X if A is understood)if, and only if, µ satisfies the following conditions (i) and (ii):

(i) µ(∅) = 0.

(ii) µ is countably additive (also called σ-additive), i.e., if (Ai)i∈N is a sequence inA consisting of (pairwise) disjoint sets, then

µ

( ∞⋃

i=1

Ai

)=

∞∑

i=1

µ(Ai). (1.1)

If µ is a measure on (X,A), then the triple (X,A, µ) is called a measure space. Inthe context of measure spaces, the sets A ∈ A are sometimes called µ-measurableinstead of A-measurable.

(b) Let (X,A, µ) be a measure space. The measure µ is called finite or bounded if, andonly if, µ(X) < ∞ (for µ(X) = 1, µ is called a probability measure, (X,A, µ) iscalled a probability space); it is called σ-finite if, and only if, there exists a sequence(Ai)i∈N in A such that X =

⋃∞i=1Ai and µ(Ai) <∞ for each i ∈ N.

Remark 1.11. In (1.1), we need to extend addition to [0,∞]. A countable sum ofsummands from [0,∞] is defined to be ∞ if at least one of the summands equals ∞; ifall summands are finite, then the partial sums either converge to some s ∈ R+

0 or theydiverge to s = ∞. In both cases, we define s to be the value of the sum. Throughoutthe class, we will sometimes need arithmetic in the so-called extended real numbersR := R ∪ −∞,∞. It is mostly straightforward to extend addition, subtraction, andmultiplication from R to R – the details are provided in Appendix A for future reference.

Example 1.12. Let (X,A) be a measurable space.

(a) The zero measure µ0 : A −→ [0,∞], µ0 ≡ 0, is clearly a measure, and so is

µ∞ : A −→ [0,∞], µ∞(A) :=

0 if A is countable,

∞ if A is uncountable.

Then µ∞ is not σ-finite if X is uncountable (if X is countable, then µ∞ = µ0). ForX 6= ∅, A = ∅, X, a variant is the measure µ defined by µ(∅) = 0, µ(X) = ∞,which is never σ-finite.

(b) Another obvious measure is the so-called counting measure, defined by

µc : A −→ [0,∞], µc(A) :=

#A if A is finite,

∞ if A is infinite.

If A = P(X), then counting measure is σ-finite if, and only if, X is countable.

1 MEASURE THEORY 10

(c) If X 6= ∅ and a ∈ X, then

δa : A −→ [0,∞], δa(A) :=

1 if a ∈ A,

0 if a /∈ A,

is called the Dirac measure concentrated in a (clearly, it is a measure).

(d) Let X be nonempty and finite, #X = n ∈ N. Let m : X −→ R+0 be some function.

Thenµm : P(X) −→ R+

0 , µm(A) :=∑

x∈Am(x),

defines a measure on X. Moreover, if M := µm(X) 6= 0, then P := M−1µm definesa probability measure on X.

1.3 Extension

The examples at the end of the previous section show that there are many measures thatare simple to define and simple to recognize as measures. However, to accomplish ouroriginal goal of defining a measure on the Borel subsets of Rn that yields the expectedgeometric size of intervals, is not so simple. We will be able to do this using the techniqueof extension, which is also useful beyond the application we have in mind. The idea isto first define a suitable [0,∞]-valued function on some set E that is not a σ-algebra(e.g. the set of intervals) and then to extend this function to a measure on the σ-algebragenerated by E . We will extend our functions from so-called semirings to σ-algebras.Useful structures in between are rings and algebras.

1.3.1 Semirings, Rings, Algebras

Definition 1.13. Let X be a set, E ⊆ P(X).

(a) E is called a semiring on X if, and only if, it satisfies the following three conditions:

(i) ∅ ∈ E .(ii) If A,B ∈ E , then A ∩ B ∈ E .(iii) If A,B ∈ E , then there exist disjoint E1, . . . , En ∈ E , n ∈ N, such that

A \B =n⋃

i=1

Ei. (1.2)

(b) E is called a ring on X if, and only if, it satisfies the following three conditions:

(i) ∅ ∈ E .(ii) If A,B ∈ E , then A ∪ B ∈ E .

1 MEASURE THEORY 11

(iii) If A,B ∈ E , then A \B ∈ E .

(c) E is called an algebra on X if, and only if, E is a ring and X ∈ E .

Remark 1.14. Let X be a set, E ⊆ P(X). Then we have the following implications:

E σ-algebra ⇒ E algebra ⇒ E ring ⇒ E semiring,

where everything is immediate from the respective definitions, except that Def. 1.13(b)implies Def. 1.13(a)(ii). However, if E is a ring and A,B ∈ E , then

A ∩ B = A \ (A \B) ∈ E . (1.3)

We also note that E is a ring in the sense of Def. 1.13(b) if, and only if, E is a ring in thesense of algebra, provided one uses the symmetric difference A∆B := (A \B)∪ (B \A)as addition and ∩ as multiplication (see Appendix B).

Remark and Definition 1.15. Analogous to Prop. 1.5, arbitrary intersections of ringsyield again rings, arbitrary intersections of algebras yield again algebras: Let X be a set,let I be a nonempty index set, and let (Ri)i∈I be a family of rings on X, R :=

⋂i∈I Ri.

Since ∅ is in each Ri, it is also in R. If A,B ∈ R, then A,B ∈ Ri for each i ∈ I, i.e.A\B ∈ Ri and A∪B ∈ Ri for each i ∈ I, implying A\B ∈ R and A∪B ∈ A, showingR to be a ring on X. Now let (Ai)i∈I be a family of algebras on X, A :=

⋂i∈I Ai. Since

X is in each Ai, it is also in A. Since each Ai is a ring on X, so is A, showing A is analgebra on X. Thus, if X is a set and E ⊆ P(X), then we can let ρ(E) := ρX(E) denotethe intersection of all rings on X that are supersets of E , and we can let α(E) := αX(E)denote the intersection of all algebras on X that are supersets of E . We then know ρ(E)to be a ring and α(E) to be an algebra, namely the smallest ring (resp. algebra) on Xcontaining E . Thus, we call ρ(E) the ring (and α(E) the algebra) generated by E . Wepoint out that there exists a (rough) conceptual analogy between the way a σ-algebracan be built from a semiring and the way a topology can be built from a subbase.

Example 1.16. (a) Let X be a set. Then S := ∅ ∪ x : x ∈ X is a semiring onX. Moreover, ρ(S) = A ⊆ X : A finite, α(S) = A ⊆ X : A finite or Ac finite,and σ(S) is the σ-algebra of Ex. 1.4(b).

(b) The sets I1 = ]a, b] : a, b ∈ R, a ≤ b and I1Q = ]a, b] : a, b ∈ Q, a ≤ b of Ex.

1.8(c) are semirings on R, since, for a, b, c, d ∈ R with a ≤ b and c ≤ d, lettingα := maxa, c, β := minb, d, one has

]a, b]∩ ]c, d] =

]α, β] if α < β,

∅ otherwise,

as well as]a, b] \ ]c, d] = ]a, c]︸︷︷︸

= ∅ for c ≤ a

∪ ]d, b]︸︷︷︸= ∅ for b ≤ d

.

—

1 MEASURE THEORY 12

If n > 1, then In and InQ are still semirings on Rn:

Proposition 1.17. (a) If X, Y are sets, S is a semiring on X and T is a semiring onY , then

U := S ∗ T := A× B : A ∈ S, B ∈ T is a semiring on X × Y .

(b) For each n ∈ N, the sets In = ]a, b] : a, b ∈ Rn, a ≤ b and InQ = ]a, b] : a, b ∈

Qn, a ≤ b of Ex. 1.8(c) are semirings on Rn.

Proof. (a): It is ∅ = ∅ × ∅ ∈ U . Now let A1, A2 ∈ S and B1, B2 ∈ T . Then

(A1 × B1) ∩ (A2 × B2) = (A1 ∩ A2)× (B1 ∩B2) ∈ U .

Moreover,

(A1 ×B1) \ (A2 × B2) =((A1 \ A2)×B1

)∪(A1 × (B1 \B2)

)

=((A1 \ A2)×B1

)∪((A1 ∩ A2)× (B1 \B2)

). (1.4)

As S and T are semirings, there are M,N ∈ N, disjoint C1, . . . , CM ∈ S, and disjointD1, . . . , DN ∈ T such that

A1 \ A2 =M⋃

i=1

Ci, B1 \B2 =N⋃

i=1

Di.

Using this in (1.4), we can write (A1 ×B1) \ (A2 ×B2) as a disjoint union of sets in U ,namely

(A1 × B1) \ (A2 ×B2) =

(⋃M

i=1(Ci ×B1)

)∪(⋃N

i=1

((A1 ∩ A2)×Di

)),

which completes the proof that U is a semiring.

(b) follows by induction on n: The base case (n = 1) is given by Ex. 1.16(b), and theinduction step follows from (a), since, for n > 1,

](a1, . . . , an), (b1, . . . , bn)] =](a1, . . . , an−1), (b1, . . . , bn−1)]× ]an, bn]

for each a, b ∈ Rn with a ≤ b.

Proposition 1.18. (a) Let S be a semiring on X and A,B1, . . . , BN ∈ S, N ∈ N.Then there exists M ∈ N and disjoint C1, . . . , CM ∈ S such that

A \N⋃

i=1

Bi =M⋃

i=1

Ci. (1.5)

(b) If S is a semiring on X, then

ρ(S) = R := ⋃Nk=1Ak : N ∈ N, A1, . . . , AN ∈ S disjoint. (1.6)

1 MEASURE THEORY 13

(c) For each n ∈ N, one has ρ(In) = InΣ and ρ(In

Q) = InΣ,Q, where In, In

Σ, InQ, and In

Σ,Q

are as defined in Ex. 1.8(c).

Proof. (a) is proved via induction on N . The base case (N = 1) holds, since S isa semiring. For the induction step, let N ∈ N, A,B1, . . . , BN , BN+1 ∈ S. By theinduction hypothesis, there exist disjoint C1, . . . , CM ∈ S such that (1.5) holds true.Then

A \N+1⋃

i=1

Bi =

(A \

N⋃

i=1

Bi

)\BN+1 =

(⋃M

i=1Ci

)\BN+1 =

⋃M

i=1(Ci \BN+1).

Since each of the Ci \ BN+1 is a disjoint finite union of sets from S, the induction iscomplete.

(b): Since rings are closed under finite unions, R ⊆ ρ(S) is clear. For the remaininginclusion, it suffices to show that R is a ring. Since ∅ ∈ S, ∅ ∈ R. Let M,N ∈ N,A1, . . . , AM ∈ S disjoint, B1, . . . , BN ∈ S disjoint, A :=

⋃Mi=1Ai, B :=

⋃Ni=1Bi. We

have to show A \B ∈ R and A ∪ B ∈ R. It is

A \B =⋃M

i=1

(Ai \

⋃M

j=1Bj

).

Since, by (a), each of the Ai \⋃M

j=1Bj is a disjoint finite union of sets from S, we haveA \B ∈ R. We also have

A ∩ B =⋃M

i=1

⋃N

j=1Ai ∩Bj ∈ R,

since each Ai∩Bj ∈ S by Def. 1.13(a)(ii). Thus, A∪B = (A\B) ∪(B\A) ∪(A∩B) ∈ R,completing the proof that R is a ring and the proof of (b).

(c) follows by combining Prop. 1.17(b) with (b).

1.3.2 Contents, Premeasures

Definition 1.19. Let X be a set and let S be a semiring on X.

(a) A map µ : S −→ [0,∞] is called a content if, and only if, µ satisfies the followingconditions (i) and (ii):

(i) µ(∅) = 0.

(ii) µ is finitely additive, i.e., if n ∈ N and A1, . . . , An ∈ S are disjoint sets suchthat

⋃ni=1Ai ∈ S, then

µ

(n⋃

i=1

Ai

)=

n∑

i=1

µ(Ai). (1.7)

1 MEASURE THEORY 14

The content µ is called finite or bounded if, and only if, µ(A) <∞ for each A ∈ S;it is called σ-finite if, and only if, there exists a sequence (Ai)i∈N in S such thatX =

⋃∞i=1Ai and µ(Ai) <∞ for each i ∈ N.

(b) A content µ : S −→ [0,∞] is called a premeasure if, and only if, it is countablyadditive (also called σ-additive), i.e., if (Ai)i∈N is a sequence in S consisting ofdisjoint sets such that

⋃∞i=1Ai ∈ S, then

µ

( ∞⋃

i=1

Ai

)=

∞∑

i=1

µ(Ai).

Example 1.20. (a) A measure is a premeasure that is defined on a σ-algebra.

(b) Among the most important premeasures is so-called Lebesgue premeasure λn, which,for n ∈ N, is defined on the semiring In on Rn from Ex. 1.8(c) by letting

λn : In −→ R+0 , λn(I) :=

∏nj=1(bj − aj) for I =]a, b], a, b ∈ Rn, a < b,

0 for I = ∅.

We will study it in detail in Sec. 1.3.3 below, where we will show in Th. 1.31 thatit is, indeed, a premeasure.

(c) The following example shows that not every content is a premeasure: ConsiderS := ]a, b] : a, b ∈ R, 0 ≤ a ≤ b ≤ 1. Then S is clearly a semiring on ]0, 1] (cf.Ex. 1.16(b)). Define

µ : S −→ [0,∞], µ(I) :=

0 for I = ∅,b− a for I =]a, b], 0 < a < b ≤ 1,

∞ for I =]0, b], 0 < b ≤ 1.

Then µ is a content, since

∀n∈N

∀0<s0<s1<···<sn≤1

n∑

i=1

µ(]si−1, si]

)=

n∑

i=1

(si − si−1) = sn − s0 = µ(]s0, sn]

)

and

∀n∈N

∀0<s0<s1<···<sn≤1

µ(]0, s0]

)+

n∑

i=1

µ(]si−1, si]

)= ∞+

n∑

i=1

(si − si−1)

= ∞ = µ(]0, sn]

).

However, µ is not a premeasure, since, letting, for each n ∈ N, In :=] 1n+1

, 1n], one

has ]0, 1] =⋃∞

n=1In, but

∞∑

n=1

µ(In) =∞∑

n=1

(1

n− 1

n+ 1

)= lim

n→∞

(1− 1

n+ 1

)= 1 6= µ

(]0, 1]

)= ∞.

—

1 MEASURE THEORY 15

Before compiling important properties of contents and premeasures in Th. 1.22 and Th.1.24 below, we show in the following Prop. 1.21 that each content on a semiring extendsin a unique way to a content on the generated ring.

Proposition 1.21. Let S be a semiring on the set X and let µ : S −→ [0,∞] be acontent. Furthermore, let R := ρ(S). Then there is a unique extension ν : R −→ [0,∞]of µ such that ν is a content. Moreover ν is a premeasure if, and only if, µ is apremeasure.

Proof. Uniqueness: Assume ν is a content on R such that νS= µ. If A ∈ R, then, byProp. 1.18(b), there exist disjoint A1, . . . , AN ∈ S, N ∈ N, such that A =

⋃Ni=1Ai. Thus,

as ν is finitely additive, ν(A) =∑N

i=1 µ(Ai), showing that ν is uniquely determined byµ.

Existence: As we have seen that ν(A) =∑N

i=1 µ(Ai) must hold if A and A1, . . . , AN areas above, we use this formula to define the function ν on R. We now need to verify thatν is well-defined, i.e. that, if we pick, possibly different, disjoint B1, . . . , BM ∈ S withA =

⋃Mi=1Bi, then

∑Ni=1 µ(Ai) =

∑Mi=1 µ(Bi). Indeed, we have, since µ is a content on

S,N∑

i=1

µ(Ai) =N∑

i=1

µ

(Ai ∩

M⋃

j=1

Bj

)=

N∑

i=1

µ

(M⋃

j=1

(Ai ∩ Bj)

)=

N∑

i=1

M∑

j=1

µ(Ai ∩ Bj)

=M∑

j=1

µ

(N⋃

i=1

(Ai ∩ Bj)

)=

M∑

j=1

µ

(Bj ∩

N⋃

i=1

Ai

)=

M∑

j=1

µ(Bj),

proving that ν is well-defined. Then νS= µ is also clear. To prove that ν is a content,we still need to check that it is finitely additive. To this end, let A,B ∈ R be disjoint.Then there are disjoint A1, . . . , AN , B1, . . . , BM ∈ S; M,N ∈ N, such that A =

⋃Ni=1Ai

and B =⋃M

i=1Bi. In consequence,

ν(A ∪ B) =N∑

i=1

µ(Ai) +M∑

i=1

µ(Bi) = ν(A) + ν(B),

as needed.

Since νS= µ, if ν is a premeasure, so is µ. For the converse, assume µ to be a premeasureand let (Ai)i∈N be a sequence of disjoint sets in R such that A :=

⋃∞i=1Ai ∈ R. Then

there are disjoint sets B1, . . . , BN ∈ S, N ∈ N, with A =⋃N

i=1Bi, and, for each i ∈ N,there are disjoint sets Ci1, . . . , CiNi

∈ S, Ni ∈ N, with Ai =⋃Ni

j=1Cij. Then

∀i∈1,...,N

Bi = Bi ∩ A =⋃∞

j=1(Bi ∩ Aj) =

⋃∞

j=1

⋃Nj

k=1(Bi ∩ Cjk︸ ︷︷ ︸

∈S

).

Thus, as µ is a premeasure,

∀i∈1,...,N

µ(Bi) =∞∑

j=1

Nj∑

k=1

µ(Bi ∩ Cjk) =∞∑

j=1

ν(Bi ∩ Aj),

1 MEASURE THEORY 16

implying

ν(A) =N∑

i=1

µ(Bi) =N∑

i=1

∞∑

j=1

ν(Bi ∩ Aj) =∞∑

j=1

ν(A ∩ Aj) =∞∑

j=1

ν(Aj),

proving ν to be a premeasure.

During the construction of a content µ, it can be helpful to first define and study it ona semiring. However, Prop. 1.21 tells us that we may assume µ to be defined on a ringwhen considering a content’s general properties, as in the following Th. 1.22 (one stillhas to use some care, though, cf. Ex. 1.25(c) below).

Theorem 1.22. Let R be a ring on the set X and let µ : R −→ [0,∞] be a content.Then the following rules hold, where we assume A,B ∈ R as well as Ai ∈ R for eachi ∈ N.

(a) Monotonicity: If B ⊆ A, then µ(B) ≤ µ(A).

(b) Subtractivity: If B ⊆ A and µ(B) <∞, then µ(A \B) = µ(A)− µ(B).

(c) µ(A) + µ(B) = µ(A ∪B) + µ(A ∩ B).

(d) If the Ai are disjoint and⋃∞

i=1Ai ⊆ A, then∑∞

i=1 µ(Ai) ≤ µ(A).

(e) Subadditivity: For each n ∈ N, one has µ(⋃n

i=1Ai) ≤∑n

i=1 µ(Ai).

(f) If µ is a premeasure, then one also has σ-subadditivity: If A ⊆ ⋃∞i=1Ai, then

µ(A) ≤∑∞i=1 µ(Ai).

Proof. (a): If B ⊆ A, then A = B ∪(A \B), implying

µ(A) = µ(B) + µ(A \B). (1.8)

Since µ(A \B) ∈ [0,∞], this shows µ(B) ≤ µ(A).

(b) is immediate from (1.8), as µ(B) is assumed finite.

(c): Finite additivity yields

µ(B) = µ(B \ A) + µ(A ∩B),

µ(A ∪ B) = µ(A) + µ(B \ A).

If µ(A∪B) <∞, then we add µ(A) = µ(A∪B)−µ(B\A) to the first equation to obtain(c). If µ(A∪B) = µ(A) = ∞, then (c) also holds. In the remaining case, µ(A∪B) = ∞and µ(A) <∞, i.e. µ(B \ A) = ∞, implying µ(B) = ∞, i.e. (c) holds once again.

(d) holds, since, by finite additivity and monotonicity,∑n

i=1 µ(Ai) ≤ µ(A) holds foreach n ∈ N.

1 MEASURE THEORY 17

(e): Since⋃n

i=1Ai =⋃n

i=1(Ai \⋃n−1

j=1 Aj), we have

µ

(n⋃

i=1

Ai

)=

n∑

i=1

µ

(Ai \

n−1⋃

j=1

Aj

)≤

n∑

i=1

µ(Ai).

(f) follows with the same kind of trick as in (e): Since A =⋃n

i=1(A ∩ (Ai \⋃n−1

j=1 Aj)),we have

µ(A) =∞∑

i=1

µ

(A ∩

(Ai \

n−1⋃

j=1

Aj

))≤

∞∑

i=1

µ(Ai),

completing the proof of (f) and the theorem.

Notation 1.23. Let A and An, n ∈ N, be sets. We introduce the following notation:

An ↑ A :⇔(A =

∞⋃

n=1

An ∧ A1 ⊆ A2 ⊆ . . .

),

An ↓ A :⇔(A =

∞⋂

n=1

An ∧ A1 ⊇ A2 ⊇ . . .

).

Theorem 1.24. Let R be a ring on the set X, let µ : R −→ [0,∞] be a content, andconsider the following statements:

(i) µ is a premeasure.

(ii) µ is continuous from below, i.e. if A ∈ R and (An)n∈N is a sequence in R, then

An ↑ A ⇒ limn→∞

µ(An) = µ(A). (1.9)

(iii) µ is continuous from above, i.e. if A ∈ R and (An)n∈N is a sequence in R, then

(µ(A1) <∞ ∧ An ↓ A

)⇒ lim

n→∞µ(An) = µ(A). (1.10)

(iv) If (An)n∈N is a sequence in R, then

(µ(A1) <∞ ∧ An ↓ ∅

)⇒ lim

n→∞µ(An) = 0. (1.11)

Then the following implications hold:

(i) ⇔ (ii) ⇒ (iii) ⇔ (iv).

Under the additional assumption that µ is finite, (i) – (iv) are equivalent.

1 MEASURE THEORY 18

Proof. (i) ⇒ (ii): Let A ∈ R and let (An)n∈N be a sequence in R such that An ↑ A.Then A = A1 ∪

⋃∞i=2(Ai \ Ai−1), implying

µ(A) = µ(A1) +∞∑

i=2

µ(Ai \ Ai−1) = limn→∞

(µ(A1) +

n∑

i=2

µ(Ai \ Ai−1)

)= lim

n→∞µ(An),

as desired.

(ii) ⇒ (i): If A ∈ R and (Bi)i∈N is a sequence of disjoint sets in R, satisfying A =⋃∞i=1Bi, then, letting, for each n ∈ N, An :=

⋃ni=1Bi ∈ R, one has An ↑ A. Then (ii)

yields

µ(A) = limn→∞

µ(An) = limn→∞

n∑

i=1

µ(Bi) =∞∑

i=1

µ(Bi),

showing µ to be a premeasure.

(ii)⇒ (iii): Let A ∈ R and let (An)n∈N be a sequence inR such that An ↓ A, µ(A1) <∞.Then µ(A) <∞ and µ(An) <∞ for each n ∈ N. Moreover,

An ↓ A ⇒(A =

∞⋂

n=1

An ∧ A1 ⊇ A2 ⊇ . . .

)

⇒(A1 \ A =

∞⋃

n=1

(A1 \ An) ∧ A1 \ A1 ⊆ A1 \ A2 ⊆ . . .

)

⇒ A1 \ An ↑ A1 \ A.

Thus, we can apply Th. 1.22(b) and (ii) to obtain

limn→∞

(µ(A1)− µ(An)

)= lim

n→∞µ(A1 \ An) = µ(A1 \ A) = µ(A1)− µ(A),

proving (iii).

(iii) ⇒ (iv) is immediate by setting A := ∅.(iv) ⇒ (iii): Let A ∈ R and let (An)n∈N be a sequence in R such that An ↓ A, µ(A1) <∞. For each n ∈ N, set Bn := An \ A ∈ R. Then Bn ↓ ∅ and µ(B1) ≤ µ(A1) < ∞, i.e.we can apply (iv) to obtain

limn→∞

(µ(An)− µ(A)

)= lim

n→∞µ(Bn) = 0,

proving (iii).

Finally, assume µ to be finite and (iv). Let A ∈ R and let (An)n∈N be a sequence in Rsuch that An ↑ A. Then A \ An ↓ ∅ and, since µ(A \ A1) <∞, (iv) yields

limn→∞

(µ(A)− µ(An)

)= lim

n→∞µ(A \ An) = 0,

proving (ii) and the theorem.

1 MEASURE THEORY 19

The following Ex. 1.25(a)–(c) provides some counterexamples related to Th. 1.24 above.

Example 1.25. (a) The following example shows that one cannot omit the conditionµ(A1) < ∞ in (iii) and (iv) of Th. 1.24: Consider the counting measure of Ex.1.12(b) with X := N. If, for each n ∈ N, An := k ∈ N : k ≥ n. Then An ↓ ∅, butµ(An) = ∞ for each n ∈ N.

(b) The following example shows statements (iii) and (iv) of Th. 1.24 are, in general,not equivalent to statements (i) and (ii): Let X be a countable infinite set (e.g.,X = N). Then A := A ⊆ X : A finite or Ac finite is clearly an algebra on X and

µ : A −→ [0,∞], µ(A) :=

0 for A finite,

∞ for A infinite,

clearly defines a content on A. Moreover, µ satisfies (iii) and (iv) of Th. 1.24, since,if A ∈ A and (An)n∈N is a sequence in A with An ↓ A, then µ(A1) <∞ implies thatall An and A are finite with µ(An) = µ(A) = 0. However, µ is not a premeasure,since

∑x∈X µ(x) = 0 6= ∞ = µ(X) (where we used that X is the countable

disjoint union of the x).

(c) The following example shows Th. 1.24 does, in general, not hold if the ring R isreplaced with a semiring S: Let X := Q and

S := I1|Q := ]a, b] ∩Q : a, b ∈ R, a ≤ b.

Then, clearly, S is a semiring on Q. We show

µ : S −→ R+0 , µ

(]a, b] ∩Q

):= b− a (a ≤ b)

defines a finite content, satisfying Th. 1.24(ii) on S (and, thus, also Th. 1.24(iii),(iv)on S, since the corresponding proofs of Th. 1.24 remain valid), but (the extension of)µ does not satisfy Th. 1.24(ii) on R := ρ(S) (in particular, µ is not a premeasure):Since

∀n∈N

∀s0<s1<···<sn

n∑

i=1

µ(]si−1, si] ∩Q

)=

n∑

i=1

(si − si−1) = sn − s0 = µ(]s0, sn] ∩Q

),

µ is a content. We now show µ satisfies Th. 1.24(ii) on S: Let A =]a, b] ∩ Q ∈ S,a, b ∈ R, a < b, and (An)n∈N a sequence in S with An ↑ A. Then, for each n ∈ N,An =]an, bn] ∩ Q, an, bn ∈ R, an < bn, with (an)n∈N decreasing, (bn)n∈N increasing,limn→∞ an = a, limn→∞ bn = b. Then

limn→∞

µ(An) = limn→∞

(bn − an) = b− a = µ(A),

showing Th. 1.24(ii) holds on S. Next, we show that (the extension of) µ doesnot satisfy Th. 1.24(ii) on R := ρ(S): Let A :=]0, 1] ∩ Q and let q1, q2, . . . bean enumeration of the (countable) set A, 0 < ǫ < 1. We recursively construct asequence (An)n∈N, An =]an, bn] ∩ Q, of disjoint sets in S, and a sequence (Bn)n∈N

1 MEASURE THEORY 20

in R, Bn :=⋃n

i=1Ai, by letting a1 := max0, q1 − ǫ 2−1, b1 := q1 (note µ(B1) =µ(A1) ≤ ǫ 2−1) and, for n > 1, let Nn := mink ∈ N : qk /∈ Bn−1, bn := qNn

,an := max

(t ∈ Bn−1 : t < bn ∪ bn − ǫ 2−n

), noting µ(An) ≤ ǫ 2−n and

µ(Bn) =n∑

i=1

µ(Ai) < ǫ

∞∑

i=1

2−i = ǫ < 1. (1.12)

We point out that Nn above is well-defined, since, if Bn−1 =⋃n−1

i=1 Ai = A, then∑n−1i=1 µ(Ai) < ǫ < 1 = µ(A) would mean that µ were not a content. Moreover,

since Bn ↑ A (note that every qk must be in precisely one of the Ai), (1.12) showsthat µ does not satisfy Th. 1.24(ii) on R.

1.3.3 Lebesgue Premeasure on Rn

At this point, we are in a position to perform the first step of the process describedat the beginning of Sec. 1.3: We can define a premeasure on a semiring that, whenextended to all Borel sets in Rn, will satisfy the conditions of Def. 1.1 (cf. Ex. 1.47).

Definition 1.26. Let n ∈ N and recall the definition of the semiring In on Rn fromEx. 1.8(c). The function

λn : In −→ R+0 , λn(I) :=

∏nj=1(bj − aj) for I =]a, b], a, b ∈ Rn, a < b,

0 for I = ∅, (1.13)

is called the (n-dimensional) Lebesgue premeasure on In (or, somewhat less precisely,on Rn, as in the section title above).

—

The main objective of this section is to show that λn is, indeed, a premeasure. It turnsout not to be as easy as one might have thought. We will first prove it to be a content inProp. 1.28 and then to be a premeasure in Th. 1.31. We introduce the following notionsthat will be useful to prove Prop. 1.28:

Definition 1.27. Let n ∈ N.

(a) Given k ∈ 1, . . . , n, α ∈ R, we define the following subsets of Rn:

Hk,α := x ∈ Rn : xk = α, (1.14a)

H+

k,α := x ∈ Rn : xk ≥ α, (1.14b)

H−k,α := x ∈ Rn : xk ≤ α, (1.14c)

where Hk,α is called the hyperplane and H+k,α (resp. H−

k,α) is called the (closed)positive (resp. negative) halfspace corresponding to (k, α).

1 MEASURE THEORY 21

(b) Given a half-open interval I =]a, b] ∈ In, a, b ∈ Rn with a < b, we call the set

C(I) := (k, α) ∈ 1, . . . , n × R : ak = α ∨ bk = α

the coordinate set of I. We also define C(∅) := ∅.

(c) Now let I, I1, . . . , IN ∈ In be half-open intervals such that I =]a, b] =⋃N

j=1Ij,a, b ∈ Rn with a < b. Moreover, let (k1, α1), . . . , (kM , αM) ∈ 1, . . . , n×R,M ∈ N.We say that the hyperplanes Hk1,α1 , . . . , HkM ,αM

cut the interval I into the intervalsI1, . . . , IN if, and only if, the following two conditions hold:

∀i∈1,...,N

∀j∈1,...,M

(Ii ⊆ H

+

kj ,αj∨ Ii ⊆ H

−kj ,αj

), (1.15a)

(kj, αj) : j ∈ 1, . . . ,M

=

(k, α) ∈

N⋃

j=1

C(Ij) : ak < α < bk

(1.15b)

(condition (1.15a) says that each small interval Ij has to lie on precisely one side ofeach of the cutting halfspaces; condition (1.15b) says that all halfspaces intersectthe interior of I and that the cutting halfspaces extend all the faces of the Ij thatintersect the interior of I to the boundary of I).

Proposition 1.28. For each n ∈ N, Lebesgue premeasure λn as defined in Def. 1.26constitutes a content on the semiring In.

Proof. We conduct the proof in several steps. First, we consider an interval I =]a, b],a, b ∈ Rn, a < b, that is cut by a hyperplane Hk,α, k ∈ 1, . . . , n and α ∈]ak, bk[, intointervals I1, I2 ∈ In in the sense of Def. 1.27. Then, clearly I1 ∪ I2, where I1 =]a1, b1],I2 =]a2, b2], with

a1 := a, b1 := (b1, . . . , bk−1, α, bk+1, . . . , bn),

a2 := (a1, . . . , ak−1, α, ak+1, . . . , an), b2 := b.

Moreover,

λn(I) =n∏

j=1

(bj − aj) =

(k−1∏

j=1

(bj − aj)

)(α− ak)

(n∏

j=k+1

(bj − aj)

)

+

(k−1∏

j=1

(bj − aj)

)(bk − α)

(n∏

j=k+1

(bj − aj)

)= λn(I1) + λn(I2).

We can now use an induction on M ∈ N to show that, if I ∈ In is cut by M hy-perplanes Hk1,α1 , . . . , HkM ,αM

, (k1, α1), . . . , (kM , αM ) ∈ 1, . . . , n × R, into intervals

I1, . . . , IN ∈ In, then λn(I) =∑N

i=1 λn(Ii): The case M = 1 has already been carried

out above. Thus, let M ∈ N. By induction, the M hyperplanes Hk1,α1 , . . . , HkM ,αM,

(k1, α1), . . . , (kM , αM) ∈ 1, . . . , n × R cut I into N intervals I1, . . . , IN ∈ In andλn(I) =

∑Ni=1 λ

n(Ii). Now consider (kM+1, αM+1) ∈ 1, . . . , n × R. Then, using the

1 MEASURE THEORY 22

caseM = 1, HkM+1,αM+1cuts each Ii either not at all or into two intervals I(i,1), I(i,2) ∈ In

such that λn(Ii) = λn(I(i,1)) + λn(I(i,2)). Letting

JM+1,1 :=i ∈ 1, . . . , N : Ii not cut by HkM+1,αM+1

,

JM+1,2 :=i ∈ 1, . . . , N : Ii cut into I(i,1), I(i,2) by HkM+1,αM+1

,

Ji := i for i ∈ JM+1,1, Ji := (i, 1), (i, 2) for i ∈ JM+1,2,

we have

λn(I) =N∑

i=1

λn(Ii) =N∑

i=1

∑

k∈Jiλn(Ik),

completing the induction. Finally, given I =]a, b] ∈ In (a, b ∈ Rn with a < b), let

K ∈ N, and let I1, . . . , IK ∈ In be disjoint intervals such that I =⋃K

j=1Ij. Set

J :=

(k, α) ∈

K⋃

j=1

C(Ij) : ak < α < bk

,

We cut I by the M := #J ∈ N hyperplanes in H(k,α) : (k, α) ∈ J (i.e. we extend allthe faces of the Ij that intersect the interior of I to the boundary of I). Then we knowthat these hyperplanes cut I into N disjoint intervals A1, . . . , AN ∈ In, N ∈ N, andλn(I) =

∑Ni=1 λ

n(Ai). Moreover, we define

∀j∈1,...,K

Jj :=(k, α) ∈ J : Ij cut by H(k,α)

.

If Jj 6= ∅, then theMj hyperplanes in H(k,α) : (k, α) ∈ Jj (where 1 ≤Mj := #Jj ∈ N)

cut Ij into Nj disjoint intervals Aj1, . . . , A

jNj

∈ In, Nj ∈ N, and λn(Ij) =∑Nj

ι=1 λn(Aj

ι ).

For Jj = ∅, set Aj1 := Ij, Nj := 1. The definition of J yields

∀i∈1,...,N

∃j∈1,...,K

Ai ⊆ Ij

and, thus,Ai : i ∈ 1, . . . , N

=⋃K

j=1

Aj

ι : ι ∈ 1, . . . , Nj.

In consequence, altogether, we obtain

λn(I) =N∑

i=1

λn(Ai) =K∑

j=1

Nj∑

ι=1

λn(Ajι ) =

K∑

j=1

λn(Ij),

completing the proof that λn is a content on In.

To show that λn is a premeasure, it remains to verify it is σ-additive. In the literature,one finds many different proofs. Here, we choose to use a method that makes use of λn

being regular in the sense of the following Def. 1.29 with respect to the norm topology onRn. Regular measures on topological spaces play an important role throughout analysisand, while we will not have time in this class to study them in depth, this gives us theopportunity to at least briefly touch on the subject.

1 MEASURE THEORY 23

Definition 1.29. Let (X, T ) be a topological space, let S be a semiring on X, and letµ : S −→ R+

0 be a finite content. Then µ is called inner regular (with respect to T ) if,and only if,

∀ǫ∈R+

∀A∈S

∃K∈S

(K compact ∧ K ⊆ A ∧ µ(A) ≤ µ(K) + ǫ

). (1.16)

Proposition 1.30. Let (X, T ) be a topological space, let S be a semiring on X, and letµ : S −→ R+

0 be an inner regular content. Then the following assertions hold true:

(a) If R := ρ(S) and µ : R −→ R+0 denotes the extension of µ to a content on R, then

µ is still inner regular.

(b) µ is a premeasure.

Proof. (a): According to Prop. 1.18(b), R consists of disjoint finite unions of sets fromS. Thus, if µ is finite on S, then µ is also finite on R. Now let A1, . . . , An ∈ S bedisjoint, n ∈ N, A :=

⋃ni=1Ai. Given ǫ ∈ R+, for each i ∈ 1, . . . , n, let Ki ∈ S be

such that Ki is compact, Ki ⊆ Ai, and µ(Ai) ≤ µ(Ki) +ǫn. Let K :=

⋃ni=1Ki. Then

K ∈ R, K =⋃n

i=1Ki is compact, K ⊆ A, and

µ(A) =n∑

i=1

µ(Ai) ≤n∑

i=1

(µ(Ki) +

ǫ

n

)= µ(K) + ǫ,

establishing µ to be inner regular on R.

(b): Using (a), it suffices to show that µ on R satisfies Th. 1.24(iv). Proceeding bycontraposition, we let (An)n∈N be a sequence in R such that A1 ⊇ A2 ⊇ . . . and

ǫ := limn→∞

µ(An) > 0,

and show that this implies⋂∞

n=1An 6= ∅. To this end, for each n ∈ N, letKn ∈ R be suchthat Kn is compact, Kn ⊆ An, and µ(An) ≤ µ(Kn) + 2−n ǫ. Also let Hn :=

⋂ni=1Kn.

Then Hn ∈ R. Moreover, Hn ⊆ Kn, i.e. Hn is a closed subset of the compact set Kn

and, thus, itself compact. If we can show

∀n∈N

Hn 6= ∅, (1.17)

then Hn =⋂n

i=1H i 6= ∅ for each n ∈ N and, by the finite intersection property of thecompact set H1,

⋂∞n=1An ⊇ ⋂∞

n=1Hn 6= ∅ as desired. It remains to show (1.17). Weaccomplish this via proving, for each n ∈ N,

µ(Hn) ≥ µ(An)− ǫ(1− 2−n) ≥ ǫ− ǫ+ ǫ 2−n = ǫ 2−n > 0 (1.18)

by induction on n. For n = 1, H1 = K1 and (1.18) holds, since µ(A1) ≤ µ(K1) + 2−1 ǫby the choice of K1. For the induction step, fix n ∈ N and estimate

µ(Hn+1) = µ(Hn ∩Kn+1)Th. 1.22(c)

= µ(Kn+1) + µ(Hn)− µ(Hn ∪Kn+1)ind.hyp

≥ µ(Kn+1) + µ(An)− ǫ(1− 2−n)− µ(Hn ∪Kn+1). (1.19)

1 MEASURE THEORY 24

Due to the choice ofKn+1, we have µ(Kn+1) ≥ µ(An+1)−2−(n+1) ǫ as well asKn+1∪Hn ⊆An+1 ∪ An = An and, thus, µ(Kn+1 ∪Hn) ≤ µ(An), which we use in (1.19) to obtain

µ(Hn+1) ≥ µ(An+1)− 2−(n+1) ǫ− ǫ(1− 2−n) = µ(An+1)− ǫ(1− 2−(n+1)),

completing the induction as well as the proof of the proposition.

Theorem 1.31. For each n ∈ N, Lebesgue premeasure λn as defined in Def. 1.26constitutes a premeasure on the semiring In.

Proof. We know λn to be a content from Prop. 1.28. Thus, the assertion will follow fromProp. 1.30, if we can show λn to be inner regular with respect to the norm topologyon Rn. To this end, let a, b, c ∈ Rn, a < c < b, A =]a, b], K =]c, b]. Then K ⊆ A iscompact and, letting Mn := maxbj − aj : j ∈ 1, . . . , n,

λn(A) ≤ λn(K) +Mn−1n

n∑

j=1

(cj − aj). (1.20)

We prove (1.20) by induction on n ∈ N: For n = 1, (1.20) follows from λ1(A) = b1−a1 =b1 − c1 + c1 − a1 = λ1(K) +M0

1 (c1 − a1). For n ∈ N, we have

λn+1(A×]an+1, bn+1]

)= (bn+1 − cn+1 + cn+1 − an+1)λ

n(A)

ind.hyp

≤ (bn+1 − cn+1)

(λn(K) +Mn−1

n

n∑

j=1

(cj − aj)

)+ (cn+1 − an+1)λ

n(A)

≤ λn+1(K×]cn+1, bn+1]

)+Mn

n+1

n∑

j=1

(cj − aj) +Mnn+1(cn+1 − an+1)

= λn+1(K×]cn+1, bn+1]

)+Mn

n+1

n+1∑

j=1

(cj − aj),

as needed to conclude the induction. Now fix n ∈ N, ǫ > 0, and let A and M := Mn

be as above. For each j ∈ 1, . . . , n, let cj := min bj+aj2, aj +

ǫnMn−1, K :=]c, b] (as

before). Then

λn(A)(1.20)

≤ λn(K) +Mn−1

n∑

j=1

(cj − aj) ≤ λn(K) +Mn−1 nǫ

nMn−1= λn(K) + ǫ,

proving λn to be inner regular and, thus, a premeasure.

1.3.4 Outer Measures, Caratheodory Extension Theorem

We would now like to extend premeasures to measures. An important tool to accomplishthis, are so-called outer measures.

Definition 1.32. Let X be a set. A map η : P(X) −→ [0,∞] is called an outer measureon X if, and only if, η satisfies the following conditions (i) – (iii):

1 MEASURE THEORY 25

(i) η(∅) = 0.

(ii) Monotonicity:∀

A⊆B⊆Xη(A) ≤ η(B). (1.21)

(iii) η is countably subadditive (also called σ-subadditive), i.e., if (Ai)i∈N is a sequencein P(X), then

η

( ∞⋃

i=1

Ai

)≤

∞∑

i=1

η(Ai). (1.22)

Remark 1.33. Since we know measures to be monotone and σ-subadditive, every mea-sure on P(X) is an outer measure. However, the converse is not true in general: Forexample,

η : P(X) −→ [0,∞], η(A) :=

0 if A = ∅,1 if A 6= ∅,

defines an outer measure that is not a measure if X has more than one element.

Definition 1.34. Let X be a set and let η : P(X) −→ [0,∞] be an outer measure onX. We call A ⊆ X η-measurable if, and only if,

∀Q⊆X

η(Q) ≥ η(Q ∩ A) + η(Q ∩ Ac). (1.23)

Lemma 1.35. Let X be a set and let η : P(X) −→ [0,∞] be an outer measure on X,A ⊆ X.

(a) If η(A) = 0 or η(Ac) = 0, then A is η-measurable.

(b) A is η-measurable if, and only if,

∀Q⊆X,

η(Q)<∞

η(Q) ≥ η(Q ∩ A) + η(Q ∩ Ac).

(c) A is η-measurable if, and only if,

∀Q⊆X

η(Q) = η(Q ∩ A) + η(Q ∩ Ac). (1.24)

Proof. (a) If η(A) = 0, then the monotonicity of η implies, for each Q ⊆ X, η(Q∩A) = 0and, thus η(Q ∩ A) + η(Q ∩ Ac) = η(Q ∩ Ac) ≤ η(Q), showing A to be η-measurable.Analogously, we see A to be η-measurable for η(Ac) = 0.

(b) is clear, since (1.23) always holds for η(Q) = ∞.

(c) is also clear since ≤ always holds in (1.23) by the subadditivity of η.

Proposition 1.36. Let X be a set and let η : P(X) −→ [0,∞] be an outer measure onX. Define

Aη := A ⊆ X : A is η-measurable. (1.25)

Then Aη is a σ-algebra and ηAηis a measure.

1 MEASURE THEORY 26

Proof. We first show Aη to be an algebra: Let Q ⊆ X. Since η(Q) = η(Q∩X)+η(Q∩∅),∅ ∈ Aη and X ∈ Aη. It is immediate from (1.23) that A ∈ Aη implies Ac ∈ Aη. Nowlet A,B ∈ Aη. Then

η(Q)(1.23)

≥ η(Q ∩ A) + η(Q ∩ Ac)(1.23)

≥ η(Q ∩ A) + η(Q ∩ Ac ∩ B) + η(Q ∩ Ac ∩ Bc)Def. 1.32(iii)

≥ η((Q ∩ A) ∪ (Q ∩ Ac ∩B)

)+ η(Q ∩ (A ∪ B)c

)

= η(Q ∩ (A ∪ B)

)+ η(Q ∩ (A ∪ B)c

),

proving A ∪ B ∈ Aη. Now

A \B = A ∩ Bc = (Ac ∪ B)c ∈ Aη, (1.26)

establishing Aη to be an algebra.

Now let (Ai)i∈N be a sequence of disjoint sets in Aη and let A :=⋃∞

i=1Ai. We claimthat A ∈ Aη and

η(A) =∞∑

i=1

η(Ai). (1.27)

To prove the claim, fix Q ∈ P(X). We first show

∀n∈N

η

(Q ∩

n⋃

i=1

Ai

)=

n∑

i=1

η(Q ∩ Ai) (1.28)

via induction on n: For n = 1, there is nothing to prove. For n > 1, we apply (1.24)with Q∩⋃n

i=1Ai instead of Q and⋃n−1

i=1 Ai instead of A (we know⋃n−1

i=1 Ai ∈ Aη, as Aη

is an algebra) to obtain

η

(Q ∩

n⋃

i=1

Ai

)= η

(Q ∩

n−1⋃

i=1

Ai

)+ η(Q ∩ An)

ind.hyp.=

n∑

i=1

η(Q ∩ Ai)

as needed. Now we prove A ∈ Aη and (1.27): For each n ∈ N, since⋃n

i=1Ai ∈ Aη, wecan apply (1.28) to estimate

η(Q)(1.23)

≥ η

(Q ∩

n⋃

i=1

Ai

)+η

(Q ∩

(n⋃

i=1

Ai

)c)(1.28),Def. 1.32(ii)

≥n∑

i=1

η(Q∩Ai)+η(Q∩Ac)

implying

η(Q) ≥∞∑

i=1

η(Q∩Ai)+η(Q∩Ac)Def. 1.32(iii)

≥ η(Q∩A)+η(Q∩Ac)Def. 1.32(iii)

≥ η(Q). (1.29)

Thus, all terms in (1.29) must be equal, proving A ∈ Aη and, for Q := A, (1.27).

1 MEASURE THEORY 27

Due to (1.27), ηAηis a measure, provided Aη is a σ-algebra. So, finally, let (Ai)i∈N be

a sequence in Aη (now the Ai do not have to be disjoint) and let A :=⋃∞

i=1Ai. Then

A =⋃∞

i=1

(Ai \

i−1⋃

k=1

Ak

)

︸ ︷︷ ︸∈Aη

∈ Aη

showing Aη to be a σ-algebra, completing the proof of the proposition.

The following Prop. 1.37 yields a useful construction method for outer measures:

Proposition 1.37. Let X be a set, S ⊆ P(X) with ∅ ∈ S, and µ : S −→ [0,∞] withµ(∅) = 0. Then

η : P(X) −→ [0,∞],

η(A) := inf

∞∑

i=1

µ(Ai) : (Ai)i∈N sequence in S,A ⊆∞⋃

i=1

Ai

, inf ∅ := ∞. (1.30)

defines an outer measure on X. This outer measure is called the outer measure inducedby µ.

Proof. Setting Ai := ∅ for each i ∈ N in (1.30) shows η(∅) = 0. If A ⊆ B ⊆ X and(Ai)i∈N is a sequence in S with B ⊆ ⋃∞

i=1Ai, then A ⊆ ⋃∞i=1Ai, showing η(A) ≤ η(B).

Only the σ-subadditivity of η requires slightly more effort: Let (Ai)i∈N be a sequence inP(X) and A :=

⋃∞i=1Ai. We need to show

η(A) ≤∞∑

i=1

η(Ai). (1.31)

Since (1.31) tivially holds if, for at least one i ∈ N, η(Ai) = ∞, we may assumeη(Ai) < ∞ for each i ∈ N. Let ǫ ∈ R+. Then, for each i ∈ N, there exists a sequence(Bik)k∈N in S such that

Ai ⊆∞⋃

k=1

Bik ∧∞∑

k=1

µ(Bik) < η(Ai) + ǫ · 2−i.

Since (Bik)(i,k)∈N2 is a countable family (i.e. a sequence) in S with A ⊆ ⋃(i,k)∈N2 Bik,

η(A) ≤∞∑

i=1

∞∑

k=1

µ(Bik) ≤∞∑

i=1

(η(Ai) + ǫ · 2−i

)= ǫ+

∞∑

i=1

η(Ai)

implies (1.31) and establishes the case.

Theorem 1.38 (Caratheodory Extension Theorem). Let S be a semiring on the set Xand let µ : S −→ [0,∞] be a content. If η is defined by (1.30), then the following holds:

1 MEASURE THEORY 28

(a) η is an outer measure and each A ∈ S is η-measurable (i.e. S ⊆ σ(S) ⊆ Aη).

(b) If µ is a premeasure, then ηS= µ (thus, in this case, ηAηis a measure that extends

the premeasure µ to a σ-algebra containing σ(S)).

(c) If µ is not a premeasure, then

∃A∈S

η(A) < µ(A). (1.32)

Proof. (a): According to Prop. 1.37, η is an outer measure. It remains to show thateach A ∈ S is η-measurable, i.e. we have to verify (1.23), given some A ∈ S. It will beconvenient to extend the content µ to a content on R := ρ(S) via Prop. 1.21 (we stillcall the extension µ). Then (1.30) and Prop. 1.18(b) imply

∀Q∈P(X)

η(Q) = inf

∞∑

i=1

µ(Ai) : (Ai)i∈N sequence in R,Q ⊆∞⋃

i=1

Ai

. (1.33)

To verify (1.23), let Q ⊆ X. As (1.23) always holds if η(Q) = ∞, we assume η(Q) <∞.For each sequence (Ai)i∈N in S with Q ⊆ ⋃∞

i=1Ai (where η(Q) < ∞ guarantees such asequence exists), the following must hold (using that µ is a content on R):

∞∑

i=1

µ(Ai) =∞∑

i=1

µ(Ai ∩ A) +∞∑

i=1

µ(Ai \ A)(1.33)

≥ η(Q ∩ A) + η(Q ∩ Ac). (1.34)

Now taking the infimum in (1.34) over all admissible sequences (Ai)i∈N in S accordingto (1.30), we obtain η(Q) ≥ η(Q ∩ A) + η(Q ∩ Ac), proving S ⊆ Aη as claimed.

(b): The inequality η S≤ µ is immediate from (1.30). It remains to show µ ≤ η S .We know from Prop. 1.21 that, if µ is a premeasure on S, then the extended µ is apremeasure on R. Thus, if (Ai)i∈N is a sequence in R such that Q ⊆ ⋃∞

i=1Ai, Q ∈ S,then, by Th. 1.22(f), µ(Q) ≤ ∑∞

i=1 µ(Ai). Taking the infimum over all admissiblesequences according to (1.33) yields µ(Q) ≤ η(Q) and µ ≤ ηS as needed.

(c): If µ is not a premeasure on S, then there exists A ∈ S and a sequence (Ai)i∈N ofdisjoint sets in S such that A =

⋃∞i=1Ai and µ(A) 6=

∑∞i=1 µ(Ai). Due to Th. 1.22(d)

(applied to µ on R),∑∞

i=1 µ(Ai) ≤ µ(A). Thus,

η(A) ≤∞∑

i=1

µ(Ai) < µ(A),

which establishes the case.

1.3.5 Dynkin Systems, Uniqueness of Extensions

The following simple Ex. 1.39 shows that, in general, one can not expect the extensionof a premeasure µ on a semiring S to a measure on σ(S) to be unique:

1 MEASURE THEORY 29

Example 1.39. Let X be a nonempty set and R := ∅. Then R is a semiring (andeven a ring) on X, and µ : R −→ [0,∞], µ(∅) := 0, defines a premeasure. ThenA := σ(R) = ∅, X and

∀α∈[0,∞]

µα : A −→ [0,∞], µα(A) :=

0 for A = ∅,α for A = X,

constitutes an extension of µ to a measure on A.

—

The problem in Ex. 1.39, as it turns out, is that the premeasure µ on the semiring S isnot σ-finite. In preparation for the proof of the uniqueness Th. 1.45, we develop someadditional technical tools that are also of measure-theoretic use beyond the applicationin the present section.

Definition 1.40. Let X be a set, E ⊆ P(X).

(a) E is called ∩-stable if, and only if,

∀A,B∈E

A ∩ B ∈ E .

(b) E is called a monotone class on X if, and only if, it satisfies the following twoconditions:

(i) If A ⊆ X and (En)n∈N is a sequence in E with En ↑ A, then A ∈ E .(ii) If A ⊆ X and (En)n∈N is a sequence in E with En ↓ A, then A ∈ E .

(c) E is called a Dynkin system on X if, and only if, it satisfies the following threeconditions:

(i) X ∈ E .(ii) If A ∈ E , then Ac ∈ E .(iii) If (En)n∈N is a sequence of disjoint sets in E , then ⋃n∈NEn ∈ E .

Remark and Definition 1.41. Arbitrary intersections of monotone classes yield againmonotone classes, arbitrary intersections of Dynkin systems yield again Dynkin systems:Let X be a set, let I be a nonempty index set, and let (Mi)i∈I be a family of monotoneclasses on X, M :=

⋂i∈I Mi. If A ⊆ X and (En)n∈N is a sequence in M with En ↑ A

(resp. En ↓ A), then, for each i ∈ I, (En)n∈N is a sequence in Mi, i.e. A ∈ Mi for eachi ∈ I, proving M to be a monotone class on X. Now let (Di)i∈I be a family of Dynkinsystems on X, D :=

⋂i∈I Di. Since X is in each Di, it is also in D. If A ∈ D, then

A ∈ Di for each i ∈ I, i.e. Ac ∈ Di for each i ∈ I, implying Ac ∈ D. If (En)n∈N isa sequence of disjoint sets in D then, for each i ∈ I, (En)n∈N is a sequence of disjointsets in Di. Thus,

⋃n∈NEn ∈ D, showing D is a Dynkin system on X. Thus, if X

is a set and E ⊆ P(X), then we can let µ(E) := µX(E) denote the intersection of all

1 MEASURE THEORY 30

monotone classes on X that are supersets of E , and we can let δ(E) := δX(E) denote theintersection of all Dynkin systems on X that are supersets of E . We then know µ(E) tobe a monotone class and δ(E) to be a Dynkin system, namely the smallest monotoneclass (resp. Dynkin system) on X containing E . Thus, we call µ(E) the monotone class(and δ(E) the Dynkin system) generated by E .

Proposition 1.42. Let X be a set, E ⊆ P(X). Then E is a Dynkin system on X if,and only if, the following conditions (i) – (iii) hold:

(i) X ∈ E .

(ii) For each A,B ∈ E , B ⊆ A implies A \B ∈ E .

(iii) E is a monotone class.

Proof. Assume (i) – (iii) hold. We have to show E is closed under complements andunder countable unions of disjoint sets. If B ∈ E , then, choosing A := X in (ii) showsBc = X\B ∈ E . If (Ei)i∈N is a sequence of disjoint sets in E , then E1∪E2 = (Ec

1\E2)c ∈ E

(if E1, E2 are disjoint, then E2 ⊆ Ec1 and Ec

1 \ E2 ∈ E by (ii)). Now an induction overn ∈ N shows An :=

⋃ni=1Ei ∈ E for each n ∈ N. Since An ↑ A :=

⋃∞i=1Ei, (iii) implies

A ∈ E , showing E to be a Dynkin system.

Conversely, assume E to be a Dynkin system. We have to show that (ii) and (iii) hold.If A,B ∈ E with B ⊆ A, then Ac, B, ∅ are disjoint sets in E , yielding

C := Ac ∪ B ∪ ∅ ∪ · · · ∈ E , A \B = Cc ∈ E ,

showing E satisfies (ii). Now let A ⊆ X and let (Ei)i∈N be a sequence of sets in E .If Ei ↑ A, then A = E1 ∪

⋃∞i=2(Ei \ Ei−1) ∈ E . If Ei ↓ A, then Ac =

⋃∞i=1E

ci , i.e.

Eci ↑ Ac ∈ E . Thus, A ∈ E as well and E is a monotone class, which completes the

proof.

Proposition 1.43. Let X be a set and let E ⊆ P(X) be a Dynkin system on X. Thenthe following statements are equivalent:

(i) E is ∩-stable.

(ii) E is a σ-algebra.

Proof. As every σ-algebra is ∩-stable, it only remains to show (i) implies (ii). If A,B ∈ Eand E is ∩-stable, then A ∪ B = (A ∩ Bc) ∪(B ∩ Ac) ∪(A ∩ B) ∈ E . Let (Ei)i∈N be asequence of sets in E . An induction over n ∈ N shows An :=

⋃ni=1Ei ∈ E for each n ∈ N.

Since An ↑ A :=⋃∞

i=1Ei and the Dynkin system E is a monotone class, A ∈ E , showingE to be a σ-algebra.

Proposition 1.44. Let X be a set and let E ⊆ P(X) be ∩-stable. Then δ(E) = σ(E).

1 MEASURE THEORY 31

Proof. Since every σ-algebra is a Dynkin system, we have D := δ(E) ⊆ σ(E). So it onlyremains to show that D is a σ-algebra. Moreover, according to Prop. 1.43, it suffices toshow D is ∩-stable. We will achieve this by showing

∀D∈D

D ⊆ D(D) := A ⊆ X : A ∩D ∈ D. (1.35)

While this strategy might seem strange at first, in various variants, it turns out to bequite powerful in measure theory, and we will use it again in the proof of Th. 1.45below. We note that, for each D ∈ D, D(D) is a Dynkin system: X ∈ D(D), sinceX ∩D = D ∈ D. If A ∈ D(D), then Ac ∈ D(D), as Ac∩D = D \ (A∩D) ∈ D by Prop.1.42(ii). If (Ai)i∈N is a sequence of disjoint sets in D(D), then A :=

⋃∞i=1Ai ∈ D(D),

since

A ∩D =⋃∞

i=1(Ai ∩D) ∈ D.

Since E is ∩-stable, we have E ⊆ D(E) for each E ∈ E . Thus, D = δ(E) ⊆ D(E) for eachE ∈ E , since D(E) is a Dynkin system. Thus, if E ∈ E and D ∈ D, then D ∈ D(E), i.e.D ∩ E ∈ D. But this means also E ∈ D(D), showing E ⊆ D(D), which proves (1.35)and the proposition.

Theorem 1.45. Let (X,A) be a measurable space and let µ, ν : A −→ [0,∞] be mea-sures. Moreover, let E ⊆ P(X) have the following properties:

(i) σ(E) = A.

(ii) µE= νE .

(iii) E is ∩-stable.

(iv) There exists a sequence (Ei)i∈N of sets in E such that

(∀i∈N

µ(Ei) = ν(Ei) <∞)

∧ X =∞⋃

i=1

Ei.

Then µ = ν.

Proof. For each E ∈ E with µ(E) = ν(E) <∞, we show

A = D(E) := A ∈ A : µ(A ∩ E) = ν(A ∩ E). (1.36)

First, we note that, for each E ∈ E with µ(E) = ν(E) < ∞, D(E) is a Dynkin system:X ∈ D(E), since X ∩ E = E ∈ E . If A ∈ D(E), then Ac ∈ D(E), since

µ(Ac ∩ E) = µ(E \ (A ∩ E)

)= µ(E)− µ(A ∩ E) (iii)

= ν(E)− ν(A ∩ E)= ν

(E \ (A ∩ E)

)= ν(Ac ∩ E).

1 MEASURE THEORY 32

If (Ai)i∈N is a sequence of disjoint sets in D(E), then A :=⋃∞

i=1Ai ∈ D(E), since

µ(A ∩ E) = µ

(⋃∞

i=1(Ai ∩ E)

)=

∞∑

i=1

µ(Ai ∩ E) =∞∑

i=1

ν(Ai ∩ E)

= ν

(⋃∞

i=1(Ai ∩ E)

)= ν(A ∩ E).

Moreover, since (iii) implies E ⊆ D(E), one obtains

A = σ(E) (iii), Prop. 1.44= δ(E) ⊆ D(E),

proving (1.36). Let (Ei)i∈N be as in (iv) and define, F0 := ∅ as well as, for each n ∈ N,Fn :=

⋃ni=1Ei. Then Fn ↑ X. For each n ∈ N, we have Fn =

⋃n

i=1(Ei ∩ F ci−1), implying

∀n∈N,A∈A

µ(A ∩ Fn) =n∑

i=1

µ(A ∩ F ci−1︸ ︷︷ ︸

∈A

∩Ei)(1.36)=

n∑

i=1

ν(A ∩ F ci−1︸ ︷︷ ︸

∈A

∩Ei) = ν(A ∩ Fn)

and∀

A∈Aµ(A) = lim

n→∞µ(A ∩ Fn) = lim

n→∞ν(A ∩ Fn) = ν(A),

which proves the theorem.

Corollary 1.46. Let X be a set and let S be a semiring on X.

(a) If µ : S −→ [0,∞] is a σ-finite premeasure, then there exists a unique measureν : σ(S) −→ [0,∞] that extends µ.

(b) One can strengthen (a) to the following comparison result: If α, β : σ(S) −→ [0,∞]are measures that satisfy αS≤ βS , where βS is σ-finite, then α ≤ β holds on allof σ(S).

Proof. (a): Existence was shown in Th. 1.38(b), whereas uniqueness is due Th. 1.45,which applies with E := S, since the semiring S is ∩-stable.(b): Since β S is σ-finite and α S≤ β S , α S must be σ-finite as well. Let α∗, β∗ :P(X) −→ [0,∞] be the outer measures induced by αS and βS , respectively (as definedin Prop. 1.37). Then the Caratheodory Extension Th. 1.38 combined with (a) yieldsα = α∗σ(S) and β = β∗σ(S). In consequence, αS≤ β S combined with (1.30) provesα ≤ β.

Example 1.47. Let n ∈ N. From Th. 1.31, we know the Lebesgue premeasure λn tobe a premeasure on the semiring In on Rn. Clearly, λn is also σ-finite. Thus, accordingto Cor. 1.46(a), λn can be uniquely extended to a measure βn on Bn = σ(In), calledLebesgue-Borel measure, where Bn denotes the σ-algebra of Borel sets in Rn (see Ex.1.8(c)). According to the Caratheodory Extension Th. 1.38, one obtains βn by firstextending λn to the induced outer measure λn : P(Rn) −→ [0,∞] (also called theLebesgue outer measure) and restricting it, first to Ln := Aλn and then to Bn. The

1 MEASURE THEORY 33

measure λn : Ln −→ [0,∞] is called Lebesgue measure. The notation βn will be usedwhen it is to be emphasized that λn is considered only on Bn. The sets in Bn are calledβn-measurable or Borel-measurable, the sets in Ln are called λn-measurable or Lebesgue-measurable. According to Cor. 1.52 below, Lebesgue measure is the so-called completionof Lebesgue-Borel measure (see Def. 1.48 and Def. 1.50) and one can show that Bn ( Ln,where Bn has the same cardinality as R and Ln has the same cardinality as P(R) (see[Els07, Sec. 2.8.3] and the last paragraph of Sec. 1.5.3 below). We will reencounter theimportant measures βn and λn throughout this class. In particular, they will be seen tosatisfy Def. 1.1, except that they are not defined on all of P(Rn).

1.3.6 Completion

Sets of measure zero play a special role in measure theory, giving rise to the followingdefinitions.

Definition 1.48. Let (X,A, µ) be a measure space.

(a) A set N ∈ A is called a µ-null set (or just a null set if µ is understood) if, and onlyif, µ(N) = 0.

(b) (X,A, µ) and µ are called complete if, and only if, each subset of a µ-null set isµ-measurable, i.e. if, and only if,

∀A∈A

∀B⊆A

(µ(A) = 0 ⇒ B ∈ A

).

—

If (X,A, µ) is a measure space, then µ can be used as the content of the CaratheodoryExtension Th. 1.38, which yields the extension ηAη

. According to Lem. 1.35(a), ηAηis

complete. Thus, every measure can be extended to a complete measure. This can alsobe seen, using a simpler and more natural construction that does not make use of outermeasures, according to the following Th. 1.49. If µ is σ-finite, then both constructionsyield the same complete extension (the so-called completion of µ), see Th. 1.51 below.

Theorem 1.49. Let (X,A, µ) be a measure space. Define N to be the set consisting ofall subsets of µ-null sets, i.e.

N :=

N ∈ P(X) : ∃

A∈A

(N ⊆ A ∧ µ(A) = 0

). (1.37a)

Also define

A := A ∪N : A ∈ A, N ∈ N, (1.37b)

µ : A −→ [0,∞], µ(A ∪N) := µ(A) for A ∈ A, N ∈ N . (1.37c)

Then the following holds:

1 MEASURE THEORY 34

(a) A is a σ-algebra, µ is well-defined by (1.37c), and (X, A, µ) is complete. Moreover,µ is the only content on A that is an extension of µ.

(b) Each complete measure ν that extends µ must also extend µ (i.e. µ is the “smallest”complete extension of µ).

Proof. (a): We verify A to be a σ-algebra: ∅ = ∅ ∪ ∅ ∈ A; if A ∈ A, N ∈ N , then thereexists B ∈ A such that N ⊆ B, µ(B) = 0. Then we compute

(A ∪N)c = Ac ∩N c = Ac ∩(Cc ∪ (C ∩N c)

)=(Ac ∩ Cc︸ ︷︷ ︸

∈A

)∪(Ac ∩ C ∩N c︸ ︷︷ ︸

⊆C

)∈ A,

showing (A ∪ N)c ∈ A. If (Ak)k∈N is a sequence in A and (Nk)k∈N is a sequencein N , then, for each k ∈ N, there exists Bk ∈ A such that Nk ⊆ Bk, µ(Bk) = 0.Then A :=

⋃∞k=1Ak ∈ A, B :=

⋃∞k=1Bk ∈ A, µ(B) = 0, and N :=

⋃∞k=1Nk ⊆ B,

showing⋃∞

k=1(Ak ∪ Nk) = A ∪ N ∈ A, proving A to be a σ-algebra. Next, we showµ to be well-defined: If A1 ∪ N1 = A2 ∪ N2 with A1, A2 ∈ A and N1, N2 ∈ N , thenthere exists B ∈ A with N2 ⊆ B, µ(B) = 0, implying A1 ⊆ A1 ∪ N1 ⊆ A2 ∪ Band µ(A1) ≤ µ(A2) + µ(B) = µ(A2). Analogously, one obtains µ(A2) ≤ µ(A1), i.e.µ(A1) = µ(A2) and µ is well-defined. Choosing N := ∅ in (1.37c) proves µ to be anextension of µ. To verify σ-additivity of µ, let (Ak)k∈N be a sequence of disjoint sets inA and (Nk)k∈N a sequence of disjoint sets in N . Then

µ

( ∞⋃

k=1

(Ak ∪Nk)

)= µ

(( ∞⋃

k=1

Ak

)∪( ∞⋃

k=1

Nk

))

= µ

( ∞⋃

k=1

Ak

)=

∞∑

k=1

µ(Ak) =∞∑

k=1

µ(Ak ∪Nk),

i.e. µ is σ-additive and, thus, a measure. To see that µ is complete, let C ⊆ A ∪ Nwith A ∈ A, N ∈ N , µ(A) = 0. Then there exits B ∈ A such that µ(B) = 0. ThenC ⊆ A ∪ B ∈ A with µ(A ∪ B) = 0. Thus, C ∈ N ⊆ A, proving µ to be complete. Letρ be a content on A that is an extension of µ. Then ρ(A) = µ(A) for each A ∈ A and,thus, ρ(N) = 0 for each N ∈ N . Thus, if A ∈ A and N ∈ N , then, by monotonicityand subadditivity of ρ, µ(A) = ρ(A) ≤ ρ(A ∪ N) ≤ ρ(A) ∪ ρ(N) = µ(A), showingρ(A ∪N) = µ(A) and ρ = µ as claimed.

(b): Let (X,B, ν) a complete measure space such that A ⊆ B and ν A= µ. As ν iscomplete, one has N ⊆ B. As ν extends µ, one has ν(N) = 0 for each N ∈ N . ThenA ⊆ B and νA= µ follows from the last part of (a).

Definition 1.50. Let (X,A, µ) be a measure space. Then the measure space (X, A, µ),where A and µ are defined as in (1.37), is called the completion of (X,A, µ); µ is calledthe completion of µ. According to Th. 1.49(a),(b), the completion of µ is the smallestcomplete measure that extends µ.

Theorem 1.51. Let S be a semiring on the set X, let µ : S −→ [0,∞] be a σ-finitepremeasure and η : P(X) −→ [0,∞] the induced outer measure defined by (1.30). Then

1 MEASURE THEORY 35

ηAηis the completion of ησ(S) and, in particular, this is the unique extension of µ to

a measure on Aη.

Proof. As we know ηAηto be complete, it remains to prove Aη ⊆ σ(S). To this end,

let B ∈ Aη. First, we assume η(B) <∞. Then, for each n ∈ N, there exists a sequence(Ank)k∈N of sets in S such that B ⊆ ⋃∞

k=1Ank and∑∞

k=1 µ(Ank) ≤ η(B) + 1n. Defining

A :=∞⋂

n=1

∞⋃

k=1

Ank,

we obtain

A ∈ σ(S), B ⊆ A, ∀n∈N

η(A) ≤ η(B) +1

n.

Thus, η(A) = η(B). We can now apply what we have just shown to A \ B instead ofB, proving the existence of C ∈ σ(S) such that A \ B ⊆ C and η(C) = η(A \ B) =

η(A) − η(B) = 0. Then B = (A \ C) ∪ (B ∩ C) ∈ σ(S), since A \ C ∈ σ(S) andB ∩ C is a subset of the (η σ(S))-null set C. It remains to consider the case B ∈ Aη

with η(B) = ∞. Since µ is assumed to be σ-finite, there exists a sequence (En)n∈N inS such that X =

⋃∞n=1En and µ(En) < ∞ for each n ∈ N. Since, for each n ∈ N,

η(B∩En) <∞ implies B∩En ∈ σ(S), we obtain B =⋃∞

n=1(B∩En) ∈ σ(S) as needed.According to Th. 1.49(a), η Aη

is the only measure on Aη that extends η σ(S). Since,by Cor. 1.46(a), η σ(S) is the only measure on σ(S) that extends µ, η Aη

is the onlymeasure on Aη that extends µ.

Corollary 1.52. Let n ∈ N. Then Lebesgue measure λn : Ln −→ [0,∞] is the comple-tion of Lebesgue-Borel measure βn : Bn −→ [0,∞]. Moreover, Lebesgue measure is theunique extension of the premeasure λn : In −→ [0,∞] to a measure on Ln.

Proof. Since λn : In −→ [0,∞] is a σ-finite premeasure, everything is immediate fromTh. 1.51.

1.4 Inverse Image, Trace

We investigate the behavior of σ-algebras when forming inverse images and subsets.

Notation 1.53. Let X, Y be sets f : X −→ Y . For E ⊆ P(Y ), we introduce thenotation

f−1(E) := f−1(E) : E ∈ E ⊆ P(X). (1.38)

Definition and Remark 1.54. Let X be a set, and let E ⊆ P(X) be a collection ofsubsets. If B ⊆ X, then let E|B := A ∩ B : A ∈ E denote the trace of E on B, alsoknown as the restriction of E to B. Consider the canonical inclusion map f : B −→ X,f(a) = a. Then,

f−1(E) = f−1(A) : A ∈ E = A ∩B : A ∈ E = E|B. (1.39)

1 MEASURE THEORY 36

Proposition 1.55. Consider sets X, Y and a map f : X −→ Y .

(a) If A is a σ-algebra on X, then B := B ⊆ Y : f−1(B) ∈ A is a σ-algebra on Y .

(b) If B is a σ-algebra on Y , then f−1(B) is a σ-algebra on X.

(c) Consider B ⊆ X. If A is a σ-algebra on X, then the trace A|B is a σ-algebra onB.

Proof. (a): Since ∅ = f−1(∅) ∈ A, we have ∅ ∈ B. If B ∈ B, then Y \ B ∈ B, sincef−1(Y \B) = X \ f−1(B) ∈ A, due to A being a σ-algebra. If (Bn)n∈N is a sequence ofsets in B, then

f−1

( ∞⋃

n=1

Bn

)=

∞⋃

n=1

f−1(Bn) ∈ A,

as A is a σ-algebra. Thus,⋃∞

n=1Bn ∈ B, which completes the proof that B is a σ-algebraon Y .

(b): As ∅ = f−1(∅), ∅ ∈ f−1(B). If B ∈ B, then X \ f−1(B) = f−1(Y \ B) ∈ f−1(B)since Y \B ∈ B due to B being a σ-algebra. If (Bn)n∈N is a sequence of sets in B, then,as B is a σ-algebra,

⋃∞n=1Bn ∈ B. Then

∞⋃

n=1

f−1(Bn) = f−1

( ∞⋃

n=1

Bn

)∈ f−1(B),

completing the proof that f−1(B) is a σ-algebra on X.

(c): Due to (1.39), (c) follows immediately from (b).

The following Th. 1.56 can be seen as a generalization of Prop. 1.55(b),(c).

Theorem 1.56. Consider sets X, Y and a map f : X −→ Y .

(a) The forming of generated σ-algebras commutes with the forming of inverse images:If E ⊆ P(Y ), then

σX(f−1(E)

)= f−1

(σY (E)

). (1.40)

(b) Let A be a σ-algebra on X, B ⊆ X, and E ⊆ A. If A = σX(E), then A|B =σB(E|B).

(c) Let T be a topology on X and let B denote the corresponding Borel σ-algebra onX. If B ⊆ X, then B|B is the Borel σ-algebra on B with respect to the relativetopology TB = T |B on B, i.e. B|B = σB(TB).

Proof. (a): Since E ⊆ σY (E), one has f−1(E) ⊆ f−1(σY (E)), which implies the ⊆-partof (1.40), as f−1

(σY (E)

)is a σ-algebra by Prop. 1.55(b). To prove the ⊇-part of (1.40),

defineB :=

B ⊆ Y : f−1(B) ∈ σX

(f−1(E)

).

1 MEASURE THEORY 37

Then E ⊆ B, B is a σ-algebra by Prop. 1.55(a), implying σY (E) ⊆ B. This, in turn,yields f−1

(σY (E)

)⊆ f−1(B) ⊆ σX

(f−1(E)

), which is precisely the ⊇-part of (1.40),

completing the proof.

(b): One merely has to apply (a) to the canonical inclusion map f : B −→ X, f(a) = a:

A|B (1.39)= f−1(A) = f−1

(σX(E)

) (1.40)= σB

(f−1(E)

) (1.39)= σB(E|B),

establishing the case.

(c): Since B = σX(T ), the statement is a special case of (b).

Please note the analogy between the restriction of a σ-algebra to a subset and theforming of the relative topology on a subset.

Proposition 1.57. If (X,A, µ) is a measure space and A ∈ A, then the restrictionν := µA|A is a measure on (A,A|A).

Proof. According to Prop. 1.55(c), A|A is a σ-algebra on A, and A ∈ A impliesA|A ⊆ Asuch that ν is actually defined for each A′ ∈ A|A. Moreover, ν(∅) = µ(∅) = 0, and, as νis the restriction of µ, ν inherits the σ-additivity from µ, proving that ν is a measure.

Example 1.58. Let n ∈ N, A ⊆ Rn. Then we can restrict the σ-algebras Bn and Ln

to A to obtain σ-algebras Bn|A and Ln|A. According to Prop. 1.57, if A ∈ Bn, thenβn Bn|A is a measure and, if A ∈ Ln, then λn Ln|A is a measure. By a slight abuse ofnotation, we also write (A,Bn, βn) (for A ∈ Bn) and (A,Ln, λn) (for A ∈ Ln) to denotethe measure spaces (A,Bn|A, βnBn|A) and (A,Ln|A, λnLn|A), respectively.

1.5 Lebesgue Measure on Rn

1.5.1 Intervals and Countable Sets

Proposition 1.59. Let n ∈ N.

(a) If a, b ∈ Rn, a ≤ b, I := [a, b], then λn(I) = βn(I) =∏n

j=1 |bj − aj|.

(b) If A ⊆ Rn is countable, then λn(A) = βn(A) = 0.

Proof. First note that the statements make sense, since all intervals and all countablesubsets of Rn are Borel-measurable.

(a): For each k ∈ N, let ak := (a1 − 1k, . . . , an − 1

k), Ik :=]ak, b] ∈ In. Then Ik ↓ I and

one computes

λn(I)(1.10)= lim

k→∞λn(Ik)

Ex. 1.20(b)= lim

k→∞

n∏

j=1

(bj − aj +

1

k

)=

n∏

j=1

(bj − aj),

proving (a).

1 MEASURE THEORY 38

(b): If a ∈ Rn, then λn(a) = λn([a, a]) = 0 by (a). Then (b) follows due to thecountable additivity of λn.

Thus, λn and βn satisfy Def. 1.1(a)(c). That they also satisfy Def. 1.1(b) will be seenin Sec. 1.6.3.

1.5.2 Approximation

The Lebesgue measure λn has the benign property of being regular in the sense that,with respect to λn, each A ∈ Ln can be approximated by open sets from the outsideand by closed and even by compact sets from the inside. Moreover, measurability ofA ⊆ Rn can be characterized by means of open and closed sets:

Theorem 1.60. Let n ∈ N.

(a) For each A ∈ Ln and each ǫ ∈ R+, there exist O ⊆ Rn open and F ⊆ Rn closedsuch that

F ⊆ A ⊆ O ∧ λn(O \ A) < ǫ ∧ λn(A \ F ) < ǫ.

Thus, for each A ∈ Ln,

λn(A) = infλn(O) : A ⊆ O, O open= supλn(F ) : F ⊆ A, F closed= supλn(C) : C ⊆ A, C compact. (1.41)

(b) A set A ⊆ Rn is λn-measurable if, and only if, for each ǫ ∈ R+, there exist O ⊆ Rn

open and F ⊆ Rn closed such that

F ⊆ A ⊆ O ∧ λn(O \ F ) < ǫ.

Proof. (a): Fix ǫ ∈ R+. First, assume λn(A) < ∞. Then, due to (1.30), there existsa sequence (Ik)k∈N in In such that A ⊆ ⋃∞

k=1 Ik and∑∞

k=1 λn(Ik) < λn(A) + ǫ

2. For

each k ∈ N, we enlarge Ik slightly to obtain some Jk ∈ In satisfying Ik ⊆ Jk and

λn(Jk) ≤ λn(Ik) + ǫ · 2−k−1. Then O :=⋃∞

k=1 Jk is open, A ⊆ O, and

λn(O \ A) = λn(O)− λn(A) ≤∞∑

k=1

λn(Ik) + ǫ∞∑

k=1

2−k−1 − λn(A) <ǫ

2+ǫ

2= ǫ.

If λn(A) = ∞, then, for each k ∈ N, set Ak := A ∩ [−k, k]n. Using what we havealready shown, for each k ∈ N, there exists an open Ok ⊆ Rn such that Ak ⊆ Ok andλn(Ok \ Ak) < ǫ · 2−k. Letting O :=

⋃∞k=1Ok, we obtain O open with A ⊆ O and

λn(O \ A) ≤∞∑

k=1

λn(Ok \ A) ≤∞∑

k=1

λn(Ok \ Ak) < ǫ∞∑

k=1

2−k = ǫ,

1 MEASURE THEORY 39

as desired. One now obtains the closed set F by taking complements: To Ac, thereexists an open set U ⊂ Rn such that Ac ⊆ U and λn(U \ Ac) < ǫ. Then F := U c isclosed, F ⊆ A, and λn(A \ F ) = λn(A ∩ U) = λn(U \ Ac) < ǫ, as needed. The first twoequalities of (1.41) are a direct consequence of what we have proved above. It remainsto show the third equality of (1.41). To this end, let 0 < α < λn(A). Then there existsa closed F ⊆ A such that λn(F ) > α. For each k ∈ N, the set Ck := F ∩ [−k, k]n iscompact and Ck ↑ F . Thus limk→∞ λn(Ck) = λn(F ) > α, i.e. λn(Ck) > α must hold foralmost all of the Ck.

(b): First, assume A ∈ Ln. Given ǫ > 0, choose O and F as in (a), except with ǫreplaced by ǫ/2. Then

λn(O \ F ) = λn(O \ A) + λn(A \ F ) < ǫ

2+ǫ

2= ǫ

proves the first implication. For the converse, given A ⊆ Rn, for each k ∈ N, choosea closed Fk and an open Ok such that Fk ⊆ A ⊆ Ok and λn(Ok \ Fk) <

1k. Then

B :=⋃∞

k=1 Fk ∈ Bn, C :=⋂∞

k=1Ok ∈ Bn, B ⊆ A ⊆ C, and λn(C \ B) = 0. Thus,A = B ∪ (A \B) is the union of the Borel set B and the set A \B, which is a subset ofthe βn-null set C \B. As λn is the completion of βn, A ∈ Ln.

1.5.3 Cantor Set

According to Prop. 1.59(b), λn(A) = 0 for each countable A ⊆ Rn. For n > 1, Prop.1.59(a) immediately provides examples of uncountable sets A ⊆ Rn with λn(A) = 0.Finding uncountable subsets of R that are λ1-null sets is not as simple. The Cantor setC is an example. The Cantor set occurs frequently in the literature due to its manyinteresting (e.g. topological) properties. The Cantor set C is what is left of the unitinterval [0, 1] after removing, in the initial step, the open middle third ]1

3, 23[ and, then,

in each successive step, the open middle third of each remaining interval (after the firststep, one has left [0, 1

3]∪[2

3, 1], after the second step, one has left [0, 1

9]∪[2

9, 39]∪[6

9, 79]∪[8

9, 99],

etc.). We now give the precise definition:

Definition and Remark 1.61. Inductively, we define two sequences (In)n∈N0 and(Kn)n∈N0 of subsets of [0, 1] such that, for each n ∈ N0, the following statements holdtrue:

(i) In consists of 2n disjoint open intervals In,1, . . . , In,2n .

(ii) Kn consists of 2n+1 disjoint compact intervals Kn,1, . . . , Kn,2n+1 , where

∀i∈1,...,2n+1

(Kn,i = [αn,i, βn,i], αn,i, βn,i ∈ [0, 1], βn,i = αn,i + 3−n−1

).

(iii) λ1(I) = λ1(K) = 3−n−1 for each I ∈ In and each K ∈ Kn.

(iv) [0, 1] is the disjoint union of all elements of Kn ∪⋃n

m=0 Im, i.e.

[0, 1] =⋃2n+1

i=1Kn,i ∪

⋃n

m=0

⋃2m

i=1Im,i.

1 MEASURE THEORY 40

(v) If n > 0, then,∀

i∈1,...,2nKn−1,i = Kn,2i−1 ∪ In,i ∪Kn,2i.

Initially, we set I0,1 :=]13, 23[, K0,1 := [0, 1

3], K0,2 := [2

3, 1], clearly satisfying the above

conditions for n = 0. Now let n ∈ N0 and assume I0, . . . , In as well as K0, . . . ,Kn to bealready defined such that (i) – (v) hold. Define, for each i ∈ 1, . . . , 2n+1,

In+1,i :=]αn,i + 3−n−2, αn,i + 2 · 3−n−2[,

Kn+1,2i−1 := [αn,i, αn,i + 3−n−2], Kn+1,2i := [αn,i + 2 · 3−n−2, αn,i + 3−n−1].

Then (ii) and (iii) for n + 1 are immediately clear as well as (v) for n + 1. Togetherwith (iv) for n, this implies (iv) for n+ 1. Finally, (v) for n+ 1 together with (ii) for nimplies (i) for n + 1, completing the inductive definition of (In)n∈N0 and (Kn)n∈N0 . Weare now in a position to define the Cantor set as

C :=∞⋂

n=0

2n+1⋃

i=1

Kn,i(iv)=

∞⋂

n=0

([0, 1] \

⋃n

m=0

⋃2m

i=1Im,i

)= [0, 1] \

⋃∞

m=0

⋃2m

i=1Im,i. (1.42)

Proposition 1.62. Let C ⊆ [0, 1] be the Cantor set as defined in (1.42). Then thefollowing holds:

(a) C is compact.

(b) C does not contain any nontrivial intervals, i.e. no intervals of positive length.

(c) λ1(C) = 0.

Proof. (a): As the intersection of compact subsets of Rn, C is compact.

(b): Using the notation of Def. and Rem. 1.61, define, for each n ∈ N, Kn :=⋃2n+1

i=1 Kn,i.Let I ⊆ R be a nontrivial interval. Then α := λ1(I) > 0. If n ∈ N is such that3−n−1 < α, then I 6⊆ Kn (since Kn is the disjoint union of intervals of length 3−n−1).Since C ⊆ Kn, I 6⊆ C.

(c): Due to (1.42), we have

λ1(C) = 1−∞∑

m=0

2m∑

i=1

λ1(Im,i) = 1−∞∑

m=0

2m 3−m−1 = 1− 1

3

1

1− 23

= 0,

proving (c).

It is not difficult to see directly that all the endpoints αn,i and βn,i of the Kn,i areelements of C, showing C must be infinite. However, there are only countably manyαn,i and βn,i, and, thus, to see that C is uncountable (Th. 1.64(b) below), we first haveto learn a bit more about the elements of C.

1 MEASURE THEORY 41

Lemma 1.63. Using the notation of Def. and Rem. 1.61, the left endpoints αn,i of Kn,i,n ∈ N0, i ∈ 1, . . . , 2n+1, are precisely the numbers x ∈ R that have a finite triadicexpansion, where all digits are either 0 or 2: Let n ∈ N0, x ∈ R. Then

(∃

i∈1,...,2n+1x = αn,i

)⇔

(∃

d1,...,dn+1∈0,2x =

n+1∑

k=1

dk · 3−k

).

Similarly, the right endpoints βn,i of Kn,i, n ∈ N0, i ∈ 1, . . . , 2n+1, are precisely thenumbers x ∈ R that have a periodic triadic expansion, where the period is 2 and alldigits are either 0 or 2: Let n ∈ N0, x ∈ R. Then

(∃

i∈1,...,2n+1x = βn,i

)⇔

(∃

d1,...,dn+1∈0,2x =

n+1∑

k=1

dk · 3−k +∞∑

k=n+2

2 · 3−k

).

Proof. Since βn,i = αn,i + 3−n−1 and∑∞

k=n+2 2 · 3−k = 2 · 3−n−2 · 32= 3−n−1, it suffices

to prove the statement for the αn,i. The statement for the αn,i is proved via inductionon n ∈ N0: For n = 0, x = α0,1 if, and only if, x = 0 = 0 · 3−1; x = α0,2 if, and only if,x = 2

3= 2 · 3−1. Thus, the statement holds for n = 0. For the induction step, assume

the statement to hold for some fixed n ∈ N0. Then x = αn+1,2i−1, i ∈ 1, . . . , 2n+1, isequivalent to x = αn,i and, by induction, this is equivalent to x =

∑n+1k=1 dk·3−k+0·3−(n+2)

with d1, . . . , dn+1 ∈ 0, 2. Similarly, x = αn+1,2i, i ∈ 1, . . . , 2n+1, is equivalent tox = αn,i+2 ·3−n−2 and, by induction, this is equivalent to x =

∑n+1k=1 dk ·3−k+2 ·3−(n+2)

with d1, . . . , dn+1 ∈ 0, 2, completing the induction and the proof of the lemma.

Theorem 1.64. Let C be the Cantor set defined in Def. and Rem. 1.61.

(a) C consists of all x ∈ [0, 1] that have a triadic expansion such that all digits areeither 0 or 2.

(b) C has the same cardinality as R; in particular, C is uncountable.

Proof. (a): Let x ∈ [0, 1]. According to the definition of C in (1.42), we have to show

(∀

n∈N0

x ∈2n+1⋃

i=1

Kn,i

)⇔

(∃

(dk)k∈N in 0,2x =

∞∑

k=1

dk · 3−k

).

“⇐”: Suppose x =∑∞

k=1 dk · 3−k with dk ∈ 0, 2 for each k ∈ N. Fix n ∈ N0.According to Lem. 1.63, there exists i ∈ 1, . . . , 2n+1 such that αn,i =

∑n+1k=1 dk · 3−k.

Then αn,i ≤ x ≤ αn,i +∑∞

k=n+2 2 · 3−k = βn,i, showing x ∈ Kn,i.

“⇒”: Suppose x ∈ ⋃2n+1

i=1 Kn,i for each n ∈ N0. If x is a right endpoint βn,i, then x hasa triadic expansion of the required form by Lem. 1.63. If x is not a right endpoint βn,i,then, for each n ∈ N0, there exists i ∈ 1, . . . , 2n+1 such that αn,i ≤ x < αn,i + 3−n−1.Thus, x− αn,i < 3−n−1, i.e. the coefficients of 3−1, . . . , 3−(n+1) in the triadic expansionsof αn,i and x must be identical. Then Lem. 1.63 yields that x =

∑∞k=1 dk · 3−k with

dk ∈ 0, 2 for each k ∈ N.

1 MEASURE THEORY 42

(b): Consider the map φ : 0, 2N −→ C, φ((dk)k∈N) :=∑∞

k=1 dk · 3−k. According to(a), φ is well-defined and surjective. It is also injective by [Phi16a, Th. 7.99], proving#0, 2N = #C. Thus

#C = #0, 2N = #0, 1N [Phi16a, Th. F.2]= #R,

completing the proof of (b).

As mentioned at the beginning of this section, for n > 1, Prop. 1.59(a) provides examplesof sets A ⊆ Rn with #A = #R and λn(A) = 0. We have now also seen that the Cantorset is a set A ⊆ R with #A = #R and λ1(A) = 0. Since each λn is complete, thisimplies #Ln = #P(R) for each n ∈ N. As #Bn = #R according to [Els07, Sec. 2.8.3],this also shows Bn ( Ln.

1.6 Measurable Maps

1.6.1 Definition, Composition

Measurable maps are for measure theory what continuous maps are for topology: Recallthat a map f : X −→ Y between topological spaces (X,S) and (Y, T ) is continuous if,and only if, f−1(B) ∈ S for each B ∈ T . If one replaces each topology with a σ-algebra,then one obtains the concept of a measurable map:

Definition 1.65. Let (X,A) and (Y,B) be measurable spaces, f : X −→ Y .

(a) f is called A-B-measurable (or often merely measurable) if, and only if, f−1(B) ∈ Afor each B ∈ B.

(b) If S and T are topologies onX and Y , respectively, then f is called Borelmeasurableif, and only if, it is σ(S)-σ(T )-measurable.

—

According to [Phi16b, Th. 2.7(iii)], to check if a map f : X −→ Y between topologicalspaces (X,S) and (Y, T ) is continuous, one only needs to verify if all preimages of setsin a subbase of T are open in X. The following Prop. 1.66 is the analogous result formeasurable maps:

Proposition 1.66. Let (X,A) and (Y,B) be measurable spaces. If E ⊆ P(Y ) is agenerator of B (i.e. σY (E) = B), then a map f : X −→ Y is A-B-measurable if, andonly if,

f−1(B) ∈ A for each B ∈ E . (1.43)

Proof. Since E ⊆ B, A-B-measurability implies (1.43). To prove the converse, note thatf−1(E) ⊆ A implies

f−1(B) = f−1(σY (E)

) Th. 1.56(a)= σX

(f−1(E)

)⊆ A,

completing the proof that f is A-B-measurable.

1 MEASURE THEORY 43

Example 1.67. (a) Constant maps are always measurable. More precisely, if (X,A)and (Y,B) are measurable spaces, f : X −→ Y , f ≡ c, c ∈ Y , then, for eachB ∈ B, there are precisely two possibilities: c ∈ B, implying f−1(B) = X ∈ A;c /∈ B, implying f−1(B) = ∅ ∈ A. Thus, f is A-B-measurable.

(b) Continuous maps are always Borel measurable: Let (X,S) and (Y, T ) be topologicalspaces with corresponding Borel σ-algebras σ(S) and σ(T ). If f : X −→ Y iscontinuous, then f−1(U) ∈ S ⊆ σ(S) for each U ∈ T and Prop. 1.66 implies f isσ(S)-σ(T ) measurable. In particular, if M ⊆ Rn and f : M −→ Rm is continuous(n,m ∈ N), then f is Borel measurable, i.e. Bn-Bm-measurable (cf. Ex. 1.58 andTh. 1.56(c)).

(c) Let X be a set and (Y,B) a measurable space, f : X −→ Y . Then A := f−1(B))is the smallest σ-algebra on X that makes f measurable (i.e. A =

⋂C ∈ P(X) :C σ-algebra on X and f C-B-measurable).

Proposition 1.68. The composition of measurable maps is measurable – more precisely,if (X,A), (Y,B), and (Z, C) are measurable spaces, f : X −→ Y is A-B-measurable,and g : Y −→ Z is B-C-measurable, then g f is A-C-measurable.

Proof. For each C ∈ C, we have A := (g f)−1(C) = f−1(g−1(C)

). Then g−1(C) ∈ B,

since g is B-C-measurable. In consequence, A ∈ A, since f is A-B-measurable, provingthe A-C measurability of g f .

1.6.2 Pushforward Measure

Theorem 1.69. Let (X,A, µ) be a measure space, let (Y,B) be a measurable space, andlet f : X −→ Y be A-B-measurable.

(a) The functionµf : B −→ [0,∞], µf (B) := µ

(f−1(B)

), (1.44)

defines a measure on B, called a pushforward measure – the measure µ is pushedforward from A to B by f . Alternatively to µf , one also finds the notation f(µ) :=µf .

(b) The forming of push forward measures commutes with the composition of maps: If(Z, C) is another measurable space, where g : Y −→ Z is B-C-measurable, then

µgf = (µf )g or (g f)(µ) = g(f(µ)). (1.45)

Proof. (a): One has µf (∅) = µ(∅) = 0 and, if (Bn)n∈N is a sequence of disjoint sets in

1 MEASURE THEORY 44

B, then (f−1(Bn))n∈N is a sequence of disjoint sets in A, implying

µf

( ∞⋃

n=1

Bn

)= µ

(f−1

( ∞⋃

n=1

Bn

))= µ

( ∞⋃

n=1

f−1(Bn)

)=

∞∑

n=1

µ(f−1(Bn)

)

=∞∑

n=1

µf (Bn),

verifying σ-additivity of µf and completing the proof of (a).

(b): First note that the hypotheses imply g f to be A-C measurable, such that µgfis well-defined by (a). Since (1.45) claims the equality of the two maps µgf and (µf )g,we have to verify µgf (C) = (µf )g(C) for each C ∈ C. To this end, for each C ∈ C, onecalculates

µgf (C) = µ((g f)−1(C)

)= µ

(f−1(g−1(C)

))= µf

(g−1(C)

)= (µf )g(C),

thereby establishing the case.

Proposition 1.70. Let (X,A, µ) and (Y,B, ν) be measure spaces with completions(X, A, µ) and (Y, B, ν). Moreover, let f : X −→ Y be A-B-measureable.

(a) If µ(f−1(C)) = 0 for each ν-null set C ∈ B, then f is A-B-measureable.

(b) If ν = f(µ), then ν = f(µ).

Proof. (a): Let B ∈ B. Then there exist B,C ∈ B and N ⊆ C such that ν(C) = 0 andB = B ∪N . Then

A := f−1(B) = f−1(B) ∪ f−1(N),

where f−1(B) ∈ A and f−1(N) ⊆ f−1(C) with µ(f−1(C)) = 0, showing A ∈ A, i.e. f isA-B-measureable.

(b): If ν = f(µ), and C ∈ B is a ν-null set, then µ(f−1(C)) = ν(C) = 0. Thus, (a)applies such that f is A-B-measureable and f(µ) is well-defined. If B, B, C,N are as inthe proof of (a), then

(f(µ)

)(B) = µ

(f−1(B) ∪ f−1(N)

)= µ

(f−1(B)

)= ν(B) = ν(B),

which establishes the case.

1.6.3 Invariance of Lebesgue Measure under Euclidean Isometries

In the present section, we will show that the measures βn and λn satisfy Def. 1.1(b), i.e.they are translation invariant. Indeed, in Th. 1.73 below, we will obtain the change ofβn and λn under arbitrary bijective affine maps, which will include the invariance of βn

and λn under Euclidean isometries, in particular translations, as a special case (also seethe paragraph before the statement of Th. 1.73 for a brief review of the relation betweenaffine maps, Euclidean isometries, and translations). For A ∈ P(Rn), x ∈ Rn, n ∈ N,we recall the notation

A+ x = a+ x : a ∈ A.

1 MEASURE THEORY 45

Definition 1.71. Let n ∈ N.

(a) If a ∈ Rn, then the affine map

Ta : Rn −→ Rn, Ta(x) := x+ a,

is called a translation, namely, the translation by a.

(b) Let (Rn,A, µ) be a measure. Then µ is called translation invariant if, and only if,for each a ∈ Rn, the translation Ta is A-A-measurable and Ta(µ) = µ, i.e. if, andonly if,

∀a∈Rn

∀A∈A

µ(A+ a) = µ(T−1−a (A)

)=(T−a(µ)

)(A) = µ(A). (1.46)

Theorem 1.72. Let n ∈ N.

(a) βn (resp. λn) is translation invariant and, moreover, it is the only translation in-variant measure µ on Bn (resp. Ln) satisfying µ([0, 1]n) = 1.

(b) If µ is a translation invariant measure on Bn (resp. Ln) such that µ([0, 1]n) = α <∞, then µ = αβn (resp. µ = αλn).

Proof. (a): Let a ∈ Rn. Since the translation Ta is continuous, it is Bn-Bn-measurable.Furthermore, if c, d ∈ Rn with c ≤ d, then ]c, d] + a =]c+ a, d+ a] and

βn(]c, d] + a) =n∏

j=1

(dj + aj − cj − aj) =n∏

j=1

(dj − cj) = βn(]c, d]).

Thus, the σ-finite measures βn and T−a(βn) agree on the semiring In. By Cor. 1.46(a),

they must also agree on Bn, proving the translation invariance of βn. Then, accordingto Prop. 1.70(a), Ta is Ln-Ln-measurable. Now apply Prop. 1.70(b) with µ := βn andν := T−a(β

n) = βn and obtain

T−a(λn) = T−a(µ) = ν = µ = λn,

proving the translation invariance of λn. Conversely, assume µ : Bn −→ [0,∞] tobe a translation invariant measure. In a first step, we assume µ(]0, 1]n) = 1. Giveng := (n1, . . . , nn) ∈ Nn, we can decompose ]0, 1]n into n1 · · ·nn translative copies of theinterval Ig :=

∏ni=1]0,

1ni]: If

Kg :=

(k1n1

, . . . ,knnn

)∈ Qn : (k1, . . . , kn) ∈ (N0)

n, ∀i∈1,...,n

0 ≤ ki < ni

,

then one has the disjoint union

]0, 1]n =⋃

k∈Kg

(Ig + k).

1 MEASURE THEORY 46

Due to the translation invariance of µ, this yields

1 = µ(]0, 1]n) = n1 · · ·nn · µ(Ig) ⇒ µ(Ig) =1

n1 · · ·nn

= βn(Ig).

If ∅ 6= I ∈ InQ, then there exist g := (n1, . . . , nn) ∈ Nn and a, b ∈ Zn with a < b, such

that

I =n∏

i=1

]aini

,bini

]=

⋃

k∈Kg,a,b

(Ig + k),

Kg,a,b :=

(k1n1

, . . . ,knnn

)∈ Qn : (k1, . . . , kn) ∈ Zn, ∀

i∈1,...,nai ≤ ki < bi

.

In this case, the translation invariance of µ yields

µ(I) =n∏

i=1

bi − aini

= βn(I).

Thus, the σ-finite measures µ and βn agree on InQ, i.e. they must agree on Bn by Cor.

1.46(a). If µ extends to a measure on Ln, then µ = λn by Th. 1.51. It remains toconsider a translation invariant measure µ on Bn (resp. on Ln) with µ([0, 1]n) = 1. Setα := µ(]0, 1]n). As µ is translation invariant, we obtain

1 = µ([0, 1]n) ≤ µ(]− 1, 1]n) = 2n · α ⇒ α ≥ 2−n > 0.

Thus, ν := α−1µ is a translation invariant measure with ν(]0, 1]n) = 1, implying ν = βn

(resp. ν = λn). Since α−1 = α−1µ([0, 1]n) = ν([0, 1]n) = βn([0, 1]n) = 1, it is µ = βn

(resp. µ = λn), completing the proof of (a).

(b): If α > 0, then α−1µ = βn (resp. α−1µ = λn) according to (a), proving µ = αβn

(resp. µ = αλn). If α = 0, then µ(]0, 1]n

)= 0 and the translation invariance of µ yields

µ(Rn) = µ

( ⋃

a∈Zn

(]0, 1]n + a

))

=∑

a∈Zn

µ(]0, 1]n

)= 0,

showing µ ≡ 0 ≡ 0 · βn (resp. µ ≡ 0 ≡ 0 · λn).

Note that Th. 1.72(b) does not hold without the condition α <∞, as counting measureis a translation invariant measure on both Bn and Ln that is not a multiple of βn (resp.λn).

In generalization of the above result on the translation invariance of βn and λn, we willnow investigate these measures’ behavior under general bijective affine maps. Recallfrom Linear Algebra that a map f : Rn −→ Rn is called affine if, and only if, thereexists a linear map L : Rn −→ Rn and a vector a ∈ Rn such that f(x) = L(x) + a foreach x ∈ Rn. Thus, using Def. 1.71(a), f : Rn −→ Rn is affine if, and only if, f = Ta Lwith a linear L and a vector a as above. Clearly, in that case, a and L are uniquelydetermined by the affine f (a = f(0), L = f − a), and one defines det f := detL. Inconsequence, one has the equivalences

f bijective ⇔ L bijective ⇔ detL 6= 0 ⇔ det f 6= 0.

1 MEASURE THEORY 47

Theorem 1.73. Let n ∈ N, f : Rn −→ Rn.

(a) If f is affine and bijective, then f is Bn-Bn-measurable as well as Ln-Ln-measurableand

f(βn) = | det f |−1 βn, f(λn) = | det f |−1 λn. (1.47a)

Furthermore, if A ∈ Bn (resp. A ∈ Ln), then f(A) ∈ Bn (resp. f(A) ∈ Ln) and

βn(f(A)

)= | det f | βn(A), λn

(f(A)

)= | det f |λn(A). (1.47b)

(b) If f is a Euclidean isometry (i.e. ‖f(x) − f(y)‖2 = ‖x − y‖2), then f is Bn-Bn-measurable as well as Ln-Ln-measurable and

f(βn) = βn, f(λn) = λn. (1.48a)

Furthermore, if A ∈ Bn (resp. A ∈ Ln), then f(A) ∈ Bn (resp. f(A) ∈ Ln) and

βn(f(A)

)= βn(A), λn

(f(A)

)= λn(A). (1.48b)

Proof. Since we know from Linear Algebra that f is a Euclidean isometry if, and onlyif, f = L + a with a linear map L : Rn −→ Rn that has | detL| = 1 (i.e. an orthogonallinear map L) and a ∈ Rn (see, e.g., [Koe03, Sec. 5.5.1]), it suffices to prove (a).

Since the affine map f is continuous, it is Bn-Bn-measurable. As f is affine and bijective,there exists a linear map L : Rn −→ Rn with detL 6= 0 and some a ∈ Rn such thatf = Ta L. The main work consists of showing

L(βn) = | detL|−1 βn. (1.49)

Let µ : Bn −→ [0,∞] be some arbitrary translation invariant measure on Bn. We claimL(µ) to be translation invariant as well: Indeed, for each A ∈ Bn and each v ∈ Rn, usingthe linearity of L and the translation invariance of µ,

(L(µ)

)(A+ v) = µ

(L−1(A+ v)

)= µ

(L−1(A) + L−1(v)

)= µ

(L−1(A)

)=(L(µ)

)(A).

We introduce the abbreviation I := [0, 1]n and, for the rest of the proof, we make theadditional assumption that αµ := µ(I) <∞. Then µ = αµ β

n by Th. 1.72(b), and

αL,µ :=(L(µ)

)(I) = µ

(L−1(I)

)= αµ β

n(L−1(I)

)<∞

(as the continuous image L−1(I) of the compact set I is compact), implying

L(µ) = αL,µ βn.

We need to show αL,βn = | detL|−1. Suppose L is orthogonal (i.e. | detL| = 1) and letB := x ∈ Rn : ‖x‖2 ≤ 1. Then, since L is a Euclidean isometry, L−1(B) = B and,thus,

αL,µ βn(B) =

(L(µ)

)(B) = µ

(L−1(B)

)= µ(B) = αµ β

n(B),

1 MEASURE THEORY 48

implying αL,µ = αµ, thereby proving (1.49) for orthogonal L. Next, assume L is repre-sented by a diagonal matrix with diagonal entries d1, . . . , dn ∈ R+ with respect to thestandard basis (e1, . . . , en) of Rn. Then

αL,βn = βn(L−1(I)

)= βn

([(0, . . . , 0), (d−1

1 , . . . , d−1n )]

)=

n∏

j=1

d−1j = | detL|−1,

proving (1.49) also for this case. Now let L be a general linear map with detL 6= 0. Weobtain this general case from the above special cases, using a bit more linear algebra:According to [Koe03, Sec. 6.3.2], P := LLt is positive definite. Using [Koe03, Sec. 6.3.2],P can be written as P = L1DL

t1, where L1 is orthogonal and D is represented by a

diagonal matrix with diagonal entries d1, . . . , dn ∈ R+ as above. If S is represented bythe diagonal matrix with diagonal entries

√d1, . . . ,

√dn, then S

2 = D and K := S−1Lt1L

is orthogonal:

KKt = S−1Lt1

P︷︸︸︷LLt L1S

−1 = S−1 Lt1L1︸ ︷︷ ︸Id

S2 Lt1L1︸ ︷︷ ︸Id

S−1 = Id .

Moreover, L1SK = L1SS−1Lt

1L = L and, thus, | detL| = | detL1| · detS · | detK| =1 · detS · 1 = detS, implying

αL,βn =(L(βn)

)(I) =

((L1 S K)(βn)

)(I)

(1.45)=(L1

(S(K(βn)

)))(I)

=(L1

(S(βn)

))(I) =

(L1

((detS)−1βn

))(I) =

(| detL|−1βn

)(I) = | detL|−1,

completing the proof of (1.49). We now come back to our bijective affine map f =Ta L. Since L(βn) is translation invariant by (1.49), f(βn) = Ta(L(β

n)) = L(βn) =| det f |−1 βn as claimed. Then Prop. 1.70(b) yield with µ := βn and ν := f(βn) =| det f |−1 βn and obtain

f(λn) = f(µ) = ν(1.37c)= | det f |−1 µ = | det f |−1 λn,

completing the proof of (1.47a). Finally, (1.47b) follows by applying (1.47a) to f−1:

∀A∈Ln

λn(f(A)

)=(f−1(λn)

)(A) = | det f |λn(A),

finishing the proof of the theorem.

1.6.4 Nonmeasurable Sets

We are now in a position to show the existence of subsets M of Rn that are not λn-measurable and, more generally, that no ideal n-dimensional volume as defined in Def.1.1 can exist (see Th. 1.74 below). The proof hinges on the validity of the axiom ofchoice AC. Without the axiom of choice the situation becomes subtle and difficult: Ithas been shown that there exist models of ZF set theory (without AC), where every

1 MEASURE THEORY 49

subset of Rn is λn-measurable. However, these constructions assume the existence ofso-called inaccessible cardinals, and if the existence of such cardinals is consistent withZF remains unknown (see the comments and references at the end of [Els07, Sec. III.3]).

In preparation for Th. 1.74, let G ⊆ Rn, n ∈ N, be an additive subgroup of Rn andrecall from Linear Algebra that the factor group (or quotient group) Rn/G consists ofall cosets

G+ a = g + a : g ∈ G, a ∈ Rn.

These are precisely the equivalence classes of the equivalence relation defined by x ∼y :⇔ x − y ∈ G. By AC, there exists a set MG ⊆ Rn of representatives of Rn/G,containing precisely one element from each coset G+ a ∈ Rn/G.

Theorem 1.74. Let n ∈ N.

(a) Let (Rn,A, µ) be a measure space such that µ extends βn (i.e. Bn ⊆ A and µBn=βn). Moreover, let G := Qn ⊆ Rn. We consider G as an additive subgroup of Rn

and let M :=MG be a set of representatives of Rn/G. If µ is translation invariantwith respect to G (i.e. A + g ∈ A and µ(A + g) = µ(A) hold for each A ∈ A andeach g ∈ G), then M /∈ A and µ(A) = 0 for each A ∈ A with A ⊆M (in particular,choosing A := Ln, µ := λn, yields the existence of a set M ⊆ Rn (by use of AC)that is not λn-measurable).

(b) There does not exist an ideal n-dimensional volume as defined in Def. 1.1.

(c) Let ηn : P(Rn) −→ [0,∞] denote the outer measure induced by λn. Then eachA ⊆ Rn with ηn(A) > 0 contains a set B that is not Lebesgue measurable (i.e.B /∈ Ln).

Proof. (a): Seeking a contradiction, assume M ∈ A. Define I :=]0, 1]n and

M0 :=⋃

g∈Zn

(− g +

(M ∩ (g + I)

)).

Clearly, M0 ⊆ I. Moreover, M0 is still a set of representatives of Rn/G (since Rn =⋃g∈Zn(g + I) and adding the element −g ∈ G to a coset does not change the coset).

Moreover, the assumed translation invariance of µ yields −g+(M∩(g+I)

)∈ A for each

g ∈ Zn and, thus, M0 ∈ A. Using the translation invariance of µ once again togetherwith the countability of G, we obtain

∞ = βn(Rn) = µ(Rn) = µ

(⋃

g∈G(g +M0)

)=∑

g∈Gµ(g +M0) =

∑

g∈Gµ(M0),

implying µ(M0) > 0. On the other hand, since G ∩ I is infinite (and countable),

∑

g∈G∩Iµ(M0) =

∑

g∈G∩Iµ(g +M0) = µ

( ⋃

g∈G∩I(g +M0)

)

≤ µ(]0, 2]n

)= βn

(]0, 2]n

)= 2n <∞,

1 MEASURE THEORY 50

implying µ(M0) = 0, in contradiction to µ(M0) > 0. Now, if A ∈ A with A ⊆ M , thenone can repeat the above argument with M replaced by A, except that one no longerobtains

∑g∈G µ(M0) = ∞, i.e., one obtains µ(A) = 0 without any contradiction.

(b): An ideal n-dimensional volume µ would satisfy all the hypotheses of (a) withA = P(Rn), in contradiction to the existence of some M ∈ P(Rn) with M /∈ A.

(c): Let G := Qn and M be as in (a). Then Rn =⋃

g∈G(g+M) and the σ-subadditivityof ηn imply

ηn(A) ≤∑

g∈Gηn(A ∩ (g +M)

). (1.50)

Since, for each g ∈ G, g + M is still a set of representatives of Rn/G, (a) impliesηn(A∩ (g+M)

)= λn

(A∩ (g+M)

)= 0 if A∩ (g+M) ∈ Ln. Thus, if A∩ (g+M) ∈ Ln

for each g ∈ G, then (1.50) implies ηn(A) = 0.

1.6.5 R-Valued Measurable Maps

Recall that, in [Phi16b, Ex. 1.52(b)], we extended the total order on R to a total orderon R = R ∪ −∞,∞. Then the resulting order topology T on R also extended thestandard topology on R, where a base for T is given by the set of open intervals

Io :=R∪]a, b[: a, b ∈ R, a < b

∪[−∞, a[: a ∈ R ∪ ∞

∪]a,∞] : a ∈ R ∪ −∞

(1.51)

(also see Def. A.2(b) and (A.5) in the Appendix).

Notation 1.75. Let B denote the Borel sets on R, i.e. the Borel sets with respect tothe order topology T on R.

Lemma 1.76. For the Borel sets on R, we have the following identities:

B =B ∪ E : B ∈ B1, E ⊆ −∞,∞

, (1.52a)

B = σ([−∞, α[: α ∈ R) = σ([−∞, α] : α ∈ R)= σ(]α,∞] : α ∈ R) = σ([α,∞] : α ∈ R), (1.52b)

B|R = B1. (1.52c)

Proof. (1.52a): Let A denote the right-hand side of (1.52a). Since −∞ and ∞ areclosed sets in R, every E ⊆ −∞,∞ is a Borel set, implying A ⊆ B. To verify theremaining inclusion, note σR(Io) = B, Io ⊆ A, and A is a σ-algebra.

(1.52b): Clearly, all four generator sets are subsets of B. Let E1 := [−∞, α[: α ∈ R,E2 := [−∞, α] : α ∈ R. We first show σ(E2) = B: If α, β ∈ R, then ]α, β] =[−∞, β] \ [−∞, α]. Thus, I1 ⊆ σ(E2), implying B1 ⊆ σ(E2). Since also −∞ =⋂

n∈N[−∞,−n] ∈ σ(E2) and ∞ =⋂

n∈N]n,∞] ∈ σ(E2), σ(E2) = B now follows from

(1.52a). If α ∈ R, then [−∞, α[=⋃

n∈N[−∞, α − 1n] ∈ σ(E1) implies σ(E1) = B. Since

the elements of ]α,∞] : α ∈ R are precisely the complements of the elements of E2

1 MEASURE THEORY 51

and the elements of [α,∞] : α ∈ R are precisely the complements of the elements ofE1, the proof of (1.52b) is complete.

(1.52c): Since the topology on R is the relative topology inherited from R, (1.52c) followsfrom Th. 1.56(c). Alternatively, (1.52c) is immediate from (1.52a).

Definition 1.77. Let (X,A) be a measurable space.

(a) A function f : X −→ R is called measurable if, and only if, it is A-B-measurable.

(b) A function f : X −→ C is called measurable if, and only if, it is A-B2-measurable,where B2 are the Borel sets on C with respect to the standard topology on C (thenotation B2 is reasonable, since C = R2 and B2 is precisely the set of Borel sets onR2).

(c) A function f : X −→ Rn, n ∈ N, is called measurable if, and only if, it is A-Bn-measurable.

(d) A function f : R −→ R is called measurable if, and only if, it is L1-B-measurable.

Notation 1.78. In the present context, it is useful to introduce a shorthand notation forcertain subsets of X: Let F := (fi)i∈I be a family of functions defined on X. Let P (F)denote a sentence (i.e. a sequence of symbols) and let P (F)(x) denote the correspondingsentence, where each occurrance of fi in P (F) was replaced by fi(x). If P (F) is suchthat P (F)(x) constitutes a statement (i.e. is either true or false) for each x ∈ X, thenwe define

P (F) := x ∈ X : P (F)(x) holds true. (1.53)

For example, consider f, g : X −→ R. If F = (f) and α ∈ R, then, e.g.,

f < α︸ ︷︷ ︸P (F)

= x ∈ X : f(x) < α = f−1([−∞, α[) (1.54a)

and, if F = (f, g), then, e.g.,

f = g︸ ︷︷ ︸P (F)

= x ∈ X : f(x) = g(x). (1.54b)

Theorem 1.79. Let (X,A) be a measurable space and f : X −→ R. Then the followingstatements are equivalent:

(i) f is measurable.

(ii) f < α ∈ A for each α ∈ R.

(iii) f ≤ α ∈ A for each α ∈ R.

(iv) f > α ∈ A for each α ∈ R.

1 MEASURE THEORY 52

(v) f ≥ α ∈ A for each α ∈ R.

Proof. One merely has to combine Prop. 1.66 with (1.52b).

Notation 1.80. Let X be a set and let (fi)i∈I be a family of functions fi : X −→ R.Then define

infi∈I

fi : X −→ R,

(infi∈I

fi

)(x) := inffi(x) : i ∈ I, (1.55a)

supi∈I

fi : X −→ R,

(supi∈I

fi

)(x) := supfi(x) : i ∈ I. (1.55b)

For I = N, we also define

limi→∞

fi : X −→ R,

(limi→∞

fi

)(x) := lim inf

i→∞fi(x), (1.55c)

limi→∞

fi : X −→ R,(limi→∞

fi

)(x) := lim sup

i→∞fi(x), (1.55d)

limi→∞

fi : X −→ R,(limi→∞

fi

)(x) := lim

i→∞fi(x), (1.55e)

where (1.55e) is defined, provided each limit limi→∞ fi(x) exists in R (of course, this isthe pointwise limit of the fi, which we have already studied previously). The followingnotation will also turn out to be useful: If f : X −→ R, then define

fi ↑ f :⇔(f = lim

i→∞fi ∧ ∀

x∈Xf1(x) ≤ f2(x) ≤ . . .

), (1.56a)

fi ↓ f :⇔(f = lim

i→∞fi ∧ ∀

x∈Xf1(x) ≥ f2(x) ≥ . . .

). (1.56b)

In the case of (1.55e) (and, in particular, (1.56)), we call the convergence uniform if,and only if,

∀ǫ∈R+

∃N∈N

∀i≥N

∀x∈X

∣∣fi(x)− f(x)∣∣ < ǫ. (1.57)

Lemma 1.81. Let X be a set and let (fk)k∈N be a sequence of functions fk : X −→ R.Then

∀α∈R

(infk∈N

fk ≥ α

=

∞⋂

k=1

fk ≥ α ∧supk∈N

fk ≤ α

=

∞⋂

k=1

fk ≤ α), (1.58a)

limk→∞

fk = supk∈N

(infi≥k

fi

)∧ lim

k→∞fk = inf

k∈N

(supi≥k

fi

). (1.58b)

If limk→∞ fk exists, then also

limk→∞

fk = limk→∞

fk = limk→∞

fk. (1.58c)

1 MEASURE THEORY 53

Proof. Let α ∈ R, x ∈ X. Then the two assertions in (1.58a) are precisely the equiva-lences

inffk(x) : k ∈ N ≥ α ⇔ ∀k∈N

fk(x) ≥ α,

supfk(x) : k ∈ N ≤ α ⇔ ∀k∈N

fk(x) ≤ α.

We prove (1.58b) next. Let x ∈ X. We have to show

α := lim infk→∞

fk(x) = supinffi(x) : i ≥ k : k ∈ N

=: β.

For each k ∈ N, let rk := inffi(x) : i ≥ k. Then α ≥ rk: If there exists i0 ≥ k suchthat α ≥ fi0(x), then α ≥ fi0(x) ≥ rk, since rk is a lower bound for the fi(x), i ≥ k.If α < fi(x) for all i ≥ k, then α < ∞ and α = rk = inffi(x) : i ≥ k, since, forα > −∞, each ]α, α+ ǫ[, ǫ > 0, must contain some fi(x), i ≥ k, and, for α = −∞, each] −∞,−n[, n ∈ N, must contain some fi(x) (as α is a cluster point). Now α ≥ rk foreach k ∈ N implies β ≤ α, since β is the smallest upper bound of the rk. It remains toshow α ≤ β. To this end, let γ ∈ R be an arbitrary upper bound of R := rk : k ∈ N.As β is an upper bound of R, it suffices to show α ≤ γ. Seeking a contradiction, assumeγ < α. For each k ∈ N, since rk ≤ γ and rk = inffi(x) : i ≥ k, there exists i ≥ k suchthat fi(x) < β, showing that [α, rk[ must contain fj(x) for infinitely many j ∈ N, incontradiction to α being the smallest cluster point of (fj(x))j∈N. Thus, we have shownα = β, as desired. One can conclude the second equation of (1.58b) from the first byapplying it to the reversed order on R. Alternatively, one can conduct an analogousproof as for the first equation: Let x ∈ X. We have to show

α := lim supk→∞

fk(x) = infsupfi(x) : i ≥ k : k ∈ N

=: β.

For each k ∈ N, let rk := supfi(x) : i ≥ k. Then α ≤ rk: If there exists i0 ≥ k suchthat α ≤ fi0(x), then α ≤ fi0(x) ≤ rk, since rk is an upper bound for the fi(x), i ≥ k.If α > fi(x) for all i ≥ k, then α > −∞ and α = rk = supfi(x) : i ≥ k, since, forα < ∞, each ]α − ǫ, α[, ǫ > 0, must contain some fi(x), i ≥ k, and, for α = ∞, each]n,∞[, n ∈ N, must contain some fi(x) (as α is a cluster point). Now α ≤ rk for eachk ∈ N implies β ≥ α, since β is the largest lower bound of the rk. It remains to showα ≥ β. To this end, let γ ∈ R be an arbitrary lower bound of R = rk : k ∈ N. Asβ is a lower bound of R, it suffices to show α ≥ γ. Seeking a contradiction, assumeγ > α. For each k ∈ N, since rk ≥ γ and rk = supfi(x) : i ≥ k, there exists i ≥ ksuch that fi(x) > β, showing that ]rk, α] must contain fj(x) for infinitely many j ∈ N,in contradiction to α being the largest cluster point of (fj(x))j∈N. Thus, we have shownα = β, as desired, finishing the proof of (1.58b). If the limit of the sequence (fk(x))k∈N,x ∈ X, exists, then it is the only cluster point of the sequence, proving (1.58c).

Theorem 1.82. Let (X,A) be a measurable space, let (fk)k∈N be a sequence of measur-able functions, fk : X −→ R.

(a) Each of the functions infk∈N fk, supk∈N fk, limk→∞ fk, limk→∞ fk is measurable. Iff = limk→∞ fk, f : X −→ R, then f is measurable as well.

1 MEASURE THEORY 54

(b) For each n ∈ N, min(f1, . . . , fn) and max(f1, . . . , fn) are measurable.

Proof. (a): If fk is measurable, then fk ≥ α ∈ A for each α ∈ R. Thus, infk∈N fkis measurable by (1.58a) and Th. 1.79(v); supk∈N fk is measurable by (1.58a) and Th.1.79(iii). In consequence, limk→∞ fk and limk→∞ fk are measurable by (1.58b), implyinglimk→∞ fk to be measurable by (1.58c).

(b) follows by applying the inf and sup parts of (a) to the sequence of measurablefunctions (f1, . . . , fn, fn, fn, . . . ).

Theorem 1.83. Let (X,A) be a measurable space and n ∈ N.

(a) A function f : X −→ Rn is measurable if, and only if, all coordinate functionsf1, . . . , fn : X −→ R are measurable.

(b) A function f : X −→ C is measurable if, and only if, the real-valued functions Re fand Im f are both measurable.

Proof. (a): Each projection πj : Rn −→ R, j ∈ 1, . . . , n, is continuous and, thus,Bn-B1-measurable. Thus, if f is A-Bn-measurable, then fj = πj f is A-B1-measurable.Conversely, if each fj, j ∈ 1, . . . , n, is measurable, then, for each I :=]a, b] ∈ In,a, b ∈ Rn, a < b, one has f−1(I) =

⋂nj=1 f

−1j (]aj, bj]) ∈ A. Since σ(In) = Bn, this proves

the A-Bn-measurability of f .

(b) is an immediate consequence of (a), as C = R2 with Re f = f1, Im f = f2.

Theorem 1.84. Let (X,A) be a measurable space.

(a) If f, g : X −→ R are measurable and α ∈ R, then f + g, αf , fg, and |f | are allmeasurable, and so is each of the sets f < g, f ≤ g, f = g, f 6= g. Ifg 6= 0,−∞,∞, then f/g is measurable as well.

(b) If f, g : X −→ C are measurable and α ∈ C, then f + g, αf , fg, and |f | are allmeasurable, and so are the sets f = g, f 6= g. If g 6= 0, then f/g is measurableas well.

Proof. (a): Let f, g be measurable. Since Q is countable, we obtain

f < g =⋃

q∈Q

(f < q ∩ q < g

)∈ A, f ≤ g = g < fc ∈ A,

f = g = f ≤ g ∩ g ≤ f ∈ A, f 6= g = f = gc ∈ A.

If α ∈ R, then f < α = −f > −α, showing −f to be measurable by Th. 1.79.Now assume f, g to be R-valued. Then h := (f, g) : X −→ R2 is measurable by Th.1.83(a). Since the continuous functions s, p : R2 −→ R, s(x, y) := x + y, p(x, y) := xy

1 MEASURE THEORY 55

are B2-B1-measurable, f + g = sh and fg = ph are measurable. If f, g are R-valued,then we have

A0 := f + g = 0 = f = −g ∈ A,A1 := f + g = ∞ =

(f = ∞ ∩ g 6= −∞

)∪(g = ∞ ∩ f 6= −∞

)∈ A,

A2 := f + g = −∞ =(f = −∞ ∩ g 6= ∞

)∪(g = −∞ ∩ f 6= ∞

)∈ A.

Let C := X \ (A0 ∪ A1 ∪ A2). Then f C and gC are R-valued and A|C-B-measurable.As C ∈ A, they are also A-B-measurable. Thus, (f + g)C is A-B-measurable as well.Now, if B ∈ B, then

(f + g)−1(B) = (f + g)−1(B ∩ 0,−∞,∞) ∪ (f + g)−1(B ∩ 0,−∞,∞c),

where the first set is in A, since A0, A1, A2 ∈ A, and the second set is in A, since(f + g)C is A-B-measurable. This shows f + g to be measurable. Now fg follows tobe measurable in an analogous manner, except this time we let

A0 := fg = 0 = f = 0 ∪ g = 0 ∈ A,A1 := fg = ∞ =

(f = ∞ ∩ g > 0

)∪(g = ∞ ∩ f > 0

)

∪(f = −∞ ∩ g < 0

)∪(g = −∞ ∩ f < 0

)∈ A,

A2 := fg = −∞ =(f = ∞ ∩ g < 0

)∪(g = ∞ ∩ f < 0

)

∪(f = −∞ ∩ g > 0

)∪(g = −∞ ∩ f > 0

)∈ A.

In particular, αf is measurable, as the constant function with value α is measurable.Next, we note |f | = max(f,−f) to be measurable. If g 6= 0,−∞,∞, then g−1 is well-defined and measurable (since g−1 = i g, where i : R \ 0 −→ R, i(x) := x−1, iscontinuous).

(b) now follows from (a) combined with Th. 1.83(b), where, for |f |, one also uses thecontinuity of the square root function.

Corollary 1.85. Let (X,A) be a measurable space.

(a) f : X −→ R is measurable if, and only if, both f+ and f− are measurable.

(b) f : X −→ C is measurable if, and only if, Re f+, Re f− Im f+, Im f− all aremeasurable.

Proof. (a): If f is measurable, then f+ = max(f, 0) and f− = max(−f, 0) are measur-able. If f+ and f− are measurable, then so is f = f+ − f−.

(b) follows by combining (a) with Th. 1.83(b).

1 MEASURE THEORY 56

Caveat 1.86. One has to use caution with the statement “The composition of measur-able functions is measurable.”: While it is correct if it is meant in the sense of Prop.1.68, it is false if applied to functions f, g : R −→ R, that are measurable in the senseof Def. 1.77(d): If f and g are L1-B1-measurable, then g f is not necessarily L1-B1-measurable (see, e.g., [RF10, Sec. 3.1, p. 57] for a counterexample). Of course, g f isL1-B1-measurable, if g is B1-B1-measurable (in particular, if g is continuous).

—

In the rest of the present section, we study so-called simple or step functions (Def. 1.87).Their importance lies in the fact that every R-valued measurable map is the pointwiselimit of a suitable sequence of simple functions (see Th. 1.89 below). We will make useof this result, when developing the integration theory of R-valued measurable maps (seeSections 2.1.1 and 2.1.2).

Definition 1.87. Let (X,A) be a measurable space. Then a real-valued measurablefunction f : X −→ R is called an A-simple function or an A-step function (often, A isunderstood and f is merely called a simple function or a step function) if, and only if,f takes only finitely many different values, i.e. if, and only if, #f(X) < ∞. Moreover,we introduce the following notation:

S := S(A) := (f : X −→ R) : f is A-simple, (1.59a)

S+ := S+(A) := f ∈ S : f ≥ 0, (1.59b)

M := M(A) := (f : X −→ R) : f is measurable, (1.59c)

M+ := M+(A) := f ∈ M : f ≥ 0. (1.59d)

Caveat: The functions in S and S+ are R-valued, but the functions in M and M+ areallowed to be R-valued.

—

We compile some basic properties of simple functions in the following proposition. Inpreparation, recall the notion of the characteristic function χA of a subset A of a set X:

χA : X −→ R, χA(x) :=

1 if x ∈ A,

0 if x /∈ A.

Proposition 1.88. Let (X,A) be a measurable space.

(a) If f, g ∈ S and α ∈ R, then f + g ∈ S, αf ∈ S (i.e. S is a vector space over R),fg ∈ S, max(f, g) ∈ S, min(f, g) ∈ S, and |f | ∈ S. Moreover, f/g ∈ S if g 6= 0.Everything remains true if S is replaced by S+ and α ∈ R+

0 (of course, S+ is nolonger a vector space (if X 6= ∅)).

(b) If f : X −→ R is a simple function, then there exist distinct numbers α1, . . . , αn ∈R, n ∈ N, and disjoint measurable sets A1, . . . , An ∈ A such that

f =n∑

i=1

αiχAi. (1.60)

1 MEASURE THEORY 57

More precisely, if f(X) = α1, . . . , αn with distinct α1, . . . , αn ∈ R, then oneobtains the representation (1.60) with disjoint A1, . . . , An ∈ A by setting Ai :=f−1(αi).

(c) A function f : X −→ R is a simple function if, and only if, it is a linear combina-tion of characteristic functions of measurable subsets of X.

Proof. (a): If f, g are measurable, then we already know f + g, αf , fg, max(f, g),min(f, g), |f |, and f/g (for g 6= 0) to be measurable. Moreover, if #f(X) = m,#g(X) = n, with m,n ∈ N, f(X) = α1, . . . , αm, g(X) = β1, . . . , βn, then

#(f + g)(X) = #αi + βj : i ∈ 1, . . . ,m, j ∈ 1, . . . , n

≤ m · n <∞,

#(αf)(X) = #ααi : i ∈ 1, . . . ,m

≤ m <∞,

#(fg)(X) = #αiβj : i ∈ 1, . . . ,m, j ∈ 1, . . . , n

≤ m · n <∞,

#max(f, g)(X) = #max(αi, βj) : i ∈ 1, . . . ,m, j ∈ 1, . . . , n

≤ m+ n <∞,

#min(f, g)(X) = #min(αi, βj) : i ∈ 1, . . . ,m, j ∈ 1, . . . , n

≤ m+ n <∞,

#|f |(X) = #|αi| : i ∈ 1, . . . ,m

≤ m <∞.

For β1, . . . , βn 6= 0, also

#(f/g)(X) = #αi/βj : i ∈ 1, . . . ,m, j ∈ 1, . . . , n

≤ m · n <∞.

Clearly, if f, g ≥ 0, then f + g ≥ 0, αf ≥ 0 for α ≥ 0, fg ≥ 0, max(f, g) ≥ 0,min(f, g) ≥ 0, |f | ≥ 0 (holds even without the hypothesis f ≥ 0) and f/g ≥ 0 for g 6= 0.

(b): If f is a simple function, then it takes finitely many distinct values α1, . . . , αn ∈ R,n ∈ N. Thus, if we let Ai := f−1(αi) for i ∈ 1, . . . , n, then each Ai is measurable(as f is measurable) and the validity of (1.60) is clear.

(c) is an immediate consequence of (a) and (b), since χA ∈ S for A ∈ A.

Theorem 1.89. Let (X,A) be a measurable space and f : X −→ R.

(a) f ∈ M+ if, and only if, there exists a sequence (φk)k∈N in S+ with φk ↑ f .

(b) f ∈ M with f bounded from below by α ∈ R (and bounded from above by β ∈ R) if,and only if, there exists a sequence (φk)k∈N in S with all φk ≥ α (and all φk ≤ β)and φk ↑ f .

(c) f ∈ M if, and only if, there exists a sequence (φk)k∈N in S with f = limk→∞ φk.

Additionally, in each case, if f is also bounded, then one can obtain the φk to convergeuniformly to f .

Proof. Note that (a) is a special case of (b) (to obtain (a) from (b), choose α := 0 and,in the bounded case, also choose β ≥ 0 to be some upper bound of f in (b)). We, thus,proceed to prove (b): If (φk)k∈N is a sequence in S that converges pointwise to f , then,

1 MEASURE THEORY 58

as the simple functions φk are measurable, f is measurable by Th. 1.82(a). Moreover, if,for x ∈ X, α ≤ φk(x) (resp. φk(x) ≤ β) for each k ∈ N, then, α ≤ f(x) (resp. f(x) ≤ β)as well. For the converse, given f , we have to construct suitable simple functions φk.To this end, we define the A-measurable sets

∀k∈N

∀j∈0,1,...,k·2k

Ak,j :=

α + j

2k≤ f < α + j+1

2k for 0 ≤ j ≤ k · 2k − 1,

f ≥ α + k for j = k · 2k

(as f is assumed to be measurable, the Ak,j are A-measurable by Th. 1.79). Clearly, foreach fixed k ∈ N, the Ak,j form a decomposition of X:

∀k∈N

X =⋃k·2k

j=0Ak,j. (1.61)

For each k ∈ N, we define the simple function

φk : X −→ R, φk :=k·2k∑

j=0

(α +

j

2k

)χAk,j

.

Since X is the disjoint union of the Ak,j by (1.61), this immediately implies

∀k∈N

α ≤ φk ≤ α + k · 2k.

We next verify that the sequence (φk)k∈N is increasing: Note that, for each k ∈ N, wehave the disjoint unions

Ak,j = Ak+1,2j ∪Ak+1,2j+1 for 0 ≤ j ≤ k · 2k − 1,

Ak,k·2k =⋃(k+1)·2k+1

l=k·2k+1Ak+1,l.

Since j2k

≤ l2k+1 for l ∈ 2j, 2j + 1, and k·2k

2k≤ l

2k+1 for l ∈ k · 2k+1, . . . , (k + 1) · 2k+1,we have φk(x) ≤ φk+1(x) for each x ∈ X. We now fix x ∈ X and show limk→∞ φk(x) =f(x): For each k ∈ N, there exists a unique j = j(k, x) such that x ∈ Ak,j. Thenφk(x) = α+ j

2k≤ f(x), showing f(x) to be an upper bound for the φk(x). If f(x) = ∞,

then, for each k ∈ N, φk(x) = α+ k → ∞. On the other hand, if f(x) <∞, then, givenǫ > 0, we can choose k0 ∈ N such that 1

2k0< ǫ and f(x) < α + k0. If k ≥ k0, then

f(x)− φk(x) < 2−k ≤ 2−k0 < ǫ,

proving limk→∞ φk(x) = f(x). Under the additional hypothesis f ≤ β ∈ R, one canchoose α+ k0 ≥ β, independently of x, showing uniform convergence. Moreover, in thiscase, the above observation φk(x) ≤ f(x) for each k ∈ N and each x ∈ X also yieldsφk(x) ≤ β for each k ∈ N, x ∈ X.

(c): As before, if f = limk→∞ φk with (φk)k∈N in S, then f ∈ M by Th. 1.82(a). Forthe converse, we write f ∈ M as f = f+ − f− and, by (a), choose (φk)k∈N in S+ withφk ↑ f+ and (ψk)k∈N in S+ with ψk ↑ f− (with uniform convergence in each case for fbounded). Then f = limk→∞(φk − ψk) (with uniform convergence for f bounded).

1 MEASURE THEORY 59

1.6.6 Products, Projections, and Borel Sets

In [Phi16b, Ex. 1.53(a)], we constructed the product topology on a Cartesian productof topological spaces as the smallest topology that makes all projections continuous.Analogously, we are now going to construct the product σ-algebra on a Cartesian productof measurable spaces as the smallest σ-algebra that makes all projections measurable(cf. Th. 1.93(b) below). As in topology, the construction is a special case of so-calledinitial constructions (cf. Sec. C in the Appedix).

Definition 1.90. Let I be a nonempty index set and, for each i ∈ I, let (Xi,Ai) bea measurable space. We consider the Cartesian product X :=

∏i∈I Xi and, for each

j ∈ I, we recall the projection

πj : X −→ Xj, πj((xi)i∈I

):= xj.

Consider the set

Ep :=

∏

i∈IAi :

(∀i∈I

Ai ∈ Ai

)∧ #i ∈ I : Ai 6= Xi ≤ 1

=π−1i (Ai) : i ∈ I, Ai ∈ Ai

.

Then Ap :=⊗

i∈I Ai := σ(Ep) (the σ-algebra on X generated by Ep), is called theproduct σ-algebra on X.

Proposition 1.91. In the situation of Def. 1.90 above, for each i ∈ I, let Ei be agenerator of Ai.

(a) The set

E∗p :=

∏

i∈IEi :

(∀i∈I

Ei ∈ Ei ∪ Xi)

∧ #i ∈ I : Ei 6= Xi ≤ 1

=π−1i (Ei) : i ∈ I, Ei ∈ Ei

∪ X

is a generator of the product σ-algebra on X, i.e. Ap = σ(E∗p).

(b) If I is finite and, for each i ∈ I, there exists a sequence (Ei,k)k∈N in Ei such that

Xi =∞⋃

k=1

Ei,k, (1.62)

then the set

Fp :=

∏

i∈IEi : ∀

i∈IEi ∈ Ei

=

⋂

i∈Iπ−1i (Ei) : ∀

i∈IEi ∈ Ei

is a generator of the product σ-algebra on X, i.e. Ap = σ(Fp).

1 MEASURE THEORY 60

Proof. (a): As E∗p ⊆ Ep, we have σ(E∗

p) ⊆ σ(Ep). For the remaining inclusion, we note

that, by the definition of E∗p , one has π−1

i (Ei) ⊆ E∗p for each i ∈ I. Moreover, by Th.

1.56(a), σ(π−1i (Ei)

)= π−1

i

(σ(Ei)

)= π−1

i (Ai), implying σ(Ep) ⊆ σ(E∗p).

(b): From (a), we know Ap = σ(E∗p). If i ∈ I and Ei ∈ Ei, then π−1

i (Ei) ∈ E∗p by the

definition of E∗p . Thus, if Ei ∈ Ei for each i ∈ I, then

⋂i∈I π

−1i (Ei) ∈ σ(E∗

p), since I isfinite. This already yields σ(Fp) ⊆ σ(E∗

p) (for this inclusion, we actually did not use(1.62)). For the remaining inclusion, we use the finiteness of I together with (1.62) towrite every element of E∗

p as a countable union of elements from Fp:

∀i∈I

∀Ei∈Ei∪Xi

π−1i (Ei) =

⋃

(kα)α∈I\i∈NI\i

Ei ×

∏

α∈I\iEα,kα

∈ σ(Fp).

Thus, σ(E∗p) ⊆ σ(Fp), completing the proof of (b).

Example 1.92. Let p, q ∈ N. For the Borel sets on Rp and Rq, we obtain Bp+q = Bp⊗Bq

(also see Th. 1.94(c) below): We know from Ex. 1.8(c), that the Borel sets are generatedby the compact intervals: Bp = σ(Ip

c ), Bq = σ(Iqc ), Bp+q = σ(Ip+q

c ). Thus, we can applyProp. 1.91(b) with E1 = Ip

c , E2 = Iqc , and Fp = Ip+q

c to obtain

Bp+q = σ(Ip+qc ) = σ(Fp) = Bp ⊗ Bq.

Theorem 1.93. Let I be a nonempty index set and, for each i ∈ I, let (Xi,Ai) be ameasurable space. We consider the Cartesian product X :=

∏i∈I Xi with the product

σ-algebra Ap = σ(Ep) according to Def. 1.90.

(a) Each projection πi : X −→ Xi, i ∈ I, is Ap-Ai-measurable.

(b) Ap is the smallest σ-algebra A on X such that all projections πi, i ∈ I, are A-Ai-measurable.

(c) If (Y,B) is a measurable space and f : Y −→ X, then the following statements areequivalent:

(i) f is B-Ap-measurable.

(ii) For each i ∈ I, the coordinate function fi = πi f is B-Ai-measurable.

Proof. (a): If i ∈ I and A ∈ Ai, then π−1i (A) ∈ Ep ⊆ Ap, showing πi to be Ap-Ai-

measurable.

(b) is immediate, since each σ-algebra A on X such that all projections πi, i ∈ I, areA-Ai-measurable must contain Ep.(c): If f is B-Ap-measurable, then each fi, i ∈ I, is B-Ai-measurable by (a) and Prop.1.68. For the converse, assume each fi, i ∈ I, to be B-Ai-measurable. According toProp. 1.66, it suffices to show f−1(π−1

i (A)) ∈ B for each i ∈ I and each A ∈ Ai. Sincef−1(π−1

i (A)) = f−1i (A) and fi is measurable, the proof is, thus, complete.

1 MEASURE THEORY 61

When considering the product (X, T ) of topological spaces (Xi, Ti), i ∈ I, where T isthe corresponding product topology, then one can obtain a σ-algebra AT on X by takingthe Borel sets with respect to T . Alternatively, one can take the Borel σ-algebras Ai

on Xi and then form the resulting product σ-algebra Ap :=⊗

i∈I Ai on X. An obviousquestion is, whether the results are the same. Unfortunately, in general, they are not:While one always has Ap ⊆ AT (see Th. 1.94(a) below), the inclusion can be strict (evenfor the product of just two spaces, see, e.g., [Els07, Rem. III.5.16, Problem III.5.3]). Asufficient condition for Ap = AT will be given in Th. 1.94(b).

Theorem 1.94. Let I be a nonempty index set and, for each i ∈ I, let (Xi, Ti) be atopological space. Let Ai denote the corresponding Borel σ-algebra on Xi. Moreover, letX :=

∏i∈I Xi, let T denote the product topology on X, let AT be the resulting Borel

σ-algebra on X, and let Ap :=⊗

i∈I Ai be the product σ-algebra of the Ai.

(a) One always has Ap ⊆ AT .

(b) If I is countable and each (Xi, Ti) is a C2 space (i.e. each Ti has a countable base),then Ap = AT .

(c) Let p, q ∈ N, X ⊆ Rp, Y ⊆ Rq. Then

Bp|X ⊗ Bq|Y = Bp+q|(X × Y ). (1.63)

In particular, Bp+q = Bp ⊗ Bq and Bp = ⊗pk=1B1.

Proof. (a): Since each projection πi, i ∈ I, is continuous, it is π−1i (A) ∈ T for each

A ∈ Ti, implying it to be Ai-AT -measurable. Then Ap ⊆ AT follows from Th. 1.93(b).

(b): Due to (a), it remains to show AT ⊆ Ap. For each i ∈ I, let Bi be a countable baseof Ti. According to [Phi16b, Ex. 1.53(a)], a base for T is given by

B :=

⋂

j∈Jπ−1j (Bj) : J ⊆ I, 0 < #J <∞, ∀

j∈JBj ∈ Bj

.

Since I and each Bi are countable, so is B. Since Bi ⊆ Ai, we have B ⊆ Ap. Now everyO ∈ T is a countable union of sets from B (since B is countable), showing T ⊆ σ(B) ⊆Ap and, thus, AT = σ(T ) ⊆ Ap, as desired.

(c): Let Tp, Tq, and T denote the norm topologies on Rp, Rq, and Rp+q, respectively.As we know the norm topologies to be C2, we can apply (b) to obtain

Bp|X ⊗ Bq|Y = σ(Tp)|X ⊗ σ(Tq)|YTh. 1.56(b)

= σX(Tp|X)⊗ σY (Tq|Y )

= σX((Tp)X)⊗ σY ((Tq)Y )(b)= σX×Y ((Tp)X × (Tq)Y )

= σX×Y (TX×Y ) = σX×Y (T |(X × Y ))Th. 1.56(b)

= σ(T )|(X × Y )

= Bp+q|(X × Y ),

as claimed.

2 INTEGRATION 62

Caveat 1.95. While we know from [Phi16b, Ex. 2.12(b)] that every projection is an openmap with respect to the product topology (i.e. projections map open sets to open sets),the analogous statement is, in general, not true for product σ-algebras: For example, itcan occur that the projection of a Borel set B ⊆ [0, 1]2 is not in B1|[0, 1] (projections ofsuch Borel sets are known as Suslin sets – the construction of a Suslin set that is not aBorel set needs some work and can be found in [Beh87, Appendix II]).

2 Integration

2.1 Integration of R-Valued Measurable Maps

We define the so-called Lebesgue integral first for nonnegative simple functions in Def.2.1, then for nonnegative measurable functions in Def. 2.4, and then for so-called inte-grable functions in Def. 2.11.

2.1.1 Simple Functions

The definition of the Lebesgue integral for nonnegative simple functions makes use ofrepresentations of the integrands, where Lem. 2.2 then states that the value of theintegral does actually not depend on the representation of the integrated function.

Definition 2.1. Let (X,A, µ) be a measure space. Moreover, let f : X −→ R+0 be a

simple function (i.e. f ∈ S+(A)), where

f =N∑

i=1

αi χAi(2.1)

with N ∈ N; α1, . . . , αN ≥ 0; and A1, . . . , AN ∈ A. Then

∫

X

f dµ :=

∫

X

f(x) dµ(x) :=N∑

i=1

αi µ(Ai) ∈ [0,∞] (2.2)

is called the Lebesgue integral of f over X with respect to µ.

Lemma 2.2. Let (X,A, µ) be a measure space. Then the value of the integral in(2.2) does not depend on the representation of the simple function f : If M,N ∈ N;α1, . . . , αN ≥ 0; β1, . . . , βM ≥ 0; A1, . . . , AN ∈ A; and B1, . . . , BM ∈ A are such that

f =N∑

i=1

αi χAi=

M∑

i=1

βi χBi, (2.3a)

thenN∑

i=1

αi µ(Ai) =M∑

i=1

βi µ(Bi). (2.3b)

2 INTEGRATION 63

Proof. We add the list of Bi to the list of Ai by letting AN+1 := B1, . . . , AN+M := BM ,and define the set of intersections

D :=

N+M⋂

i=1

Mi : ∀i∈1,...,N+M

Mi ∈ Ai, Aci.

Note that two distinct elements of D are always disjoint: Indeed, if A,B ∈ D withA 6= B, then there exists i0 ∈ 1, . . . , N +M such that A ⊆ Ai0 and B ⊆ Ac

i0. Next,

we consider i ∈ 1, . . . , N +M and x ∈ Ai. Then, for each j ∈ 1, . . . , N +M, eitherx ∈ Aj or x ∈ Ac

j, implying

Ai =⋃

N+M⋂

j=1

Mj : Mj = Ai ∧ ∀j∈1,...,N+M

Mj ∈ Aj, Acj.

Thus, if C1, . . . , CK , K ∈ N, is an enumeration of the distinct elements of D, then wehave the disjoint union

Ai =⋃

j∈1,...,K:Cj⊆Ai

Cj,

implying

f =K∑

i=1

γi χCi, where γi =

∑

j∈1,...,N:Ci⊆Aj

αj =∑

j∈1,...,M:Ci⊆Bj

βj

and

N∑

i=1

αi µ(Ai) =N∑

i=1

αi

∑

j∈1,...,K:Cj⊆Ai

µ(Cj) =K∑

j=1

γj µ(Cj)

=M∑

i=1

βi∑

j∈1,...,K:Cj⊆Bi

µ(Cj) =M∑

i=1

βi µ(Bi),

proving (2.3b).

Lemma 2.3. Let (X,A, µ) be a measure space.

(a) For each A ∈ A, one has∫AχA dµ = µ(A).

(b) For each f, g ∈ S+(A) and each α, β ∈ R+0 , one has

∫

X

(αf + βg) dµ = α

∫

X

f dµ + β

∫

X

g dµ .

(c) For each f, g ∈ S+(A), one has

f ≤ g ⇒∫

X

f dµ ≤∫

X

g dµ .

2 INTEGRATION 64

Proof. (a) is immediate from Def. 2.1.

(b): If f =∑N

i=1 αi χAiand g =

∑Mi=1 βi χBi

with N,M ∈ N; α1, . . . , αN , β1, . . . , βM ≥ 0;and A1, . . . , AN , B1, . . . , BM ∈ A, then

αf + βg =N∑

i=1

ααi χAi+

M∑

i=1

ββi χBi,

implying

α

∫

X

f dµ + β

∫

X

g dµ =N∑

i=1

ααi µ(Ai) +M∑

i=1

ββi µ(Bi) =

∫

X

(αf + βg) dµ ,

proving (b).

(c): If f ≤ g, then g − f ∈ S+(A). Thus, (b) applies to g = f + (g − f), yielding

∫

X

g dµ =

∫

X

f dµ +

∫

X

(g − f) dµ ≥∫

X

f dµ ,

as desired.

Note that, so far, we have not used the σ-additivity of the measure µ – it would havesufficed to take µ to be a content on an algebra. However, this will change in thefollowing section.

2.1.2 Nonnegative Functions

As for nonnegative simple functions above, the definition of the Lebesgue integral fornonnegative measurable functions also makes use of representations of the integrand,where, once again, a subsequent lemma (Lem. 2.5) then states that the value of theintegral does actually not depend on the representation of the integrated function.

Definition 2.4. Let (X,A, µ) be a measure space. Moreover, let f : X −→ [0,∞] bemeasurable (i.e. f ∈ M+(A)), where (φi)i∈N is a sequence in S+(A) such that φi ↑ f .Then ∫

X

f dµ :=

∫

X

f(x) dµ(x) := limi→∞

∫

X

φi dµ ∈ [0,∞] (2.4)

is called the Lebesgue integral of f over X with respect to µ (note that the limit in (2.4)exists, since

∫Xφ dµ ≥

∫Xψ dµ for simple functions φ ≥ ψ due to Lem. 2.3(c)).

Lemma 2.5. Let (X,A, µ) be a measure space.

(a) If (ui)i∈N is an increasing sequence in S+(A) and v ∈ S+(A), then

v ≤ limi→∞

ui ⇒∫

X

v dµ ≤ limi→∞

∫

X

ui dµ . (2.5)

2 INTEGRATION 65

(b) The value of the integral in (2.4) does not depend on the representation of the non-negative measurable function f ∈ M+(A): If (φi)i∈N and (ψi)i∈N both are sequencesin S+(A) with φi ↑ f and ψi ↑ f , then

limi→∞

∫

X

φi dµ = limi→∞

∫

X

ψi dµ . (2.6)

In particular, (2.2) and (2.4) are consistently defined.

Proof. (a): Since (ui)i∈N is increasing, limi→∞∫Xui dµ exists in [0,∞] due to Lem.

2.3(c). Fix β ∈]1,∞[ and, for each n ∈ N, define Bn := βun ≥ v. The measurabilityof v and un implies Bn ∈ A. Let x ∈ X. If v(x) = 0, then x ∈ Bn for each n ∈ N.If v(x) > 0, then v(x) ≤ limi→∞ ui(x) and β > 1 yield the existence of N ∈ N suchthat x ∈ Bn for each n ≥ N . Thus, Bn ↑ X (since (ui)i∈N is increasing). We now writev =

∑Ni=1 αi χAi

with αi ∈ R+0 , Ai ∈ A, N ∈ N, and use µ being continuous from below

to compute

∫

X

v dµ =N∑

i=1

αiµ(Ai) = limn→∞

N∑

i=1

αiµ(Ai ∩ Bn) = limn→∞

∫

X

v · χBndµ

Lem. 2.3(c)

≤ limn→∞

∫

X

β un · χBndµ

Lem. 2.3(c)

≤ β limn→∞

∫

X

un dµ . (2.7)

As (2.7) holds for each β > 1, it must still hold for β = 1, proving (2.5).

(b): For each k ∈ N, we have φk ≤ f = limi→∞ ψi. Thus, by (a),

∀k∈N

∫

X

φk dµ ≤ limi→∞

∫

X

ψi dµ ,

implying

limk→∞

∫

X

φk dµ ≤ limi→∞

∫

X

ψi dµ . (2.8)

Interchanging the roles of φi and ψi above, yields (2.8) with the inequality reversed,proving (2.6).

Lemma 2.6. Let (X,A, µ) be a measure space.

(a) For each f ∈ M+(A), one has

∫

X

f dµ = 0 ⇔ µ(f > 0

)= 0.

(b) For each f, g ∈ M+(A) and each α, β ∈ [0,∞], one has

∫

X

(αf + βg) dµ = α

∫

X

f dµ + β

∫

X

g dµ .

2 INTEGRATION 66

(c) For each f, g ∈ M+(A), one has

f ≤ g ⇒∫

X

f dµ ≤∫

X

g dµ .

(d) For each f ∈ M+(A), one has

∫

X

f dµ = sup

∫

X

φ dµ : φ ∈ S+(A), φ ≤ f

.

Proof. (a): Choose a sequence (φn)n∈N in S+(A) such that φn ↑ f . Set A := f > 0.As f is measurable, we have A ∈ A. If µ(A) > 0, then, letting An := f > 1

n, we have

An ↑ A and, thus, µ(An0) > 0 for some n0 ∈ N. In consequence,

∫

X

f dµ = limn→∞

∫

X

φn dµLem. 2.5(a)

≥ limn→∞

∫

X

1

nχAn

dµ ≥ 1

n0

µ(An0) > 0.

showing ∞ ·∫Xf dµ = ∞ =

∫X∞ · f dµ . If µ(A) = 0, then µ(φn > 0) ≤ µ(A) = 0

for each n ∈ N, implying∫Xf dµ = limn→∞

∫Xφn dµ = 0.

(b): Choose sequences (φi)i∈N and (ψi)i∈N in S+(A) such that φi ↑ f and ψi ↑ f . Then(φi + ψi) ↑ (f + g), implying

∫

X

(f + g) dµ = limi→∞

∫

X

(φi + ψi) dµLem. 2.3(b)

= limi→∞

∫

X

φi dµ + limi→∞

∫

X

ψi dµ

=

∫

X

f dµ +

∫

X

g dµ .

Similarly, if α ∈ R+0 , then αφi ∈ S+(A) for each i ∈ N, (αφi) ↑ (αf), and

∫

X

α f dµ = limi→∞

∫

X

αφi dµLem. 2.3(b)

= α limi→∞

∫

X

φi dµ = α

∫

X

f dµ .

It remains to consider the case α = ∞. As in the proof of (a), set A := f > 0. Thenn · χA ↑ ∞ · χA, implying

∫

X

∞ · f dµ = limn→∞

(n · µ(A)) =∞ for µ(A) > 0,

0 for µ(A) = 0.

If µ(A) > 0, then,∫Xf dµ > 0 by (a), showing ∞ ·

∫Xf dµ = ∞ =

∫X∞ · f dµ . If

µ(A) = 0, then∫Xf dµ = 0 by (a). Thus, using (A.2e), ∞·

∫Xf dµ = 0 =

∫X∞· f dµ .

(c): If f ≤ g, then g − f ∈ M+(A). Thus, (b) applies to g = f + (g − f), yielding∫

X

g dµ =

∫

X

f dµ +

∫

X

(g − f) dµ ≥∫

X

f dµ ,

as desired.

(d): While “≤” is immediate from Def. 2.4, “≥” is due to (c).

2 INTEGRATION 67

Theorem 2.7 (Monotone Convergence). If (X,A, µ) is a measure space and (fi)i∈N isan increasing sequence in M+(A), then

∫

X

(limi→∞

fi

)dµ = lim

i→∞

∫

X

fi dµ . (2.9)

Proof. Letting f := limi→∞ fi, we have fi ↑ f and f ∈ M+(A) by Th. 1.82(a). Now “≥”in (2.9) is due to Lem. 2.6(c). To prove the remaining inequality, consider φ ∈ S+(A),φ ≤ f , and β ∈]1,∞[. Then, for each n ∈ N, we obtain Bn := βfn ≥ φ ∈ A andφn := φ · χBn

∈ S+(A). Also, clearly, Bn ↑ X and φn ↑ φ. Then Lem. 2.5(a) yields

∫

X

φ dµ ≤ limn→∞

∫

X

φ · χBndµ ≤ β lim

n→∞

∫

X

fn dµ

and∫Xφ dµ ≤ limn→∞

∫Xfn dµ , since β > 1 was arbitrary. Now Lem. 2.6(d) yields

“≤” in (2.9) and completes the proof.

Corollary 2.8. If (X,A, µ) is a measure space and (gi)i∈N is a sequence in M+(A),then ∫

X

( ∞∑

i=1

gi

)dµ =

∞∑

i=1

∫

X

gi dµ . (2.10)

Proof. Letting, for each n ∈ N, fn :=∑n

i=1 gi, (2.10) is precisely (2.9).

Example 2.9. (a) Theorem 2.7 does, in general, not hold without the monotonicityassumption of the fn: Consider (X,A, µ) := (R,B1, β1) with fn := 1

nχ[0,n]. Then

fn → 0 (even uniformly), but

∫

X

fn dµ =1

n· n = 1 6→ 0 =

∫

X

0 dµ .

(b) Let X := N, A := P(X), and let µ be counting meaure. Then M+(A) =F(N, [0,∞]) = [0,∞]N (all functions are measurable), and we claim that

∀f∈M+(A)

∫

X

f dµ =∞∑

n=1

f(n) :

Indeed, letting gn := f(n) · χn and using Cor. 2.8, we have

∫

X

f dµ =∞∑

n=1

∫

X

gn dµ =∞∑

n=1

(f(n) · µ(n)

)=

∞∑

n=1

f(n).

2.1.3 Integrable Functions

We are already familiar with the notation K to denote either R or C. Similarly, it willsometimes be useful to treat R and C in parallel, giving rise to the following notation:

2 INTEGRATION 68

Notation 2.10. We use the symbol K to denote either R or C; we write (K, B) todenote either (R,B) or (C,B2).

Definition 2.11. Let (X,A, µ) be a measure space and let f : X −→ K be measurable.

(a) f is called integrable (or, more precisely, µ-integrable) if, and only if,

∫

X

(Re f)+ dµ <∞,

∫

X

(Re f)− dµ <∞,

∫

X

(Im f)+ dµ <∞,

∫

X

(Im f)− dµ <∞.

(2.11)

For integrable functions f ,∫

X

f dµ :=

∫

X

f(x) dµ(x)

:=

∫

X

(Re f)+ dµ −∫

X

(Re f)− dµ

+ i

∫

X

(Im f)+ dµ − i

∫

X

(Im f)− dµ ∈ K (2.12)

is called the Lebesgue integral of f over X with respect to µ.

(b) Let A ∈ A. If f ∈ M+(A) or f is integrable, then define

∫

A

f dµ :=

∫

A

f(x) dµ(x) :=

∫

X

f χA dµ . (2.13)

Remark 2.12. Let (X,A, µ) be a measure space.

(a) A measurable function f : X −→ R is integrable if, and only if, both f+ and f−

are integrable and, in that case,∫

X

f dµ =

∫

X

f+ dµ −∫

X

f− dµ .

(b) A measurable function f : X −→ C is integrable if, and only if, both Re f andIm f are integrable and, in that case,

∫

X

f dµ =

∫

X

Re f dµ + i

∫

X

Im f dµ ,

Re

(∫

X

f dµ

)=

∫

X

Re f dµ , Im

(∫

X

f dµ

)=

∫

X

Im f dµ .

Lemma 2.13. Let (X,A, µ) be a measure space, let A ∈ A, and let f : X −→ K.

(a) f · χA is A-measurable if, and only if, f A is (A|A)-measurable.

2 INTEGRATION 69

(b) It is f · χA ∈ M+(A) if, and only if, f A∈ M+(A|A), and, in that case,

∫

A

f dµ =

∫

X

f · χA dµ =

∫

A

f A d(µA|A) (2.14)

(note that we can restrict µ to A|A, since A ∈ A).

(c) It is f · χA µ-integrable if, and only if, f A is (µA|A)-integrable, and, in that case,(2.14) holds.

Proof. As A ∈ A, we have A|A = C ∈ A : C ⊆ A.(a): If B ∈ B, then

(f A)−1(B) = A ∩ f−1(B), (f · χA)

−1(B) =

Ac ∪ (A ∩ f−1(B)) if 0 ∈ B,

A ∩ f−1(B) if 0 /∈ B.

Since Ac ∈ A, we obtain (f · χA)−1(B) ∈ A if, and only if, (f A)

−1(B) ∈ A|A, proving(a).

(b): Since f · χA ≡ 0 on Ac, the equivalence claimed in (b) is clear. If f · χA ∈ M+(A),then let (φn) be a sequence in S+(A) with φn ↑ (f · χA). Then (φn A) ↑ (f A) (withφnA∈ S+(A|A)) and

∫

A

f dµ =

∫

X

f · χA dµ = limn→∞

∫

X

φn dµφnAc≡0= lim

n→∞

∫

A

(φnA) d(µA|A)

=

∫

A

f A d(µA|A),

proving (b).

(c) is a direct consequence of (b).

Theorem 2.14. Let (X,A, µ) be a measure space, f : X −→ K. Then the followingstatements are equivalent:

(i) f is integrable.

(ii) Re f and Im f are both integrable.

(iii) (Re f)+, (Re f)−, (Im f)+, (Im f)− all are integrable.

(iv) There exist integrable functions p, q, r, s ∈ M+(A) such that f = p− q + i (r − s).

(v) f is measurable and there exists an integrable g ∈ M+(A) such that |f | ≤ g (i.e.|f | is dominated by g).

(vi) f is measurable and |f | is integrable.

2 INTEGRATION 70

Proof. The equivalences of (i) – (iii) are immediate from Def. 2.11(a).

(iii) implies (iv) by setting p := (Re f)+, q := (Re f)−, r := (Im f)+, s := (Im f)−.

(iv) ⇒ (v): If f = p − q + i (r − s) with integrable p, q, r, s ∈ M+(A), then |f | ≤ g :=p+ q + r + s, where g is integrable by Lem. 2.6(b).

(v) implies (vi), since∫X|f | dµ ≤

∫Xg dµ <∞.

(vi) implies (iii): Let h ∈ (Re f)+, (Re f)−, (Im f)+, (Im f)−. Then h is measurable,since f is measurable, and h is integrable, since

∫Xh dµ ≤

∫X|f | dµ <∞.

Example 2.15. It can happen that |f | is integrable, but f is not: Let (X,A, µ) :=([0, 1],B1, β1), A ⊆ X, A /∈ B1. Then f := χA − χAc is not measurable (in particular,not integrable), but |f | ≡ 1 is integrable.

Theorem 2.16. Let (X,A, µ) be a measure space, let f, g : X −→ K be integrable, letA,B ∈ A, and let α, β ∈ K.

(a) If f : X −→ R is integrable, then

µ(|f | = ∞

)= 0.

(b) The Lebesgue integral is linear: αf + βg is integrable and∫

A

(αf + βg) dµ = α

∫

A

f dµ + β

∫

A

g dµ . (2.15a)

(c) If A,B are disjoint, then∫

A∪Bf dµ =

∫

A

f dµ +

∫

B

f dµ (2.15b)

(this also holds for nonintegrable f ∈ M+(A)).

(d) The Lebesgue integral is isotone (here, let f, g be R-valued):

f ≤ g ⇒∫

A

f dµ ≤∫

A

g dµ . (2.15c)

(e) The Lebesgue integral satisfies the triangle inequality:∣∣∣∣∫

A

f dµ

∣∣∣∣ ≤∫

A

|f | dµ . (2.15d)

(f) Mean Value Theorem for Integration: Let f be R-valued and assume there existnumbers m,M ∈ R such that m ≤ f ≤M on A. Moreover, let p : X −→ [0,∞] besuch that χA p and χA fp are integrable. Then

m

∫

A

p dµ ≤∫

A

fp dµ ≤M

∫

A

p dµ . (2.15e)

2 INTEGRATION 71

In particular, if f(A) = [m,M ] (e.g., if f is R-valued and continuous, A is compactand connected, m = min f(A), M := max f(A)), then

∃ξ∈A

∫

A

fp dµ = f(ξ)

∫

A

p dµ . (2.15f)

Returning to a general integrable f , if p ≡ 1 and µ(A) < ∞, then we obtain thetheorem’s classical form:

mµ(A) ≤∫

A

f dµ ≤M µ(A), (2.15g)

where the theorem’s name comes from the fact that, for 0 < µ(A) <∞,

−∫

A

f dµ := µ(A)−1

∫

A

f dµ (2.15h)

is sometimes referred to as the mean value of f on A.

(g) If f, g : X −→ R are integrable, then max(f, g) and min(f, g) are integrable.

Proof. (a): Letting A∞ := |f | = ∞, we have A∞ ∈ A and h := ∞ · χA∞ ≤ |f |. Sinceh, |f | ∈ M+(A), we apply Lem. 2.6(c) to obtain ∞·µ(A∞) =

∫Xh dµ ≤

∫X|f | dµ <∞,

implying µ(A∞) = 0.

(b): It suffices to consider A = X due to Lem. 2.13. Since |αf + βg| ≤ |α||f | + |β||g|,αf + βg is integrable by Th. 2.14(v). If f, g are R-valued, then h := f + g = f+ − f− +g+ − g−, i.e. h+ + f− + g− = h− + f+ + g+. Thus, due to Lem. 2.6(b),

∫

X

h+ dµ +

∫

X

f− dµ +

∫

X

g− dµ =

∫

X

h− dµ +

∫

X

f+ dµ +

∫

X

g+ dµ . (2.16)

Due to integrability, all integrals in (2.16) are finite, implying

∫

X

(f + g) dµ =

∫

X

h+ dµ −∫

X

h− dµ

=

∫

X

f+ dµ +

∫

X

g+ dµ −∫

X

f− dµ −∫

X

g− dµ =

∫

X

f dµ +

∫

X

g dµ .

If f, g are C-valued, then, since Re(f + g) = Re f + Re g and Im(f + g) = Im f + Im g,∫X(f + g) dµ =

∫Xf dµ +

∫Xg dµ follows from the R-valued case. If f is R-valued and

α ≥ 0, then (αf)+ = αf+, (αf)− = αf− and we obtain∫Xαf dµ = α

∫Xf dµ due to

Lem. 2.6(b). If α < 0, then (αf)+ = |α|f−, (αf)− = |α|f+, and one computes

∫

X

αf dµ = |α|(∫

X

f− dµ −∫

X

f+ dµ

)= α

∫

X

f dµ .

2 INTEGRATION 72

Finally, using the R-valued case, one computes for C-valued f and α ∈ C,∫

X

(αf) =

∫

X

(ReαRe f − Imα Im f) dµ + i

∫

X

(Reα Im f + ImαRe f) dµ

= Reα

∫

X

Re f dµ − Imα

∫

X

Im f dµ + i Reα

∫

X

Im f dµ + i Imα

∫

X

Re f dµ

= α

∫

X

f dµ ,

proving (b).

(c): Since fχA∪B = fχA + fχB, the case f ∈ M+(A) follows from Lem. 2.6(b) and theintegrable case follows from (b).

(d): If f ≤ g, then g − f ≥ 0. Thus, (b) applies to g = f + (g − f), yielding∫

A

g dµ =

∫

A

f dµ +

∫

A

(g − f) dµ ≥∫

A

f dµ ,

as desired.

(e) follows from (d): Due to the existence of polar coordinates for complex numbers,there exists ζ ∈ K with |ζ| = 1 and

∣∣∣∣∫

A

f dµ

∣∣∣∣ = ζ

∫

A

f dµ =

∫

A

ζf dµ .

In the above equality, all terms are real numbers, implying∣∣∣∣∫

A

f dµ

∣∣∣∣ = Re

(∫

A

ζf dµ

)=

∫

A

Re(ζf) dµ ≤∫

A

|f | dµ ,

proving (e).

(f): Since mp ≤ fp ≤Mp, we compute

m

∫

A

p dµ(d)

≤∫

A

fp dµ(d)

≤ M

∫

A

p dµ .

If∫Ap dµ = 0, then (2.15e) implies

∫Afp dµ = 0 and (2.15f) is immediate. It remains

to consider that∫Ap dµ > 0. If f(A) = [m,M ], then (2.15e) shows there is ξ ∈ A such

that f(ξ) =∫

Afp dµ

∫

Ap dµ

, i.e. (2.15f) holds.

(g) follows from Th. 2.14(v), since max(f, g) and min(f, g) are measurable and both|max(f, g)| and |min(f, g)| are dominated by |f |+ |g|.

2.2 Null Sets, Properties Holding Almost Everywhere

Definition 2.17. Let (X,A, µ) be a measure space. Let P (x) be a predicate (informally,a property) of x ∈ X, i.e. a sentence that, for each x ∈ X, is either true or false. Then we

2 INTEGRATION 73

say that P holds µ-almost everywhere (abbreviated µ-a.e.) if, and only if, there exists aµ-null set N ∈ A such that P (x) is true for each x ∈ N c. One writes just a.e. insteadof µ-a.e. if the measure µ is understood.

—

Here are some examples of potentially useful properties that might hold µ-almost ev-erywhere: Functions f, g : X −→ Y are equal µ-a.e. if, and only if, there exists a µ-nullset N ∈ A such that f Nc= gNc . A function f : X −→ R is finite µ-a.e. if, and onlyif, there exists a µ-null set N such that f(N c) ⊆ R. The sequence (fn)n∈N of functionsfn : X −→ K converges a.e. to f : X −→ K if, and only if, there exists a µ-null set Nsuch that limn→∞ fnNc= f Nc .

Note that “P holds µ-a.e.” does not imply that M := x ∈ X : P (x) is false ∈ A. Itmerely implies that there exists a µ-null set N ∈ A such that M ⊆ N (of course, if µ iscomplete, then we do know M ∈ A).

Theorem 2.18. Let (X,A, µ) be a measure space.

(a) If f, g : X −→ R are integrable or nonnegative measurable, then f ≤ g µ-a.e.implies ∫

X

f dµ ≤∫

X

g dµ . (2.17a)

In particular, if f = g µ-a.e., then

∫

X

f dµ =

∫

X

g dµ . (2.17b)

(b) If f, g : X −→ K are measurable, g is integrable, and f = g µ-a.e., then f isintegrable and (2.17b) holds.

(c) If f : X −→ K is measurable and there exists an integrable g ∈ M+(A) such that|f | ≤ g µ-a.e., then f is integrable.

(d) If f, g : X −→ R are integrable and

∀A∈A

∫

A

f dµ ≤∫

A

g dµ , (2.18)

then f ≤ g µ-a.e. In particular, if (2.18) holds with equality, then f = g µ-a.e.

Proof. (a): As f, g are both measurable, we have N := f > g ∈ A with µ(N) = 0.Thus, ∫

X

f+ dµ =

∫

N

f+ dµ +

∫

Nc

f+ dµ ≤ 0 +

∫

X

g+ dµ .

Analogously, we obtain∫Xf− dµ ≥

∫Xg− dµ , proving (a).

2 INTEGRATION 74

(b): One applies (a) to the positive and negative parts of the real and imaginary partsof f and g: If g is integrable, then (Re g)+ is integrable, implying

∫X(Re f)+ dµ =∫

X(Re g)+ dµ <∞ by (a), i.e. (Re f)+ is integrable etc. Now (2.17b) also follows.

(c): As in the proof of (b), one obtains (Re f)+, (Re f)−, (Im f)+, (Im f)− to be inte-grable: If g is integrable, then

∫X(Re f)+ dµ ≤

∫X|f | dµ ≤

∫Xg dµ < ∞ by (a), i.e.

(Re f)+ is integrable etc.

(d): The measurability of f, g implies M := f > g ∈ A as well as Mn := f >g + 1

n ∈ A for each n ∈ N. We estimate∫

Mn

f dµ ≥∫

Mn

(g +

1

n

)dµ =

∫

Mn

g dµ +µ(Mn)

n

(2.18)

≥∫

Mn

f dµ +µ(Mn)

n,

implying µ(Mn) = 0 for each n ∈ N. SinceMn ↑M , we obtain µ(M) = 0 as claimed.

2.3 Convergence Theorems

2.3.1 Fatou’s Lemma and Dominated Convergence

Proposition 2.19 (Fatou’s Lemma). Let (X,A, µ) be a measure space. Then, for eachsequence (fn)n∈N in M+(A), one has

∫

X

limn→∞

fn dµ ≤ lim infn→∞

∫

X

fn dµ . (2.19)

Proof. Letting f := limn→∞ fn and, for each n ∈ N, gn := infk≥n fk, we obtain f, gn ∈M+(A) by Th. 1.82(a). Clearly, gn ↑ f , such that the monotone convergence Th. 2.7applies, yielding ∫

X

f dµ = limn→∞

∫

X

gn dµ .

Since (∀

k≥ngn ≤ fk

)⇒∫

X

gn dµ ≤ infk≥n

∫

X

fk dµ ,

we obtain ∫

X

f dµ ≤ limn→∞

infk≥n

∫

X

fk dµ = lim infn→∞

∫

X

fn dµ ,

which establishes the case.

Theorem 2.20 (Dominated Convergence). Let (X,A, µ) be a measure space. Let f, fn :X −→ K be measurable functions, n ∈ N, such that f = limn→∞ fn µ-a.e. Moreover,assume the fn to be dominated by an integrable function, i.e. assume there exists anintegrable g ∈ M+(A) such that, for each n ∈ N, |fn| ≤ g holds µ-a.e. Then all fn andf are integrable with

limn→∞

∫

X

fn dµ =

∫

X

f dµ (2.20a)

and

limn→∞

∫

X

|fn − f | dµ = 0. (2.20b)

2 INTEGRATION 75

Proof. The fn are integrable by Th. 2.18(c) and then |f | = limn→∞ |fn| ≤ g µ-a.e.implies f to be integrable as well. Then we know that outside of a µ-null set N ∈ A,all the functions g, f, fn, n ∈ N, must be K-valued and limn→∞ fn = f . Since all theinvolved integrals over N equal 0, we may assume, without loss of generality, that N = ∅.Then, for each n ∈ N, gn := |f |+ g − |fn − f | ∈ M+(A), and Fatou (Prop. 2.19) yields

∫

X

(|f |+ g) dµ =

∫

X

limn→∞

gn dµ ≤ lim infn→∞

∫

X

gn dµ

=

∫

X

(|f |+ g) dµ − lim supn→∞

∫

X

|fn − f | dµ .

Since∫X(|f |+ g) dµ ∈ R+

0 , we obtain (2.20b). As

∀n∈N

∣∣∣∣∫

X

fn dµ −∫

X

f dµ

∣∣∣∣ ≤∫

X

|fn − f | dµ ,

(2.20a) is also proved.

Note that we have already seen in Ex. 2.9(a), in the context of the monotone convergencetheorem, that one can not expect (2.20a) to hold without additional hypotheses (suchas monotonicity or domination).

Corollary 2.21. Let (X,A, µ) be a measure space.

(a) Let f, fn : X −→ K be measurable functions, n ∈ N, such that f =∑∞

n=1 fn µ-a.e.Moreover, assume there exists an integrable g ∈ M+(A) such that, for each n ∈ N,|∑n

k=1 fk| ≤ g holds µ-a.e. Then all fn and f are integrable with

∫

X

f dµ =∞∑

n=1

∫

X

fn dµ .

(b) Let f : X −→ K be integrable and A,An ∈ A such that A =⋃∞

n=1An and, fork 6= l, µ(Ak ∩ Al) = 0. Then

∫

A

f dµ =∞∑

n=1

∫

An

f dµ .

Proof. (a): The integrability of the fn follows since fn =∑n

k=1 fk−∑n−1

k=1 fk, where bothsums on the right-hand side are integrable by hypothesis. Now the claimed formula for∫Xf dµ is obtained from applying Th. 2.20 with fn replaced by

∑nk=1 fk.

(b): Since |f | = ∞ and⋃

(k,l)∈N2, k 6=l(Ak ∩ Al) are both µ-null sets, we may assume,without loss of generality, that A is the disjoint union of the An and that f · χA isK-valued. Then f · χA =

∑∞n=1 f · χAn

with |∑nk=1 f · χAk

| ≤∑nk=1 |f | · χAk

≤ |f | · χA,i.e. (b) now follows from (a).

2 INTEGRATION 76

2.3.2 Parameter-Dependent Integrals: Continuity, Differentiation

Theorem 2.22. Let (Z, d) be a metric space and let (X,A, µ) be a measure space. Letf : Z ×X −→ K. Fix z0 ∈ Z and assume f has the following properties:

(a) For each z ∈ Z, the function x 7→ f(z, x) is integrable.

(b) For µ-a.e. x ∈ X, the function z 7→ f(z, x) is continuous in z0.

(c) There exists ǫ > 0 and an integrable h : X −→ [0,∞] such that

∀z∈Bǫ(z0)

|f(z, x)| ≤ h(x) for µ-a.e. x ∈ X (2.21)

(i.e. f is uniformly dominated by an integrable h in some neighborhood of z0).

Then the function

φ : Z −→ K, φ(z) :=

∫

X

f(z, x) dµ(x) , (2.22)

is continuous in z0.

Proof. Let (zm)m∈N be a sequence in Bǫ(z0) such that limm→∞ zm = z0. According to(a), this gives rise to a sequence (gm)m∈N of integrable functions

gm : X −→ K, gm(x) := f(zm, x),

such that, using (b), limm→∞ gm = g µ-a.e., where

g : X −→ K, g(x) := f(z0, x).

Since, by (c), all gm are dominated by the integrable function h, we can apply thedominated convergence Th. 2.20 to obtain

limm→∞

φ(zm) = limm→∞

∫

X

f(zm, x) dµ(x) = limm→∞

∫

X

gm(x) dµ(x)Th. 2.20=

∫

X

g(x) dµ(x)

=

∫

X

f(z0, x) dµ(x) = φ(z0),

proving φ is continuous at z0.

Theorem 2.23. Let I ⊆ R be a nontrivial interval and let (X,A, µ) be a measure space.Let f : I ×X −→ K. Fix t0 ∈ I and assume f has the following properties:

(a) For each t ∈ I, the function x 7→ f(t, x) is integrable.

(b) For µ-a.e. x ∈ X, the function fx : I −→ K, fx(t) := f(t, x), is differentiable at t0.

(c) There exists an integrable h : X −→ [0,∞] such that

∀t∈I\t0

∣∣∣∣f(t, x)− f(t0, x)

t− t0

∣∣∣∣ ≤ h(x) for µ-a.e. x ∈ X. (2.23)

2 INTEGRATION 77

Then the function

φ : I −→ K, φ(t) :=

∫

X

f(t, x) dµ(x) , (2.24)

is differentiable at t0 (this means the existence of the one-sided derivative if t0 ∈ ∂I),x 7→ f ′(t0, x) (with value 0 on the µ-null set N ⊆ X, where f ′(t0, x) does not exist), isintegrable, and

φ′(t0) =

∫

X

f ′(t0, x) dµ(x) . (2.25)

Proof. Let (tm)m∈N be a sequence in I \ t0 such that limm→∞ tm = t0. According to(a), this gives rise to a sequence (gm)m∈N of integrable functions

gm : X −→ K, gm(x) :=f(tm, x)− f(t0, x)

tm − t0,

such that, using (b), limm→∞ gm = g µ-a.e., where

g : X −→ K, g(x) :=

f ′(t0, x) if x 6∈ N,

0 if x ∈ N.

Since, by (c), all gm are dominated by the integrable function h, we can apply thedominated convergence Th. 2.20 to obtain the integrability of the gm and g, as well as

limm→∞

φ(tm, x)− φ(t0, x)

tm − t0= lim

m→∞

∫

X

gm(x) dµ(x)Th. 2.20=

∫

X

g(x) dµ(x)

=

∫

X

f ′(t0, x) dµ(x) ,

proving φ is differentiable at t0 with derivative according to (2.25).

Corollary 2.24. The statement of Th. 2.23 remains true if (b) and (c) are replaced by

(b′) For µ-a.e. x ∈ X, the function fx : I −→ K, fx(t) := f(t, x), is differentiable on I.

(c′) There exists an integrable h : X −→ [0,∞] such that, for µ-a.e. x ∈ X,

∀t∈I

|f ′x(t)| ≤ h(x)

(note that, this time, the null set, where the condition does not hold, must notdepend on t).

Proof. Since (b′) immediately implies (b), it remains to verify that (c′) implies (c). Tothis end, let N be a µ-null set such that, for each x ∈ X \ N , |f ′

x(t)| ≤ h(x) holds foreach t ∈ I. Then, for each x ∈ X \N , by the mean value theorem of diffential calculus[Phi16a, Th. 9.18], for each t ∈ I \ t0, there exists τ(t, x) ∈ I such that

∣∣∣∣f(t, x)− f(t0, x)

t− t0

∣∣∣∣ =∣∣f ′

x

(τ(t, x)

)∣∣ ≤ h(x),

implying the validity of (2.23).

2 INTEGRATION 78

2.4 Riemann versus Lebesgue

In the present section, we investigate the obvious question regarding the relationsshipbetween the Riemann integral of [Phi16a, Sec. 10] and the Lebesgue integral for L1-measurable functions f : [a, b] −→ K.

Definition 2.25. Let n ∈ N, A ∈ Ln. We call f : A −→ K Lebesgue integrable if, andonly if, f is (λnA)-integrable.

—

We will now see in Th. 2.26(a) that every Riemann integrable function is also Lebesgueintegrable with identical values for the integral. Thus, the Lebesgue integral can beseen as a generalization of the Riemann integral. In [Phi16a, Sec. 10], we only studiedthe Riemann integral for functions on one-dimensional intervals. However, an analogousconstruction can be performed for functions on intervals in Rn, n ∈ N. It is provided inSec. D of the Appendix. While the following Th. 2.26 is provided for arbitrary n ∈ N,one can understand its contents and structure already from the case n = 1 (in particular,without having studied Sec. D). The main difference between n = 1 and n > 1 is thatthe notation gets somewhat more complicated.

As a consequence of Th. 2.26(a), for functions on R, one will, in most cases, still computeLebesgue integrals as Riemann integrals, using the techniques introduced in [Phi16a,Sec. 10]. The most important technique for computing integrals on (subsets of) Rn isthe Fubini Th. 2.36(b) below, which, if it applies, allows to compute an integral of afunction on Rn by evaluating n one-dimensional integrals.

Theorem 2.26. Let n ∈ N and I := [a, b] ⊆ Rn, where a, b ∈ Rn with a < b. Moreover,let f : I −→ C be bounded.

(a) If f is Riemann integrable, then f is Lebesgue integrable and∫

I

f dλn = R-

∫

I

f, (2.26)

where R-∫

denotes the Riemann integral.

(b) f is Riemann integrable if, and only if, the set of points where f is discontinuousis a λn-null set (i.e. if, and only if, f is continuous λn-a.e.).

Proof. We begin with some general preparations: As we may consider Re f and Im f ,we consider f to be real-valued without loss of generality. For each N ∈ N and eachk ∈ 1, . . . , n, consider the partition ∆N

k = (xNk,0, . . . , xNk,2N ) of [ak, bk], x

Nk,j := ak +

j · (bk − ak)2−N for each j ∈ 0, . . . , 2N, into the disjoint intervals INk,1 := [xNk,0, x

Nk,1],

INk,2 :=]xNk,1, xNk,2], . . . , I

Nk,2N :=]xNk,2N−1, x

Nk,2N ], and the resulting partition ∆N of I (cf.

Def. D.2). Analogous to Def. D.2, for each (k1, . . . , kn) ∈ P (∆N) = 1, . . . , 2Nn, define

IN(j1,...,jn) :=n∏

k=1

INk,jk ,

2 INTEGRATION 79

which is as in (D.4), except that we are using the (disjoint) halfopen intervals here.Then, analogous to Rem. D.3, we obtain the (now disjoint) unions

I =⋃

p∈P (∆N )

INp .

According to Def. D.4, if we let, for each p ∈ P (∆N),

mNp := inff(x) : x ∈ cl(INp ), MN

p := supf(x) : x ∈ cl(INp ),

then we obtain the lower and upper Riemann sums

r(∆N , f) =∑

p∈P (∆N )

mNp · λn(INp ), R(∆N , f) =

∑

p∈P (∆N )

MNp · λn(INp ).

Moreover, defining the simple functions

gN : I −→ R, gN :=∑

p∈P (∆N )

mNp χINp

,

hN : I −→ R, hN :=∑

p∈P (∆N )

MNp χINp

,

we obtain

r(∆N , f) =

∫

I

gN dλn , R(∆N , f) =

∫

I

hN dλn .

Moreover, clearly, gN ≤ f ≤ hN , (gN)N∈N is increasing, and (hN)N∈N is decreas-ing. Thus, the pointwise limits g := limN→∞ gN and h := limN→∞ hN exist, are βn-measurable, bounded, and, hence, Lebesgue integrable. They also satisfy g ≤ f ≤ h.

(a): Suppose f is Riemann integrable. Since

∀N∈N

|gN |, |hN | ≤ |g1|+ |h1|,

we can apply the dominated convergence Th. 2.20 to compute∫

I

g dλn = limN→∞

∫

I

gN dλn = limN→∞

r(∆N , f) = R-

∫

I

f

= limN→∞

R(∆N , f) = limN→∞

∫

I

hN dλn =

∫

I

h dλn .

Thus,∫I(h− g) dλn = 0, such that h = g λn-a.e. on I by Lem. 2.6(a). Since g ≤ f ≤ h,

this also yields f = g λn-a.e. on I. In consequence, there exists a λn-null set M ∈ Ln|Isuch that f = g on M c. If B ∈ B1, then

f−1(B) = (f−1(B) ∩M c) ∪(f−1(B) ∩M) = (g−1(B) ∩M c) ∪(f−1(B) ∩M) ∈ Ln|I,

since g is measurable and λn is complete. Thus, f is measurable as well. Moreover,∫

I

f dλn =

∫

Mc

f dλn +

∫

M

f dλn =

∫

Mc

g dλn + 0 =

∫

I

g dλn = R-

∫

I

f,

2 INTEGRATION 80

proving (a).

(b): Let D := x ∈ I : f not continuous at x and let

R :=⋃

N∈N

⋃

p∈P (∆N )

∂INp

be the set of boundary points occurring in any of the considered partitions of I, which,as a countable union of λn-null sets is itself a λn-null set. We show that D ⊆ R∪g < hby proving that (R∪g < h)c = Rc∩g = h ⊆ Dc: Let x ∈ Rc∩g = h and let ǫ > 0.Note f(x) = g(x) = h(x). ChooseN ∈ N such that f(x)−ǫ < gN(x) < hN(x) < f(x)+ǫ.Since x /∈ R, there exists pN ∈ P (∆N) such that x ∈ int(INpN ). Set δ := dist(x, ∂IpN ).Then δ > 0 and, for each y ∈ Rn with ‖y − x‖ < δ, one has y ∈ INpN , implying

f(x)− ǫ < gN(x) = mNpN

≤ f(y) ≤MNpN

= hN(x) < f(x) + ǫ,

showing |f(y) − f(x)| < ǫ and the continuity of f at x, i.e. x ∈ Dc as claimed. Now,if f is Riemann integrable, then, as shown in the proof of (a), g = h λn-a.e., i.e.λn(g < h) = 0. Since D ⊆ R∪ g < h, this implies λn(D) = 0. For the converse, weshow g < h ⊆ D: Indeed, for x ∈ I with g(x) < h(x) let δ := (h(x)− g(x))/3. Then,for each ǫ > 0, there exist x1, x2 ∈ Bǫ(x) such that f(x1) ≤ g(x)+δ and f(x2) ≥ h(x)−δ,showing x ∈ D and g < h ⊆ D. Thus, if D ∈ Ln with λn(D) = 0, then g = h λn-a.e.,yielding

limN→∞

r(∆N , f) = limN→∞

∫

I

gN dλn =

∫

I

g dλn

=

∫

I

h dλn = limN→∞

∫

I

hN dλn = limN→∞

R(∆N , f),

thereby proving f to be Riemann integrable by Th. D.11.

Example 2.27. (a) Let I = [a, b] ⊆ Rn be an interval, a, b ∈ Rn, n ∈ N, a < b. TheDirichlet function

f : I −→ R, f(x) :=

0 for x ∈ I \Qn,

1 for x ∈ I ∩Qn,

is everywhere discontinuous and, thus, not Riemann integrable (cf. Ex. D.7(b) and[Phi16a, Ex. 10.7(b)]). However, f is the characteristic function of the βn-null setI ∩ Qn, i.e. f is Lebesgue integrable with

∫If dλn = 0. Now let (r1, r2, . . . ) be an

enumeration of I ∩Qn and define

∀k∈N

φk : I −→ R, φk(x) :=

1 for x ∈ r1, . . . , rk,0 otherwise.

Then each φk is Riemann integrable, as it has only finitely many discontinuities.Since, clearly, φk ↑ f , this shows that there does not exist a “Riemann measure”, i.e.no measure space (I,A, ρ) such that a function g : I −→ R is Riemann integrableif, and only if, g is ρ-integrable (otherwise, we had φk ∈ S+(A) for each k ∈ N, andf had to be ρ-integrable as well).

2 INTEGRATION 81

(b) The following example shows that there exist Riemann integrable functions thatare not Borel measurable: Let C ⊆ I := [0, 1] be the Cantor set of Def. and Rem.1.61 and let A ⊆ C be such that A /∈ B1 (such an A exists since #P(C) = #P(R),but #B1 = #R). Then f : I −→ R, f := χA is continuous on the open set I \ C.Since λ1(C) = 0, f is Riemann integrable with

∫If = 0.

—

While we saw in Th. 2.26(a) that every Riemann integrable function is Lebesgue in-tegrable, for improperly Riemann integrable functions this only remains true if theimproper Riemann integral converges absolutely:

Theorem 2.28. Let a, b, c ∈ R, a < c < b. Let I ⊆]a, b[ be one of the following threekinds of intervals: I = [c, b[, I =]a, c], or I =]a, b[. Moreover, assume f : I −→ R to belocally Riemann integrable, i.e. Riemann integrable over each compact subinterval of I.Then f is Lebesgue integrable if, and only if, |f | is improperly Riemann integrable. Inthat case, both integrals agree, i.e. (2.26) holds, where R-

∫now denotes the improper

Riemann integral.

Proof. Let m := inf I, M := sup I. Moreover, let (an)n∈N and (bn)n∈N be sequences inI such that m < an < bn < M and an ↓ m as well as bn ↑ M . If, for each n ∈ N,fn := f · χ[an,bn], then fn ↑ f on ]m,M [. Since f is locally Riemann integrable, eachfn is Lebesgue integrable by Th. 2.26(a), implying f to be Lebesgue measurable. Thus,the monotone convergence Th. 2.7 yields

limn→∞

R-

∫ bn

an

|f | = limn→∞

∫

[an,bn]

|f | dλ1 =

∫

I

|f | dλ1 .

If |f | is improperly Riemann integrable, then the left-hand side is finite, i.e. |f | (and,thus, f) is Lebesgue integrable. Conversely, if f is Lebesgue integrable, then so is|f |, i.e. the right-hand side is finite, implying |f | to be improperly Riemann integrable.Finally, if |f | is improperly Riemann integrable, then so is f by [Phi16a, Prop. 10.39(b)],implying, together with Th. 2.26(a) and the dominated convergence Th. 2.20,

R-

∫ M

m

f = limn→∞

R-

∫ bn

an

f = limn→∞

∫

[an,bn]

f dλ1 =

∫

I

f dλ1 ,

proving the validity of (2.26).

In [Phi16a, Ex. 10.40(c)], we saw that the function

f : [0,∞[−→ R, f(t) :=

(−1)n+1 for n ≤ t ≤ n+ 1

n, n ∈ N,

0 otherwise,(2.27)

is improperly Riemann integrable, but that the improper Riemann integral does notconverge absolutely. Therefore, by Th. 2.28, this function is not Lebesgue integrable,even though it is improperly Riemann integrable.

In the next example, we introduce the important gamma function, which can be seenas a continuous extension of the factorial function (cf. (2.29b) below).

2 INTEGRATION 82

Example 2.29. The gamma function is defined by

Γ : R+ −→ R+, Γ(x) :=

∫ ∞

0

e−ttx−1 dt . (2.28)

(a) The gamma function is well-defined by (2.28), i.e. the integral exists as an improperRiemann integral and as a Lebesgue integral with values in R+: Since, for eacht, x > 0, one has e−ttx−1 > 0, and fx : R+ −→ R+, fx(t) := e−ttx−1, is continuous,it suffices to show fx is dominated by some improperly Riemann integrable functiongx. In anticipation of (c) below, it will actually be useful to estimate fx uniformlyfor each x in some bounded interval. Thus, let α, β ∈ R+ such that 0 < α < β.Then, for each x ∈ [α, β],

∀0<t≤1

0 < fx(t) ≤ tx−1 ≤ tα−1,

∀t≥1

0 < fx(t) = tx−1 e−t/2 e−t/2 ≤Mβ e−t/2,

where Mβ ∈ R+ is an upper bound for the bounded function φ : [1,∞[−→ R+,φ(t) := tβ−1 e−t/2 (note limt→∞ φ(t) = 0). Thus, the function

gα,β : R+ −→ R+, gα,β(t) :=

tα−1 for t ≤ 1,

Mβ e−t/2 for t > 1,

dominates fx for each x ∈ [α, β]. It is also improperly Riemann integrable, since∫ 1

0tα−1 dt = 1

αby [Phi16a, Ex. 10.35(a)], and, if (tk)k∈N is a sequence in ]1,∞[ such

that limk→∞ tk = ∞, then

∫ ∞

1

e−t/2 dt = limk→∞

∫ tk

1

e−t/2 dt

= limk→∞

[− 2e−t/2

]tk1= lim

k→∞(2e−1/2 − 2e−tk/2) = 2e−1/2.

(b) The gamma function satisfies

∀x∈R+

Γ(x+ 1) = xΓ(x), (2.29a)

∀n∈N0

Γ(n+ 1) = n! : (2.29b)

To prove (2.29a), we fix x ∈ R+ and integrate by parts to obtain

Γ(x+ 1) =

∫ ∞

0

e−ttx dt = [−e−ttx]∞0 + x

∫ ∞

0

e−ttx−1 dt = −0 + 0 + xΓ(x),

establishing (2.29a). Since, using [Phi16a, Ex. 10.35(c)],

Γ(1) =

∫ ∞

0

e−t dt = 1, (2.30)

(2.29a) implies (2.29b) via induction on n ∈ N0.

2 INTEGRATION 83

(c) The gamma function is C∞ and

∀k∈N0

Γ(k) : R+ −→ R, Γ(k)(x) =

∫ ∞

0

(ln t)k e−ttx−1 dt : (2.31)

Using [Phi16a, Ex. 9.5(b)], we differentiate the integrand of (2.28) k times withrespect to x to obtain, for each x ∈ R+,

∀k∈N

∂kx(e−ttx−1) = (ln t)k e−ttx−1.

Thus, according to Cor. 2.24, (2.31) is proved if we can estimate each integrand of(2.31) by an integrable function, uniformly for x ∈ [α, β], where 0 < α < β as in(a). For each k ∈ N and x ∈ [α, β], we estimate

∀0<t≤1

0 < | ln t|k e−ttx−1 ≤ | ln t|k tα/2t−α/2tx−1 ≤ Cα,k tα/2−1,

∀t≥1

0 < | ln t|k e−ttx−1 = | ln t|k tx−1 e−t/2 e−t/2 ≤Mβ,k e−t/2,

where Cα,k ∈ R+ is an upper bound for the bounded function ψk : ]0, 1] −→ R+,ψ(t) := | ln t|k tα/2 (note limt→0 ψk(t) = 0), andMβ,k ∈ R+ is an upper bound for thebounded function φk : [1,∞[−→ R+, φ(t) := | ln t|k tβ−1 e−t/2 (note limt→∞ φk(t) =0). Thus, the function

hα,β,k : R+ −→ R+, hα,β,k(t) :=

Cα,k t

α/2−1 for t ≤ 1,

Mβ,k e−t/2 for t > 1,

dominates | ln t|k e−ttx−1 for each x ∈ [α, β]. That it is improperly Riemann inte-grable follows analogously to the integrability of gα,β in (a).

(d) One has the identities

∀x∈R+

Γ(x) = limn→∞

∫ n

0

(1− t

n

)n

tx−1 dt = limn→∞

nx n!

x(x+ 1) · · · (x+ n)(2.32)

and ∫ ∞

−∞e−x2

dx = Γ

(1

2

)=

√π : (2.33)

For each x ∈ R, one has 1 + x ≤ ∑∞i=0

xi

i!= ex. Thus, for each x ∈] − 1,∞[,

applying ln yields ln(1 + x) ≤ x. If 0 ≤ t < n ∈ N, then −1 < − tnand we can

apply ln(1 + x) ≤ x with x := − tnto obtain (1 − t

n)n ≤ e−t. In consequence, we

can apply the dominated convergence Th. 2.20 to obtain, for each x ∈ R+,

Γ(x) =

∫ ∞

0

e−ttx−1 dt = limn→∞

∫ ∞

0

(1− t

n

)n

tx−1 · χ]0,n[ dt

= limn→∞

∫ n

0

(1− t

n

)n

tx−1 dt ,

2 INTEGRATION 84

proving the first identity of (2.32). To obtain the second identity of (2.32), for eachn ∈ N, we now use integration by parts n times:

∫ n

0

(1− t

n

)n

tx−1 dt =

[(1− t

n

)ntx

x

]n

0

+n

nx

∫ n

0

(1− t

n

)n−1

tx dt

= . . .

= 0 +n!

nn x(x+ 1) · · · (x+ n− 1)

∫ n

0

(1− t

n

)0

tx+n−1 dt

=n!

nn x(x+ 1) · · · (x+ n)

[tx+n

]n0=

nx n!

x(x+ 1) · · · (x+ n).

The first identiy in (2.33) can be proved using the change of variables t := x2:∫ ∞

0

e−x2

dx =

∫ ∞

0

e−t

2t−1/2 dt =

1

2Γ

(1

2

).

The second identiy in (2.33) is a consequence of (2.32) and the so-called Wallisproduct (see Prop. E.1(c) of the Appendix for an elementary proof of the Wallisproduct)

π

2=

∞∏

k=1

(2k

2k − 1· 2k

2k + 1

)= lim

n→∞

n∏

k=1

(2k

2k − 1· 2k

2k + 1

)=

2

1· 23· 43· 45· 65· 67· · · :

(2.34)

√π = lim

n→∞

√√√√2 (2n+ 1)n∏

k=1

(2k

2k + 1· 2k

2k + 1

)

= limn→∞

(√2 (2n+ 1)

n·

√nn! 2n

1 · 3 · · · (2n+ 1)

)

= limn→∞

√nn! 2n+1

1 · 3 · · · (2n+ 1)= Γ

(1

2

),

where we also used the continuity of the square root function.

2.5 Products

2.5.1 Product Measure

Definition 2.30. Let n ∈ N and let (Xi,Ai, µi)ni=1 be a finite family of measure spaces,

X :=∏n

i=1Xi, A := ⊗ni=1Ai. Then a measure µ on (X,A) is called a product measure

if, and only if,

∀(A1,...,An)∈

∏ni=1 Ai

µ

(n∏

i=1

Ai

)=

n∏

i=1

µi(Ai). (2.35)

2 INTEGRATION 85

Lemma 2.31. Let n ∈ N, n ≥ 2, and let (Xi,Ai)ni=1 be a finite family of measurable

spaces, X :=∏n

i=1Xi, A := ⊗ni=1Ai. Define the sections

∀M⊆X

∀xn∈Xn

Mxn:=

(x1, . . . , xn−1) ∈

n−1∏

i=1

Xi : (x1, . . . , xn) ∈M

. (2.36)

(a) Forming sections commutes with set-theoretic rules: Let M ⊆ X and let (Mj)j∈Jbe a family of subsets of X, J 6= ∅. Also let Ei ⊆ Xi for i ∈ 1, . . . , n. Then, foreach xn ∈ Xn, (⋂

j∈JMj

)

xn

=⋂

j∈J(Mj)xn

, (2.37a)

(⋃

j∈JMj

)

xn

=⋃

j∈J(Mj)xn

, (2.37b)

(M c)xn= (Mxn

)c, (2.37c)(

n∏

i=1

Ei

)

xn

=

∅ if xn /∈ En,∏n−1

i=1 Ei if xn ∈ En.(2.37d)

Moreover, if f : X −→ Y , f(·, xn) :∏n−1

i=1 Xi −→ Y , and B ⊆ Y , then

f(·, xn)−1(B) =(f−1(B)

)xn. (2.37e)

(b) If M is measurable, then so is each section:

∀M⊆X

∀xn∈Xn

(M ∈ A ⇒ Mxn

∈ ⊗n−1i=1 Ai

).

(c) If (Y,B) is another measurable space and f : X −→ Y is A-B-measurable, then,for each xn ∈ Xn, the map f(·, xn) :

∏n−1i=1 Xi −→ Y is ⊗n−1

i=1 Ai-B-measurable.

Proof. Let Q :=∏n−1

i=1 Xi, Q := ⊗ni=1Ai.

(a): One has

q ∈(⋂

j∈JMj

)

xn

⇔ (q, xn) ∈⋂

j∈JMj ⇔ ∀

j∈J(q, xn) ∈Mj ⇔ ∀

j∈Jq ∈ (Mj)xn

⇔ q ∈⋂

j∈J(Mj)xn

,

proving (2.37a). One has

q ∈(⋃

j∈JMj

)

xn

⇔ (q, xn) ∈⋃

j∈JMj ⇔ ∃

j∈J(q, xn) ∈Mj ⇔ ∃

j∈Jq ∈ (Mj)xn

⇔ q ∈⋃

j∈J(Mj)xn

,

2 INTEGRATION 86

proving (2.37b). One has

q ∈ (M c)xn⇔ (q, xn) ∈M c ⇔ (q, xn) ∈ X \M ⇔ q ∈ Q \Mxn

⇔ q ∈ (Mxn)c,

proving (2.37c). One has

q ∈(

n∏

i=1

Ei

)

xn

⇔ (q, xn) ∈(

n−1∏

i=1

Ei

)× En,

proving (2.37d). One has

q ∈ f(·, xn)−1(B) ⇔ f(q, xn) ∈ B ⇔ (q, xn) ∈ f−1(B) ⇔ q ∈(f−1(B)

)xn,

proving (2.37e).

(b): LetM := M ⊆ X : Mxn

∈ Q for each xn ∈ Xn.We show M to e a σ-algebra: It is ∅ ∈ Q, since ∅ ∈ An. Let M ∈ M and xn ∈ Xn.Then, by (2.37c), (M c)xn

= (Mxn)c ∈ Q, since Mxn

∈ Q. Thus, M c ∈ M. Let (Mk)k∈Nbe a sequence in M, xn ∈ Xn. Then, by (2.37b), (

⋃k∈NMk)xn

=⋃

k∈N(Mk)xn∈ Q, since

each (Mk)xn∈ Q. Thus,

⋃k∈NMk ∈ M, showing M to be a σ-algebra. If M =

∏ni=1Ai

with each Ai ∈ Ai, then (2.37d) impliesM ∈ M. Thus, A ⊆ M, sinceM is a σ-algebra.Since A ⊆ M is precisely the claim of (b), the proof is done.

(c): If B ∈ B, then, by (2.37e), f(·, xn)−1(B) =(f−1(B)

)xn

∈ Q by (b), since f−1(B) ∈A by the measurability of f .

Theorem 2.32. Let n ∈ N and let (Xi,Ai, µi)ni=1 be a finite family of measure spaces,

X :=∏n

i=1Xi, A := ⊗ni=1Ai.

(a) There always exists at least one product measure on (X,A).

(b) Let n = 2. If µ1 is σ-finite, then, using the notation of (2.36), one has

∀M∈A

x2 7→ µ1(Mx2) is A2-B-measurable (2.38a)

and

µ : A −→ [0,∞], µ(M) :=

∫

X2

µ1(Mx2) dµ2(x2), (2.38b)

defines a product measure.

(c) Let n ≥ 2. If each measure µi is σ-finite, then there exists a unique product measureµ on (X,A), denoted by ⊗n

i=1µi := µ. Moreover, ⊗ni=1µi is itself σ-finite and, using

the notation of (2.36), one has

∀M∈A

xn 7→(⊗n−1

i=1 µi

)(Mxn

) is An-B-measurable (2.39a)

2 INTEGRATION 87

and

∀M∈A

(⊗ni=1µi) (M) =

∫

Xn

(⊗n−1

i=1 µi

)(Mxn

) dµn(xn). (2.39b)

In terms of the canonical identification (∏n−1

i=1 Xi)×Xn∼=∏n

i=1Xi, one has

(⊗n−1i=1 µi)⊗ µn = ⊗n

i=1µi. (2.39c)

Proof. (a): For n = 1, there is nothing to prove. We consider the case n = 2, fromwhich the case n > 2 follows by induction. Let n = 2. We know from Prop. 1.17(a) thatE := A1 ∗ A2 = A1 × A2 : Ai ∈ Ai, i = 1, 2 is a semiring. We also know A = σ(E).We now define µ on E by (2.35), i.e. µ(A1×A2) := µ1(A1)µ2(A2) for A1 ∈ A1, A2 ∈ A2.We show that this defines a premeasure on E : µ(∅) = 0 is clear. Let A ∈ A1, B ∈ A2,and let (Ai)i∈N and (Bi)i∈N be sequences in A1 and A2, respectively, such that A×B is

the disjoint union of the Ai ×Bi, i.e. A×B =⋃

i∈N(Ai ×Bi). Then

µ(A× B) =

∫

X2

µ1(A)χB(x) dµ2(x)(2.37d)=

∫

X2

µ1

((A×B)x

)dµ2(x)

(2.37b)=

∫

X2

µ1

(⋃

i∈N(Ai ×Bi)x

)dµ2(x) =

∫

X2

∞∑

i=1

µ1

((Ai ×Bi)x

)dµ2(x)

Cor. 2.8=

∞∑

i=1

∫

X2

µ1

((Ai ×Bi)x

)dµ2(x) =

∞∑

i=1

µ(Ai ×Bi),

showing µ to be a premeasure. Then, using the Caratheodory extension Th. 1.38, µextends to a measure on (X,A) that constitutes a product measure.

(b): For each M ∈ A = A1 ⊗ A2, we introduce the notation fM : X2 −→ [0,∞],fM(x) := µ1(Mx), for the map defined in (2.38a). The bulk of the proof will consist ofshowing the measurability of fM . Defining

M := M ∈ A : fM is measurable,we have to show M = A. If A ∈ A1, B ∈ A2, then

∀x∈X2

fA×B(x) = µ1((A×B)x)(2.37d)= µ1(A)χB(x),

showing fA×B to be measurable. Thus, E ⊆ M, where E := A1 ∗ A2 as in the proof of(a). We now consider the case, where µ1 is finite, i.e. µ1(X1) < ∞. We show M to bea Dynkin system: X ∈ M, since X ∈ E . If M ∈ M, then, due to µ1(X1) < ∞, we canuse subtractivity of µ1 to obtain

∀x∈X2

fMc(x) = µ1((Mc)x)

(2.37c)= µ1((Mx)

c) = µ1(X1)− fM(x),

showing fMc to be measurable and M c ∈ M. Now let (Mk)k∈N be a sequence of disjointsets in M, M :=

⋃k∈NMk. Then

∀x∈X2

fM(x) = µ1(Mx)(2.37b)= µ1

(⋃

k∈N(Mk)x

)=

∞∑

k=1

µ1((Mk)x) =∞∑

k=1

fMk(x),

2 INTEGRATION 88

showing fM to be measurable by Th. 1.82(a), i.e. M ∈ M and M is a Dynkin system.Since E is ∩-stable (by [Phi16b, (A.1)]), we have δ(E) = σ(E) = A by Prop. 1.44. Sincealso δ(E) ⊆ M, we have A ⊆ M ⊆ A, i.e. M = A, as desired. Now let µ1 be σ-finiteand choose a sequence (Ak)k∈N in A1 such that Ak ↑ X1 and µ1(Ak) <∞ for each k ∈ N.Then, clearly, for each k ∈ N, νk : A1 −→ R+

0 , νk(A) := µ1(Ak ∩ A), defines a finitemeasure on X1, and, for each M ∈ A, k ∈ N, we know

fk,M : X2 −→ [0,∞[, fk,M(x) := νk(Mx),

to be measurable. Then fM = limk→∞ fk,M must also be measurable. In consequence,the definition of µ by (2.38b) makes sense and it merely remains to verify µ is a productmeasure. We clearly have µ(∅) = 0 and if (Mk)k∈N is a sequence of disjoint sets in A,M :=

⋃k∈NMk, then

µ(M)(2.37b)=

∫

X2

µ1

(⋃

k∈N(Mk)x

)dµ2(x)

Cor. 2.8=

∞∑

k=1

∫

X2

µ1

((Mk)x

)dµ2(x) =

∞∑

k=1

µ(Mk),

showing µ to be a measure. If A ∈ A1, B ∈ A2, then

µ(A× B) =

∫

X2

µ1

((A× B)x

)dµ2(x) =

∫

X2

µ1(A)χB(x) dµ2(x) = µ1(A)µ2(B),

showing µ to be a product measure.

(c): It suffices to consider the case n = 2, since, then, the general case n ≥ 2 follows byinduction: For the induction step, one merely applies the case n = 2 to the two measurespaces (

∏n−1i=1 Xi,⊗n−1

i=1 Ai,⊗n−1i=1 µi) (which is uniquely determined and σ-finite by the

induction hypothesis) and (Xn,An, µn). Thus, let n = 2. It now suffices to show thatthe product measure of µ1 and µ2 is unique and σ-finite, since, then, the representationof (2.39b) is due to (b). In the proof of (a), we defined a premeasure µ on the semiringE = A1 ∗ A2 by setting µ(A1 × A2) := µ1(A1)µ2(A2) for A1 ∈ A1, A2 ∈ A2. In ourpresent situation, we have µ1, µ2 σ-finite and can choose sequences (Ak)k∈N and (Bk)k∈Nin A1 and A2, respectively, such that Ak ↑ X1, Bk ↑ X2, and µ1(Ak) <∞, µ2(Bk) <∞for each k ∈ N. Then (Ak ×Bk) ↑ X with µ(Ak ×Bk) <∞, showing µ to be σ-finite onE . According to Cor. 1.46(a), µ uniquely extends to a measure on σ(E) = A, provingthe existence of a unique product measure. Since X ∈ E , the extension is still σ-finite,completing the proof.

Caveat 2.33. If µ1 in Th. 2.32(b) is not σ-finite, then it can, indeed, occur that themap of (2.38a) is nonmeasurable so that (2.38b) does not even make sense (see [Beh87,p. 96] for an example). Even if µ1 is σ-finite (but µ2 is not), there can exist severaldifferent product measures on X: Let X1 := X2 := R, A1 := A2 := B1, µ1 := β1

(which is σ-finite), and let µ2 be counting measure (which is not σ-finite). Let α be theproduct measure constructed according to the proof of Th. 2.32(a), i.e. by defining αon B1 ∗ B1 by letting α(A×B) := µ1(A)µ2(B) for A,B ∈ B1 and extending it to B2 viaTh. 1.38: Then α on B2 is the restriction of the corresponding outer measure on P(R2).Set D := (x, x) ∈ R2 : x ∈ [0, 1]. Then D ∈ B2 and we claim α(D) = ∞: Seeking a

2 INTEGRATION 89

contradiction, assume α(D) < ∞. Then there exist sequences (Ak)k∈N and (Bk)k∈N inB1 such that D ⊆ ⋃∞

k=1(Ak × Bk) and∑∞

k=1 α(Ak × Bk) =∑∞

k=1 β1(Ak)µ2(Bk) < ∞.

This implies

∀k∈N

(β1(Ak) = 0 ∨ Bk finite

).

Thus, letting A :=⋃Ak : β

1(Ak) = 0, B :=⋃Bk : Bk finite, we haveD ⊆ (A×R)∪

(R×B). Note π1(M) = π2(M) for each M ⊆ D. We obtain π1(D \ (A×R)) = [0, 1] \Aand π1(D \ (A × R)) = π2(D \ (A × R)) ⊆ B. Since β1([0, 1] \ A) = 1, but β1(B) = 0,this is a contradiction, proving α(D) = ∞. Now let β be the product measure given byTh. 2.32(b). Then

β(D) =

∫

R

β1(Dx) dµ2(x) =

∫

[0,1]

β1(x) dµ2(x) =

∫

[0,1]

0 dµ2(x) = 0.

It is shown in [Beh87, pp. 94–96] that, even though µ2 is not σ-finite, in our consideredsituation, for each M ∈ B2, the map x 7→ µ2(Mx), Mx := y ∈ R : (x, y) ∈ M isλ1-measurable and then

γ : B2 −→ [0,∞], γ(M) :=

∫

R

µ2(Mx) dλ1(x),

again, defines a product measure. Since

γ(D) =

∫

R

µ2(Dx) dλ1(x) =

∫

[0,1]

µ2(x) dλ1(x) =∫

[0,1]

1 dλ1(x) = 1,

we see that the three product measures α, β, γ must all be different.

Example 2.34. (a) Let p, q ∈ N. We already know Bp+q = Bp ⊗ Bq from Th. 1.94(c).This σ-algebra is generated by the semiring E := Ip+q = Ip ∗ Iq. Since βp ⊗ βq andβp+q agree on E , we obtain βp⊗βq = βp+q. Now let X ∈ Bp, Y ∈ Bq, βp

X := βpBp|X ,βqY := βq Bq |Y . We already know Bp|X ⊗ Bq|Y = Bp+q|(X × Y ) from Th. 1.94(c).

This σ-algebra is generated by the semiring F := Bp|X ∗ Bq|Y ⊆ Bp+q. Since wehave (βp

X⊗βqY )F= (βp⊗βq)F= βp+qF= βp+q

X×Y F , we also obtain βpX⊗βq

Y = βp+qX×Y .

(b) Let p, q ∈ N. ThenLp ⊗ Lq ( Lp+q : (2.40)

The inclusion in (2.40) holds: Letting E := Lp ∗ Lq, we have E ⊆ Lp+q: Indeed, ifE ∈ E , then E = (A ∪M) × (B ∪N) with M ⊆ M1, N ⊆ N1, where A,M1 ∈ Bp,B,N1 ∈ Bq, βp(M1) = βq(N1) = 0. Then E = (A × B) ∪ P , where P = (A ×N) ∪ (M ×B) ∪ (M ×N) ⊆ P1 := (A×N1) ∪ (M1 ×B) ∪ (M1 ×N1) ∈ Bp+q withβp+q(P1) = 0, showing E ∈ Lp+q. Thus Lp ⊗ Lq = σ(E) ⊆ Lp+q. The inclusionin (2.40) is strict: Let M ⊆ [0, 1]p such that M /∈ Lp (such a set M exists by Th.1.74(c)). Also let y ∈ Rq. Then N := M × y ∈ Lp+q, since N is a subset ofthe βp+q-null set [0, 1]p × y. However, N /∈ Lp ⊗Lq, since, otherwise, the sectionNy = M had to be in Lp by Lem. 2.31(b). In consequence, the complete measureλp+q is a strict extension of the incomplete measure λp ⊗ λq.

2 INTEGRATION 90

(c) Let n ∈ N. We compute the volume of n-dimensional balls with respect to theEuclidean norm: Let x ∈ Rn, r > 0, Br(x) := y ∈ Rn : ‖y − x‖2 < r.

vol(Br(x)) := βn(Br(x)) = ωn rn, where ωn := βn(B1(0)) =

πn/2

Γ(n2+ 1)

, (2.41)

where Γ denotes the Gamma function from Ex. 2.29: It suffices to prove the formulafor ωn: Indeed, consider the bijective affine map A : Rn −→ Rn, A(v) := rv + x.Then A(B1(0)) = Br(x), detA = rn, and βn(Br(x)) = rn βn(B1(0)) by (1.47b).We now prove the formula for ωn via induction on n ∈ N: For n = 1, the claimis ω1 = β1(] − 1, 1[) = 2 =

√π

Γ( 32), which holds, as Γ(3

2) = 1

2Γ(1

2) = 1

2

√π by (2.29a)

and (2.33). For the induction step, let n > 1, write Rn = Rn−1 × R and useβn = βn−1 ⊗ β1. For x ∈ R, we obtain the sections

(B1(0)

)x=

y ∈ Rn−1 : x2 +

n−1∑

i=1

y2i < 1

=

∅ for |x| ≥ 1,

Bn−1√1−x2(0) for −1 < x < 1,

where Bn−1√1−x2(0) denotes a ball in Rn−1. Thus,

ωn(2.39b)=

∫ 1

−1

βn−1(B1(0)

)xdβ1(x) = ωn−1

∫ 1

−1

(1− x2)(n−1)/2 dx .

As we may consider the last integral as a 1-dimensional Riemann integral, we mayuse the change of variables t := arccos x (recall arccos′(x) = −(1−x2)−1/2) to obtain

ωn = −ωn−1

∫ 0

π

(1− (cos t)2

)n/2dt = ωn−1

∫ π

0

(sin t)n dt .

In particular, for n = 2, ω2 = ω1

∫ π

0(sin t)2 dt

(E.3a)= 2 · π

2= π = π

Γ(2), proving (2.41)

for n = 2. For n > 2,

ωn = ωn−2

∫ π

0

(sin t)n−1 dt

∫ π

0

(sin t)n dt .

Finally, using the induction hypothesis and (E.3), we obtain, for n even

ωn =π(n−2)/2

Γ(n2)

2π

n/2−1∏

k=1

2k

2k + 1

n/2∏

k=1

2k − 1

2k=

2πn/2

nΓ(n2)

(2.29a)=

πn/2

Γ(n2+ 1)

and, for n odd,

ωn =π(n−2)/2

Γ(n2)

2π

(n−1)/2∏

k=1

2k − 1

2k

(n−1)/2∏

k=1

2k

2k + 1=

2πn/2

nΓ(n2)

(2.29a)=

πn/2

Γ(n2+ 1)

,

completing the induction.

2 INTEGRATION 91

2.5.2 Theorems of Tonelli and Fubini

For the actual computation of higher-dimensional integrals, the theorems of Tonelli andFubini, which we provide in Th. 2.36 below, are among the most powerful tools, as theyallow, under suitable hypotheses, to compute an n-dimensional integral as a sequenceof n 1-dimensional integrals.

The literature knows many variants of the theorems of Tonelli and Fubini, where differentvariants can have subtle differences, and one has to use some care in applications toalways verify that the variant one desires to use does, indeed, apply. The most naturalvariant in our setting is probably the one that considers two σ-finite measure spaces(X1,A1, µ1) and (X2,A2, µ2) and a function f that is (µ1⊗µ2)-measurable or (µ1⊗µ2)-integrable. However, one would also like to apply the theorem to functions that areλp+q-measurable or λp+q-integrable and, recalling Ex. 2.34(b), Lp ⊗Lq ( Lp+q. For thisreason, we also provide a version that assumes f to be µ-measurable or µ-integrable,where µ is the completion of µ1 ⊗ µ2.

The following convention is sometimes useful, especially in the context of the theoremsof Tonelli and Fubini:

Convention 2.35. Let (X,A, µ) be a measure space, N ∈ A with µ(N) = 0, f :N c −→ K, z ∈ f(X). Define

g : X −→ K, g(x) :=

f(x) for x ∈ N c,

z for x ∈ N.

If f is integrable or nonnegative measurable (with respect to A and µ restricted to N c),then, clearly, g is µ-integrable or nonnegative µ-measurable. In this case, one defines

∫

X

f dµ :=

∫

X

g dµ

and observes that this definition does not depend on the choice of z ∈ f(X) (since Nis a µ-null set). In practise, one will usually still denote g by f and one often does noteven specify z explicitly.

Theorem 2.36. Let n ∈ N, n ≥ 2. For each i ∈ 1, . . . , n, let (Xi,Ai, µi) be a σ-finitemeasure space, X :=

∏ni=1Xi. We consider two cases:

(i) (X,A, µ) is the corresponding product measure space.

(ii) Each µi is complete and (X,A, µ) is the completion of the corresponding productmeasure space.

In both cases, the following holds:

(a) Tonelli’s Theorem: Let f : X −→ [0,∞] be A-measurable. For n = 2, the functionf(x1, ·) is A2-measurable (for every x1 ∈ X1 in Case (i), for µ1-almost every x1 ∈

2 INTEGRATION 92

X1 in Case (ii)), the function f(·, x2) is A1-measurable (for every x2 ∈ X2 in Case(i), for µ2-almost every x2 ∈ X2 in Case (ii)), the functions given by

x1 7→∫

X2

f(x1, x2) dµ2(x2), x2 7→∫

X1

f(x1, x2) dµ1(x1), (2.42a)

are A1-measurable and A2-measurable, respectively (in Case (ii), they are definedaccording to Convention 2.35), and

∫

X1×X2

f(x1, x2) dµ(x1, x2)

=

∫

X2

∫

X1

f(x1, x2) dµ1(x1) dµ2(x2)

=

∫

X1

∫

X2

f(x1, x2) dµ2(x2) dµ1(x1). (2.42b)

For n ≥ 2, by applying (2.42b) inductively, one obtains

∫

X

f dµ =

∫

X1

. . .

∫

Xn

f(x1, . . . , xn) dµn(xn) · · · dµ1(x1) , (2.42c)

where one can also arbitrarily permute the order of the inner integrals withoutchanging the overall value (as above, the inner integrals are defined according toConvention 2.35 in Case (ii)).

(b) Fubini’s Theorem: Let f : X −→ K be µ-integrable. For n = 2, the function f(x1, ·)is µ2-integrable for µ1-almost every x1 ∈ X1 (in both cases), the function f(·, x2)is µ1-integrable for µ2-almost every x2 ∈ X2 (in both cases), the functions given by(2.42a) are µ1-integrable and µ2-integrable, respectively (in the sense of Convention2.35), and (2.42b) holds. As in (a), for n ≥ 2, by applying (2.42b) inductively, oneobtains the validity of (2.42c), where, once again, one can also arbitrarily permutethe order of the inner integrals without changing the overall value (of course, asbefore, the inner integrals are defined according to Convention 2.35).

(c) If f : X −→ K is µ-measurable and one of the integrals

∫

X

|f | dµ ,∫

Xπ(1)

. . .

∫

Xπ(n)

|f(x1, . . . , xn)| dµπ(n)(xπ(n)) · · · dµπ(1)(xπ(1)) ,(2.43)

where π is a permutation of 1, . . . , n, is finite, then all the integrals (for anarbitrary permutation π of 1, . . . , n) are finite, f is µ-integrable and, in particular,all the assertions of (b) hold true.

Proof. It suffices to prove the case n = 2 – then the case n > 2 follows by induction(and using that every permutation is a finite composition of transpositions).

2 INTEGRATION 93

(a): Case (i): Let M ∈ A = A1 ⊗A2. If x2 ∈ X2, then, for the section Mx2 ⊆ X1, onehas Mx2 ∈ A1 by Lem. 2.31(b), and x2 7→ µ1(Mx2) is A2-B-measurable by Th. 2.32(b).One computes

∫

X

χM(x1, x2) dµ(x1, x2) = µ(M)(2.38b)=

∫

X2

µ1(Mx2) dµ2(x2)

=

∫

X2

∫

X1

χMx2(x1) dµ1(x1) dµ2(x2) =

∫

X2

∫

X1

χM(x1, x2) dµ1(x1) dµ2(x2)

proving the first equality of (2.42b) for characteristic functions of measurable sets. Nowlet f be a simple function f ∈ S+(A), i.e. f =

∑Nk=1 αkχMk

with α1, . . . , αN ∈ R+0 ;

M1, . . . ,Mn ∈ A; N ∈ N. Then the linearity of the integrals yields

∫

X

f dµ =N∑

k=1

αk

∫

X

χMk(x1, x2) dµ(x1, x2)

=N∑

k=1

αk

∫

X2

∫

X1

χMk(x1, x2) dµ1(x1) dµ2(x2)

=

∫

X2

∫

X1

f(x1, x2) dµ1(x1) dµ2(x2) , (2.44)

proving the first equality of (2.42b) for nonnegative simple functions. Next, we considerf ∈ M+(A) and choose a sequence (φk)k∈N in S+(A) such that φk ↑ f . According toLem. 2.31(c), for each x2 ∈ X2, f(·, x2) ∈ M+(A1) and, for each k ∈ N, φk(·, x2) ∈S+(A1). Since φk(·, x2) ↑ f(·, x2), we obtain, according to (2.4)

∫

X1

φk(x1, x2) dµ1(x1) ↑∫

X1

f(x1, x2) dµ1(x1) . (2.45)

Moreover, each function x2 7→∫X1φk(x1, x2) dµ1(x1) on the left-hand side of (2.45) is

A2-measurable, implying the function x2 7→∫X1f(x1, x2) dµ1(x1) on the right-hand side

to be A2-measurable as well. We now compute∫

X

f dµ(2.4)= lim

k→∞

∫

X

φk dµ = limk→∞

∫

X2

∫

X1

φk(x1, x2) dµ1(x1) dµ2(x2)

MCT=

∫

X2

∫

X1

limk→∞

φk(x1, x2) dµ1(x1) dµ2(x2)

(2.4)=

∫

X2

∫

X1

f(x1, x2) dµ1(x1) dµ2(x2) , (2.46)

proving the first equality of (2.42b) for each f ∈ M+(A) as claimed. To prove thesecond equality of (2.42b) as well as all the remaining claims of (a) (in Case (i)), onemerely exchanges the roles of (X1,A1, µ1) and (X2,A2, µ2) in the above argument.

Case (ii): LetM ∈ A = (A1⊗A2)∼. Then there exist A,C ∈ A1⊗A2 and N ⊆ C, such

that (µ1⊗µ2)(C) = 0 and M = A∪N . If x2 ∈ X2, then, for the section Mx2 ⊆ X1, one

2 INTEGRATION 94

has Mx2 = Ax2 ∪Nx2 , Nx2 ⊆ Cx2 . Since (µ1 ⊗ µ2)(C) = 0, (2.38b) implies µ1(Cx2) = 0for µ2-a.e. x2 ∈ X2. Thus, since µ1 is complete, Mx2 ∈ A1 and χM(·, x2) = χMx2

isµ1-measurable for µ2-a.e. x2 ∈ X2. Using Convention 2.35, one computes

∫

X

χM(x1, x2) dµ(x1, x2) = µ(M) = µ(A)(2.38b)=

∫

X2

µ1(Ax2) dµ2(x2)

=

∫

X2

∫

X1

χAx2(x1) dµ1(x1) dµ2(x2) =

∫

X2

∫

X1

χM(x1, x2) dµ1(x1) dµ2(x2) ,

proving the first equality of (2.42b) for characteristic functions of measurable sets. Iff ∈ S+(A), f =

∑Nk=1 αkχMk

as in the proof of Case (i), then the linearity of theintegrals, once again, yields the validity of (2.44) proving the first equality of (2.42b)for f ∈ S+(A). As in Case (i), we now consider f ∈ M+(A) and choose a sequence(φk)k∈N in S+(A) such that φk ↑ f . We have already seen above that, if M ∈ A,then χM(·, x2) is µ1-measurable for µ2-a.e. x2 ∈ X2. Since a countable union of µ2-nullsets is still a µ2-null set, there exists a µ2-null N ∈ A2 such that all φk(·, x2), k ∈ N,are µ1-measurable for each x2 ∈ N c. This shows f(·, x2) to be µ1-measurable for eachx2 ∈ N c (i.e. for µ2-a.e. x2 ∈ X2) as well. Thus, for each x2 ∈ N c, f(·, x2) ∈ M+(A1),φk(·, x2) ∈ S+(A1), φk(·, x2) ↑ f(·, x2), and (2.45) holds. As in Case (i), each functionx2 7→

∫X1φk(x1, x2) dµ1(x1) on the left-hand side of (2.45) (extended to N by 0) is A2-

measurable, implying the function x2 7→∫X1f(x1, x2) dµ1(x1) on the right-hand side to

be A2-measurable as well. Then, in terms of Convention 2.35, (2.46) still holds, provingthe first equality of (2.42b) for each f ∈ M+(A). Once again, exchanging the roles of(X1,A1, µ1) and (X2,A2, µ2) proves the remaining claims of (a).

(b): If f is µ-integrable, then so is |f | and (a) yields

∫

X2

∫

X1

|f(x1, x2)| dµ1(x1) dµ2(x2) =

∫

X

|f | dµ <∞,

implying ∫

X1

|f(x1, x2)| dµ1(x1) <∞ µ2-a.e.

In other words, f(·, x2) is µ1-integrable for each x2 ∈ N c, where N is a µ2-null set, i.e.,letting f1 := (Re f)+, f2 := −(Re f)−, f3 := i (Im f)+, f4 := −i (Im f)−,

∀x2∈Nc

∫

X1

f(x1, x2) dµ1(x1) =4∑

k=1

∫

X1

fk(x1, x2) dµ1(x1) .

Moreover, for each k ∈ 1, 2, 3, 4, x2 7→∫X1fk(x1, x2) dµ1(x1) (extended to N by 0) is

µ2-integrable, since

∫

X2

∫

X1

|fk(x1, x2)| dµ1(x1) dµ2(x2) ≤∫

X2

∫

X1

|f(x1, x2)| dµ1(x1) dµ2(x2) <∞.

2 INTEGRATION 95

We apply (a) to each |fk| to obtain

∫

X2

∫

X1

f(x1, x2) dµ1(x1) dµ2(x2) =4∑

k=1

∫

X2

∫

X1

fk(x1, x2) dµ1(x1) dµ2(x2)

=4∑

k=1

∫

X

fk dµ =

∫

X

f dµ ,

i.e. the first equality of (2.42b). Exchanging the roles of (X1,A1, µ1) and (X2,A2, µ2)completes the proof of (b).

(c) is just (a) combined with (b).

The technique employed in the proof of Th. 2.36 can often be employed to prove an asser-tion about integrable funtions: One first proves the assertion for characteristic functionsof measurable sets, then for simple functions (linear combinations of characteristic func-tions of measurable sets), then for nonnegative measurable functions (pointwise limitsof simple functions), then, finally, for integrable functions f (whose parts (Re f)± and(Im f)± are nonnegative measurable).

Example 2.37. (a) Let I := [1, 2] × [2, 3] × [0, 2]. We use the Fubini theorem tocompute

∫If :=

∫If dλ3 for

f : I −→ R, f(x, y, z) :=2z

(x+ y)2:

Since f is continuous on the compact interval I, it is integrable and Fubini applies,yielding

∫

I

f =

∫ 2

1

∫ 3

2

∫ 2

0

2z

(x+ y)2dz dy dx =

∫ 2

1

∫ 3

2

[z2

(x+ y)2

]z=2

z=0

dy dx

=

∫ 2

1

∫ 3

2

4

(x+ y)2dy dx =

∫ 2

1

[− 4

x+ y

]y=3

y=2

dx =

∫ 2

1

(− 4

x+ 3+

4

x+ 2

)dx

= 4[ln(x+ 2)− ln(x+ 3)

]21= 4 ln

(4 · 4 · 1

3· 15

)= 4 ln

16

15.

(b) Using the Tonelli theorem, we provide another proof for∫∞−∞ e−x2

dx =√π of

(2.33): We let I := [0,∞[2 and consider

f : I −→ R+0 , f(x, y) := y e−(1+x2)y2 .

Since f is continuous, it is measurable. It is also nonnegative, i.e. f ∈ M+ and theTonelli theorem applies. Thus, we can compute

∫If :=

∫If dλ2 in two different

2 INTEGRATION 96

ways, where, for the second way, we also use the change of variable t := xy,

∫

I

f =

∫ ∞

0

∫ ∞

0

y e−(1+x2)y2 dy dx =

∫ ∞

0

[−e

−(1+x2)y2

2(1 + x2)

]y=∞

y=0

dx

=1

2

∫ ∞

0

1

1 + x2dx =

1

2[arctan x]∞0 =

π

4,

∫

I

f =

∫ ∞

0

e−y2∫ ∞

0

y e−x2y2 dx dy =

∫ ∞

0

e−y2∫ ∞

0

e−t2 dt dy =

(∫ ∞

0

e−t2 dt

)2

,

proving∫∞−∞ e−x2

dx =√π.

(c) The following example shows that, in general, the Fubini theorem does not holdfor nonintegrable functions: It can happen that all iterated integrals exist, but theresult depends on the order in which they are executed: Consider X1 := X2 :=]0, 1[,X :=]0, 1[2, A1 := A2 := B1, µ1 := µ2 := β1. Consider the continuous function

f : X −→ R, f(x, y) :=x2 − y2

(x2 + y2)2.

We show∫ 1

0

∫ 1

0

f(x, y) dy dx =π

46= −π

4=

∫ 1

0

∫ 1

0

f(x, y) dx dy :

Observe that, on X,

∂y arctanx

y= − 1

1 + x2

y2

· xy2

= − x

x2 + y2,

∂x∂y arctanx

y= −x

2 + y2 − 2x2

(x2 + y2)2=

x2 − y2

(x2 + y2)2,

∂x arctany

x= − y

x2 + y2,

∂y∂x arctany

x=

y2 − x2

(x2 + y2)2.

Thus,

∫ 1

0

∫ 1

0

f(x, y) dy dx =

∫ 1

0

[y

x2 + y2

]y=1

y=0

dx =

∫ 1

0

1

1 + x2dx = [arctan x]10 =

π

4,

∫ 1

0

∫ 1

0

f(x, y) dx dy =

∫ 1

0

[− x

x2 + y2

]x=1

x=0

dy = −∫ 1

0

1

1 + y2dy = −π

4.

In particular, f cannot be β2-integrable over X.

2 INTEGRATION 97

2.6 Lp-Spaces

2.6.1 Definition, Norm, Completeness

Notation 2.38. Let (X,A, µ) be a measure space. Using the convention

∀p∈R+

∞p := ∞, (2.47)

we define for every measurable f : X −→ K:

∀p∈R+

Np(f) :=

(∫

X

|f |p dµ)1/p

∈ [0,∞] (2.48a)

and

N∞(f) := inf

sup

|f(x)| : x ∈ X \N

: N ∈ A, µ(N) = 0

∈ [0,∞], (2.48b)

where the number N∞(f) is also known as the essential supremum of f (in (2.48b), it isuseful to set sup ∅ := inf[0,∞] = 0 – this setting is only relevant if µ(X) = 0, resultingin N∞(f) = 0). Note that |f | ≤ N∞(f) µ-a.e., since, for N∞(f) <∞,

|f | > N∞(f)

=⋃

n∈N

|f | > N∞(f) +

1

n

is a µ-null set.

Theorem 2.39 (Holder Inequality). Let (X,A, µ) be a measure space and p, q ∈ [1,∞]such that 1

p+ 1

q= 1 (note 1/∞ = 0). If f, g : X −→ K are measurable, then

N1(fg) ≤ Np(f)Nq(g), (2.49)

where (2.49) is known as Holder inequality. We point out the following special cases:(a) For p = q = 2, one obtains the Cauchy-Schwarz inequality. (b) If µ is countingmeasure on 1, . . . , n, n ∈ N, then one recovers the Holder inequality of [Phi16b, Th.1.12]. (c) If µ is counting measure on N, then one obtains the Holder (and, in particular,Cauchy-Schwarz) inequality for series of real (or complex) numbers.

Proof. If p = 1, q = ∞, then

N1(fg) =

∫

X

|fg| dµ ≤ N∞(g)

∫

X

|f | dµ

proves (2.49), and the case p = ∞, q = 1 is analogous. Thus, let 1 < p, q < ∞. IfNp(f) = 0 or Nq(g) = 0, then fg = 0 µ-a.e., implying N1(fg) = 0 and (2.49). As (2.49)is also immediate for Np(f)Nq(g) = ∞, it merely remains to consider the case 0 <Np(f), Nq(g) <∞. Recall that, according to [Phi16a, Th. 9.35], if n ∈ N, x1, . . . , xn ≥ 0

2 INTEGRATION 98

and λ1, . . . , λn > 0 such that λ1 + · · · + λn = 1, then xλ11 · · · xλn

n ≤ λ1x1 + · · · + λnxn.Using this with n = 2, λ1 :=

1p, λ2 :=

1q, x1 :=

(|f |/Np(f)

)p, x2 :=

(|g|/Nq(g)

)q, yields

|f |Np(f)

|g|Nq(g)

≤ 1

p

( |f |Np(f)

)p

+1

q

( |g|Nq(g)

)q

(which, trivially, also holds for |f | = ∞ or |g| = ∞). Integrating this inequality over Xproves (2.49).

As in the special case of Kn in Analysis II, we obtain the Minkowski inequality (triangleinequality for Np) from the Holder inequality:

Theorem 2.40. Let (X,A, µ) be a measure space and let f, g : X −→ K be measurablefunctions.

(a) Minkowski Inequality: If p ∈ [1,∞], then

Np(f + g) ≤ Np(f) +Np(g). (2.50)

We point out the following special cases: (a) If µ is counting measure on 1, . . . , n,n ∈ N, then one recovers the Minkowski inequality of [Phi16b, Th. 1.13]. (b) If µis counting measure on N, then one obtains the Minkowski inequality for series ofreal (or complex) numbers.

(b) If p ∈]0, 1], then one has the following inequalities:

Npp (f + g) ≤ Np

p (f) +Npp (g), (2.51a)

Np(f + g) ≤ 21p−1(Np(f) +Np(g)

). (2.51b)

Proof. (a): If Np(f + g) = 0 or Np(f) = ∞ or Np(g) = ∞, then (2.50) is trivially true.If p = 1, then

N1(f + g) =

∫

X

|f + g| dµ ≤∫

X

|f | dµ +

∫

X

|g| dµ = N1(f) +N1(g).

If p = ∞, then

∀A⊆X

sup|f(x) + g(x)| : x ∈ A

≤ sup

|f(x)| : x ∈ A

+ sup

|g(x)| : x ∈ A

,

proves N∞(f + g) ≤ N∞(f) +N∞(g). For the remaining case, consider 1 < p <∞ andf, g such that Np(f + g) > 0 as well as 0 ≤ Np(f), Np(g) <∞. Set q := (1− 1

p)−1 = p

p−1

2 INTEGRATION 99

and apply the Holder inequality (2.49) to obtain

Npp (f + g) =

∫

X

|f + g|p dµ ≤∫

X

|f | |f + g|p−1 dµ +

∫

X

|g| |f + g|p−1 dµ

(2.49)

≤(Np(f) +Np(g)

)Nq

(|f + g|p−1

)

=(Np(f) +Np(g)

) (∫

X

|f + g|(p−1)q dµ

)1/q

(p−1)q=p=

(Np(f) +Np(g)

) (Np(f + g)

)p/q

p/q=p−1=

(Np(f) +Np(g)

) (Np(f + g)

)p−1. (2.52)

Next, we use|f + g|p ≤

(2 max(|f |, |g|)

)p ≤ 2p(|f |p + |g|p

)

and Np(f), Np(g) < ∞ to conclude Np(f + g) ∈ R+. Thus, we can devide (2.52) by(Np(f + g)

)p−1to prove (2.50).

(b): For p = 1, (2.51) is just (2.50). Thus, let 0 < p < 1. Fix a ∈ R+ and consider thefunction

φ : R+0 −→ R, φ(t) := ap + tp − (a+ t)p,

which is, clearly, continuous and, for t > 0, differentiable. Moreover,

∀t>0

φ′(t) = p tp−1 − p (a+ t)p−1 > 0

(since p − 1 < 0, it is tp−1 > (a + t)p−1). Thus, φ is strictly increasing and, notingφ(0) = 0, we have ap + tp − (a+ t)p ≥ 0 for each t ≥ 0. We restate our findings as

∀a,b∈R+

0

(a+ b)p ≤ ap + bp.

Applying this with a := |f | and b := |g| yields (2.51a) after an integration over X. Toprove (2.51b), we, once again, fix a ∈ R+, this time considering the function

ψ : R+0 −→ R, ψ(t) := (a1/p + t1/p) (a+ t)−1/p,

which is also, clearly, continuous and, for t > 0, differentiable, where

∀t>0

ψ′(t) =1

pt1p−1 (a+ t)−1/p − 1

p(a1/p + t1/p) (a+ t)−

1p−1

=(a+ t)−

1p−1

p

(t1p−1 (a+ t)− (a1/p + t1/p)

)

=a (a+ t)−

1p−1

p(t

1p−1 − a

1p−1),

2 INTEGRATION 100

showing ψ to be strictly decreasing on [0, a] and strictly increasing on [a,∞[ with a strict

global min at t = a, ψ(a) = 2 a1/p 2−1/p a−1/p = 21−1p . Taking reciprocals in ψ(b) ≥ ψ(a)

proves

∀a,b∈[0,∞]

(a+ b)1/p ≤ 21p−1 (a1/p + b1/p).

We apply this with a := Npp (f) and b := Np

p (g) to obtain

Np(f + g)(2.51a)

≤(Np

p (f) +Npp (g)

)1/p ≤ 21p−1(Np(f) +Np(g)

),

proving (2.51b).

Definition and Remark 2.41. Let (X,A, µ) be a measure space and p ∈]0,∞].

(a) Let Lp := LpK := Lp

K(µ) := LpK(X,A, µ) denote the set of all µ-measurable functions

f : X −→ K such that Np(f) <∞ (i.e. such that |f |p is µ-integrable (for p <∞)).For p ≥ 1, it is common to call Np the p-norm (caveat: on Lp, Np is, in general,only a seminorm, see below) and to introduce the notation

∀p∈[1,∞]

∀f∈Lp

‖f‖p := Np(f). (2.53a)

We also define

∀p∈[1,∞]

dp : Lp × Lp −→ R+0 , dp(f, g) := ‖f − g‖p, (2.53b)

∀p∈]0,1[

dp : Lp × Lp −→ R+0 , dp(f, g) := Np

p (f − g). (2.53c)

Clearly, each Lp forms a vector space over K (here we use that Lp contains onlyK-valued functions and not K-valued functions). Moreover, for p ∈ [1,∞], Np

constitutes a seminorm on Lp (i.e.Np(0) = 0, Np is homogeneous of degree 1, andNp

satisfies the triangle inequality due to the Minkowski inequality (2.50), cf. [Phi16b,Def. E.2]) and, in consequence, dp forms a pseudometric on Lp (i.e. dp(f, f) = 0,dp is symmetric, and dp satisfies the triangle inequality, cf. [Phi16b, Def. E.1]). Forp ∈]0, 1[, dp with the modified definition (2.53c) still forms a pseudometric on Lp:Indeed, dp(f, f) = 0 and dp(f, g) = dp(g, f) are clear,

∀p∈]0,1[

∀f,g,h∈Lp

dp(f, g) = Npp (f − g) = Np

p (f − h+ h− g)

≤ Npp (f − h) +Np

p (h− g) = dp(f, h) + dp(h, g)

is due to (2.51a). As described in [Phi16b, Sec. E], pseudometrics induce topologiesin precisely the same way that metrics do (the open balls Br(x) still form the baseof a topology). However, if the pseudometric d is not a metric (resp., the seminormis not a norm), i.e. if it is not positive definite and there are f 6= g such thatd(f, g) = 0, then the induced topology is not Hausdorff (not even T1, since thereexists no open set separating f and g). In the present case, Np and dp are positivedefinite if, and only if, there does not exist a nonempty µ-null set N (otherwise,f ≡ 0 and g := χN satisfy f 6= g, but dp(f, g) = 0). Due to this reason, one cannot be completely happy with Lp, but needs to proceed to (b):

2 INTEGRATION 101

(b) Let N denote the set of measurable f : X −→ K that vanish µ-almost everywhere.Clearly, N is a vector space over K, namely a subspace of Lp. Thus, we can definethe factor vector space (a.k.a. quotient vector space):

Lp := LpK := Lp

K(µ) := LpK(X,A, µ) := Lp

K(X,A, µ)/N .

If X = N, A = P(N), and µ is counting measure, then it is customary to writelp := Lp (i.e. lp is the space of p-summable sequences; in this case N = 0 andlp = Lp = Lp). Coming back to the general case, the elements of Lp are preciselythe equivalence classes of the equivalence relation defined by

f ∼ g :⇔ f − g ∈ N ⇔ dp(f, g) = 0,

Lp = [f ] : f ∈ Lp. If f, g ∈ Lp represent the same element of Lp, i.e. [f ] = [g] ∈Lp, then Np(f) = Np(g). Thus, it makes sense to define

∀p∈]0,∞]

Np : Lp −→ R+

0 , Np([f ]) := Np(f),

∀p∈[1,∞]

‖ · ‖ : Lp −→ R+0 , ‖[f ]‖p := ‖f‖p,

∀p∈]0,∞]

dp : Lp × Lp −→ R+

0 , dp([f ], [g]) := dp(f, g).

It is then straightforward to verify that ‖ · ‖p constitutes a norm on Lp for eachp ∈ [1,∞], and dp constitutes a metric on Lp for each p ∈]0,∞], where the mapι : Lp −→ Lp, ι(f) := [f ], is surjective and continuous (see [Phi16b, Th. E.8, Th.E.9]). In practise, it is very common not to properly distinguish between elements[f ] ∈ Lp and their representatives f ∈ Lp (subsequently, we will also make use of thispractise, where suitable). In most situations, using representatives of [f ] does notcause any problems. Some caution is, however, advisable: There are circumstances(e.g., restricting elements of Lp or Lp to µ-null sets, typically to obtain traces onboundaries), where one does have to properly distinguish between f and [f ].

—

An obvious question is if, for p < q, one of the spaces Lp and Lq contains the other?The answer is “no” in general (e.g., f ≡ 1 is in L∞(λ1), but not in L1(λ1), g withg(t) := |t|−1/2 · χ]0,1](t) is in L1(λ1), but not in L∞(λ1)), but the answer is affirmativefor spaces with finite measure. The latter result is a simple consequence of the Holderinequality (2.49) and is provided as Th. 2.42. The complete answer is then given (withproof omitted) in Rem. 2.43 below.

Theorem 2.42. Let (X,A, µ) be a finite measure space, i.e. µ(X) <∞. Then, for eachp, q ∈]0,∞] with 0 < p < q ≤ ∞, one has Lq ⊆ Lp and

∀f∈Lq

Np(f) ≤ µ(X)1p− 1

q Nq(f).

2 INTEGRATION 102

Proof. It suffices to prove the estimate, which, clearly, implies Lq ⊆ Lp. Thus, letf ∈ Lq. For q = ∞, we compute

Np(f) =

(∫

X

|f |p dµ) 1

p

≤(∫

X

(N∞(f)

)pdµ

) 1p

= N∞(f)(µ(X)

) 1p ,

thereby establishing the case. For q <∞, set r := qp> 1 and s := (1− 1

r)−1 = r

r−1> 1.

Then 1r+ 1

s= 1. Since |f |p ∈ Lr (since |f |pr = |f |q ∈ L1) and 1 ∈ Ls, we obtain

Npp (f) =

∫

X

|f |p · 1 dµ(2.49)

≤ Nr(|f |p)Ns(1) =

(∫

X

|f |pr dµ) 1

r

µ(X)1s = Np

q (f)µ(X)1−pq ,

which, after taking pth roots, completes the proof.

Remark 2.43. Let (X,A, µ) be a measure space. In generalization of Th. 2.42, onecan show (cf. Problems 6.2.5 and 6.2.6 in [Els07]) the equivalence of the statements

(i) there exist p, q ∈]0,∞] such that 0 < p < q ≤ ∞ and Lq ⊆ Lp,

(ii) supµ(A) : A ∈ A, µ(A) <∞

<∞,

(iii) for each p, q ∈]0,∞] such that 0 < p < q ≤ ∞, one has Lq ⊆ Lp;

as well as the equivalence of the statements

(i′) there exist p, q ∈]0,∞] such that 0 < p < q ≤ ∞ and Lp ⊆ Lq,

(ii′) infµ(A) : A ∈ A, µ(A) > 0

> 0,

(iii′) for each p, q ∈]0,∞] such that 0 < p < q ≤ ∞, one has Lp ⊆ Lq.

In particular, a space with µ(X) = ∞ can only have the properties (i) – (iii) if itis not σ-finite. As a concrete simple example, consider X := 0, 1, A := P(X),µ(∅) := µ(0) := 0, µ(1) := µ(X) := ∞. Then every function f : X −→ Kis measurable, properties (ii) and (ii′) are clearly satisfied and, for each p ∈]0,∞],Lp = (f : X −→ K) : f(1) = 0.

—

The following Th. 2.44(a), regarding the completeness of the Lp spaces, is sometimesknown as the Riesz-Fischer theorem. However, part of the literature reserves the namefor a different theorem and, thus, we leave Th. 2.44(a) unnamed.

Theorem 2.44. Let (X,A, µ) be a measure space and p ∈]0,∞].

(a) All the spaces LpK(µ) are complete (with respect to the pseudometric dp), where the

definition of Cauchy sequence and completeness in pseudometric spaces is preciselythe same as in metric spaces. In particular, all Lp

K(µ), p ∈ [1,∞], are Banachspaces; all Lp

K(µ), p ∈]0, 1[, are complete metric spaces.

2 INTEGRATION 103

(b) If (fk)k∈N is a sequence in LpK(µ), converging to g ∈ Lp

K(µ), then (fk)k∈N has asubsequence that converges to g pointwise µ-almost everywhere. For p = ∞, (fk)k∈Nitself converges to g, even uniformly µ-almost everywhere.

Proof. (a): Let (fk)k∈N be a Cauchy sequence in Lp. First, consider 0 < p < ∞. Thenthere exists a subsequence (fkl)l∈N such that

∀l∈N

∀m>kl

dp(fm, fkl) < 2−l.

For each l ∈ N, set gl := fkl − fkl+1. Then

∀n∈N

∀p≥1

Np

(n∑

l=1

|gl|)

≤n∑

l=1

Np(gl) =n∑

l=1

dp(fkl , fkl+1) <

n∑

l=1

2−l < 1,

∀n∈N

∀p<1

Npp

(n∑

l=1

|gl|)

≤n∑

l=1

Npp (gl) =

n∑

l=1

dp(fkl , fkl+1) <

n∑

l=1

2−l < 1.

The monotone convergence Th. 2.7 allows to take the limit for n → ∞, resulting inNp(∑∞

l=1 |gl|) ≤ 1. In particular, for µ-a.e. x ∈ X, the series∑∞

l=1 gl(x) convergesabsolutely and, in consequence, converges to some h(x) ∈ K. However,

∀x∈X

∀n∈N

n∑

l=1

gl(x) = fk1(x)− fkn+1(x),

showing there exists a µ-null set N such that (fkl)l∈N converges on N c (i.e. µ-a.e.) tothe measurable function

f : X −→ K, f(x) :=

fk1(x)− h(x) for x ∈ N c,

0 for x ∈ N.

We still need to verify f ∈ Lp and limk→∞ dp(fk, f) = 0. To this end, let ǫ > 0. Thenthere exists k0 ∈ N such that Np

p (fk − fm) < ǫ for each k,m > k0. Fixing m > k0 and

applying Fatou (Prop. 2.19) to the sequence(|fkl − fm|p

)l∈N yields

Npp (fm − f) =

∫

X

|f − fm|p dµ =

∫

X

liml→∞

|fkl − fm|p dµ ≤ lim infl→∞

∫

X

|fkl − fm|p dµ

= lim infl→∞

Npp (fkl − fm) ≤ ǫ.

Since f = f − fm + fm, this shows both f ∈ Lp and limk→∞ dp(fk, f) = 0. It remains toconsider the case p = ∞. In this case, we have N ∈ A with µ(N) = 0 for the set

N :=⋃

(k,l)∈N2

|fk − fl| > N∞(fk − fl)

.

If x ∈ N c, then (fk(x))k∈N is a Cauchy sequence in K (since (fk)k∈N is a Cauchy sequencein L∞). Thus, there exists f(x) ∈ K such that f(x) = limk→∞ fk(x). On N c, the

2 INTEGRATION 104

convergence fk → f is even uniform: Given ǫ > 0, choose n0 ∈ N such thatN∞(fk−fl) <ǫ2for each k, l > n0. Also choose m > n0 such that |fm(x)− f(x)| < ǫ

2. Then, for each

k > n0 (and independently of x ∈ N c), |fk(x)−f(x)| ≤ |fk(x)−fm(x)|+|fm(x)−f(x)| <ǫ2+ ǫ

2= ǫ. Extending f by 0 to X, we obtain f = limk→∞ fk χNc . In particular, f is

measurable as the pointwise limit of measurable functions. The uniform convergencefk → f on N c also yields limk→∞N∞(fk−f) = 0 and N∞(f) ≤ N∞(f−fk)+N∞(fk) <∞, showing f ∈ L∞.

(b): If (fk)k∈N converges in Lp, then it is a Cauchy sequence in Lp. Thus, the proofof (a) shows the statement of (b) to be true if g is replaced by the function f ∈ Lp

constructed in the proof of (a). Then dp(f, g) ≤ dp(f, fk) + dp(fk, g) → 0 for k → ∞,showing dp(f, g) = 0, i.e. f = g µ-a.e., thereby establishing the case.

In Th. 2.44(b), for p < ∞, one can, in general, not expect for the sequence (fk)k∈∞ toconverge pointwise µ-a.e. to some f ∈ Lp, as is demonstrated by the following example:Consider the measure space ([0, 1],B1, β1) and the sequence of intervals

(Ik)k∈N :=

([0, 1],

[0,

1

2

],

[1

2, 1

],

[0,

1

3

],

[1

3,2

3

],

[2

3, 1

],

[0,

1

4

], . . .

).

If, for each k ∈ N, fk := χIk , then, for each p ∈]0,∞[, limk→∞ dp(fk, 0) = 0: Indeed,

limk→∞

Npp (fk) = lim

k→∞

∫ 1

0

χpIkdβ1 = lim

k→∞β1(Ik) = lim

k→∞

1

k= 0.

However, for no x ∈ [0, 1] does (fk(x))k∈N converge.

2.6.2 Integration with Respect to Pushforward Measures

Given a measure space (X,A, µ), a measurable space (Y,B), and an A-B-measurableφ : X −→ Y , recall the pushforward measure

µφ : B −→ [0,∞], µφ(B) = µ(φ−1(B)

),

from Th. 1.69(a).

Theorem 2.45. (a) Let (X,A, µ) be a measure space, (Y,B) a measurable space, φ :X −→ Y A-B-measurable, and µφ the resulting pushforward measure on (Y,B).Then, for each f ∈ M+(B), one has

∫

Y

f dµφ =

∫

X

(f φ) dµ . (2.54)

Moreover, a B-measurable function f : Y −→ K is µφ-integrable if, and only if,f φ is µ-integrable, and then (2.54) holds.

2 INTEGRATION 105

(b) Let n ∈ N and consider (Rn,A, µn), where (A, µn) is either (Bn, βn) or (Ln, λn).Moreover, let φ : Rn −→ Rn be bijective and affine. If f : Rn −→ K is nonnegativeµn-measurable or µn-integrable, then f φ has the same property and

∫

Rn

f dµn = | detφ|∫

Rn

(f φ) dµn . (2.55)

Proof. (a): We note that f ∈ M+(B) implies f φ ∈ M+(A) and then use the techniquesummarized after the proof of Th. 2.36: First, let f = χB, B ∈ B. Then

∫

Y

χB dµφ = µφ(B) = µ(φ−1(B)

)=

∫

X

χφ−1(B) dµ =

∫

X

(χB φ) dµ ,

such that (2.54) holds. Due to the linearity of the integral, (2.54) then also holds foreach f ∈ S+(B). Now let f ∈ M+(B) and choose a sequence (uk)k∈N in S+(B) suchthat uk ↑ f . Then, clearly, (uk φ)k∈N is a sequence in S+(A) such that (uk φ) ↑ (f φ),implying

∫

Y

f dµφ = limk→∞

∫

Y

uk dµφ = limk→∞

∫

X

(uk φ) dµ =

∫

X

(f φ) dµ ,

proving (2.54) for each f ∈ M+(B). The remaining assertion of (a) follows by applying(2.54) to (Re f)± and (Im f)±.

(b): According to Th. 1.73(a), φ is µn-µn-measurable and (µn)φ = | detφ|−1 µn. Thus,

if f : Rn −→ K is µn-measurable, then so is f φ. Using (a), if f : Rn −→ K isnonnegative µn-measurable or µn-integrable, then f φ has the same property and

∫

Rn

(f φ) dµn (2.54)=

∫

Rn

f d(µn)φ = | detφ|−1

∫

Rn

f dµn ,

completing the proof of (b).

2.6.3 Dense Subsets, Separability

Below, we provide some dense subsets of Lp-spaces (for p < ∞ – L∞ turns out tohave a less benign structure in most situations). Dense subsets are, e.g., useful, sincecontinuous functions are already uniquely determined by values on dense subsets. Recallthat a topological space is called separable if, and only if, it has a countable dense subset.Separable spaces are typically much easier to handle than nonseparable spaces. We willprovide some sufficient conditions for Lp-spaces to be separable.

Remark 2.46. Let (X,A, µ) be a measure space, p ∈]0,∞], and let N denote the setof measurable f : X −→ K that vanish µ-almost everywhere. Let ι : Lp

K(µ) −→ LpK(µ),

ι(f) := [f ], be the quotient map. Consider F ⊆ Lp and F := ι(F) ⊆ Lp. Then F isdense in Lp if, and only if, F is dense in Lp (in particular, Lp is separable if, and only if,Lp is separable): Note that dp(f, g) = dp(ι(f), ι(g)) for each f, g ∈ Lp. Let F be densein Lp. Given ι(g) ∈ Lp and ǫ > 0, choose f ∈ F such that dp(f, g) < ǫ. Then ι(f) ∈ F

2 INTEGRATION 106

and dp(ι(f), ι(g)) < ǫ, showing F to be dense in Lp. Conversely, let F be dense in Lp.Given g ∈ Lp and ǫ > 0, choose f ∈ F such that dp(ι(f), ι(g)) < ǫ. Then dp(f, g) < ǫ,showing F to be dense in Lp.

Theorem 2.47. Let (X,A, µ) be a measure space, p ∈]0,∞[.

(a) The vector space over K, defined by

S0 := S0(A, µ) := spanχA : A ∈ A, µ(A) <∞,

i.e. the vector space of simple functions that are 0 outside a set of finite measure,is dense in Lp: More precisely,

∀f∈Lp

∀ǫ∈R+

∃φ∈S0

(|φ| ≤ |f | ∧ dp(f, φ) < ǫ

).

(b) If S ⊆ A is a semiring on X such that σ(S) = A and µ S is σ-finite, then thevector space over K, defined by

D := spanχA : A ∈ S, µ(A) <∞,

is dense in Lp.

(c) If there exists a countable semiring S on X such that A = σ(S) and µS is σ-finite,then Lp is separable.

(d) If there exists a countable semiring S on X such that E = σ(S), E ⊆ A ⊆ E, where(X, E , α) is the completion of (X, E , α), α := µE , µ = α A, and µS is σ-finite,then Lp(µ) is separable.

(e) Let n ∈ N and consider (Rn,B, µn), where (B, µn) = (Bn, βn) or (B, µn) = (Ln, λn).If B ∈ B and (X,A, µ) = (B,B|B, µnA), then Lp(µ) is separable.

Proof. (a): Clearly, S0(A, µ) ⊆ Lp. Let f ∈ Lp. First, we also assume f ≥ 0. Thenthere exists a sequence (φk)k∈N in S+(A) such that φk ↑ f . Then Np(φk) ≤ Np(f) <∞,showing φk ∈ S0. Since |f − φk|p ≤ |f |p, we can apply the dominated convergence Th.2.20 to obtain,

limk→∞

dp(f, φk) = limk→∞

Np(f − φk) = limk→∞

∫

X

|f − φk|p dµ Th. 2.20= 0,

proving the assertion for f ≥ 0. For a general f ∈ Lp, we proceed in the usual way,letting f1 := (Re f)+, f2 := (Re f)−, f3 := (Im f)+, f4 := (Im f)−. Then each fk ∈ Lp

and fk ≥ 0, k = 1, . . . , 4. Given ǫ > 0, choose φk ∈ S0 such that φk ≤ fk anddp(fk, φk) <

ǫ4and let φ := φ1 − φ2 + i(φ3 − φ4) ∈ S0. Then

|φ|2 = (φ1 + φ2)2 + (φ3 + φ4)

2 ≤ (f1 + f2)2 + (f3 + f4)

2 = |f |2

2 INTEGRATION 107

and, for p ≥ 1,

Np(f − φ) = Np

(f1 − φ1 − (f2 − φ2) + i((f3 − φ3)− (f4 − φ4))

)≤

4∑

k=1

Np(fk − φk) < ǫ.

For p < 1, one replaces Np by Npp in the above estimate, thereby completing the proof

of (a).

(b): According to (a), it suffices to show that, given E ∈ A with µ(E) < ∞ and ǫ > 0,there exist disjoint sets A1, . . . , AN ∈ S, N ∈ N, such that µ(Ak) <∞ and

Npp

(χE −

N∑

k=1

χAk

)=

∫

X

∣∣∣χE − χ⋃Nk=1 Ak

∣∣∣p

dµ

= µ

(E \

N⋃

k=1

Ak

)+ µ

((N⋃

k=1

Ak

)\ E)< ǫ.

To obtain the Ak, we use the Caratheodory extension Th. 1.38 together with the unique-ness result of Cor. 1.46(a) (which holds, since µS is σ-finite) to see that µ on A = σ(S)must be the restriction of the outer measure on P(X) induced by µ on S. Thus, ifR := ρ(S) is the generated ring, then

µ(E)(1.30)= inf

∞∑

k=1

µ(Ak) : (Ak)k∈N sequence in S,E ⊆∞⋃

k=1

Ak

Prop. 1.18(b)= inf

∞∑

k=1

µ(Ak) : (Ak)k∈N sequence in R,E ⊆∞⋃

k=1

Ak

= inf

∞∑

k=1

µ(Ak) : (Ak)k∈N disjoint sequence in R,E ⊆∞⋃

k=1

Ak

Prop. 1.18(a)= inf

∞∑

k=1

µ(Ak) : (Ak)k∈N disjoint sequence in S,E ⊆∞⋃

k=1

Ak

.

In consequence, since µ(E) < ∞, there exists a sequence of disjoint sets (Ak)k∈N ∈ S,E ⊆ ⋃∞

k=1Ak and µ(⋃∞

k=1Ak

)< µ(E) + ǫ

2. Then

limN→∞

µ

(E \

N⋃

k=1

Ak

)= 0 and lim

N→∞µ

((N⋃

k=1

Ak

)\ E)

≤ µ

( ∞⋃

k=1

Ak

)− µ(E) <

ǫ

2.

Thus, there exists N ∈ N such that

µ

(E \

N⋃

k=1

Ak

)+ µ

((N⋃

k=1

Ak

)\ E)<ǫ

2+ǫ

2= ǫ,

and the proof of (b) is complete.

2 INTEGRATION 108

(c): According to (b), D = spanKχA : A ∈ S, µ(A) < ∞ is dense in LpK(µ). Since Q

is dense in R and Q(i) := q+ ir : q, r ∈ Q is dense in C, it follows that the countableset spanQχA : A ∈ S, µ(A) <∞, where Q := Q for K = R and Q := Q(i) for K = C,is dense in Lp

K(µ).

(d): According to (c), Lp(α) is separable. Let D be a countable dense subset of Lp(α).If f ∈ Lp(µ), then, by Prop. G.3, there exists g ∈ Lp(α) such that f = g α-a.e. Thus,given ǫ > 0, there exists φ ∈ D, satisfying dp(f, φ) = dp(g, φ) < ǫ, showing D to bedense in Lp(µ).

(e): First, consider the case (B, µn) = (Bn, βn). We know that Bn is generated by thecountable semiring In

Q: Bn = σ(InQ). If B ⊆ Rn, then S := In

Q|B is a countable semiringon B by Prop. G.2. For B ∈ Bn, according to Th. 1.56(b), A = Bn|B = σB(S) and,for µ = βnA, Lp(µ) is separable by (c). The case (B, µn) = (Ln, λn) now follows from(d).

It turns out that, in many respects, L∞ is typically less benign than Lp for p <∞. Forexample, in general, the space S0 of Th. 2.47(a) is not dense in L∞ and, in the situationsof Th. 2.47(c),(d),(e), L∞ is typically not separable (cf. Rem. 2.55 below).

Definition 2.48. (a) Let (X, T ) be a topological space. Given a function f : X −→ K,we call the set

supp f := x ∈ X : f(x) 6= 0the support of f . Continuous functions whose support is a compact subset of X areof particular importance, giving rise to the notation

Cc(X) := Cc(X,K) :=f ∈ C(X,K) : supp f is compact

.

If O ⊆ Rn is open, n ∈ N, then we also define

∀k∈N0∪∞

Ckc (O) := Ck

c (O,K) := Cc(O,K) ∩ Ck(O,K).

(b) If O ⊆ Rn is open, n ∈ N, then a function f : O −→ K is said to vanish atinfinity if, and only if, for each ǫ ∈ R+, there exists a compact set K ⊆ O such that|f(x)| < ǫ for each x ∈ O \K (for O = Rn, this is, clearly, equivalent to

limx→∞

f(x) = 0, (2.56)

where (2.56) is defined to mean that, for each sequence (xk)k∈N in Rn such thatlimk→∞ ‖xk‖ = ∞, one has limk→∞ f(xk) = 0). Define

C0(O) := C0(O,K) :=(f ∈ C(O,K) : f vanishes at infinity

.

Clearly, Cc(O) ( C0(O) (if O = Rn, then f(x) := e−‖x‖22 defines f ∈ C0(O) \Cc(O);if O 6= Rn, then f(x) := dist(x,Oc) · e−‖x‖22 defines f ∈ C0(O) \ Cc(O)).

Theorem 2.49. Let n ∈ N and let O ⊆ Rn be open and consider (O,A, µn), where(A, µn) is either (Bn, βn) or (Ln, λn).

2 INTEGRATION 109

(a) For each p ∈]0,∞[, Cc(O,K) is dense in LpK(µ

n).

(b) For each p ∈]0,∞[ and f ∈ LpK(R

n,A, µn), one has

limt→0

∫

Rn

∣∣f(x+ t)− f(x)∣∣p dµn(x) = 0. (2.57)

Proof. (a): It suffices to consider (A, µn) = (Ln, λn). Due to Th. 2.47(a), it suffices toshow that, for each A ∈ A with λn(A) <∞ and each ǫ > 0, there exists g ∈ Cc(O) suchthat Np

p (χA − g) < ǫ. Let δ := min ǫ2, ( ǫ

2)p. Choose k ∈ N sufficiently large such that,

for B := A∩ [−k, k]n, one has λn(A) ≤ λn(B)+δ. Then Npp (χA−χB) < δ. According to

(1.41), there exists a closed set K ⊆ Rn and an open set V ⊆ Rn such that K ⊆ B ⊆ Vand λn(V \K) < δ. If we let U := V ∩O ∩ (]− k− 1, k+1[)n, then U is still open withB ⊆ U and λn(U \K) < δ, but U is also bounded and U ⊆ O. Moreover, since K is asubset of the bounded set B, K is compact. If K = ∅, then λn(B) < δ and λn(A) < ǫ,such that we can choose g ≡ 0. Thus, consider K 6= ∅. Then, using the continuity ofdist (cf. [Phi16b, Ex. 2.6(b)]), we observe dist(K,U c) > 0 and define the continuousfunction

g : O −→ R, g(x) := 1−min

1,

dist(x,K)

dist(K,U c)

. (2.58)

Then, clearly, 0 ≤ g ≤ 1, gK≡ 1, and gUc≡ 0. The support of g is a closed subset ofthe compact set U , i.e. supp g is compact and g ∈ Cc(O). Moreover, noting

|χB(x)− g(x)| = 1− 1 = 1− 1 = |χU(x)− χK(x)| for x ∈ K,

|χB(x)− g(x)| = 1− g(x) ≤ 1− 0 = |χU(x)− χK(x)| for x ∈ B \K,|χB(x)− g(x)| = |0− g(x)| ≤ 1− 0 = |χU(x)− χK(x)| for x ∈ U \B,|χB(x)− g(x)| = 0− 0 = 0− 0 = |χU(x)− χK(x)| for x ∈ U c,

we obtain

Npp (χB − g) =

∫

O

|χB − g|p dλn ≤∫

O

|χU − χK |p dλn < δ.

Then, for p < 1, Npp (χA − g) ≤ Np

p (χA − χB) + Npp (χB − g) < ǫ

2+ ǫ

2= ǫ. For p ≥ 1,

Np(χA− g) ≤ Np(χA−χB)+Np(χB − g) < 2δ1/p ≤ 2 ǫ2= ǫ, completing the proof of (a).

(b): Seeking a contradiction, assume there exists ǫ > 0 such that for each s > 0, thereexists s > t > 0 with

∫Rn

∣∣f(x+ t)−f(x)∣∣p dµn(x) ≥ ǫ. Using (a), choose g ∈ Cc(Rn,K)

such that∫Rn

∣∣f(x) − g(x)∣∣p dµn(x) < ǫ

3. Since g is continuous and, thus, bounded on

the compact set supp g, there exists M ∈ R+ such that 2p|g|p ≤ M . Choosing r > 0such that supp g ⊆ Br(0) and letting B := Br(0), M χB is integrable and dominatesx 7→ |g(x+t)−g(x)|p for all sufficiently small t ≥ 0. Thus, by the dominated convergenceTh. 2.20,

limt→0

∫

Rn

∣∣g(x+ t)− g(x)∣∣p dµn(x)

Th. 2.20=

∫

Rn

limt→0

∣∣g(x+ t)− g(x)∣∣p dµn(x)

g cont.= 0.

In consequence,

∃s>0

∀0≤t<s

∫

Rn

∣∣g(x+ t)− g(x)∣∣p dµn(x) <

ǫ

3

2 INTEGRATION 110

and, for each 0 ≤ t < s,

∫

Rn

∣∣f(x+ t)− f(x)∣∣p dµn(x)

≤∫

Rn

∣∣f(x+ t)− g(x+ t)∣∣p dµn(x) +

∫

Rn

∣∣g(x+ t)− g(x)∣∣p dµn(x)

+

∫

Rn

∣∣g(x)− f(x)∣∣p dµn(x)

(2.55)<

ǫ

3+ǫ

3+ǫ

3= ǫ.

This contradiction proves (2.57).

We will see in Th. 2.54(d) below that even C∞c (O,K) is dense in Lp

K(µn) for p ∈]0,∞[.

One can actually extend Th. 2.49(a) to measure spaces (X,A, µ), where µ is a suitablemeasure on a locally compact Hausdorff space (see, e.g., [Rud87, Th. 3.14]).

2.6.4 Convolution and Fourier Transform

We will now define the so-called convolution of two integrable functions on Rn.

Remark and Definition 2.50. Let n ∈ N and consider (Rn,A, µn), where (A, µn) iseither (Bn, βn) or (Ln, λn). Consider f, g ∈ L1

K(µn). Then

h : R2n −→ K, h(x, y) := f(x− y)g(y),

is µ2n-measurable: Indeed, the function A : R2n −→ R2n, A(x, y) := (x− y, y), is linearand bijective and, thus, µ2n-µ2n-measurable. The projections π1, π2 : R2n −→ Rn,π1(x, y) = x, π2(x, y) = y are µ2n-µn-measurable. Thus, as f, g are µn-measurable, andh = (f π1 A) · (g π2), h is µ2n-measurable. Next, we compute

∫

Rn

∫

Rn

∣∣f(x− y)g(y)∣∣ dµn(x) dµn(y) =

∫

Rn

∣∣g(y)∣∣∫

Rn

∣∣f(x− y)∣∣ dµn(x) dµn(y)

(2.55)= ‖f‖1

∫

Rn

∣∣g(y)∣∣ dµn(y) = ‖f‖1 ‖g‖1 <∞.

Since h is measurable, we may apply Th. 2.36(c) to conclude that h is µ2n-integrableand there exists a µn-null set N ⊆ Rn such that y 7→ h(x, y) is µn-integrable for eachx ∈ N c. We now define

f ∗ g : Rn −→ K, (f ∗ g)(x) :=∫

Rn f(x− y)g(y) dµn(y) for x ∈ N c,

0 for x ∈ N,(2.59)

and call this function the convolution of f and g. From our considerations above, wethen know f ∗ g ∈ L1

K(µn). If f , g ∈ L1

K(µn) with f = f µn-a.e. and g = g µn-a.e.,

then, for each x ∈ Rn, we have f(x − y)g(y) = f(x − y)g(y) for a.e. y ∈ Rn. Thus,y 7→ f(x − y)g(y) is µn-integrable for each x ∈ N c, and f ∗ g = f ∗ g. In particular,convolution “∗” is well-defined as a map from L1

K(µn)× L1

K(µn) to L1

K(µn).

2 INTEGRATION 111

Proposition 2.51. Let n ∈ N and consider (Rn,A, µn), where (A, µn) is either (Bn, βn)or (Ln, λn). Let

∗ : L1K(µ

n)× L1K(µ

n) −→ L1K(µ

n)

be the convolution map as defined in Rem. and Def. 2.50 above.

(a) Convolution is commutative:

∀f,g∈L1

K(µn)

f ∗ g = g ∗ f.

(b) Convolution is distributive:

∀f,g,h∈L1

K(µn)

(f + g) ∗ h = f ∗ h+ g ∗ h µn-a.e.

(c) Convolution is associative:

∀f,g,h∈L1

K(µn)

(f ∗ g) ∗ h = f ∗ (g ∗ h) µn-a.e.

(d) supp(f ∗ g) ⊆ supp f + supp g.

Proof. (a): Since, for each x ∈ Rn, the map φx : Rn −→ Rn, φx(y) := x − y, isaffine and bijective with determinant | detφx| = 1, (2.55) implies, for each x ∈ Rn,y 7→ f(x− y)g(y) to be µn-integrable if, and only if, y 7→ f(y)g(x− y) is µn-integrableand, in that case,

(f ∗ g)(x) =∫

Rn

f(x− y)g(y) dµn(y) =

∫

Rn

f(y)g(x− y) dµn(y) = (g ∗ f)(x).

(b) is due to the fact that, for each x, y ∈ Rn,

(f + g)(x− y)h(y) =(f(x− y) + g(x− y)

)h(y) = f(x− y)h(y) + g(x− y)h(y),

which is applied for each x ∈ Rn such that y 7→ (f + g)(x− y)h(y) is µn-integrable.

(c): We apply Fubini for each x ∈ Rn such that y 7→ (f ∗ g)(x− y)h(y) is µn-integrableto obtain

((f ∗ g) ∗ h

)(x) =

∫

Rn

(f ∗ g)(x− y)h(y) dµn(y)(2.55)=

∫

Rn

(f ∗ g)(y)h(x− y) dµn(y)

(a)=

∫

Rn

(∫

Rn

f(z)g(y − z) dµn(z)

)h(x− y) dµn(y)

Fubini=

∫

Rn

f(z)

∫

Rn

g(y − z)h(x− y) dµn(y) dµn(z)

(2.55)=

∫

Rn

f(z)

∫

Rn

g(x− y − z)h(y) dµn(y) dµn(z)

=

∫

Rn

f(z)(g ∗ h)(x− z) dµn(z)(2.55)=(f ∗ (g ∗ h)

)(x),

2 INTEGRATION 112

proving associativity.

(d): If (f ∗ g)(x) 6= 0, then∫Rn f(x− y)g(y) dµn(y) 6= 0, i.e. there exists y ∈ supp g such

that x− y ∈ supp f , showing x = y + (x− y) ∈ supp f + supp g and proving (d).

Definition 2.52. Let n ∈ N. We call a sequence (ϕk)k∈N in L1(βn) a Dirac sequence if,and only if, it has the following properties (i) – (iii):

(i) ϕk ≥ 0 for each k ∈ N,

(ii) ‖ϕk‖1 = 1 for each k ∈ N,

(iii) limk→∞ diam(0 ∪ suppϕk

)= 0.

A Dirac sequence can be seen as an approximation of the Dirac measure δ0 : P(Rn) −→[0,∞] (cf. Ex. 1.12(c)) which, in turn, constitutes a representation of the so-called Diracdistribution (concentrated at 0).

Example 2.53. Let n ∈ N.

(a) Clearly,

ϕk : Rn −→ K, ϕk(x) :=

(2/k)−n for x ∈ [− 1

k, 1k]n,

0 otherwise,

defines a (discontinuous) Dirac sequence.

(b) The sequence defined by

ϕk : Rn −→ K, ϕk(x) :=

n∏

j=1

max0, k − k2|xj|,

defines a continuous (but not differentiable) Dirac sequence: Clearly, ϕk ≥ 0,suppϕk = [− 1

k, 1k]n, and, using Fubini,

‖ϕk‖1 =∫

Rn

ϕk =n∏

j=1

(2

∫ 1/k

0

(k − k2 xj) dxj

)=

n∏

j=1

(2

[kxj −

k2x2j2

]1/k

0

)= 1.

(c) We show the sequence defined by

ϕk : Rn −→ K, ϕk(x) :=

ck exp

(− 1

k−2−‖x‖22

)for ‖x‖2 < 1

k,

0 otherwise,

where

ck :=

(∫

B

exp

(− 1

k−2 − ‖x‖22

)dx

)−1

, B := x ∈ Rn : ‖x‖2 < 1/k,

2 INTEGRATION 113

defines a Dirac sequence in C∞c (Rn): Clearly, each ϕk maps into R+

0 , ‖ϕk‖1 = 1by the definition of ck, and suppϕk = Bk−1,‖·‖2(0), showing (ϕk)k∈N to be a Diracsequence. It remains to verify that each ϕk is C∞. To this end, we show that,for each p = (p1, . . . , pm) ∈ 1, . . . , nm, m ∈ N, ∂pϕk exists. Let x ∈ Rn. For‖x‖2 > 1

k, ∂pϕk(x) = 0 is immediate from the definition of ϕk. We claim that, for

‖x‖2 < 1k,

∂pϕk(x) = ck

(Np∑

j=1

lp,j Qp,j(x)(α(k, x)

)j

)exp

(− 1

α(k, x)

), (2.60)

where α(k, x) := (k−2 − ‖x‖22), Np ∈ N, each lp,j ∈ Z, and each Qp,j is a monomialin x1, . . . , xn: To prove (2.60) by induction, fix p = (p1, . . . , pm) ∈ 1, . . . , nm,m ∈ N, assume (2.60) to hold for this p, and let ν ∈ 1, . . . , n. Then we have todifferentiate ∂pϕk with respect to xν using the product rule, resulting in

∂ν∂pϕk(x) = ck

(Np∑

j=1

lp,j ∂νQp,j(x)(α(k, x)

)j

)exp

(− 1

α(k, x)

)

+ ck

(Np∑

j=1

(−j) (−2xν) lp,j Qp,j(x)(α(k, x)

)j+1

)exp

(− 1

α(k, x)

)

+ ck

(Np∑

j=1

lp,j Qp,j(x)(α(k, x)

)j

) (− 2xν(

α(k, x))2

)exp

(− 1

α(k, x)

)

= ck

(Nq∑

j=1

lq,j Qq,j(x)(α(k, x)

)j

)exp

(− 1

α(k, x)

),

where q := (ν, p1, . . . , pm), Nq ∈ N, each lq,j ∈ Z, and each Qq,j a monomial inx1, . . . , xn, completing the induction. Finally, suppose ‖x‖2 = k−1. We prove viainduction that, for each p = (p1, . . . , pm) ∈ 1, . . . , nm, m ∈ N, ∂pϕk(x) = 0: Onceagain, we assume the claim for a fixed p and let ν ∈ 1, . . . , n. Let (tγ)γ∈N bea sequence in R \ 0 such that, with xγ := x + tγ eν , (xγ)γ∈N is a sequence inBk−1,‖·‖2(0), satisfying limγ→∞ xγ = x. Then, for each γ ∈ N,

α(k, xγ) = k−2 − ‖xγ‖22‖x‖22 − ‖xγ‖22 = x2ν − (xν + tγ)2 = −2xνtγ − t2γ

= |tγ| (2|xν | − |tγ|) > 0

and

limγ→∞

α(k, xγ) = 0 ⇒ limγ→∞

exp

(− 1

α(k, xγ)

)= 0

⇒ ∀j∈N

limγ→∞

((α(k, xγ)

)−jexp

(− 1

α(k, xγ)

))= 0.

This, together with (2.60), t−1γ = (−2xν − tγ)

(α(k, xγ)

)−1, and limγ→∞Qp,j(xγ) =

Qp,j(x) ∈ R, implies limγ→∞(t−1γ ∂pϕk(xγ)

)= 0 and, thus,

∂ν∂pϕk(x) = limγ→∞

∂pϕk(xγ)− 0

tγ= 0,

2 INTEGRATION 114

completing the proof that ϕk is C∞.

Theorem 2.54. Let n ∈ N.

(a) If the sequence (ϕk)k∈N in L1(βn) is a Dirac sequence according to Def. 2.52, then

∀f∈L1(λn)

limk→∞

‖ϕk ∗ f − f‖1 = 0. (2.61)

(b) If f, g ∈ L1(λn) and at least one of the functions f, g is bounded, then f ∗ g isuniformly continuous.

(c) If f ∈ L1(λn) and g ∈ C∞c (Rn), then f ∗ g ∈ C∞(Rn), and, for each p =

(p1, . . . , pk) ∈ 1, . . . , nk, k ∈ N,

∂p(f ∗ g) = ∂p1 . . . ∂pk(f ∗ g) = f ∗ (∂p g). (2.62)

(d) If O ⊆ Rn is open, then C∞c (O,K) is dense in Lp

K(O,Ln, λn) for each p ∈]0,∞[.

Proof. (a): Let f ∈ L1(λn) and ǫ > 0. According to (2.57),

∃δ>0

∀t∈Bδ(0)

∫

Rn

∣∣f(x+ t)− f(x)∣∣ dλn(x) < ǫ.

Since (ϕk)k∈N is a Dirac sequence, there exists N ∈ N such that, for each k > N ,suppϕk ⊆ Bδ(0). Thus, for each k > N ,

‖ϕk ∗ f − f‖1 =

∫

Rn

∣∣∣∣∫

Rn

ϕk(x− y)f(y) dλn(y) − f(x)

∣∣∣∣ dλn(x)

=

∫

Rn

∣∣∣∣∫

Rn

ϕk(y)(f(x− y)− f(x)

)dλn(y)

∣∣∣∣ dλn(x)

Fubini

≤∫

Rn

ϕk(y)

∫

Rn

∣∣f(x− y)− f(x)∣∣ dλn(x) dλn(y)

≤∫

Rn

ϕk(y) ǫ dλn(y) = ǫ,

proving (a).

(b): Let f, g ∈ L1(λn). Since convolution is commutative, it suffices to consider thecase, where g is bounded. If g is bounded, then, for each x ∈ Rn, y 7→ f(y)g(x − y) isλn-integrable, implying y 7→ f(x− y)g(y) to be λn-integrable as well. If x, h ∈ Rn, then

|(f ∗ g)(x+ h)− (f ∗ g)(x)| ≤∫

Rn

∣∣f(x+ h− y)− f(x− y)∣∣ ∣∣g(y)

∣∣ dµn(y)

=

∫

Rn

∣∣f(y + h)− f(y)∣∣ ∣∣g(x− y)

∣∣ dµn(y)

≤ ‖g‖∞∫

Rn

∣∣f(y + h)− f(y)∣∣ dµn(y) .

2 INTEGRATION 115

Thus, according to (2.57), for each ǫ > 0, there exists δ > 0 (not depending on x ∈ Rn),such that |(f ∗g)(x+h)−(f ∗g)(x)| < ǫ for each h ∈ Bδ(0), proving f ∗g to be uniformlycontinuous.

(c): Let f ∈ L1(λn), g ∈ C∞c (Rn), j ∈ 1, . . . , n. Clearly, ∂jg ∈ C∞

c (Rn) as well,and ∂jg is uniformly continuous by [Phi16b, Th. 3.20]. Thus, given ǫ > 0, there existsδ > 0, such that |∂jg(u) − ∂jg(v)| < ǫ for each u, v ∈ Rn with ‖u − v‖1 < δ. Let ejdenote the jth standard unit vector of Rn. Then, for each x ∈ Rn and each real number0 6= t ∈]− δ, δ[, one computes

∣∣∣∣(f ∗ g)(x+ tej)− (f ∗ g)(x)

t−(f ∗ (∂jg)

)(x)

∣∣∣∣

=

∣∣∣∣∫

Rn

f(y)

(g(x+ tej − y)− g(x− y)

t− ∂jg(x− y)

)dλn(y)

∣∣∣∣

=

∣∣∣∣∫

Rn

f(y)

t

∫ t

0

(∂jg(x+ sej − y)− ∂jg(x− y)

)dλ1(s) dλn(y)

∣∣∣∣

≤ ‖f‖1 ǫ,proving ∂j(f ∗ g) = f ∗ (∂jg). Now (2.62) follows by induction, implying f ∗ g ∈ C∞(Rn)as well.

(d): According to Th. 2.49(a), Cc(O,K) is dense in LpK(λ

n) and it merely remains toshow that C∞

c (O) is dense in Cc(O). Thus, given f ∈ Cc(O) and ǫ > 0, we need to findg ∈ C∞

c (O) such that Np(f − g) < ǫ. If supp f = ∅, then f ≡ 0 and one sets g := f .Thus, let supp f 6= ∅. Extending f by 0, we regard f as an element of C∞

c (Rn). We alsomake use of the Dirac sequence (ϕk)k∈N from Ex. 2.53(c), which is in C∞

c (Rn). For eachk ∈ N, we have f ∗ ϕk ∈ C∞

c (Rn): Indeed, f ∗ ϕk ∈ C∞(Rn) by (c) and, since f and ϕk

both have compact support, so has f ∗ ϕk by Prop. 2.51(d). Let

α :=

dist(supp f,Oc) ∈ R+ for Oc 6= ∅,∞ for Oc = ∅.

Using the uniform continuity of f , we choose δ > 0 such that |f(u) − f(v)| < ǫ ·(βn(supp f) + 1))−1/p for each u, v ∈ Rn with ‖u− v‖ < δ. As

supp f =⋂

ν∈NKν , Kν := x ∈ Rn : dist(x, supp f) ≤ ν−1,

we can choose ν0 ∈ N such that βn(Kν0) < βn(supp f)+1 and ν−10 < α. Since (ϕk)k∈N is

a Dirac sequence, we can choose k0 ∈ N, such that suppϕk0 ⊆ Bβ(0), β := minδ, ν−10 .

Then, using∫Rn ϕk0 dλ

n = 1,

∀x∈Rn

∣∣(f ∗ ϕk0)(x)− f(x)∣∣ ≤

∫

Rn

|f(x− y)− f(x)|ϕk0(y) dλn(y)

=

∫

suppϕk0

|f(x− y)− f(x)|ϕk0(y) dλn(y)

≤ ǫ · (βn(supp f) + 1)−1/p.

2 INTEGRATION 116

We note, using Prop. 2.51(d), supp(f ∗ ϕk0) ⊆ supp f + suppϕk0 ⊆ Kν0 ⊆ O, sinceβ ≤ ν−1

0 < α. Thus, the estimate

Np((f ∗ ϕk0)− f) ≤ βn(Kν−10)1/p · ǫ · (βn(supp f) + 1)−1/p

< (βn(supp f) + 1)1/p · ǫ · (βn(supp f) + 1)−1/p = ǫ

concludes the proof.

Remark 2.55. As mentioned after the proof of Th. 2.47, L∞ is typically less benignthan Lp for p <∞:

(a) If µ(X) = ∞, then the space S0 of Th. 2.47(a) is not dense in L∞: If f ∈ S0, thenthere exists A ∈ A such that µ(A) <∞ and f Ac= 0, i.e. ‖f−χX‖∞ = ‖f−1‖∞ ≥ 1(note µ(Ac) = ∞).

(b) Consider (Rn,A, µn), n ∈ N, where (A, µn) is either (Bn, βn) or (Ln, λn). If B ∈ Ais such that µn(B) > 0, then L∞(B,A|B, µnA|B) is not separable (see [Alt06, Ex.2.17〈4〉]) – whereas the corresponding Lp with p <∞ is separable according to Th.2.47(e).

(c) Let O ⊆ Rn be open. In contrast to the case p < ∞ of Th. 2.54(d), Cc(O,K) isnot dense in L∞

K (O,A, µn) (using the same notation as in (b) above): Indeed, theclosure of C∞

c (O) with respect to ‖ · ‖∞ turns out to be the space C0(O) as definedin Def. 2.48(b) (see [Rud87, Th. 3.17]).

Definition 2.56. Consider (Rn,Ln, λn), n ∈ N. For each f ∈ L1C(λ

n), define

f : Rn −→ C, f(t) := (2π)−n/2

∫

Rn

e−i 〈t,x〉 f(x) dλn(x) , (2.63a)

f : Rn −→ C, f(t) := f(−t) = (2π)−n/2

∫

Rn

ei 〈t,x〉 f(x) dλn(x) , (2.63b)

where 〈·, ·〉 denotes the Euclidean scalar product on Rn (note that, since |e−i 〈t,x〉| =|ei 〈t,x〉| = 1, the integrability of f implies the integrability of both integrands in (2.63)).We call f the Fourier transform of f and f the inverse Fourier transform of f (thename for f is due to Th. 2.59 below). Note that, if f = g λn-a.e., then f = g and f = g,i.e. f and f are also well-defined for f ∈ L1

C(λn).

Theorem 2.57. Let n ∈ N and let f, g ∈ L1C(λ

n).

(a) Both Fourier transform and inverse Fourier transform are linear, i.e. if α, β ∈ C,then (αf + βg)∧ = αf + βg and (αf + βg)∨ = αf + βg.

(b) Riemann-Lebesgue Lemma: f ∈ C0(Rn), i.e. f is continuous with limt→∞ f(t) = 0.Moreover f is bounded: |f | ≤ (2π)−n/2 ‖f‖1.

(c) f ∗ g = (2π)n/2 f · g.

2 INTEGRATION 117

(d) For each a ∈ Rn and r ∈ R+, let

f ∗ : Rn −→ C, f ∗(x) := f(−x),fa : R

n −→ C, fa(x) := f(a+ x),

(Mrf) : Rn −→ C, (Mrf)(x) := rn f(rx).

Then, for each t ∈ Rn, one has

f ∗ (t) = f(t), (2.64a)

fa(t) = ei 〈a,t〉f(t), (2.64b)

Mrf(t) = f

(t

r

), (2.64c)

(e−i 〈a,x〉 f)∧(t) = (f)a(t). (2.64d)

(e) Let p = (p1, . . . , pk) ∈ 1, . . . , nk, k ∈ N. Using the notation xp :=∏k

j=1 xpj for

each x ∈ Rn, and assuming ∂pf exists as well as xqf ∈ L1 for each q = (pα, . . . , pk)with α ∈ 1, . . . , k, we have

∂pf = (−i)k(xpf)∧.

Proof. (a) is immediate from the linearity of the integral.

(b): One has

∀t∈Rn

(2π)n/2 |f(t)| =∣∣∣∣∫

Rn

e−i 〈t,x〉 f(x) dλn(x)

∣∣∣∣ ≤∫

Rn

|f(x)| dλn(x) = ‖f‖1.

Since, for each t ∈ Rn, the integrands in (2.63a) are dominated by the integrable |f |, fis continuous by Th. 2.22. We note that, for each t, x ∈ Rn, t 6= 0,

⟨t, x+

π

‖t‖22t

⟩= 〈t, x〉+ π,

implying (since e−iπ = −1)

(2π)n/2 f(t) =

∫

Rn

e−i 〈t,x〉 f(x) dλn(x)(2.55)= −

∫

Rn

e−i 〈t,x〉 f

(x+

π

‖t‖22t

)dλn(x)

and

2 (2π)n/2 |f(t)| ≤∫

Rn

∣∣∣∣f(x)− f

(x+

π

‖t‖22t

)∣∣∣∣ dλn(x) → 0 for t→ ∞

by (2.57).

2 INTEGRATION 118

(c): For each t ∈ Rn, we compute

f ∗ g(t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉∫

Rn

f(y)g(x− y) dλn(y) dλn(x)

Fubini= (2π)−n/2

∫

Rn

(∫

Rn

e−i 〈t,x−y〉 g(x− y) dλn(x)

)e−i 〈t,y〉 f(y) dλn(y)

= (2π)n/2 f(t) g(t),

proving (c).

(d): Let t ∈ Rn. Then

f ∗ (t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉 f(−x) dλn(x) (2.55)= (2π)−n/2

∫

Rn

ei 〈t,x〉 f(x) dλn(x)

= (2π)−n/2

∫

Rn

e−i 〈t,x〉 f(x) dλn(x) = f(t),

proving (2.64a);

fa(t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉 f(a+ x) dλn(x)(2.55)= ei 〈a,t〉f(t),

proving (2.64b);

Mrf(t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉 rn f(rx) dλn(x) = f

(t

r

),

proving (2.64c);

(e−i 〈a,x〉 f)∧(t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉 e−i 〈a,x〉 f(x) dλn(x) = f(t+ a) = (f)a(t),

proving (2.64d).

(e): Let j ∈ 1, . . . , n. According to the hypothesis, |xjf | is integrable and dominatesthe corresponding derivative of the integrand of (2.63a) and we can apply Cor. 2.24 toobtain

∂j f(t) = ∂tj f(t) = (2π)−n/2 (−i)∫

Rn

e−i 〈t,x〉 xj f(x) dλn(x) = (−i) (xjf)∧(t).

From this, the general formula of (e) follows by induction.

Example 2.58. (a) Let a ∈ R+ and consider

f : R −→ C, f := χ]−a,a[.

Then f ∈ L1 and, for each t ∈ R,

f(t) = (2π)−1/2

∫ a

−a

e−itx dx = (2π)−1/2

[−e

−itx

it

]a

−a

= (2π)−1/2

(−sin(−ta)

t+

sin(ta)

t

)= 2 (2π)−1/2 sin(ta)

t.

2 INTEGRATION 119

Note that f /∈ L1, since

(2 (2π)−1/2)−1

∫

R

|f | ≥∫ ∞

0

| sin(ta)|t

dt =

∫ ∞

0

| sin x|x

dx

≥∞∑

k=1

(1

kπ

∫ kπ

(k−1)π

| sin x| dx)

=2

π

∞∑

k=1

1

k= ∞.

(b) Considerφ : R −→ C, φ(x) := max0, 1− |x|. (2.65a)

Then φ ∈ L1 and we show

φ : R −→ C, φ(t) = (2π)−1/2 21− cos t

t2[Phi16a, (I.1c)]

= (2π)−1/2

(sin(t/2)

t/2

)2

.

(2.65b)We compute, for each t ∈ R,

φ(t) = (2π)−1/2

∫ 1

−1

e−itx(1− |x|) dx .

According to (a) (with a = 1),

∫ 1

−1

e−itx dx = 2sin t

t.

Moreover,

∫ 1

−1

e−itx|x| dx =

∫ 1

−1

|x| cos(tx) dx − i

∫ 1

−1

|x| sin(tx) dx = 2

∫ 1

0

x cos(tx) dx

= 2

[cos(tx)

t2+x sin(tx)

t

]1

0

= 2

(cos t

t2+

sin t

t− 1

t2

).

Putting everything together, we obtain (2.65b). By (2.65b), φ ∈ L1, since

∫

R

|φ| dλ1 = 2

∫ 1

0

|φ| dλ1 + 2

∫ ∞

1

|φ| dλ1 <∞ (2.65c)

(the first integral is finite, since |φ| is continuous on the bounded set [0, 1], thesecond integral is also finite, as

∫∞1

|φ| dλ1 ≤ 4(2π)−1/2∫∞1t−2 dt = 4(2π)−1/2). We

now show(φ)∨ = φ : (2.65d)

For each x ∈ R, we obtain

π (φ)∨(x) =1

2

∫

R

ei tx(sin(t/2)

t/2

)2

dt =

∫

R

t−2(1− cos t) cos(tx) dt .

2 INTEGRATION 120

Note

cos(tx+ t) + cos(tx− t) = cos(tx) cos t− sin(tx) sin t+ cos(tx) cos t− sin(tx) sin(−t)= 2 cos(tx) cos t.

Thus,

π (φ)∨(x) = 2

∫ ∞

0

t−2

(cos(tx)− 1

2

(cos(t(x+ 1)) + cos(t(x− 1))

))dt

= 2[− t−1(1− cos t) cos(tx)

]∞0

− 2

∫ ∞

0

t−1

(x sin(tx)− 1

2

((x+ 1) sin(t(x+ 1)) + (x− 1) sin(t(x− 1))

))dt .

We proceed to evaluate each term:

[− t−1(1− cos t) cos(tx)

]∞0

=

[t1− cos t

t2cos(tx)

]∞

0

= 0.

For x = 0, we have∫∞0t−1x sin(tx) = 0. For x > 0, we have

∫ ∞

0

t−1x sin(tx) = x

∫ ∞

0

sin u

udu

(F.2)= x

π

2.

For x < 0, we have

∫ ∞

0

t−1x sin(tx) = x

∫ −∞

0

sin u

udu

(F.2)= −x π

2.

Altogether, we obtain

∀x≤−1

π (φ)∨(x) = −π2

(−x+ x+ 1

2+x− 1

2

)= 0,

∀−1≤x≤0

π (φ)∨(x) = −π(−x− x+ 1

2+x− 1

2

)= π (x+ 1),

∀0≤x≤1

π (φ)∨(x) = −π(x− x+ 1

2+x− 1

2

)= π (−x+ 1),

∀1≤x

π (φ)∨(x) = −π2

(x− x+ 1

2− x− 1

2

)= 0,

showing (2.65d).

(c) Let n ∈ N. We will use the results of (b) to compute the Fourier transforms ofthe members of the Dirac sequence (ϕk)k∈N of Ex. 2.53(b), and we will show that(2.65d) still holds for φ replaced by ϕk. Recall that

ϕk : Rn −→ C, ϕk(x) :=

n∏

j=1

max0, k − k2|xj|.

2 INTEGRATION 121

Using φ from (2.65a) and the notation from Th. 2.57(d), we define, for each k ∈ N,

φk : R −→ C, φk(x) := (Mkφ)(x) = kφ(kx) = max0, k − k2|x|.Then (2.65b) together with (2.64c) yields

φk : R −→ C, φk(t) = φ

(t

k

)= (2π)−1/2

(sin(t/(2k))

t/(2k)

)2

.

In other words, φk = k (M1/kφ) and we can apply (2.64c) again to obtain, for eachx ∈ R,

(φk)∨(x) = (φk)(−x) = k

φ(−kx) = k(φ)∨(kx)

(2.65d)= kφ(kx) = φk(x),

showing∀

k∈N(φk)

∨ = φk.

Finally, for ϕk, we obtain, for each t ∈ Rn,

ϕk(t) = (2π)−n/2

∫

Rn

e−i 〈t,x〉 ϕk(x) dλn(x)

= (2π)−n/2

∫

Rn

n∏

j=1

(e−i tjxjφk(xj)

)dλn(x)

Fubini=

n∏

j=1

φk(tj) = (2π)−n/2

n∏

j=1

(sin(tj/(2k))

tj/(2k)

)2

, (2.66a)

and, for each x ∈ Rn,

(ϕk)∨(x) = (2π)−n/2

∫

Rn

ei 〈t,x〉 ϕk(t) dλn(t)

= (2π)−n/2

∫

Rn

n∏

j=1

(ei tjxj

(sin(tj/(2k))

tj/(2k)

)2)

dλn(t)

Fubini=

n∏

j=1

(φk)∨(xj) =

n∏

j=1

φk(xj) = ϕk(x),

proving∀

k∈N(ϕk)

∨ = ϕk. (2.66b)

—

Due to Th. 2.57(b), not every function in L1 can be the Fourier transform of an L1

function. On the other hand, we know from Ex. 2.58(a) that the Fourier transform ofan L1 function is not necessarily in L1. Still one has the following inversion Th. 2.59(a)for the Fourier transform of L1 functions. But the point is that, due to the mentionedshortcomings of the Fourier transform for L1 functions, one has to assume f ∈ L1 in Th.2.59(a). There is a subspace S ( L1 ∩L2, the space of so-called Schwartz functions, onwhich the Fourier transform is, indeed, bijective, but the details are beyond the scopeof the present class.

2 INTEGRATION 122

Theorem 2.59 (Inversion Theorem). Let n ∈ N.

(a) If f ∈ L1C(λ

n) and f ∈ L1C(λ

n), then

f = (f)∨ λn-a.e.

(the equality becomes exact for f, f ∈ L1C(λ

n)).

(b) If f, g ∈ L1C(λ

n), then f = g implies f = g λn-a.e. (i.e. the Fourier transform isinjective as a map on L1

C(λn)).

Proof. (a): We will make use of the fact that, from (2.66b) of Ex. 2.58(c), we alreadyknow ϕk = (ϕk)

∨ for each member of the Dirac sequence (ϕk)k∈N of Ex. 2.53(b). Sinceϕk is bounded by Th. 2.57(b) and f ∈ L1, we have ϕkf ∈ L1 as well. Thus, we obtain,for each x ∈ Rn,

(ϕkf)∨(x) = (2π)−n/2

∫

Rn

ei 〈x,t〉 ϕk(t)f(t) dλn(t)

= (2π)−n/2 (2π)−n/2

∫

Rn

ei 〈x,t〉 ϕk(t)

∫

Rn

e−i 〈t,z〉 f(z) dλn(z) dλn(t)

Fubini= (2π)−n/2 (2π)−n/2

∫

Rn

f(z)

∫

Rn

ei 〈t,x−z〉 ϕk(t) dλn(t) dλn(z)

= (2π)−n/2

∫

Rn

f(z) (ϕk)∨(x− z) dλn(z)

(2.66b)= (2π)−n/2

∫

Rn

f(z)ϕk(x− z) dλn(z) = (2π)−n/2 (f ∗ ϕk)(x). (2.67)

Next, we note that limk→∞ ϕk(t) = (2π)−n/2 for each t ∈ Rn due to (2.66a), and

‖ϕk‖∞ ≤ (2π)−n/2 ‖ϕk‖1 = (2π)−n/2

by Th. 2.57(b) and Def. 2.52(ii). Thus, we can apply the dominated convergence Th.2.20 to obtain

limk→∞

‖(2π)n/2 ϕkf − f‖1 = limk→∞

∫

Rn

∣∣(2π)n/2 ϕk(t)f(t)− f(t)∣∣ dλn(t) DCT

= 0.

Recalling that f(t) = f(−t), we can apply Th. 2.57(b) again to conclude

limk→∞

∥∥∥(2π)n/2 (ϕkf)∨ − (f)∨

∥∥∥∞

= limk→∞

∥∥∥(2π)n/2 (ϕkf)∧ − (f)∧

∥∥∥∞

≤ (2π)−n/2 limk→∞

∥∥∥(2π)n/2 ϕkf − f∥∥∥1= 0,

i.e. uniform convergence (2π)n/2 (ϕkf)∨ → (f)∨. Applying Prop. G.4 of the Appendix

then yields, for each r ∈ R+,∫

Br(0)

|(f)∨ − f | dλn (G.4a)= lim

k→∞

∫

Br(0)

|(2π)n/2 (ϕkf)∨ − f | dλn

(2.67)= lim

k→∞

∫

Br(0)

|(f ∗ ϕk)− f | dλn

≤ limk→∞

‖ϕk ∗ f − f‖1(2.61)= 0,

2 INTEGRATION 123

proving (f)∨ = f λn-a.e. as claimed.

(b): If f = g, then (f − g)∧ = 0. Then, according to (a), f − g = 0 λn-a.e., proving(b).

Theorem 2.60 (Plancherel). Let n ∈ N.

(a) Fourier transform is an L2-isometry. More precisely, if f ∈ L1C(λ

n)∩L2C(λ

n), then

f ∈ L2C(λ

n) and ∫

Rn

|f |2 dλn =

∫

Rn

|f |2 dλn .

(b) If f, g ∈ L1C(λ

n) ∩ L2C(λ

n), then∫

Rn

f g dλn =

∫

Rn

f g dλn .

Proof. (a): Consider f ∈ L1 ∩ L2. According to Th. 2.57(d), if we let f ∗ : Rn −→ C,

f ∗(x) := f(−x), then f ∗ = f . Since f ∈ L1, we have f ∗ ∈ L1, and we can formg := f ∗ f ∗ ∈ L1. For a, b ∈ R, we know (a − b)2 = a2 − 2ab + b2 ≥ 0, which we canapply with a := |f(x− y)|, b := |f(−y)| to obtain

∀x,y∈Rn

∣∣∣f(x− y) f(−y)∣∣∣ ≤ 1

2

(|f(x− y)|2 + |f(−y)|2

).

Since f ∈ L2, this implies y 7→ f(x − y) f(−y) to be integrable for each x ∈ Rn and,thus,

∀x∈Rn

g(x) =

∫

Rn

f(x− y) f(−y) dλn(y) =

∫

Rn

f(x+ y) f(y) dλn(y) .

We use this equation for g together with the Cauchy-Schwarz inequality (i.e. (2.49) withp = q = 2) to estimate, for each x, x′ ∈ Rn,

|g(x)− g(x′)|2 ≤(∫

Rn

∣∣f(x+ y)− f(x′ + y)∣∣ |f(y)| dλn(y)

)2

(2.49)

≤(∫

Rn

∣∣f(x+ y)− f(x′ + y)∣∣2 dλn(y)

)·(∫

Rn

|f(y)|2 dλn(y)),

where the right-hand side converges to 0 for x → x′ (due to Th. 2.49(b) and f ∈ L2).In other words, the function g is continuous and

∀ǫ∈R+

∃δ∈R+

∀x∈Bδ(0)

|g(x)− g(0)| < ǫ.

Once again, we consider the Dirac sequence (ϕk)k∈N of Ex. 2.53(b). We choose N ∈ Nsuch that suppϕk ⊆ Bδ(0) for each k ≥ N . Then,

∀k≥N

|(g ∗ ϕk)(0)− g(0)| ϕk(−x)=ϕk(x)=

∣∣∣∣∫

Rn

(g(x)− g(0)

)ϕk(x) dλ

n(x)

∣∣∣∣ < ǫ

2 INTEGRATION 124

and we have shown

limk→∞

(g ∗ ϕk)(0) = g(0) =

∫

Rn

|f |2 dλn .

Since g ∗ ϕk ∈ L1, we may form its Fourier transform and use Th. 2.57(c) to concludeg ∗ ϕk = (2π)n/2 g · ϕk ∈ L1 (since g is bounded by Th. 2.57(b) and ϕk ∈ L1 (e.g. due to(2.65c), (2.66a), and Tonelli)). Thus, according to the inversion Th. 2.59(a),

((2π)n/2 g · ϕk

)∨=(g ∗ ϕk

)∨= g ∗ ϕk,

where the equality even holds exactly (not merely λn-a.e.), since g ∗ϕk is continuous byTh. 2.54(b). Next, we apply Fatou (Prop. 2.19) together with limk→∞ ϕk(t) = (2π)−n/2

and g = f ∗ f ∗ = (2π)n/2 f f = (2π)n/2 |f |2 to obtain

∫

Rn

|f |2 dλn = (2π)n/2∫

Rn

limk→∞

ϕk |f |2 dλn ≤ lim infk→∞

∫

Rn

ϕk g dλn

= (2π)n/2 lim infk→∞

(ϕk g)∨(0) = lim inf

k→∞(g ∗ ϕk)(0) =

∫

Rn

|f |2 dλn . (2.68)

In (2.68), the right-hand side is finite, showing f ∈ L2. Since ‖ϕk‖∞ ≤ 1, this alsoshows that, instead of applying Fatou, one can actually use the dominated convergenceTh. 2.20 to obtain (2.68) with limits instead of limit inferiors and with equality insteadof the estimate, thereby proving completing the proof of (a).

(b): Since, for each u, v ∈ C, one has

uv =1

4

((u+ v)(u+ v)− (u− v)(u− v) + i (u+ iv)(u− iv)− i (u− iv)(u+ iv)

)

=1

4

(|u+ v|2 − |u− v|2 + i |u+ iv|2 − i |u− iv|2

),

(b) is a direct consequence of (a).

Example 2.61. The Plancherel Th. 2.60 can sometimes be used to evaluate integralsin a simple manner: According to Ex. 2.58(a), for

fa : R −→ C, fa := χ]−a,a[, a ∈ R+,

we have

fa : R −→ C, fa(t) = 2 (2π)−1/2 sin(ta)

t.

Thus, for each a, b ∈ R+,

∫ ∞

−∞

sin(at) sin(bt)

t2dt =

2π

4

∫

R

fa fb dλ1 Th. 2.60(b)

=π

2

∫

R

fa fb dλ1 = π mina, b.

Here fa, fb ∈ L1 ∩ L2 and, even though fa, fb are not integrable (according to Ex.

2.58(a)), fa fb turns out to be integrable.

2 INTEGRATION 125

2.7 Change of Variables

Our mail goal in this section is to prove the change of variables theorem for (subsets of)Rn (Th. 2.66 below). It generalizes Th. 1.73 and (2.55) regarding the transformation ofβn and λn (and corresponding integrals) with respect to affine bijective transformationsφ to general diffeomorphisms φ (see Def. and Rem. 2.64). It also partially general-izes the one-dimensional version of [Phi16a, Th. 10.25]. The n-dimensional version issignificantly harder to prove and we will need some preparation. Still, for simplicity,the hypothesis in Th. 2.66 will be stronger than necessary. The hypothesis that thetransformation is a diffeomorphism can be relaxed, see, e.g., [Els07, Th. V.4.10] and[Rud87, Th. 7.26]. As shown in [Els07], one obtains [Els07, Th. V.4.10] from Th. 2.66with not too much extra work. However, the version [Rud87, Th. 7.26] makes use of theBrouwer fixed-point theorem of algebraic topology (see, e.g., [Oss09, 5.6.10]). We beginour preparations:

Proposition 2.62. If (X,A, µ) is a measure space and f : X −→ [0,∞] is measurable,then

fµ : A −→ [0,∞], (fµ)(A) :=

∫

A

f dµ , (2.69)

defines a measure on (X,A).

Proof. We have (fµ)(∅) =∫∅ f dµ = 0 and, if (Ai)i∈N is a sequence of disjoint sets in

A, then

(fµ)

( ∞⋃

i=1

Ai

)=

∫

X

(f

∞∑

i=1

χAi

)dµ

Cor. 2.8=

∞∑

i=1

∫

X

f χAidµ

=∞∑

i=1

∫

Ai

f dµ =∞∑

i=1

(fµ)(Ai), (2.70)

thereby establishing the case.

Definition 2.63. If µ, ν are measures on the measurable space (X,A), then a measur-able function f : X −→ [0,∞] is called a density of ν with respect to µ if, and only if,ν = fµ with fµ as in (2.69).

Definition and Remark 2.64. Let n ∈ N and let O,U ⊆ Rn be open. A bijectivemap φ : O −→ U is called a C1-diffeomorphism if, and only if, both φ and φ−1 arecontinuously differentiable. If φ and φ−1 are differentiable, then, by the chain rule,

∀x∈O

D(φ−1)(φ(x)

)Dφ(x) = Id,

showing all Dφ(x) and all D(φ−1)(φ(x)

)to be invertible (i.e. they have nonzero de-

terminants). Thus, by the inverse function theorem ([Phi16b, Th. 4.50]), a bijectivecontinuously differentiable map φ : O −→ U is a C1-diffeomorphism if, and only if,detDφ has no zeros.

2 INTEGRATION 126

Definition and Remark 2.65. Let n ∈ N. Given a measure space (Rn,A, µ) andA ∈ A, we introduce the abbreviations

AA := A|A, µA := µAA.

Let O,U ⊆ Rn be open. If φ : O −→ U is a homeomorphism (in particular, if φ is aC1-diffeomorphism), then

BnU = φ(A) : A ∈ Bn

O :

Indeed, both φ and φ−1 : U −→ O are continuous and, thus, Borel-measurable. Thus,if A ∈ Bn

O, then φ(A) = (φ−1)−1(A) ∈ BnU . On the other hand, if B ∈ Bn

U , thenA := φ−1(B) ∈ Bn

O, implying B = φ(A).

Theorem 2.66 (Change of Variables). Let n ∈ N. Consider (Rn,A, µn), n ∈ N, where(A, µn) is either (Bn, βn) or (Ln, λn). Let O,U ⊆ Rn be open and let φ : O −→ U be aC1-diffeomorphism.

(a) |det(Dφ)|−1 is a density of φ(µnO) with respect to µn

U , i.e.

φ(µnO) = |det(Dφ)|−1 µn

U , (2.71a)

∀B∈AU

µn(φ−1(B)

)=

∫

B

|det(Dφ)|−1 dµn (2.71b)

∀A∈AO

µn(φ(A)

)=

∫

A

| det(Dφ)| dµn . (2.71c)

(b) f ∈ M+(U,AU) or f : U −→ K is µn-integrable if, and only if, (f φ)| det(Dφ)|has the corresponding property with respect to O and, in that case,

∫

U

f dµn =

∫

O

(f φ)| det(Dφ)| dµn . (2.72)

(c) Let A ∈ AO. If f ∈ M+(φ(A),Aφ(A)) or f : φ(A) −→ K is µn-integrable, then(2.72) holds with (O,U) replaced by (A, φ(A)).

Proof. We consider the countable semiring

S :=

]a, b] : a, b ∈

∞⋃

k=1

2−kZn, a ≤ b, [a, b] ⊆ O

(S is countable as the set of admissible interval endpoints is countable; that S is asemiring follows, for n = 1, in the same manner as in Ex. 1.16(b) and, then, for n > 1,from Prop. 1.17(a)). Since every open subset of O is the (countable) union of sets fromS, we have σ(S) = Bn

O. We devide the proof into several steps: The first step consistsof showing that (2.71c) holds with ≤ for each A ∈ S. This already contains the mainwork of the proof.

2 INTEGRATION 127

Claim 1.

∀I∈S

βn(φ(I)

)≤∫

I

| det(Dφ)| dβn .

Proof. Let ǫ > 0. The key idea is to write I ∈ S as the union of sufficiently smallintervals Iν and, on each Iν , to approximate φ by a suitable affine map xν + Lν . It willbe convenient to use the max-norm ‖·‖max = ‖·‖∞ on Rn. RecallingDφ(x), D(φ−1)(x) ∈Rn2

, it would also be convenient to use the corresponding operator norm ‖ · ‖∞ on Rn2,

since it satisfies ‖T (x)‖∞ ≤ ‖T‖∞ ‖x‖∞ for each T ∈ Rn2and each x ∈ Rn. But

if one is not familiar with the concept of an operator norm, one can avoid it, onecan recall from [Phi16b, Lem. 5.4] that there exists a norm ‖ · ‖HS on Rn2

, satisfying‖T (x)‖2 ≤ ‖T‖HS ‖x‖2 for each T ∈ Rn2

and each x ∈ Rn. Thus, given an arbitrarynorm ‖ · ‖ on Rn2

, by the equivalence of norms, there exists a constant c0 ∈ R+ suchthat

∀T∈Rn2

∀x∈Rn

‖T (x)‖∞ ≤ c0 ‖T‖ ‖x‖∞.

In other words, using balls with respect to ‖ · ‖∞,

∀T∈Rn2

∀ρ>0

T (Bρ(0)) ⊆ Bc0‖T‖ρ(0). (2.73)

For a ∈ I, consider the map

ha : O −→ Rn, ha(x) := φ(x)−(Dφ(a)

)(x),

Dha : O −→ Rn2

, Dha(x) := Dφ(x)−Dφ(a).

Due to [Phi16b, Th. 4.38], there exists a constant c1 ∈ R+ such that, given an openconvex set C ⊆ O and α ∈ R+ with

∀a∈I

∀x∈C

∥∥Dha(x)∥∥ ≤ α,

each ha is (c1α)-Lipschitz on C with respect to ‖ · ‖∞, i.e.

∀a∈I

∀x,y∈C

‖ha(x)− ha(y)‖∞ ≤ c1 α ‖x− y‖∞. (2.74)

On the compact set I ⊆ O, the continuous map x 7→ ‖(Dφ(x))−1‖ attains its max

M := max‖(Dφ(x))−1‖ : x ∈ I

∈ R+. (2.75)

Since I ⊆ O, we have dist(I, Oc) > 0 and we may choose s > 0 such that

Ks :=x ∈ Rn : dist(x, I) ≤ s

⊆ O

(as we compute dist with respect to ‖ · ‖∞, Ks is even an interval). Since Ks is a closedand bounded subset of Rn, Ks is compact. Since the continuous map Dφ : O −→ Rn2

,x 7→ Dφ(x), is uniformly continuous on the compact set Ks (cf. [Phi16b, Th. 3.20]),there exists 0 < r ≤ s such that

∀a∈I

∀x∈Br(a)

∥∥Dφ(x)−Dφ(a)∥∥ ≤ ǫ

c1M. (2.76)

2 INTEGRATION 128

Since r ≤ s, we have Br(a) ⊆ Ks for each a ∈ I. According to (2.76),

∀a∈I

∀x∈Br(a)

∥∥Dha(x)∥∥ ≤ ǫ

c1M

and (2.74) yields

∀a∈I

∀x,y∈Br(a)

‖ha(x)− ha(y)‖∞ ≤ ǫ

M‖x− y‖∞. (2.77)

We now choose m ∈ N sufficiently large such that I =]a, b] with a, b ∈ 2−mZn. We canthen write

I =⋃N

ν=1Iν , N ∈ N,

with each Iν ∈ S being a hypercube with side length d := 2−m. Then βn(Iν) = 2−mn andβn(I) = N dn. By possibly enlarging m, we may assume diam Iν < r. Since det(Dφ) iscontinuous on the compact set Iν , for each ν ∈ 1, . . . , N, we may choose aν ∈ Iν suchthat ∣∣ det(Dφ)(aν)

∣∣ = min∣∣ det(Dφ)(x)

∣∣ : x ∈ Iν

. (2.78)

We apply (2.77) with y := aν to obtain, for each ν ∈ 1, . . . , N and each x ∈ Iν (usingIν ⊆ Br(aν) due to diam Iν < r)

∥∥φ(x)− φ(aν)−(Dφ(aν)

)(x− aν)

∥∥∞ = ‖haν (x)− haν (aν)‖∞ ≤ ǫ

M‖x− aν‖∞ <

ǫ d

M.

Letting Tν := Dφ(aν) and y := φ(x)− φ(aν)− Tν(x− aν) ∈ B ǫ dM(0), this shows

φ(x) = φ(aν) + Tν(x− aν) + y and φ(Iν) ⊆ φ(aν) + Tν(Iν − aν) + B ǫ dM(0).

Moreover,

B ǫ dM(0) = Tν

(T−1ν (B ǫ d

M(0))

) (2.73), (2.75)

⊆ Tν(Bc0 ǫ d(0)),

implyingφ(Iν) ⊆ φ(aν) + Tν

(Iν − aν +Bc0 ǫ d(0)

).

Now note that the set Iν − aν + Bc0 ǫ d(0) is contained in a hypercube with side lengthd+ 2 c0 ǫ d = d(1 + 2 c0 ǫ). In consequence, we obtain

βn(φ(Iν)

) (1.47b)

≤ | detTν | dn (1 + 2 c0 ǫ)n = | detTν | (1 + 2 c0 ǫ)

n βn(Iν)

and

βn(φ(I)) =N∑

ν=1

βn(φ(Iν)

)≤ (1 + 2 c0 ǫ)

n

N∑

ν=1

| detTν | βn(Iν)

(2.78)

≤ (1 + 2 c0 ǫ)n

∫

I

| det(Dφ)| dβn .

Since ǫ > 0 was arbitrary, this concludes the proof of Cl. 1. N

2 INTEGRATION 129

Claim 2.

∀A∈Bn

O

βn(φ(A)

)≤∫

A

| det(Dφ)| dβn .

Proof. The statement of Cl. 2 is precisely that (βn)φ−1 ≤ | det(Dφ)|βn holds for the twomeasures on Bn

O. However, from Cl. 1, we already know that the measures’ respectiverestrictions to the generating semiring S satisfy the inequality. Since the measures arealso σ-finite on S, the validity of Cl. 2 is due to Cor. 1.46(b). N

Claim 3.

∀f∈M+(U,Bn

U)

∫

U

f dβn ≤∫

O

(f φ)| det(Dφ)| dβn .

Proof. Let f = χB for B ∈ BnU , A := φ−1(B). Then

∫

U

χB dβn = βn(B) = βn(φ(A)

) Cl. 2

≤∫

A

| det(Dφ)| dβn =

∫

O

(χB φ)| det(Dφ)| dβn ,

proving Cl. 3 for measurable characteristic functions. It then also holds for each simplefunction f ∈ S+(U,Bn

U ) by the linearity of the integral. For a general f ∈ M+(U,BnU ), we

choose a sequence (fk)k∈N in S+(U,BnU) with fk ↑ f and use the monotone convergence

Th. 2.7 to estimate∫

U

f dβn = limk→∞

∫

U

fk dβn ≤ lim

k→∞

∫

O

(fk φ)| det(Dφ)| dβn MCT=

∫

O

(f φ)| det(Dφ)| dβn

concluding the proof of Cl. 3. N

Claim 4.

∀f∈M+(U,Bn

U )

∫

U

f dβn =

∫

O

(f φ)| det(Dφ)| dβn .

Proof. We apply Cl. 3 with φ−1 instead of φ and (f φ)| det(Dφ)| instead of f to obtain∫

O

(f φ)| det(Dφ)| dβn ≤∫

U

(((f φ)| det(Dφ)|

) φ−1

)| det(Dφ−1)| dβn =

∫

U

f dβn ,

since((Dφ)

(φ−1))

Dφ−1 = Id by the chain rule. The inequality just proved, togetherwith the one of Cl. 3, establishes Cl. 4. N

Applying Cl. 4 to f = χA with A ∈ BnO, we obtain (2.71c) for (A, µn) = (Bn, βn). In

particular, (2.71c) then says that βn(A) = 0 if, and only if, βn(φ(A)) = 0, showing thatφ provides a bijection between βn-null sets, also implying φ to be λn-measurable. Inconsequence, (2.71c) also holds for (A, µn) = (Ln, λn). We then obtain (2.71b) (which isequivalent to (2.71a)) by applying (2.71c) with φ−1 instead of φ, completing the proof of(a). According to Cl. 4, (b) holds for (A, µn) = (Bn, βn) and f ≥ 0. If f ∈ M+(U,Ln

U ),then, by Prop. G.3, there exists g ∈ M+(U,Bn

U), such that g = f βn-a.e. Then∫

U

f dλn =

∫

U

g dβn =

∫

O

(g φ)| det(Dφ)| dβn =

∫

O

(f φ)| det(Dφ)| dλn ,

3 INTEGRATION OVER SUBMANIFOLDS OF RN 130

where the last equality makes, again, use of the fact that φ maps βn-null sets to βn-nullsets. We have, thus, proved (2.72) for f ∈ M+(U,Ln

U). The µn-integrable case thenalso follows, in the usual way, by decomposing f into (Re f)± and (Im f)±, completingthe proof of (b). Finally, (c) is obtained by applying (b) with f replaced by g with

g : U −→ K, g(x) :=

f(x) for x ∈ φ(A),

0 otherwise.

This concludes the proof of the theorem.

Example 2.67. Let

O := R+×]0, 2π[, U := R2 \ (x, 0) : x ≥ 0. (2.79)

As shown in Sec. H of the Appendix, the function defining polar coordinates, i.e.

φ : O −→ U, φ(r, ϕ) := (r cosϕ, r sinϕ), (2.80)

constitutes a C1-diffeomorphism with detDφ(r, ϕ) = r for each (r, ϕ) ∈ O. Since(x, 0) : x ≥ 0, 0 × [0, 2π], R+

0 × 0, and R+0 × 2π all are λ2-null sets, we can use

Th. 2.66(b) to conclude, for each f ∈ M+(R2,L2),∫

R2

f dλ2 =

∫

O

f(r cosϕ, r sinϕ) r dλ2(r, ϕ) . (2.81)

Moreover, f : R2 −→ K is λ2-integrable if, and only if, (r, ϕ) 7→ r f(r cosϕ, r sinϕ) isλ2-integrable, and then (2.81) holds. We can use (2.81) in combination with Fubini togive a third proof of ∫ ∞

−∞e−x2

dx =√π : (2.82)

We compute

(∫ ∞

−∞e−x2

dx

)2

=

∫

R2

e−x2−y2 dλ2 (x, y)(2.81)=

∫

O

r e−r2 dλ2(r, ϕ)

=

∫ ∞

0

∫ 2π

0

r e−r2 dϕ dr = 2π

∫ ∞

0

r e−r2 dr = 2π

[−1

2e−r2

]∞

0

= π,

proving (2.82).

3 Integration Over Submanifolds of Rn

3.1 Submanifolds of Rn

Roughly speaking, a k-dimensional submanifold of Rn is a subset that, locally, lookslike an open subset of Rk. There are a number of equivalent ways of making this ideaprecise, see Def. 3.1 and Th. 3.4 below.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 131

Definition 3.1. Let k, n ∈ N, k < n, α ∈ N ∪ ∞. A set M ⊆ Rn is called a k-dimensional submanifold of class Cα if, and only if, for each a ∈M , there exists an openneighborhood Oa ⊆ Rn of a and a map fa ∈ Cα(Oa,Rn−k), satisfying the following twoconditions

(a) M ∩Oa = (fa)−1(0),

(b) rkDfa(a) = n − k, i.e. the n − k vectors ∇ fa1 (a), . . . ,∇ fa

n−k(a) are linearly inde-pendent elements of Rn.

Submanifolds of dimension k = 1 are called paths, curves or lines; submanifolds ofdimension k = 2 are called surfaces, submanifolds of dimension k = n − 1 are calledhypersurfaces.

Example 3.2. Let n ∈ N, n ≥ 2.

(a) Perhaps the most simple submanifolds of Rn are its affine subspaces: Let k ∈ N,k < n, and let V ⊆ Rn be a k-dimensional vector space over R. Moreover, letxM ∈ Rn and M := V +xM . Then M is a k-dimensional submanifold of Rn of classC∞: Let v1, . . . , vn be a basis of Rn such that v1, . . . , vk is a basis of V . LetA : Rn −→ Rn−k be the linear map defined by

A(vi) :=

0 for i ∈ 1, . . . , k,vi for i ∈ k + 1, . . . , n,

andf : Rn −→ Rn−k, f(x) := A(x− xM).

We can now let, for each a ∈M , Oa := Rn, fa := f . We verify Def. 3.1(a),(b) to besatisfied: First note that f is C∞, since f is the composition of a translation witha linear map, f = A T−xM

. Moreover, for each x ∈ Rn,

f(x) = A(x− xM) = 0 ⇔ x− xM ∈ V ⇔ x ∈ V + xM =M,

proving Def. 3.1(a); and

rkDf(x) = rkA = n− k,

proving Def. 3.1(b).

(b) Each Euclidean sphere is an (n− 1)-dimensional submanifold of Rn: Let xM ∈ Rn

and r > 0, M := x ∈ Rn : ‖x − xM‖2 = r. As in (a), for each a ∈ M , we maytake Oa := Rn and the same fa := f for each a ∈M , where

f : Rn −→ R, f(x) := ‖x− xM‖22 − r2 =n∑

i=1

(xi − xM,i)2 − r2.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 132

Since f is a polynomial, it is C∞. Since

M = x ∈ Rn : f(x) = 0,Def. 3.1(a) is satisfied. Since

∀x∈Rn

Df(x) = ∇ f(x) = 2(x1 − xM,1, . . . , xn − xM,n

),

Df(x) 6= 0 for each x 6= xM . In particular, rkDf(a) = 1 for each a ∈M .

Definition 3.3. Let n ∈ N, α ∈ N ∪ ∞.

(a) Let O,U ⊆ Rn be open. A bijective map F : O −→ U is called a Cα-diffeomorphismif, and only if, F ∈ Cα(O,Rn) and F−1 ∈ Cα(U,Rn).

(b) Let k ∈ N, k ≤ n, and let O ⊆ Rk be open. A map φ ∈ Cα(O,Rn) is called aCα-immersion if, and only if,

∀x∈O

rkDφ(x) = k (3.1)

(i.e., for each x ∈ O, the k vectors ∂1φ(x), . . . , ∂kφ(x) are linearly independentelements of Rn).

(c) Let π ∈ Sn be a permutation of 1, . . . , n. Then, clearly,Pπ : Rn −→ Rn, Pπ(x1, . . . , xn) := (xπ(1), . . . , xπ(n)), (3.2)

the map that reorders the variables according to π, is a linear isomorphism (inparticular, a C∞-diffeomorphism).

Theorem 3.4. Given k, n ∈ N with k < n, α ∈ N ∪ ∞, and M ⊆ Rn, the followingstatements are equivalent:

(i) M is a k-dimensional submanifold of class Cα.

(ii) M can, locally, be represented as the graph of a Cα-function, depending on k vari-ables, i.e., for each a ∈ M , there exists a permutation πa ∈ Sn and open neigh-borhoods Aa and Ba of αa and βa, respectively, where αa := (aπa(1), . . . , aπa(k)),βa := (aπa(k+1), . . . , aπa(n)), i.e.

αa ∈ Aa ⊆ Rk, βa ∈ Ba ⊆ Rn−k,

plus a Cα-map ga : Aa −→ Ba, satisfying

M ∩ Pτa(Aa ×Ba) = Pτa(

(x, y) ∈ Aa ×Ba : y = ga(x)), τa := (πa)−1. (3.3)

(iii) For each a ∈M , there exists an open neighborhood Va ⊆ Rn, an open set Ua ⊆ Rn,and a Cα-diffeomorphism F a : Va −→ Ua, satisfying

F a(M ∩ Va) = Ek ∩ Ua, (3.4)

whereEk := x ∈ Rn : xk+1 = · · · = xn = 0. (3.5)

3 INTEGRATION OVER SUBMANIFOLDS OF RN 133

(iv) For each a ∈ M , there exists an M-open neighborhood Ma ⊆ M (open with re-spect to the relative topology on M), an open set Wa ⊆ Rk, and a Cα-immersionφa : Wa −→ Rn, such that Ma = φa(Wa) and φa : Wa −→ Ma is a homeomor-phism. The maps φa are called (local) charts or (local) parametrizations or (local)coordinates of M .

Proof. (i) ⇒ (ii): Given a ∈ M , let fa : Oa −→ Rn−k be according to Def. 3.1. SincerkDfa(a) = n− k, there exists πa ∈ Sn such that

det

∂πa(k+1)fa1 · · · ∂πa(n)f

a1

......

∂πa(k+1)fan−k · · · ∂πa(n)f

an−k

(a) 6= 0.

Let Ga := Pπa(Oa). Then Ga ⊆ Rn is open by [Phi16b, Cor. 4.51]. Let τa := (πa)−1.Then Pτa(Ga) = Oa, Pτa(α

a, βa) = a. If we let

ha : Ga −→ Rn−k, ha(x, y) := fa(Pτa(x, y)

), x ∈ Rk, y ∈ Rn−k,

then, by the chain rule,

Dha(αa, βa) = Dfa(Pτa(α

a, βa))Pτa = Dfa(a)Pτa

=

∂πa(1)fa1 · · · ∂πa(k+1)f

a1 · · · ∂πa(n)f

a1

......

...∂πa(1)f

an−k · · · ∂πa(k+1)f

an−k · · · ∂πa(n)f

an−k

(a)

(in matrix form, Pτa is a permutation matrix that such that right multiplication of A byPτa permutes the columns of A according to (τa)−1 = πa). Thus, ha(αa, βa) = fa(a) = 0,detDyh

a(αa, βa) 6= 0, and the implicit function theorem [Phi16b, Th. 4.49] provides openneighborhoods Aa ⊆ Rk of αa and Ba ⊆ Rn−k of βa, and a Cα-map ga : Aa −→ Ba suchthat

(Aa ×Ba) ∩ (ha)−10 =(x, y) ∈ Aa ×Ba : y = ga(x)

.

Then

z ∈M ∩ Pτa(Aa ×Ba) ⇔ z ∈ (fa)−1(0) ∩ P(πa)−1(Aa ×Ba)

⇔ Pπa(z) ∈ (ha)−1(0) ∩ (Aa ×Ba)

⇔ Pπa(z) ∈(x, y) ∈ Aa × Ba : y = ga(x)

⇔ z ∈ Pτa(x, y) ∈ Aa × Ba : y = ga(x)

,

proving (ii).

(ii) ⇒ (iii): Given a ∈ M , let πa ∈ Sn, (αa, βa) = Pπa(a), Aa ⊆ Rk, Ba ⊆ Rn−k, and

ga : Aa −→ Ba be according to (ii). Define τa := (πa)−1, Va := Pτa(Aa × Ba). Then Vais an open neighborhood of a and we define

F a : Va −→ Rn, F a(z) :=(x, y − ga(x)

), where (x, y) := Pπa(z).

3 INTEGRATION OVER SUBMANIFOLDS OF RN 134

Then ga being Cα implies F a to be Cα. Moreover, detF a ≡ 1, since

DF a =

Idk 0

−Dga Idn−k

Pπa .

Then Ua := F a(Va) is open by [Phi16b, Cor. 4.51] and F a : V a −→ Ua constitutes aCα-diffeomorphism. Finally,

z ∈ F a(M ∩ Va) ⇔ z ∈ F a(M ∩ Pτa(Aa × Ba)

)

(3.3)⇔ z ∈ F a(Pτa(x, y) ∈ Aa × Ba : y = ga(x)

)

⇔ z ∈ Ek ∩ Ua,

proving (iii).

(iii) ⇒ (i): For a ∈ M , let F a : Va −→ Ua be given by (iii) such that (3.4) holds. Weset Oa := Va and define

fa : Oa −→ Rn−k, ∀j∈1,...,n−k

faj := F a

k+j. (3.6)

Then fa is Cα, since F a is Cα; rkDfa(a) = n− k, since rkDF a(a) = n (using that F a

is a diffeomorphism). Moreover, for each x ∈ Rn,

fa(x) = 0 ⇔ F ak+1(x) = · · · = F a

n (x) = 0 ⇔ F (x) ∈ Ek ∩ Ua ⇔ x ∈M ∩Oa,

proving M ∩ Oa = (fa)−1(0) and, therefore, that M is a k-dimensional submanifoldof class Cα.

(ii) ⇒ (iv): Given a ∈ M , let πa ∈ Sn, (αa, βa) = Pπa(a), Aa ⊆ Rk, Ba ⊆ Rn−k, and

ga : Aa −→ Ba be according to (ii). Define τa := (πa)−1, Wa := Aa, and

φa : Wa −→ Rn, φa(x) := Pτa(x, ga(x)).

Then φa is Cα, since ga is Cα. As, clearly, rkDφa = k, φa is a Cα-immersion. SetMa := φa(Wa). Then φ

a(αa) = Pτa(αa, βa) = a, shows a ∈Ma. Due to (3.3), we have

Ma = Pτa(

(x, y) ∈ Aa ×Ba : y = ga(x))

=M ∩ Pτa(Aa × Ba),

showing Ma to be M -open. Since

(φa)−1 : Ma −→ Wa, (φa)−1(z) = x, where (x, y) := Pπa(z),

is, clearly, continuous, φa : Wa −→Ma is a homeomorphism, proving (iv).

3 INTEGRATION OVER SUBMANIFOLDS OF RN 135

(iv) ⇒ (iii): Given a ∈ M , let Wa ⊆ Rk, φa : Wa −→ Rn, Ma = φa(Wa) be accordingto (iv). Since Ma is M -open, there exists Oa ⊆ Rn such that Ma =M ∩Oa. Let c ∈ Wa

be such that φa(c) = a. Since rkDφa(c) = k, there exists πa ∈ Sn such that

det

∂1φ

aπa(1) · · · ∂kφ

aπa(1)

......

∂1φaπa(k) · · · ∂kφ

aπa(k)

(c) 6= 0.

According to the inverse function theorem [Phi16b, Th. 4.50] there exist open setsAa, Ba ⊆ Rk such that c ∈ Aa ⊆ Wa and

ϕa : Aa −→ Ba, ϕa(x) :=(φaπa(1)(x), . . . , φ

aπa(k)(x)

),

is a Cα-diffeomorphism. Let Ua := Aa × Rn−k and define, with (x, y) ∈ Aa × Rn−k,

ϕa : Ua −→ Ba × Rn−k,

ϕaj (x, y) :=

φaπa(j)(x) for 1 ≤ j ≤ k,

φaπa(j)(x) + yi for j = k + i, 1 ≤ i ≤ n− k,

τa := (πa)−1, andGa : Ua −→ Rn, Ga

j (z) := ϕaτa(j)(z).

Then Ga is Cα, since φa is Cα. Moreover, detDGa = detDϕa 6= 0, since

DGa = Pτa

Dϕa 0

Dx(ϕak+1, . . . , ϕ

an) Idn−k

.

In particular, Va := Ga(Ua) is open by [Phi16b, Cor. 4.51] and Ga : Ua −→ Va is aCα-diffeomorphism. Then F a := (Ga)−1 : Va −→ Ua is a Cα-diffeomorphism as well.Since Ga(c, 0) = φa(c) = a, it only remains to show (3.4), i.e.

Ga(Ek ∩ Ua) =M ∩ Va.To prove this equality, note

z ∈ Ga(Ek ∩ Ua) ⇔ z ∈ Ga(Aa × 0) ⇔ z ∈ φa(Wa) ∩ Va ⇔ z ∈M ∩ Va,showing (iii) and completing the proof of the theorem.

Theorem 3.5 (Chart Transition). Let k, n ∈ N with k < n, α ∈ N∪∞, and let M ⊆Rn be a k-dimensional submanifold of class Cα. For j ∈ 1, 2, let φj : Wj −→Mj ⊆Mbe local charts according to Th. 3.4(iv), i.e. Wj ⊆ Rk open, the Mj are M-open, andboth φj are Cα-immersions as well as homeomorphisms. Suppose V := M1 ∩M2 6= ∅.Then Oj := φ−1

j (V ) ⊆ Wj are open and

τ := φ−12 φ1 : O1 −→ O2 (3.7)

is a Cα-diffeomorphism. The map τ is then also called a transition between the chartsφ1 and φ2 (note φ1 = φ2 τ on O1).

3 INTEGRATION OVER SUBMANIFOLDS OF RN 136

Proof. Since V isM -open and φj are continuous, both Oj are open. Clearly, τ is bijective(injective, since the φj are injective, surjective by the choice of the Oj). Since φ

−12 is not

a Cα-map defined on an open subset of Rn, we can not directly apply the chain rule,but have to work slightly harder. Fix an arbitrary c1 ∈ O1 and set

a := φ1(c1), c2 := φ−12 (a) = τ(c1).

Since V is M -open, there exists U ⊆ Rn open such that V = U ∩M . Since a ∈ V , byTh. 3.4(iii), there exist open sets Va, Ua ⊆ Rn such that a ∈ Va ⊆ U and there exists aCα-diffeomorphism F a : Va −→ Ua, satisfying

F a(M ∩ Va) = Ek ∩ Ua.

We also have M ∩ Va ⊆M ∩ U = V . Let Wj := φ−1j (M ∩ Va) and define

g : W1 −→ Rk, g(x) :=((F a φ1)1, . . . , (F

a φ1)k)(x),

h : W2 −→ Rk, h(x) :=((F a φ2)1, . . . , (F

a φ2)k)(x).

Now we are in a situation, where the chain rule applies (to g and to h), yielding

∀x∈W1

Dg(x) = P DF a(φ1(x)) Dφ1(x),

∀x∈W2

Dh(x) = P DF a(φ2(x)) Dφ2(x),

where P denotes the projection from Rn to Rk. Since rkDφj(x) = k, Dφj must beinjective. Since F a is a diffeomorphism, D(F a φj) must still be injective. Since φj

maps intoM ∩Va and F a mapsM ∩Va into Ek (i.e. the last n−k components of F a φj

are indentically 0 on W1), both Dg(x) and Dh(x) must be surjective as well as injective,showing g : W1 −→ g(W1) and h : W2 −→ g(W2) to be Cα-diffeomorphisms. Thus,

τ W1= (φ−1

2 φ1)W1= h−1 g

must be a Cα-diffeomorphisms as well. We have shown that, for each c1 ∈ O1, thereexists an open neighborhood W1 of c1 such that τ is a Cα-diffeomorphism on W1. Sinceτ is also bijective on all of O1, it must be a Cα-diffeomorphism on all of O1 by Def. andRem. 2.64 (detDτ has no zeros on O1).

Theorem 3.6. Let k, n ∈ N with k < n and letM ⊆ Rn be a k-dimensional submanifoldof class C1. Then the following holds:

(a) M is the union of countably many compact sets.

(b) M ∈ Bn, i.e. M is βn-measurable (one can also show that M is even a βn-null set).

Proof. (a): Recall that each open setO ⊆ Rn is the countable union of compact intervals,for example of intervals from the set C := In

c,Q (defined in Ex. 1.8(c)). Let a ∈M and let

3 INTEGRATION OVER SUBMANIFOLDS OF RN 137

Oa, fa ∈ C1(Oa,Rn−k) be according to Def. 3.1. Then, there exists a sequence (Iai )i∈N

in C such that Oa =⋃

i∈N Iai . Thus,

M ∩Oa =M ∩Oa ∩⋃

i∈NIai .

Since C is countable, so is CM := Iai : a ∈M, i ∈ N. Let (Ii)i∈N be an enumeration ofall elements in CM . For each i ∈ N, choose ai ∈M such that Ii = Iaii . Then

M =M ∩⋃

i∈NIaii =

⋃

i∈N(M ∩ Iaii ) =

⋃

i∈N(M ∩Oai ∩ Iaii )

Def. 3.1(a)=

⋃

i∈N

((fa)−1(0) ∩ Iaii

).

Since fa is continuous, (fa)−1(0) is closed. As Iaii is compact, (fa)−1(0) ∩ Iaii iscompact as well, proving (a).

(b) follows from (a), as each compact set is in Bn.

3.2 Metric Tensor, Gram Determinant

Before we can integrate over submanifolds, we still need to introduce the metric tensorand the Gram determinant as technical tools.

Definition 3.7. Let k, n ∈ N with k < n and let M ⊆ Rn be a k-dimensional sub-manifold of class C1. Moreover, let φ : W −→ V ⊆ M be a local chart according toTh. 3.4(iv), i.e. W ⊆ Rk open, V is M -open, and φ is a C1-immersion as well as ahomeomorphism. Then the matrix-valued function

G := G(φ) : W −→ Rk2 , G(x) := (Dφ(x))tDφ(x), (3.8)

is called the chart’s Gram matrix or its metric tensor. If we denote the columns of Dφby ∂iφ, then we can write the components gij of G as follows:

∀x∈W

∀i,j∈1,...,k

gij(x) = 〈∂iφ(x), ∂jφ(x)〉 =n∑

ν=1

∂iφν(x) ∂jφν(x). (3.9)

The mapγ := γ(φ) : W −→ R, γ(x) := detG(x), (3.10)

is called the chart’s Gram determinant.

Theorem 3.8. Let k, n ∈ N with k < n and letM ⊆ Rn be a k-dimensional submanifoldof class C1.

(a) Let φ : W −→ V ⊆ M be a local chart according to Th. 3.4(iv), i.e. W ⊆ Rk

open, V is M-open, and φ is a C1-immersion as well as a homeomorphism. Thenits Gram matrix G = G(φ) : W −→ Rk2 is symmetric positive definite and, inconsequence, its Gram determinant γ = γ(φ) : W −→ R is actually R+-valued.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 138

(b) For j ∈ 1, 2, let φj : Oj −→ V ⊆ M be local charts, i.e. Oj ⊆ Rk open,V is M-open, and both φj are C1-immersions as well as homeomorphisms. Letγj := γj(φj) : Oj −→ R+ be the corresponding Gram determinants. Moreover, letτ = φ−1

2 φ1 : O1 −→ O2 be the corresponding chart transition according to Th.3.5 (i.e. φ1 = φ2 τ). Then the following formula for the transition of the Gramdeterminant holds:

∀x∈O1

γ1(x) = | detDτ(x)|2 γ2(τ(x)

). (3.11)

Proof. (a): Let x ∈ W and set A := Dφ(x). Then G(x) = AtA, i.e. G(x)t = (AtA)t =AtA = G(x), showing G(x) to be symmetric. If 0 6= y ∈ Rk, then ytG(x)y = ytAtAy =‖Ay‖2 > 0, since y 6= 0 and A injective (since rkA = k) imply Ay 6= 0. Thus, G(x) ispositive definite. In particular, γ(x) = detG(x) > 0.

(b): For each x ∈ O1, we have, by the chain rule,

Dφ1(x) = Dφ2(τ(x))Dτ(x).

Thus,

G(φ1)(x) =(Dφ2(τ(x))Dτ(x)

)t(Dφ2(τ(x))Dτ(x)

)=(Dτ(x)

)tG(φ2)(τ(x))Dτ(x),

implying (3.11).

3.3 Integration Over Submanifolds

If M ⊆ Rn is a k-dimensional submanifold, then the goal is to integrate f : M −→ Kover M . Since, locally, M “looks like” an open set of Rk, we would like to integratewith respect to a measure that, locally, “looks like” βk (or λk) on Rk. The natural ideais to transport the measure from Rk to M via local charts. While this does work, somesubtleties arise: One has to make sure that the measure does not depend on the choiceof local chart. To achieve this, the Gram determinant of the previous section comesinto play. The other issue one has to deal with are overlapping charts (M might notbe coverable with just one chart), and here one makes use of a so-called partition ofunity, see Def. 3.9(b) below. For simplicity, we will only consider submanifolds that arecoverable by finitely many local charts. The treatment in [Kon04, Sec. 11.4, 11.5] showsthat this assumption is not really necessary.

Definition 3.9. Let k, n ∈ N with k < n and let M ⊆ Rn be a k-dimensional subman-ifold of class C1.

(a) A family (φi)i∈I of local charts φi : Wi −→ Mi ⊆ M is called a (finite) cover of Mif, and only if, M =

⋃i∈I Mi (with I finite). Moreover, M is called finitely coverable

if, and only if, there exists a finite cover of local charts for M .

3 INTEGRATION OVER SUBMANIFOLDS OF RN 139

(b) Given a finite cover C := (φ1, . . . , φN), N ∈ N, of local charts for M , a family offunctions (η1, . . . , ηN), ηi : M −→ R, is called a partition of unity subordinate to Cif, and only if, the following conditions are satisfied:

(i) 0 ≤ ηi ≤ 1 for each i ∈ 1, . . . , N.(ii) ηiM\Mi

≡ 0 for each i ∈ 1, . . . , N.(iii)

∑Ni=1 ηi ≡ 1.

We call the partition of unity measurable if, and only if, each function ηi φi isλk-measurable.

(c) Let C := (φ1, . . . , φN), N ∈ N, be a finite cover of M via local charts. A functionf : M −→ K is called σ-measurable (or merely measurable if the context is clear) if,and only if, (f φi) is λ

k-measurable for each i ∈ 1, . . . , N. If f : M −→ [0,∞] ismeasurable and (η1, . . . , ηN) constitutes a measurable partition of unity subordinateto C, then we define the integral

∫

M

f dσ :=N∑

i=1

∫

Wi

((ηif) φi)√γ(φi) dλ

k , (3.12)

where γ(φi) denotes the Gram determinant of φi (recall that γ(φi) is R+-valued).If A ⊆M and χA is σ-measurable, then we call A σ-measurable and define

σ(A) :=

∫

M

χA dσ . (3.13)

We call f : M −→ K σ-integrable if, and only if, f is σ-measurable and∫Mg dσ <

∞ for each g ∈ (Re f)+, (Re f)−, (Im f)+, (Im f−). In that case, we define∫

M

f dσ :=

∫

M

(Re f)+ dσ −∫

M

(Re f)− dσ + i

∫

M

(Im f)+ dσ − i

∫

M

(Im f)− dσ .

(3.14)

Example 3.10. Let k, n ∈ N with k < n and let M ⊆ Rn be a k-dimensional subman-ifold of class C1.

(a) If M is the homeomorphic image of a C1-immersion φ : W −→ M , W ⊆ Rk, thenM is finitely coverable (coverable by the single local chart φ). Special cases are nowconsidered in (b) and (c).

(b) Let I ⊆ R be a nonempty open interval and let φ : I −→ Rn, n ≥ 2, be C1 withDφ(x) 6= 0 for each x ∈ I. If φ : I −→ φ(I) =M is a homeomorphism, thenM is a1-dimensional submanifold (i.e. a curve) coverable by the single local chart φ. Here,since Dφ(x) is represented by a column vector, the corresponding Gram matrix is1 × 1 and its entry as well as its determinant is given by the squared Euclideannorm of Dφ(x):

γ(φ) : I −→ R+, γ(φ)(x) =∥∥Dφ(x)

∥∥22=∥∥φ′(x)

∥∥22=

n∑

i=1

|φ′i(x)|2. (3.15)

3 INTEGRATION OVER SUBMANIFOLDS OF RN 140

(c) Graphs of C1-functions are submanifolds coverable by a single local chart: LetA ⊆ Rk be open and let g : A −→ Rn−k be C1. Then the k-dimensional submanifold

M = (x, y) ∈ A× Rn−k : y = g(x)

is coverable by the single local chart

φ : A −→M, φ(x) := (x, g(x)).

Thus, the Gram matrix is

G(φ) : A −→ Rk2 , G(φ)(x) = Idk +(Dg(x))tDg(x).

For k = n− 1, this simplifies to

G(φ) : A −→ Rk2 , G(φ)(x) = Idk +(∇ g(x))t ∇ g(x),

and one can show (exercise) that, in this case, the Gram determinant is

γ(φ) : A −→ R+, γ(φ)(x) = 1 + ‖∇ g(x)‖22 = 1 +n−1∑

i=1

|∂ig(x)|2. (3.16)

(d) If M is compact, then M is finitely coverable: Indeed, for each a ∈ M , let φa :Wa −→ Rn be a local chart according to Th. 3.4(iv). Then (Ma)a∈M is an opencover of M . As M is compact, it must have a finite subcover, showing M to befinitely coverable.

Theorem 3.11. Let k, n ∈ N with k < n and let M ⊆ Rn be a k-dimensional subman-ifold of class C1. Assume M to be finitely coverable.

(a) Given a finite cover C := (φ1, . . . , φN), N ∈ N, of local charts for M , there exists ameasurable partition of unity subordinate to C.

(b) The measurability of f : M −→ K, as defined in Def. 3.9(c), does not depend on thechosen local charts covering M . The integral defined in Def. 3.9(c) is well-defined,i.e. the integrals on the right-hand side of (3.12) all make sense and the value of∫Mf dσ does neither depend on the choice of covering local charts nor on the choice

of partition of unity. In consequence, the integral defined in (3.14) is well-definedas well.

(c) The set Aσ := A ⊆ M : A σ-measurable constitutes a σ-algebra on M ; σ :A −→ [0,∞] defined by (3.13) constitutes a measure on M , called the surface mea-sure. Then σ-measurability (resp. σ-integrability) as defined in Def. 3.9(c) coincideswith the usual measurability (resp. integrability) with respect to the measure σ; theintegral defined in Def. 3.9(c) is the usual integral with respect to the measure σ.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 141

Proof. (a): If φi : Wi −→Mi, i ∈ 1, . . . , N, then define

∀i∈1,...,N

ηi : M −→ R, ηi := χBi, Bi :=Mi \

i−1⋃

j=1

Mj. (3.17)

Then Def. 3.9(b)(i)–(iii) are, clearly, satisfied ((iii), since M =⋃N

i=1Mi). It remainsto show that the ηi are measurable, i.e. that the Bi are measurable. As the Mi areM -open, there exists Oi ⊆ Rn open such that Mi = M ∩ Oi. Since M ∈ Bn by Th.3.6(b), Mi ∈ Bn, showing Bi ∈ Bn as well.

(b): If f : M −→ [0,∞], then all integrands on the right-hand side of (3.12) arenonnegative. If f is σ-measurable, then the integrands are λk-measurable as they arethe product of the λk-measurable functions f φi, ηi φi, and x 7→

√γ(φi)(x), the

last function even being continuous. It remains to show the independence of the chosencharts and partition of unity. Thus, in addition to the chart cover C = (φ1, . . . , φN),φi : Wi −→ Mi and the ηi, let D := (ψ1, . . . , ψM), M ∈ N, also be a finite cover of Mvia local charts, ψi : Ui −→ Ni, where (α1, . . . , αM ) constitutes a measurable partitionof unity subordinate to D. Let Vij :=Mi∩Nj and K := (i, j) : Vij 6= ∅. According toTh. 3.5, for each (i, j) ∈ K, there exists a C1-diffeomorphism τij : Uij −→ Wij such that

ψj = φiτij on Uij, where Uij := ψ−1j (Vij),Wij := φ−1

j (Vij). Then, for each f : M −→ Kand j ∈ 1, . . . ,M,

f ψj =N∑

i=1

(ηi f) ψj =∑

i: (i,j)∈K(ηi f) φi τij.

Since each τij is λk-measurable, if f is σ-measurable, then each summand is λk-measur-

able, showing f ψj to be λk-measurable as well. In consequence, the σ-measurabilityof f does not depend on the chosen charts. Now let f : M −→ [0,∞] be σ-measurable.One computes

M∑

j=1

∫

Uj

((αjf) ψj)√γ(ψj) dλ

k =M∑

j=1

N∑

i=1

∫

Uj

((ηiαjf) ψj)√γ(ψj) dλ

k

=∑

(i,j)∈K

∫

Uij

((ηiαjf) ψj)√γ(ψj) dλ

k

(3.11)=

∑

(i,j)∈K

∫

Uij

((ηiαjf) φi τij) | detDτij|√γ(φi) τij dλk

(2.72)=

∑

(i,j)∈K

∫

Wij

((ηiαjf) φi)√γ(φi) dλ

k

=N∑

i=1

M∑

j=1

∫

Wi

((ηiαjf) φi)√γ(φi) dλ

k =N∑

i=1

∫

Wi

((ηif) φi)√γ(φi) dλ

k ,

showing the integral in (3.12) to be well-defined.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 142

(c): Let φ1, . . . , φN be as in the proof of (b). Let A ⊆M . Then A ∈ Aσ if, and only if,φ−1i (A) ∈ Lk for each i ∈ 1, . . . , N. Thus,

Aσ =N⋂

i=1

Bi, Bi := B ∈M : φ−1i (B) ∈ Lk. (3.18)

Since each Bi is a σ-algebra by Prop. 1.55(a), Aσ is a σ-algebra as well. We show σ tobe a measure on Aσ: Clearly, σ(∅) = 0. Let (Ai)i∈N be a sequence of disjoint sets in Aσ

and let A :=⋃

j∈NAj. Then, for each i ∈ 1, . . . , N,∫

Wi

((ηiχA) φi)√γ(φi) dλ

k =

∫

φ−1i (A)

(ηiφi)√γ(φi) dλ

k

=∞∑

j=1

∫

φ−1i (Aj)

(ηiφi)√γ(φi) dλ

k =∞∑

j=1

∫

Wi

((ηiχAj) φi)

√γ(φi) dλ

k ,

implying

σ(A) =∞∑

j=1

σ(Aj),

i.e. σ is σ-additive and a measure. Next, we show f : M −→ K to be σ-measurable asdefined in Def. 3.9(c) if, and only if, f is Aσ-measurable, i.e. if, and only if, f−1(B) ∈ Aσ

for each B ∈ B:

f σ-measurable ⇔ ∀i∈1,...,N

f φi λk-measurable

⇔ ∀B∈B

∀i∈1,...,N

(f φi)−1(B) = φ−1

i (f−1(B)) ∈ Lk

(3.18)⇔ ∀B∈B

f−1(B) ∈ Aσ ⇔ f Aσ-measurable.

It is then immediate from Def. 3.9(c) that the respective notions of integrability are alsoidentical. It remains to show that the integral defined in Def. 3.9(c) is the usual integralwith respect to the measure σ. To this end, for the remainder of the proof, denote themeasure σ by λk (to notationally distinguish the two integrals – the notation is alsoreasonable as one may interpret the measure σ as a k-dimensional Lebesgue measureon M). Let f ∈ S+(Aσ) be a simple function. Then there exist A1, . . . , Am ∈ Aσ ands1, . . . , sm ∈ R+

0 such that f =∑m

j=1 sjχAj. Thus,

∫

M

f dσ =N∑

i=1

∫

Wi

((ηif) φi)√γ(φi) dλ

k =N∑

i=1

m∑

j=1

sj

∫

Wi

((ηiχAj) φi)

√γ(φi) dλ

k

=m∑

j=1

sj

∫

M

χAjdσ

(3.13)=

m∑

j=1

sjσ(Aj) =m∑

j=1

sjλk(Aj) =

∫

M

f dλk .

3 INTEGRATION OVER SUBMANIFOLDS OF RN 143

Now let f ∈ M+(Aσ) and let (fj)j∈N be a sequence in S+(Aσ) such that fj ↑ f . Then,using the monotone convergence Th. 2.7,

∫

M

f dσ =N∑

i=1

∫

Wi

((ηif) φi)√γ(φi) dλ

k MCT= lim

j→∞

N∑

i=1

∫

Wi

((ηifj) φi)√γ(φi) dλ

k

= limj→∞

∫

M

fj dλk =

∫

M

f dλk .

Finally,∫Mf dσ =

∫Mf dλk for integrable f : M −→ K follows in the usual way by

decomposing f into (Re f)±, (Im f)±.

Remark 3.12. The main benefit of Th. 3.11(c) is that we have identified the integrals∫Mf dσ as measure-theoretic integrals in the sense of Sec. 2. In consequence all the

abstract results of Sec. 2 regarding such integrals must hold and are available for integralsover submanifolds.

Example 3.13. (a) As in Ex. 3.10(b), let I ⊆ R be a nonempty open interval andlet φ : I −→ Rn, n ≥ 2, be C1 with Dφ(x) 6= 0 for each x ∈ I and such thatφ : I −→ φ(I) is a homeomorphism. We know from Ex. 3.10(b) that M := φ(I) isa 1-dimensional submanifold covered by the single chart φ and that

γ(φ) : I −→ R+, γ(φ)(x) =∥∥φ′(x)

∥∥22.

Thus, if f : M −→ K is nonnegative measurable or integrable, then

∫

M

f dσ =

∫

I

f(φ(x))∥∥φ′(x)

∥∥2dλ1 (x). (3.19)

In particular,

l(φ) := σ(M) =

∫

M

1 dσ =

∫

I

∥∥φ′(x)∥∥2dλ1 (x) (3.20)

is called the arc length of φ (or M). Moreover, φ and M are called rectifyable if,and only if, l(φ) < ∞. We also extend these notions to C1-paths φ : I −→ Rn

on general (not necessarily open) intervals by setting l(φ) := l(φ I) (where I

denotes the interval’s interior). It is then clear from (3.20) that C1-paths definedon compact intervals are always rectifyable.

(b) Let x0 ∈ [0, 2π] and consider

φ : [0, x0] −→ R2, φ(x) := (cos x, sin x),

φ′ : [0, x0] −→ R2, φ′(x) = (− sin x, cos x) 6= (0, 0).

Then ‖φ′(x)‖2 =√(sin x)2 + (cos x)2 = 1, implying

l(φ) =

∫

[0,x0]

∥∥φ′(x)∥∥2dλ1 (x) =

∫ x0

0

1 dx = x0.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 144

This finally proves that the definition of sine and cosine as given in [Phi16a, Def.and Rem. 8.21] is the same as the high school definition, where cos x and sin x aredefined to be the coordinates of the point p = (p1, p2) ∈ R2 on the unit circle, suchthat x is the angle measured in radian (i.e. the arc length of the corresponding circlesection as computed above) between the line segment between (0, 0) and (1, 0) andthe line segment between (0, 0) and p (cf. remarks at the beginning of [Phi16a, Sec.8.4]).

Example 3.14. (a) Let n ∈ N, n ≥ 2. As in Ex. 3.10(c) (with k := n − 1) letA ⊆ Rn−1 be open and let g : A −→ R be C1. We know from Ex. 3.10(c) that the(n− 1)-dimensional submanifold

M := (x, y) ∈ A× R : y = g(x)

is coverable by the single local chart

φ : A −→M, φ(x) := (x, g(x))

and that

γ(φ) : A −→ R+, γ(φ)(x) = 1 + ‖∇ g(x)‖22 = 1 +n−1∑

i=1

|∂ig(x)|2.

Thus, if f : M −→ K is nonnegative measurable or integrable, then∫

M

f dσ =

∫

A

f(x, g(x)

)√1 + ‖∇ g(x)‖22 dλn−1 (x). (3.21)

In particular,

σ(M) =

∫

M

1 dσ =

∫

A

√1 + ‖∇ g(x)‖22 dλn−1 (x). (3.22)

(b) Let n ∈ N, n ≥ 2. As a concrete example of the situation in (a), we consider the(n− 1)-dimensional upper hemisphere with center 0 and radius r > 0:

M := x ∈ Rn : ‖x‖2 = r, xn > 0. (3.23)

Letting A := x ∈ Rn−1 : ‖x‖2 < r, we obtain M as the graph of the C1-function

g : A −→ R, g(x1, . . . , xn−1) :=√r2 − x21 − · · · − x2n−1, (3.24)

where

∀i∈1,...,n−1

∂ig(x) =−xig(x)

. (3.25)

We have M = φ(A) if φ is defined as in (a). Thus, computing γ(φ) according to(a) yields

γ(φ) : A −→ R+, γ(φ)(x) = 1 +n−1∑

i=1

x2i(g(x))2

= 1 +‖x‖22

(g(x))2=

r2

(g(x))2

3 INTEGRATION OVER SUBMANIFOLDS OF RN 145

and, if f : M −→ K is nonnegative measurable or integrable, then, using ballswith respect to the 2-norm and the (linear) change of variables T : Rn−1 −→ Rn−1,t := T (x) := rx with detT ≡ rn−1,

∫

M

f dσ =

∫

A

f(x, g(x)

)√1 + ‖∇ g(x)‖22 dλn−1 (x)

=

∫

Br(0)

f

(x,√r2 − ‖x‖22

)r√

r2 − ‖x‖22dλn−1 (x)

=

∫

B1(0)

f

(rt, r

√1− ‖t‖22

)rn−1

√1− ‖t‖22

dλn−1 (t). (3.26)

Theorem 3.15. Let k, n ∈ N with k < n, α ∈ N ∪ ∞, and let M ⊆ Rn be ak-dimensional submanifold of class Cα. Given r ∈ R+, consider the linear isomorphism

Tr : Rn −→ Rn, Tr(x) := rx. (3.27)

Then Tr(M) is also a k-dimensional submanifold of class Cα. If M is finitely coverable,then so is Tr(M), and f : Tr(M) −→ K is nonnegative σ-measurable (resp. σ-integrable)if, and only if, (f Tr) : M −→ K has the corresponding property. In that case,

∫

Tr(M)

f dσ =

∫

M

(f Tr) rk dσ . (3.28)

Proof. If the Cα-immersion φ : W −→ Rn, W ⊆ Rk, is a local chart for M , thenthe Cα-immersion ψ := Tr φ : W −→ Rn is a local chart for Tr(M). Moreover,if C := (φ1, . . . , φN), N ∈ N, is a finite cover of M via local charts and (η1, . . . , ηN),ηi : M −→ R, is a measurable partition of unity subordinate to C, then D := (Tr φ1, . . . , Tr φN), is a finite cover via local charts of Tr(M), (η1 T−1

r , . . . , ηN T−1r ),

ηi T−1r : Tr(M) −→ R, is a measurable partition of unity subordinate to D. The

σ-measurability of both f and f Tr is then equivalent to the λk-measurability of eachf Trφi. Let gνij and gνij denote the entries of the Gram matrices G(φν) and G(Trφν),respectively. If φν : Wν −→ Rn, then

∀x∈Wν

gνij = 〈∂i(Tr φν)(x), ∂j(Tr φν)(x)〉 = 〈r ∂iφν(x), r ∂jφν(x)〉 = r2 gνij,

showing∀

x∈Wν

γ(Tr φν)(x) = r2k γ(φν)(x).

Thus, if f is nonnegative σ-measurable, then

∫

Tr(M)

f dσ =N∑

i=1

∫

Wi

(((ηi T−1

r )f) Tr φi

)√γ(Tr φi) dλ

k

=N∑

i=1

∫

Wi

((ηi(f Tr)

) φi

)rk√γ(φi) dλ

k =

∫

M

(f Tr) rk dσ ,

proving (3.28). The integrable case now follows in the usual way by decomposition of finto (Re f)± and (Im f)±.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 146

Theorem 3.16. Let n ∈ N, n ≥ 2. Let f : Rn −→ K be λn-integrable. Then, forλ1-a.e. r ∈ R+, f is σ-integrable over the sphere Sr(0) = x ∈ Rn : ‖x‖2 = r and∫

Rn

f dλn =

∫ ∞

0

∫

Sr(0)

f(x) dσ (x) dλ1 (r)(3.28)=

∫ ∞

0

(∫

S1(0)

f(rt) dσ (t)

)rn−1 dλ1 (r).

(3.29)This formula is sometimes called the coarea formula. It is actually a special case of themore general coarea formula of geometric measure theory.

Proof. Denote the (open) upper and lower half spaces by H± := x ∈ Rn : ±xn > 0and U := x ∈ Rn−1 : ‖x‖2 < 1. We will apply the change of variables given by theC1-diffeomorphism

φ : U × R+ −→ H+, φ(x1, . . . , xn−1, r) :=

(rx1, . . . , rxn−1, r

√1− ‖x‖22

),

φ−1 : H+ −→ U × R+, φ−1(x1, . . . , xn) :=

(x1‖x‖2

, . . . ,xn−1

‖x‖2, ‖x‖2

),

Dφ : U × R+ −→ H+,

Dφ(x1, . . . , xn−1, r) :=

r 0 · · · 0 x1

0 r · · · 0 x2...

. . ....

0 · · · 0 r xn−1

− rx1√1−‖x‖22

− rx2√1−‖x‖22

· · · − rxn−1√1−‖x‖22

√1− ‖x‖22

.

We compute detDφ(x, r) via eliminating the nondiagonal elements of the last row bysuccessively adding, for i = 1, . . . , n − 1, the i-th row multiplied by − xi√

1−‖x‖22to the

last row, ending up with an upper triangular matrix, where the upper n − 1 diagonalelements are r and the last diagonal element is

√1− ‖x‖22 −

n−1∑

i=1

x2i√1− ‖x‖22

=√

1− ‖x‖22 −‖x‖22√1− ‖x‖22

=1√

1− ‖x‖22.

Thus,

detDφ : U × R+ −→ R+, detDφ(x, r) =rn−1

√1− ‖x‖22

.

We now apply the change of variables formula (2.72) followed by Fubini to the integrablefunction f to obtain

∫

H+

f dλn =

∫ ∞

0

∫

U

f

(rx, r

√1− ‖x‖22

)rn−1

√1− ‖x‖22

dλn−1 (x) dλ1 (r)

(3.26)=

∫ ∞

0

∫

H+∩Sr(0)

f dσ dλ1 (r). (3.30)

3 INTEGRATION OVER SUBMANIFOLDS OF RN 147

As f is integrable, Fubini also yields that the inner integral must be finite λ1-a.e. asclaimed. Next, we note that the result of (3.30) still holds with H+ replaced by H−: ForH−, one merely puts a negative sign in front of the last component of φ, then | detDφ|remains the same as before and in the middle term of (3.30), one has a “−” in thesecond argument of f , which is consistent with what one obtains if one does Ex. 3.14(b)for the lower hemisphere (i.e. with g replaced by −g). Since Sr(0) ∩ x ∈ Rn : xn = 0is a σ-null set (exercise), the proof is complete.

Example 3.17. Let n ∈ N, n ≥ 2. We would like to compute the surface measure ofthe Euclidean sphere Sr(0) = x ∈ Rn : ‖x‖2 = r, r ∈ R+. We recall the formula forthe corresponding Euclidean ball Br(0) from Ex. 2.34(c):

βn(Br(0)) = ωn rn, ωn = βn(B1(0)) =

πn/2

Γ(n2+ 1)

.

Applying (3.29) to f := χB1(0) yields

ωn = βn(B1(0)) =

∫

Rn

χB1(0) dλn (3.29)

=

∫ 1

0

(∫

S1(0)

1 dσ

)rn−1 dλ1 (r) =

σ(S1(0))

n,

implying

σn := σ(S1(0)) = nωn =nπn/2

Γ(n2+ 1)

=2 πn/2

Γ(n2). (3.31)

An application of Th. 3.15 now yields the formula for σ(Sr(0)) = σ(Tr(S1(0))):

σ(Sr(0)) =

∫

Tr(S1(0))

1 dσ(3.28)=

∫

S1(0)

rn−1 dσ = σn rn−1. (3.32)

Example 3.18. We apply Th. 3.16 to rotationally symmetric functions: Let f : R+0 −→

K be such thatg : Rn −→ K, g(x) := f(‖x‖2),

is nonnegative measurable or integrable. Then

∫

Rn

g dλn =

∫

Rn

f(‖x‖2) dλn (x) =∫ ∞

0

(∫

S1(0)

f(r) dσ (t)

)rn−1 dλ1 (r)

(3.31)= σn

∫ ∞

0

f(r) rn−1 dλ1 (r). (3.33)

3.4 Tangent Space and Normal Space

A tangent vector to a submanifoldM is a vector that is tangent to a curve inM . At eachpoint a ∈ M , the set of all tangent vectors forms a vector space, the so-called tangentspace TaM . Normal vectors at a are vectors that are perpendicular to the tangent spaceTaM :

3 INTEGRATION OVER SUBMANIFOLDS OF RN 148

Definition 3.19. Let k, n ∈ N with k < n, and let M ⊆ Rn be a k-dimensionalsubmanifold of class C1, a ∈M .

(a) A vector τ ∈ Rn is called a tangent vector to M at a if, and only if, there existsǫ ∈ R+ and a C1-function

ψ : ]− ǫ, ǫ[−→M,

such thatψ(0) = a and ψ′(0) = τ. (3.34)

The set (we will see in the next theorem it is, indeed, a k-dimensional vector spaceover R)

TaM := τ ∈ Rn : τ is tangent vector to M at a (3.35)

is called the tangent space to M at a.

(b) A vector ν ∈ Rn is called a normal vector toM at a if, and only if, it is perpendicularto all tangent vectors at a (with respect to the Euclidean scalar product), i.e. if,and only if,

∀τ∈TaM

〈τ, ν〉 = 0. (3.36)

We also define

NaM := ν ∈ Rn : ν is normal vector to M at a (3.37)

to be the normal space to M at a.

Definition and Remark 3.20. Let n, k ∈ N0 and let V ⊆ Rn be a vector subspace ofRn, dimV = k. Then

V ⊥ := x ∈ Rn : 〈x, v〉 = 0 for each v ∈ V (3.38)

is the space perpendicular to V . We then know from Linear Algebra that V ⊥ is a vectorsubspace of Rn, Rn = V + V ⊥, V ∩ V ⊥ = 0, and, in particular, dimV ⊥ = n− k.

Theorem 3.21. Let k, n ∈ N with k < n, and let M ⊆ Rn be a k-dimensional subman-ifold of class C1, a ∈M .

(a) The tangent space TaM is a k-dimensional vector space over R. Moreover, if φ :W −→ V ⊆ M , W ⊆ Rk open, V M-open, is a local chart for M with φ(c) = a,c ∈ W , then

∂1φ(c), . . . , ∂kφ(c)⊆ Rn

forms a basis of TaM .

(b) The normal space NaM is a (n − k)-dimensional vector space over R. Moreover,if f ∈ C1(O,Rn−k) with a ∈ O, O ⊆ Rn open, according to Def. 3.1, i.e. such thatM ∩O = f−1(0) and rkDfa(a) = n− k, then

∇ f1(a), . . . ,∇ fn−k(a)

⊆ Rn

forms a basis of NaM .

3 INTEGRATION OVER SUBMANIFOLDS OF RN 149

(c) TaM ⊥ NaM , i.e. if τ ∈ TaM and ν ∈ NaM , then 〈τ, ν〉 = 0.

Proof. Set

V1 := span∂1φ(c), . . . , ∂kφ(c)

,

U := span∇ f1(a), . . . ,∇ fn−k(a)

,

V2 := U⊥.

Then V1, U, V2 all are vector subspaces of Rn. Since ∂1φ(c), . . . , ∂kφ(c) are linearly in-dependent, we have dimV1 = k. Since ∇ f1(a), . . . ,∇ fn−k(a) are linearly independent,we have dimU = n− k. Then we know from Def. and Rem. 3.20 that dimV2 = k. Wewill show

V1 ⊆ TaM ⊆ V2. (3.39)

Then dimV1 = dimV2 implies V1 = V2 and V1 = TaM = V2, proving (a). ThenNaM = V ⊥

2 = (TaM)⊥, proving (b) and (c). It remains to show (3.39).

“V1 ⊆ TaM”: Let τ ∈ V1. Then

∃α1,...,αk∈R

τ =k∑

i=1

αi ∂iφ(c).

Since c ∈ W with W open, there exists ǫ ∈ R+ such thatx ∈ Rk : ∀

i∈1,...,k|xi − ci| < ǫ |αi|

⊆ W.

We consider the C1-function

ψ : ]− ǫ, ǫ[−→M, ψ(s) := φ(c1 + sα1, . . . , ck + sαk).

Then ψ(0) = φ(c) = a and, by the chain rule,

ψ′(0) = Dφ(c)

α1...αk

=

k∑

i=1

αi ∂iφ(c) = τ,

showing τ ∈ TaM .

“TaM ⊆ V2”: Let τ ∈ TaM . Then there exists ǫ > 0 and a C1-function ψ : ]−ǫ, ǫ[−→Msuch that a = ψ(0) and τ = ψ′(0). Since a ∈ O, O open, we may actually assume thatψ maps into M ∩O = f−1(0). Thus,

∀i∈1,...,n−k

fi ψ ≡ 0,

implying, for each i ∈ 1, . . . , n− k,

0 = (fi ψ)′(0) = ∇ fi(ψ(0))ψ′(0) =

⟨∇ fi(a), τ

⟩,

showing τ ∈ U⊥ = V2.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 150

3.5 Gauss-Green Theorem and Green’s Identities

The Gauss-Green Th. 3.30 allows, under suitable hypotheses, to transform an integralover an open subset Ω of Rn into an integral over the boundary M := ∂Ω, where M isan (n − 1)-dimensional submanifold. The Gauss-Green Th. 3.30 can be seen as an n-dimensional generalization of the fundamental theorem of calculus, [Phi16a, Th. 10.20].Before we can state and prove Th. 3.30, we still need some preparation.

Definition 3.22. Let n ∈ N, n ≥ 2, and let Ω ⊆ Rn be open. We define Ω to have aC1-boundary if, and only if, for each a ∈ M := ∂Ω, there exists an open neighborhoodOa ⊆ Rn of a and a map fa ∈ C1(Oa,R), satisfying the following two conditions

(a) Ω ∩Oa = x ∈ Oa : fa(x) ≤ 0,

(b) ∇ fa(x) 6= 0 for each x ∈ Oa.

Theorem 3.23. Let n ∈ N, n ≥ 2, and let Ω ⊆ Rn be open, having a C1-boundary. Ifa ∈M := ∂Ω and fa ∈ C1(Oa,R) is according to Def. 3.22, then

M ∩Oa = (fa)−1(0). (3.40)

In particular, ∂Ω is an (n− 1)-dimensional submanifold of class C1.

Moreover, there exists ǫ ∈ R+ such that

∀s∈]0,ǫ[

a+ s ∇ fa(a) /∈ Ω. (3.41)

Proof. “⊆ of (3.40)”: Recall Ω = Ω∪∂Ω, where the union is disjoint, as Ω is open. Thus,according to Def. 3.22(a), we have to show that fa(x) < 0 implies x /∈ ∂Ω. However,if x ∈ Ω ∩ Oa with fa(x) < 0, then, by the continuity of fa, there exists an open setU ⊆ Oa such that fa < 0 on U , showing x /∈ ∂Ω.

“⊇ of (3.40)”: Let x ∈ Oa with fa(x) = 0. We need to verify x ∈ M . Consider theauxiliary function

ψ : ]− ǫ, ǫ[−→ R, ψ(s) := fa(x+ s∇ fa(x)

),

which, as x ∈ Oa, Oa open, is defined for sufficiently small ǫ > 0. Moreover, ψ is C1,since fa is C1. We obtain

ψ′ : ]− ǫ, ǫ[−→ R, ψ′(s) = ∇ fa(x+ s∇ fa(x)

)· ∇ fa(x).

In particular, ψ′(0) = ‖∇ fa(x)‖22 > 0. Thus, as ψ′ is continuous, ψ is strictly increasingin an entire neighborhood of 0. Since ψ(0) = fa(x) = 0, this shows every neighborhoodof x contains points from Ω and points from Ωc, showing x ∈ M . Setting x = a, wehave also shown (3.41).

That M = ∂Ω is an (n− 1)-dimensional submanifold of class C1 is now immediate fromDef. 3.1.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 151

Corollary 3.24 (Existence of Outer Unit Normal). Let n ∈ N, n ≥ 2, and let Ω ⊆ Rn

be open, having a C1-boundary. Then, for each a ∈ ∂Ω, there exists a unique normalvector ν(a) ∈ Na∂Ω, satisfying the following two conditions:

(i) ‖ν(a)‖2 = 1.

(ii) There exists ǫ ∈ R+ such that

∀s∈]0,ǫ[

a+ s ν(a) /∈ Ω.

The vector ν(a) is called the outer unit normal to ∂Ω at a. Moreover, the function

ν : ∂Ω −→ Rn, a 7→ ν(a),

is continuous.

Proof. Let a ∈ ∂Ω and let fa ∈ C1(Oa,R) be according to Def. 3.22. Then ∇ fa 6= 0and we may define

ν(a) :=∇ fa(a)

‖∇ fa(a)‖2. (3.42)

Then ‖ν(a)‖2 is clear, condition (ii) holds according to (3.41), and ν(a) ∈ Na∂Ω by Th.3.21(b). To show ν as a function on ∂Ω is continuous, it suffices to show ν is continuousat each a ∈ ∂Ω. However, given a ∈ ∂Ω, on Oa ∩ ∂Ω, ν is given by the continuousfunction x 7→ ∇ fa(x)

‖∇ fa(x)‖2 .

Example 3.25. Let n ∈ N, n ≥ 2. For xM ∈ Rn and r > 0, consider the Euclidean ballΩ := Br(xM). Then ∂Ω = x ∈ Rn : ‖x− xM‖2 = r and, according to Ex. 3.2(b), foreach a ∈ ∂Ω, we may choose fa := f with

f : Rn −→ R, f(x) := ‖x− xM‖22 − r2 =n∑

i=1

(xi − xM,i)2 − r2,

∇ f : Rn −→ Rn, ∇ f(x) = 2(x1 − xM,1, . . . , xn − xM,n

).

Thus,

∀a∈∂Ω

ν(a) =∇ f(a)

‖∇ f(a)‖2=a− xM

r. (3.43)

Proposition 3.26. Let n ∈ N, n ≥ 2, and let Ω ⊆ Rn be open, having a C1-boundary.Then, for each a ∈ M := ∂Ω, there exists a permutation πa ∈ Sn and open intervalsAa ⊆ Rn−1 and Ia ⊆ R, such that

αa ∈ Aa ⊆ Rn−1, βa ∈ Ia ⊆ R,

where αa := (aπa(1), . . . , aπa(n−1)), βa := aπa(n), and there exists a C1-map ga : Aa −→

Ia, satisfying

M ∩ Pτa(Aa × Ia) = Pτa(

(x, y) ∈ Aa × Ia : y = ga(x)), τa := (πa)−1, (3.44a)

3 INTEGRATION OVER SUBMANIFOLDS OF RN 152

and either

Ω ∩ Pτa(Aa × Ia) = Pτa(

(x, y) ∈ Aa × Ia : y < ga(x))

(3.44b)

or

Ω ∩ Pτa(Aa × Ia) = Pτa(

(x, y) ∈ Aa × Ia : y > ga(x)). (3.44c)

Then the outer unit normal ν(a) is given by

ν(a) = ±(−∇ g(αa), 1

)Pπa√

1 + ‖∇ g(αa)‖22, (αa, βa) := Pπa(a) (3.45)

(with “+” for (3.44b) and “−” for (3.44c)).

Proof. Let fa ∈ C1(Oa,R), a ∈ Oa ⊆ Rn open, be according to Def. 3.22. Then, byDef. 3.22 and Th. 3.23, fa satisfies the conditions of Def. 3.1. In particular, M = ∂Ωis an (n − 1)-dimensional submanifold of class C1, and we can then choose πa, Aa,Ia := Ba, and ga according to Th. 3.4(ii). If necessary, we make the sets Aa and Ba

of Th. 3.4(ii) smaller to obtain them to be open intervals and we make Oa smallerto have Oa = Pτa(Aa × Ia). By possibly shrinking Aa again, we may assume thatdist(ga(Aa), (Ia)

c) := ǫ > 0 (i.e. ga maps Aa into a compact subinterval of Ia). We mayalso assume ga to be obtained from ha := fa Pτa via the implicit function theorem asin the proof of “(i) ⇒ (ii)” of Th. 3.4(ii). Then, for in := τa(n), we have ∂inf

a(a) 6= 0.By the continuity of ∂inf

a, we may assume that ∂infa has only one sign (everywhere

positive or everywhere negative) on Oa = Pτa(Aa × Ia) (i.e., on Oa, fa is either strictly

increasing or strictly decreasing with respect to the in-th variable). Now (3.44a) holdsdue to Th. 3.4(ii) and, by Def. 3.22 and Th. 3.23, we have

M ∩Oa = (fa)−1(0),Ω ∩Oa = x ∈ Oa : f

a(x) < 0.

We consider the open sets

U+ := Pτa(

(x, y) ∈ Aa × Ia : 0 < ga(x)− y),

U− := Pτa(

(x, y) ∈ Aa × Ia : 0 > ga(x)− y),

Oa = U+ ∪U− ∪(M ∩Oa).

Then Ω∩Oa ⊆ U+∪U−. However Ω∩Oa cannot have nonempty intersection with both U+

and U−: Suppose z1 ∈ Ω∩Oa∩U+ and z2 ∈ Ω∩Oa∩U−. Let (x1, y1) := Pπa(z1) ∈ Aa×Ia,(x2, y2) := Pπa(z2) ∈ Aa × Ia. Then ga(x1) > y1, g

a(x2) < y2. Since ha is continuous,ha(x1, y1) = fa(z1) < 0, and ha does not have a zero on (x1, y) : y1 ≤ y < ga(x1), wecan assume ǫ > c := ga(x1)− y1 > 0. Then, since Aa is convex, the map

ψ1 : [0, 1] −→ Aa × Ia, ψ1(s) :=(x1 + s(x2 − x1), g

a(x1 + s(x2 − x1))− c)

3 INTEGRATION OVER SUBMANIFOLDS OF RN 153

is well-defined with ψ1(0) = (x1, y1), ψ1(1) = (x2, ga(x2)−c). Since haψ1 is continuous,

(haψ1)(0) < 0 and haψ1 does not have any zeros on [0, 1] (since c > 0), (haψ1)(1) < 0,i.e. ha(x2, g

a(x2)− c) < 0. However, due to the choice of in above, the map

ψ2 : [ga(x2)− c, y2] −→ R, ψ2(s) := ha(x2, s),

either strictly increasing or strictly decreasing. Since ψ2(ga(x2)−c) = ha(x2, g

a(x2)−c) <0, ψ2(g

a(x2)) = ha(x2, ga(x2)) = 0, ψ2(y2) = ha(x2, y2) = fa(z2) < 0, we obtain a

contradiction, proving Ω ∩ Oa ⊆ U+ (i.e. (3.44b)) or Ω ∩ Oa ⊆ U− (i.e. (3.44c)). For(3.44b), we now consider

fa : Pτa(Aa × Ia) −→ R, fa(z) := xn − g(x1, . . . , xn−1), x := Pπa(z)

(this might not be the same fa as before, but it also satisfies Def. 3.22). We then obtain

ν(a) =∇ fa(a)

‖∇ fa(a)‖2=

(−∇ g(αa), 1

)Pπa√

1 + ‖∇ g(αa)‖22, (αa, βa) := Pπa(a),

proving (3.45). For (3.44c), one uses fa(z) := xn + g(x1, . . . , xn−1), x := Pπa(z), againproving (3.45).

Lemma 3.27. Let n ∈ N, O ⊆ Rn open, and η ∈ C1c (O,R). Then

∀i∈1,...,n

∫

O

∂iη dλn = 0. (3.46)

Proof. Extending η by 0, we may assume O = Rn. Then there exists r ∈ R+ such thatsupp η ⊆ [−r, r]n. Then, using Fubini,

∫

O

∂iη dλn =

∫

[−r,r]n∂iη dλ

n

=

∫

[−r,r]n−1

((η(x1, . . . , xi = r, . . . , xn)

− η(x1, . . . , xi = −r, . . . , xn))dλn−1 (x1, . . . , xi, . . . , xn) = 0,

where xi means that xi is omitted.

In preparation for the proof of the Gauss-Green Th. 3.30, we prove a special case in thefollowing lemma:

Lemma 3.28. Let n ∈ N, n ≥ 2. Let A ⊆ Rn−1 be open and let I :=]α, β[ with α, β ∈ R,α < β, be an open interval. Given a C1-function

g : A −→ I,

set

Ω := (x, xn) ∈ A× I : xn < g(x),M := (x, xn) ∈ A× I : xn = g(x) ⊆ ∂Ω.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 154

Then, for each f ∈ C1c (A× I,R), one has

∀i∈1,...,n

∫

Ω

∂if dλn =

∫

M

fνi dσ , (3.47)

where

ν : M −→ Rn, ν(x, xn) :=

(−∇ g(x), 1

)√

1 + ‖∇ g(x)‖22. (3.48)

Moreover, (3.47) still holds if one has “>” instead of “<” in the definition of Ω and a“−” in front of the right-hand side of (3.48).

Proof. First, let 1 ≤ i ≤ n− 1. We define the auxiliary function

F : A× I −→ R, F (x, xn) :=

∫ xn

α

f(x, s) ds .

Then

∀(x,xn)∈A×I

(∂nF (x, xn) = f(x, xn), ∂iF (x, xn) =

∫ xn

α

∂if(x, s) ds

)

and the chain rule yields, for each (x, xn) ∈ A× I,

∂i

∫ g(x)

α

f(x, s) ds = ∂iF (x, g(x)) =

∫ g(x)

α

∂if(x, s) ds + f(x, g(x)) ∂ig(x). (3.49)

The function x 7→∫ g(x)

αf(x, s) ds is in C1

c (A,R) and, thus, Lem. 3.27 yields

∫

A

∂i

(∫ g(x)

α

f(x, s) ds

)dλn−1 (x) = 0. (3.50)

Using Fubini, we obtain

∫

Ω

∂if dλn =

∫

A

∫ g(x)

α

∂if(x, s) ds dλn−1 (x)

(3.49)=

∫

A

(∂i

∫ g(x)

α

f(x, s) ds

)dλn−1 (x)−

∫

A

f(x, g(x)) ∂ig(x) dλn−1 (x)

(3.50)= −

∫

A

f(x, g(x))∂ig(x)√

1 + ‖∇ g(x)‖22

√1 + ‖∇ g(x)‖22 dλn−1 (x)

(3.48), (3.21)=

∫

M

fνi dσ .

It remains to consider the case i = n. For each x ∈ A, the function s 7→ f(x, s) is inC1

c (I,R), implying

∫ g(x)

α

∂nf(x, s) ds = f(x, g(x))− 0 = f(x, g(x))

3 INTEGRATION OVER SUBMANIFOLDS OF RN 155

and, thus,

∫

Ω

∂nf dλn =

∫

A

∫ g(x)

α

∂nf(x, s) ds dλn−1 (x) =

∫

A

f(x, g(x)) dλn−1 (x)

(3.48), (3.21)= =

∫

M

fνi dσ ,

thereby completing the proof of the first case. One now obtains the second case byapplying the first to −Ω, −M , −g : −A −→ −I, followed by a change of variablesx 7→ −x.

As a last preparation for the proof of the Gauss-Green Th. 3.30, we construct a C∞

partition of unity. Here, the partition of unity is subordinate to an open cover of Rn,not to an open cover of a submanifold. For ǫ > 0, we consider the cover given by

(B2ǫ(ǫ p))p∈Zn ,

where the balls are defined with respect to the∞-norm on Rn (i.e. they are open intervalsof side length 4ǫ).

Proposition 3.29. Let n ∈ N, ǫ ∈ R+. Then there exists a family (ηp)p∈Zn of functionsin C∞

c (Rn,R), satisfying the following conditions (i) – (iii) (the balls are defined withrespect to the ∞-norm):

(i) 0 ≤ ηp ≤ 1 for each p ∈ Zn.

(ii) supp ηp = Bǫ(ǫ p) for each p ∈ Zn.

(iii)∑

p∈Zn ηp ≡ 1.

Proof. According to Ex. 2.53(c), the function

ϕ : R −→ [0, 1], ϕ(t) :=

exp

(− 1

ǫ2−t2

)for |t| < ǫ,

0 otherwise,

is in C∞c (R,R), suppϕ = [−ǫ, ǫ]. Define

G : R −→ R+, G(t) :=∑

p∈Zϕ(t− ǫ p). (3.51)

Since the support of t 7→ ϕ(t− ǫ p) is [ǫ p− ǫ, ǫ p+ ǫ], for each t ∈ R, at least one and atmost two summands in (3.51) are nonzero. In particular, G is well-defined as a functioninto R+. Thus,

g : R −→ [0, 1], g(t) :=ϕ(t)

G(t),

is well-defined, g ∈ C∞c (R,R), supp g = [−ǫ, ǫ]. We claim that letting

∀p∈Z

ηp : R −→ [0, 1], ηp(t) := g(t− ǫ p),

3 INTEGRATION OVER SUBMANIFOLDS OF RN 156

proves the proposition for n = 1: Clearly, (i) and (ii) are satisfied. Since, for each t ∈ R,

∑

p∈Zηp(t) =

∑

p∈Z

ϕ(t− ǫ p)

G(t− ǫ p)=∑

p∈Z

ϕ(t− ǫ p)

G(t)= 1,

(iii) holds as well. Now, for an arbitrary n ∈ N, define

∀p∈Zn

ηp : Rn −→ [0, 1], ηp(x) :=

n∏

i=1

g(xi − ǫ pi).

As before, ηp ∈ C∞c (Rn,R) and (i) and (ii) are immediate consequences of the properties

of g. To obtain (iii), we carry out and induction, assuming the case n− 1 by inductionhypothesis, obtaining, for each x ∈ Rn,

∑

p∈Zn

ηp(x) =∑

k∈Z

∑

q∈Zn−1

g(xn − ǫ k)n−1∏

i=1

g(xi − ǫ qi)

=∑

k∈Zg(xn − ǫ k)

∑

q∈Zn−1

n−1∏

i=1

g(xi − ǫ qi)

=∑

k∈Zg(xn − ǫ k) = 1,

completing the proof.

The following version of the Gauss-Green theorem is certainly not the most general. Forexample, it still holds if, instead of a C1-boundary, one just has a so-called Lipschitzboundary (which, e.g., allows corners if they are sufficiently benign, including polyhedralsets), see [Kon64].

Theorem 3.30 (Gauss-Green). Let n ∈ N, n ≥ 2. Let Ω ⊆ Rn be open and bounded,having a C1-boundary (then Ω and ∂Ω are compact; in particular, ∂Ω is a finitely cov-erable (n − 1)-dimensional submanifold of class C1). If Ω ⊆ U ⊆ Rn with U open andF ∈ C1(U,Rn) (in this context, F is often called a vector field), then

∫

Ω

divF dλn =

∫

∂Ω

〈F, ν〉 dσ (3.52)

(as before, ν : ∂Ω −→ Rn denotes the outer unit normal to Ω, also recall that divF =∑ni=1 ∂iFi).

Proof. Without loss of generality, we may assume U to be bounded, i.e. U to be compact.For each a ∈ ∂Ω, we let ga : Aa −→ Ia, and πa, τa ∈ Sn be as in Prop. 3.26. Then(Pτa(Aa × Ia))a∈∂Ω together with Ω provide an open cover O := (Uj)j∈J of the compactset Ω, where, by making the Aa, Ia smaller if necessary, we may also assume Uj ⊆ U ,i.e. V :=

⋃j∈J Uj is open and bounded, α := dist(Ω, V c) > 0 (here, and in the following,

we consider Rn with the ∞-norm). Moreover, O still constitutes an open cover of the

3 INTEGRATION OVER SUBMANIFOLDS OF RN 157

compact set C := x ∈ Rn : dist(x,Ω ≤ α2. According to [Phi16b, Th. 3.21], we can

choose a Lebesgue number δ > 0 for this cover. As we choose it with respect to the∞-norm on Rn, then each hypercube I with side length < δ, we have I∩∂Ω is containedin one of the open sets of the cover. We choose ǫ > 0 such that 2ǫ < δ, 2ǫ < α

2, and we

let (ηp)p∈Zn be the corresponding family of functions in C∞c (Rn,R) given by Prop. 3.29.

Since, for each p ∈ Zn, supp ηp = Bǫ(ǫ p), if supp ηp ∩ Ω 6= ∅, then supp ηp ⊆ C and, bythe choice of δ, there exists j ∈ J such that supp ηp ⊆ Uj. If supp ηp ⊆ Ω, then

∫

Ω

div(ηpF ) dλn (3.46)

= 0 =

∫

∂Ω

〈ηpF, ν〉 dσ .

If supp ηp ⊆ Pτa(Aa × Ia), then we apply Lem. 3.28 to obtain, for each i ∈ 1, . . . , n,∫

Ω

∂i(ηpFi) dλn =

∫

Ω∩Pτa (Aa×Ia)

∂i(ηpFi) dλn =

∫

(PπaΩ)∩(Aa×Ia)

∂i(ηpFi) Pτa dλn

(3.47)=

∫

(Pπa∂Ω)∩(Aa×Ia)

((ηpFi) Pτa

)νi dσ

(3.45)=

∫

∂Ω∩Pτa (Aa×Ia)

ηpFiνi dσ =

∫

∂Ω

ηpFiνi dσ .

Now letZ := p ∈ Zn : supp ηp ∩ Ω 6= ∅.

Then∫

Ω

divF dλn =∑

p∈Zn

∫

Ω

div(ηpF ) dλn =

∑

p∈Z

∫

Ω

div(ηpF ) dλn =

∑

p∈Z

∫

∂Ω

〈ηpF, ν〉 dσ

=∑

p∈Zn

∫

∂Ω

〈ηpF, ν〉 dσ =

∫

∂Ω

〈F, ν〉 dσ ,

establishing the theorem.

Example 3.31. (a) Let n ∈ N, n ≥ 2, and consider

F : Rn −→ Rn, F (x) := x,

divF : Rn −→ R, divF (x) =n∑

i=1

∂iFi(x) = n.

Thus, for each Ω ⊆ Rn be open and bounded, having a C1-boundary, Th. 3.30yields

nβn(Ω) =

∫

Ω

divF dλn =

∫

∂Ω

〈x, ν(x)〉 dσ (x). (3.53)

In particular, for the Euclidean ball Br(0), r > 0, we obtain (writing ωn = βn(B1(0))as before)

nωn rn = nβn(Br(0)) =

∫

Sr(0)

〈x, ν(x)〉 dσ (x) (3.43)=

∫

Sr(0)

‖x‖22r

dσ (x) = r σ(Sr(0)),

recovering (3.31) and (3.32), i.e. σ(S1(0)) = nωn and σ(Sr(0)) = σ(S1(0)) rn−1.

3 INTEGRATION OVER SUBMANIFOLDS OF RN 158

(b) To evaluate the integral∫S1(0)

(x4 + y4) dσ (x, y), we apply (3.52) with

F : R2 −→ R2, F (x, y) = (x3, y3),

divF : R2 −→ R, divF (x, y) = 3x2 + 3y2.

Since S1(0) = ∂B1(0), we can use (3.43) to obtain ν(x, y) = (x, y) for (x, y) ∈ S1(0).Thus,

∫

S1(0)

(x4 + y4) dσ (x, y) =

∫

S1(0)

⟨(x3, y3), (x, y)

⟩dσ (x, y)

(3.52)=

∫

B1(0)

(3x2 + 3y2) dλ2 (x, y)

= 3

∫ 1

0

∫ 2π

0

r3 dϕ dr =3 · 2π4

=3π

2,

where we used polar coordinates to evaluate the integral over B1(0).

—

Finally, as another application of Th. 3.30, we will prove Green’s identities, which, forexample, are of use in the theory of partial differential equations.

Theorem 3.32 (Green’s Identities). Let n ∈ N, n ≥ 2. Let Ω ⊆ Rn be open andbounded, having a C1-boundary (then Ω and ∂Ω are compact; in particular, ∂Ω is afinitely coverable (n− 1)-dimensional submanifold of class C1). If Ω ⊆ U ⊆ Rn with Uopen and f, g ∈ C2(U,R), then the following identities hold:

∫

Ω

〈∇ f,∇ g〉 dλn =

∫

∂Ω

f 〈∇ g, ν〉 dσ −∫

Ω

f ∆g dλn , (3.54a)∫

Ω

(f ∆g − g∆f) dλn =

∫

∂Ω

(f 〈∇ g, ν〉 − g 〈∇ f, ν〉

)dσ , (3.54b)

recalling ∆f =∑n

i=1 ∂i∂if .

Proof. To prove (3.54a), apply Th. 3.30 to

F : U −→ Rn, F (x) := f(x) ∇ g(x) :

One has

divF : U −→ R, divF (x) = 〈∇ f(x),∇ g(x)〉+ f(x)∆g(x) : (3.55)

Indeed,

〈∇ f,∇ g〉+ f ∆g =n∑

i=1

(∂if ∂ig + f ∂i∂ig

)=

n∑

i=1

∂i(f∂ig

)= divF (x).

A ORDER ON, ARITHMETIC IN, AND TOPOLOGY OF R 159

Thus,∫

Ω

(〈∇ f,∇ g〉+ f ∆g

)dλn =

∫

Ω

divF dλn(3.52)=

∫

∂Ω

〈F, ν〉 dσ =

∫

∂Ω

f 〈∇ g, ν〉 dσ ,

proving (3.54a).

If one interchanges the roles of f and g in (3.54a) and subtracts the resulting equationfrom the original one, then one obtains (3.54b).

A Order on, Arithmetic in, and Topology of R

Notation A.1. By R := R ∪ −∞,∞, we denote the set of extended real numbers.

—

One defines the order on, arithmetic in, and topology on R such that they, as far aspossible, constitute extensions of the respective (standard) definitions on R. In [Phi16b,Ex. 1.52(b)], we had actually already considered the order and (resulting order) topologyon R. Here, we include it again for convenience.

Definition A.2. (a) The (total) order on R is extended to R by setting −∞ < a <∞for each a ∈ R. The absolute value function is extended from R to R by defining|∞| := | −∞| := ∞.

(b) The set of open intervals Io in R is defined to consist of R plus all open intervalsin R plus all intervals of the form [−∞, a[ or ]a,∞], a ∈ R:

Io :=R∪]a, b[: a, b ∈ R, a < b

∪[−∞, a[: a ∈ R ∪ ∞

∪]a,∞] : a ∈ R ∪ −∞

. (A.1)

Then the (standard) topology on R is defined by calling a set O ⊆ R open if, andonly if, each element x ∈ O is contained in an open interval I ∈ Io that is containedin O, i.e. x ∈ I ⊆ O. In other words, Io is defined to be a base for the topology onR.

(c) Addition, subtraction, and multiplication are extended from R to R by defining:

∀a∈R

a+ (±∞) := (±∞) + a := ±∞, (A.2a)

∀a∈R

a− (±∞) := −(±∞) + a := ∓∞, (A.2b)

∞+∞ := ∞, −∞+ (−∞) := −∞, −(±∞) := ∓∞, (A.2c)

∀a∈R

a · (±∞) := (±∞) · a :=

±∞ for a ∈]0,∞],

∓∞ for a ∈ [−∞, 0[,(A.2d)

0 · (±∞) := (±∞) · 0 := 0, (A.2e)

∞−∞ := −∞+∞ := 0. (A.2f)

A ORDER ON, ARITHMETIC IN, AND TOPOLOGY OF R 160

Caveat A.3. Addition is not associative on R:

(−∞+∞) + 1 = 0 + 1 = 1 6= 0 = −∞+∞ = −∞+ (∞+ 1). (A.3)

Addition and multiplication are not distributive on R:

(−1 + 2) · ∞ = 1 · ∞ = ∞ 6= 0 = −∞+∞ = (−1) · ∞+ 2 · ∞. (A.4)

However, the following Lem. A.4 shows that commutativity of addition and multiplica-tion hold on R as well as associativity of multiplication; associativity of addition alsoholds if one removes one of the infinities. Moreover, (R,+) is not a group, as + is notassociative; (R \ 0, ·) is not a group, as the infinities do not have multiplicative in-verses; (R\∞,+) and (R\−∞,+) are not groups, as, in both cases, the remaininginfinity does not have an additive inverse.

Lemma A.4. Consider R with addition and multiplication as defined in Def. A.2(c).

(a) Addition is associative on ] − ∞,∞] and on [−∞,∞[: If a, b, c ∈] − ∞,∞] ora, b, c ∈ [−∞,∞[, then

(a+ b) + c = a+ (b+ c). (A.5)

(b) Addition is commutative on R: If a, b ∈ R, then

a+ b = b+ a. (A.6)

(c) Multiplication is associative on R: If a, b, c ∈ R, then

(ab)c = a(bc). (A.7)

(d) Multiplication is commutative on R. If a, b ∈ R, then

ab = ba. (A.8)

Proof. (a): If a, b, c ∈] − ∞,∞], then both sides of (A.5) equal ∞ if, and only if,∞ ∈ a, b, c; if a, b, c ∈ [−∞,∞[, then both sides of (A.5) equal −∞ if, and only if,−∞ ∈ a, b, c, proving (a).

(b): Commutativity of addition is immediate from (A.2a) and (A.2f).

(c): If 0 ∈ a, b, c, then (A.7) holds, since both sides are 0. If a, b, c ∈ R, then (A.7)holds, since · is associative on R. If 0 /∈ a, b, c, but −∞,∞ ∩ a, b, c 6= ∅, then(ab)c = ±∞ and a(bc) = ±∞. However, letting

sgn(±∞) := ±1, (A.9)

due to (A.2d), we obtain

(ab)c =((sgn a · sgn b) · sgn c

)· ∞ =

(sgn a · (sgn b · sgn c

)· ∞ = a(bc),

proving (c).

(d): Commutativity of multiplicationis immediate from (A.2d) and (A.2e).

B ALGEBRAIC STRUCTURE ON RINGS OF SUBSETS 161

Remark A.5. R with the topology defined in Def. A.2(b) constitutes a so-called com-pactification of R, i.e. it is a compact topological space such that the topology on R isrecovered as the relative topology when considering R as a subset of R. Actually, it isstraightforward to verify R is homeomorphic to [−1, 1], where

φ : R −→ [−1, 1], φ(x) :=

−1 for x = −∞,x

|x|+1for x ∈ R,

1 for x = ∞,

(A.10)

provides a homeomorphism.

B Algebraic Structure on Rings of Subsets

Notation B.1. For sets A,B, we let

A∆B := (A \B) ∪ (B \ A) = (A ∪ B) \ (A ∩B) (B.1)

denote the symmetric difference of A and B.

—

Let X be a set. Recall that, for A ⊆ X, we have the characteristic function

χA : X −→ 0, 1, χA(x) :=

1 if x ∈ A,

0 if x /∈ A.(B.2)

Moreover, we know from [Phi16b, Prop. 2.18] that

χ : P(X) −→ 0, 1X , χ(A) := χA, (B.3)

is bijective. Also recall from Linear Algebra that, with addition und multiplicationmodulo 2, Z2 = 0, 1 constitutes a field.

Proposition B.2. If X 6= ∅, then (0, 1X ,+, ·) constitutes a commutative ring withunity, if + and · denote pointwise addition and multiplication, respectively.

Proof. First note that + and · are associative, commutative, and distributive on 0, 1X ,as they are associative, commutative, and distributive on 0, 1. The neutral elementof addition is χ∅ ≡ 0; the neutral element of multiplication is χX ≡ 1. If f ∈ 0, 1X ,then f is its own additive inverse:

∀x∈X

(f + f)(x) =

0 + 0 = 0 if f(x) = 0,

1 + 1 = 0 if f(x) = 1,

completing the proof that (0, 1X ,+, ·) constitutes a commutative ring with unity.

B ALGEBRAIC STRUCTURE ON RINGS OF SUBSETS 162

Proposition B.3. Let X 6= ∅ and A,B ⊆ X. Then the following holds:

(a) χA + χB = χA∆B.

(b) χA · χB = χA∩B.

Proof. (a): Let x ∈ X. Then

(χA + χB)(x) = 0 ⇔ (χA(x) = 0 ∧ χB(x) = 0) ∨ (χA(x) = 1 ∧ χB(x) = 1)

⇔ (x /∈ A ∧ x /∈ B) ∨ (x ∈ A ∧ x ∈ B)

⇔ x /∈ A ∪ B ∨ x ∈ A ∩ B⇔ x /∈ A∆B ⇔ χA∆B(x) = 0.

(b): Let x ∈ X. Then

(χA · χB)(x) = 0 ⇔ χA(x) = 0 ∨ χB(x) = 0 ⇔ x /∈ A ∨ x /∈ B

⇔ x /∈ A ∩ B ⇔ χA∩B(x) = 0,

completing the proof.

Lemma B.4. Let A,B,C be sets. Then one has the following rules:

(a) The symmetric difference is commutative: A∆B = B∆A.

(b) The symmetric difference is associative: (A∆B)∆C = A∆(B∆C).

(c) The following law of distributivity holds: (A∆B) ∩ C = (A ∩ C)∆(B ∩ C).

(d) A ∪ B = (A∆B)∆(A ∩ B).

(e) A \B = A∆(A ∩B).

Proof. (a) is immediate from (B.1), since ∪ is commutative. Rules (b) – (d) can beproved concisely by making use of Prop. B.2 and Prop. B.3: If we let X := A ∪ B ∪ Cthen A,B,C ∈ P(X) and we are in the situation of the propositions. Let x ∈ X. For(b), we compute

x ∈ (A∆B)∆C ⇔ χ(A∆B)∆C(x) = 1 ⇔ ((χA + χB) + χC)(x) = 1

⇔ (χA + (χB + χC))(x) = 1 ⇔ χA∆(B∆C)(x) = 1

⇔ x ∈ A∆(B∆C).

For (c), we compute

x ∈ (A∆B) ∩ C ⇔ χ(A∆B)∩C(x) = 1 ⇔ ((χA + χB) · χC)(x) = 1

⇔ ((χA · χC) + (χB · χC))(x) = 1 ⇔ χ(A∩C)∆(B∩C)(x) = 1

⇔ x ∈ (A ∩ C)∆(B ∩ C).

C INITIAL AND FINAL σ-ALGEBRAS 163

For (d), we compute

x ∈ (A∆B)∆(A ∩ B) ⇔ χ(A∆B)∆(A∩B)(x) = 1 ⇔ ((χA + χB) + χA · χB)(x) = 1

⇔ χA(x) = 1 ∨ χB(x) = 1 ⇔ x ∈ A ∪ B.

Finally, (e) holds, since

A∆(A ∩ B) = (A \ (A ∩ B)) ∪ ((A ∩ B) \ A) = (A \B) ∪ ∅ = A \B,

which completes the proof.

Proposition B.5. Let X be a set, E ⊆ P(X). Then (E ,∆,∩) is a ring in the senseof Def. 1.13(b) if, and only if, it is a ring in the sense of algebra (i.e. in the sense of[Phi16a, Def. C.7]).

Proof. Let (E ,∆,∩) be a ring in the sense of Def. 1.13(b). If A,B ∈ E , then A∆B ∈ Eand A ∩B ∈ E . By Lem. B.4(a),(b), ∆ is commutative and associative. The empty setis neutral for ∆, and A∆A = ∅ shows that every set is inverse to itself with respectto ∆. Thus, (E ,∆) is a commutative group. Since ∩ is also associative and the law ofdistributivity holds by Lem. B.4(c), (E ,∆,∩) is a ring in the sense of algebra. Conversely,let (E ,∆,∩) be a ring in the sense of algebra. Then E 6= ∅, i.e. there is A ∈ E . Then∅ = A∆A ∈ E . If A,B ∈ E , then A ∪ B = (A∆B)∆(A ∩ B) ∈ E by Lem. B.4(d) andA \B = A∆(A∩B) by Lem. B.4(e), showing (E ,∆,∩) to be a ring in the sense of Def.1.13(b).

Remark B.6. The reason that one can not replace ∆ by ∪ in Prop. B.5 lies in the factof missing inverse elements: While ∅ is still neutral for ∪, if A 6= ∅, then there does notexist a set B such that A ∪ B = ∅.

C Initial and Final σ-Algebras

In [Phi16b, Sec. F], the construction of initial and final topological spaces was intro-duced. Analogous constructions exist for measurable spaces and we will briefly studythem in the present section (analogous to topology, initial σ-algebras include the con-struction of product σ-algebras and trace σ-algebras (cf. Ex. C.4(a),(b) below)).

Definition C.1. Let X be a set and let (Xi,Ai)i∈I be a family of measurable spaces,I 6= ∅.

(a) Given a family of functions (fi)i∈I , fi : X −→ Xi, the initial σ-algebra on X withrespect to the family (fi)i∈I is the smallest σ-algebra A on X that makes all fiA-Ai-measurable (i.e. A is the intersection of all σ-algebras on X that make allfi measurable – this intersection is well-defined, since the σ-algebra P(X) on Xalways makes all fi measurable).

C INITIAL AND FINAL σ-ALGEBRAS 164

(b) Given a family of functions (fi)i∈I , fi : Xi −→ X, the final σ-algebra on X withrespect to the family (fi)i∈I is

A :=

A ⊆ X : ∀

i∈If−1i (A) ∈ Ai

. (C.1)

We will see in Lem. C.2(b) below (in generalization of Prop. 1.55(a)) that A is,indeed, a σ-algebra, and that it is the largest σ-algebra on X that makes all fimeasurable.1

Lemma C.2. Let X be a set and let (Xi,Ai)i∈I be a family of measurable spaces, I 6= ∅.

(a) Given a family of functions (fi)i∈I , fi : X −→ Xi, the set

E :=f−1i (Ai) : Ai ∈ Ai, i ∈ I

(C.2)

is a generator of the initial σ-algebra A on X with respect to the family (fi)i∈I .

(b) Given a family of functions (fi)i∈I , fi : Xi −→ X, the final σ-algebra A on X withrespect to the family (fi)i∈I as defined in (C.1) is, indeed, a σ-algebra on X, and itis the largest σ-algebra on X that makes all fi measurable.

Proof. (a) is immediate, since each σ-algebra A on X such that all fi, i ∈ I, are A-Ai-measurable must contain E .(b): For each i ∈ I, since ∅ = f−1

i (∅) ∈ Ai, one has ∅ ∈ A. If A ∈ A, then X \ A ∈ A,since, for each i ∈ I, f−1

i (X \ A) = Xi \ f−1i (A) ∈ Ai, due to Ai being a σ-algebra. If

(An)n∈N is a sequence of sets in A, then, for each i ∈ I,

f−1i

( ∞⋃

n=1

An

)=

∞⋃

n=1

f−1i (An) ∈ Ai,

as Ai is a σ-algebra. Thus,⋃∞

n=1An ∈ A, which completes the proof that A is aσ-algebra on X. It is immediate from the definition of A that each fi, i ∈ I, is Ai-A-measurable. To see that A is the largest σ-algebra on X with this property, we stillneed to show that every σ-algebra B on X making all fi Ai-B-measurable is containedin A. To this end, let B be such a σ-algebra on X. If B ∈ B, then, for each i ∈ I,f−1i (B) ∈ Ai, i.e. B ∈ A, showing B ⊆ A.

Proposition C.3. Let X be a set and let (Xi,Ai)i∈I be a family of measurable spaces,I 6= ∅.

(a) Given a family of functions (fi)i∈I , fi : X −→ Xi, let A denote the initial σ-algebra on X with respect to the family (fi)i∈I . Then A has the property that eachmap g : Z −→ X from a measurable space (Z,AZ) into X is measurable if, andonly if, each map (fi g) : Z −→ Xi is measurable. Moreover, A is the onlyσ-algebra on X with this property.

1In the language of so-called Category Theory, we can say that the category of measurable spaces(analogous to the category of topological spaces) has initial and final objects.

C INITIAL AND FINAL σ-ALGEBRAS 165

(b) Given a family of functions (fi)i∈I , fi : Xi −→ X, let A denote the final σ-algebraon X with respect to the family (fi)i∈I . Then A has the property that each mapg : X −→ Z from X into a measurable space (Z,AZ) is measurable if, and only if,each map (g fi) : Xi −→ Z is measurable. Moreover, A is the only σ-algebra onX with this property.

Proof. (a): If g is AZ-A-measurable, then each composition fi g, i ∈ I, is AZ-Ai-measurable. For the converse, assume each fi g, i ∈ I, to be AZ-Ai-measurable.According to Lem. C.2(a), it suffices to show g−1(f−1

i (A)) ∈ AZ for each i ∈ I and eachA ∈ Ai. Since g

−1(f−1i (A)) = (fi g)−1(A) and fi g is measurable, the proof is, thus,

complete. Now let B be an arbitrary σ-algebra on X with the property stated in thehypothesis. Letting (Z,AZ) := (X,B) and g := IdX , we see that each fi = fi g isB-Ai-measurable, implying A ⊆ B. Now let A′ be an arbitrary σ-algebra on X thatmakes all fi A′-Ai-measurable. Letting (Z,AZ) := (X,A′), we see that g := IdX isA′-B-measurable (since each fi = IdX fi is A′-Ai-measurable) i.e., for each B ∈ B,we have g−1(B) = B ∈ A′, showing B ⊆ A′ and B ⊆ A, also completing the proof ofB = A.

(b): If g is A-AZ-measurable, then each composition g fi, i ∈ I, is Ai-AZ-measurable.For the converse, assume each g fi, i ∈ I, to be Ai-AZ-measurable. If A ∈ AZ , then,for each i ∈ I, f−1

i (g−1(A)) ∈ Ai, showing g−1(A) ∈ A according to (C.1). Thus, g is

A-AZ-measurable. Now let B be an arbitrary σ-algebra on X with the property statedin the hypothesis. Letting (Z,AZ) := (X,B) and g := IdX , we see that each fi = g fiis Ai-AZ-measurable, implying B ⊆ A. Now let A′ be an arbitrary topology on X thatmakes all fi Ai-A′-measurable. Letting (Z,AZ) := (X,A′), we see that g := IdX isB-A′-measurable (since each fi = fi IdX is Ai-A′-measurable) i.e., for each A ∈ A′,we have g−1(A) = A ∈ B, showing A′ ⊆ B and A ⊆ B, also completing the proof ofB = A.

Example C.4. (a) The product σ-algebra on X =∏

i∈I Xi (cf. Def. 1.90) is the initialσ-algebra with respect to the projections (πi)i∈I , πi : X −→ Xi (as is clear fromLem. C.2(a)).

(b) The trace σ-algebra A|B on B ⊆ X, where (X,A) is a measurable space (cf.Prop. 1.55(c)), is the initial σ-algebra with respect to the identity inclusion mapι : B −→ X, ι(x) := x: This is also clear from Lem. C.2(a), since

A|B =A ∩ B : A ∈ A

=ι−1(A) : A ∈ A

.

(c) In [Phi16b, Ex. F.4(c)], we considered the quotient topology as an example for afinal topology. Analogously, we consider the quotient σ-algebra as an example fora final σ-algebra (however, in contrast to quotient topologies, quotient σ-algebrasdo not seem to play a particularly important role in the literature): Let (X,A)be a measurable space and let ∼ be an equivalence relation on X. Moreover, letY := X/ ∼= [x] : x ∈ X be the corresponding quotient set (i.e. the set ofcorresponding equivalence classes). Then the quotient σ-algebra on Y with respect

D RIEMANN INTEGRAL ON INTERVALS IN RN 166

to ∼, denoted A/ ∼, is defined as the final σ-algebra with respect to the canonicalprojection

π : X −→ Y, π(x) := [x].

Thus, by (C.1),A/ ∼=

A ⊆ Y : π−1(A) ∈ A

.

D Riemann Integral on Intervals in Rn

In generalization of [Phi16a, Sec. 10], we will define Riemann integrals for suitablefunctions f : I −→ R, where I = [a, b] is a subset of Rn, a, b ∈ Rn, a ≤ b.

In generalization of [Phi16a, Sec. 10], given a nonnegative function f : I −→ R+0 , we

aim to compute the (n+ 1)-dimensional volume∫If of the set “under the graph” of f ,

i.e. of the set

(x1, . . . , xn, xn+1) ∈ Rn+1 : (x1, . . . , xn) ∈ I and 0 ≤ xn+1 ≤ f(x1, . . . , xn)

. (D.1)

Moreover, for functions f : I −→ R that are not necessarily nonnegative, we would liketo count volumes of sets of the form (D.1) (which are below the graph of f and abovethe set I ∼=

(x1, . . . , xn, 0) ∈ Rn+1 : (x1, . . . , xn) ∈ I

⊆ Rn+1) with a positive sign,

whereas we would like to count volumes of sets above the graph of f and below the setI with a negative sign. In other words, making use of the positive and negative partsf+ = max(f, 0) and f− = max(−f, 0) of f = f+ − f−, the integral needs to satisfy

∫

I

f =

∫

I

f+ −∫

I

f−. (D.2)

As in [Phi16a, Sec. 10], we restrict ourselves to bounded functions f : I −→ R.

As in [Phi16a, Sec. 10.1], the basic idea for the definition of the Riemann integral∫If is

to decompose the interval I into small intervals I1, . . . , IN and approximate∫If by the

finite sum∑N

j=1 f(xj)|Ij|, where xj ∈ Ij and |Ij| denotes the volume of the interval Ij.

Define∫If as the limit of such sums as the size of the Ij tends to zero (if the limit exists).

However, to carry out this idea precisely and rigorously is somewhat cumbersome, forexample due to the required notation.

As each n-dimensional interval I is a product of one-dimensional intervals, we will obtainour decompositions of I from decompositions of one-dimensional intervals (the sides ofI).

Notation D.1. In generalization of [Phi16a, Def. 10.2], if a, b ∈ Rn, n ∈ N, a ≤ b, andI := [a, b] = [a1, b1]× · · · × [an, bn], then we define

|I| :=n∏

j=1

(bj − aj) =n∏

j=1

|aj − bj| =n∏

j=1

|Ij| (Ij := [aj, bj ]). (D.3)

D RIEMANN INTEGRAL ON INTERVALS IN RN 167

Definition D.2. Recall the notion of partition for a 1-dimensional interval [a, b] ⊆ Rfrom [Phi16a, Def. 10.3]. We now generalize this notion to n-dimensional intervals,n ∈ N: Given an interval I := [a, b] ⊆ Rn, a, b ∈ Rn, a < b, i.e. I = [a, b] = [a1, b1]×· · ·×[an, bn], a partition ∆ of I is given by (1-dimensional) partitions ∆k = (xk,0, . . . , xk,Nk

)of [ak, bk], k ∈ 1, . . . , n, Nk ∈ N. Given such a partition ∆ of I, for each (k1, . . . , kn) ∈P (∆) :=

∏nk=11, . . . , Nk, define

I(j1,...,jn) :=n∏

k=1

[xk,jk−1, xk,jk ] = [x1,j1−1, x1,j1 ]× · · · × [xn,jn−1, xn,jn ]. (D.4)

The number

|∆| := max|∆k| : k ∈ 1, . . . , n

= maxxk,l − xk,l−1 : k ∈ 1, . . . , n, l ∈ 1, . . . , Nk

, (D.5)

is called the mesh size of ∆.

Moreover, if each ∆k is tagged by (tk,1, . . . , tk,Nk) ∈ RNk such that tk,j ∈ [xk,j−1, xk,j] for

each j ∈ 1, . . . , Nj, then ∆ is tagged by (tp)p∈P (∆), where

t(j1,...,jn) := (t1,j1 , . . . , tn,jn) ∈ I(j1,...,jn) for each (j1, . . . , jn) ∈ P (∆). (D.6)

Remark D.3. If ∆ is a partition of I = [a, b] ⊆ Rn, n ∈ N, a, b ∈ Rn, a < b, as in Def.D.2 above, then

I =⋃

p∈P (∆)

Ip (D.7)

and|Ip ∩ Iq| = 0 for each p, q ∈ P (∆) such that p 6= q, (D.8)

since, for p 6= q, Ip ∩ Iq is either empty or it is an interval such that one side consists ofprecisely one point. Moreover, as a consequence of (D.7) and (D.8):

|I| =∑

p∈P (∆)

|Ip|. (D.9)

Definition D.4. Consider an interval I := [a, b] ⊆ Rn, n ∈ N, a, b ∈ Rn, a < b, with apartition ∆ of I as in Def. D.2. In generalization of [Phi16a, Def. 10.4], given a functionf : I −→ R that is bounded, define, for each p ∈ P (∆),

mp := mp(f) := inff(x) : x ∈ Ip, Mp :=Mp(f) := supf(x) : x ∈ Ip, (D.10)

and

r(∆, f) :=∑

p∈P (∆)

mp|Ip|, (D.11a)

R(∆, f) :=∑

p∈P (∆)

Mp|Ip|, (D.11b)

D RIEMANN INTEGRAL ON INTERVALS IN RN 168

where r(∆, f) is called the lower Riemann sum and R(∆, f) is called the upper Riemannsum associated with ∆ and f . If ∆ is tagged by τ := (tp)p∈P (∆), then we also define theintermediate Riemann sum

ρ(∆, f) :=∑

p∈P (∆)

f(tp)|Ip|. (D.11c)

Definition D.5. Let I = [a, b] ⊆ Rn be an interval, a, b ∈ Rn, n ∈ N, a < b, andsuppose f : I −→ R is bounded.

(a) Define

J∗(f, I) := supr(∆, f) : ∆ is a partition of I

, (D.12a)

J∗(f, I) := infR(∆, f) : ∆ is a partition of I

. (D.12b)

We call J∗(f, I) the lower Riemann integral of f over I and J∗(f, I) the upperRiemann integral of f over I.

(b) The function f is called Riemann integrable over I if, and only if, J∗(f, I) = J∗(f, I).If f is Riemann integrable over I, then

∫

I

f(x) dx :=

∫

I

f := J∗(f, I) = J∗(f, I) (D.13)

is called the Riemann integral of f over I. The set of all functions f : I −→ R thatare Riemann integrable over I is denoted by R(I,R) or just by R(I).

(c) The function g : I −→ C is called Riemann integrable over I if, and only if, bothRe g and Im g are Riemann integrable. The set of all Riemann integrable functionsg : I −→ C is denoted by R(I,C). If g ∈ R(I,C), then

∫

I

g :=

(∫

I

Re g,

∫

I

Im g

)=

∫

I

Re g + i

∫

I

Im g ∈ C (D.14)

is called the Riemann integral of g over I.

Remark D.6. If I and f are as before, then (D.10) implies

mp(f) = −Mp(−f) and mp(−f) = −Mp(f), (D.15a)

(D.11) implies

r(∆, f) = −R(∆,−f) and r(∆,−f) = −R(∆, f), (D.15b)

and (D.12) implies

J∗(f, I) = −J∗(−f, I) and J∗(−f, I) = −J∗(f, I). (D.15c)

Example D.7. Let I = [a, b] ⊆ Rn be an interval, a, b ∈ Rn, n ∈ N, a < b.

D RIEMANN INTEGRAL ON INTERVALS IN RN 169

(a) Analogous to [Phi16a, Ex. 10.7(a)], if f : I −→ R is constant, i.e. f ≡ c with c ∈ R,then f ∈ R(I) and ∫

I

f = c |I| : (D.16)

We have, for each partition ∆ of I,

r(∆, f) =∑

p∈P (∆)

mp|Ip| = c∑

p∈P (∆)

|Ip|(D.9)= c |I| =

∑

p∈P (∆)

Mp|Ip| = R(∆, f), (D.17)

proving J∗(f, I) = c |I| = J∗(f, I).

(b) An example of a function that is not Riemann integrable is given by the n-dimen-sional version of the Dirichlet function of [Phi16a, Ex. 10.7(b)], i.e. by

f : I −→ R, f(x) :=

0 for x ∈ I \Qn,

1 for x ∈ I ∩Qn.(D.18)

Since r(∆, f) = 0 and R(∆, f) =∑

p∈P (∆) |Ip| = |I| for every partition ∆ of I, one

obtains J∗(f, I) = 0 6= |I| = J∗(f, I), showing that f /∈ R(I).

—

For a general characterization of Riemann integrable functions, see Th. 2.26(b).

Definition D.8. Recall the notions of refinement and superposition of partitions ofa 1-dimensional interval [a, b] ⊆ R from [Phi16a, Def. 10.8]. We now generalize bothnotions to partitions of n-dimensional intervals, n ∈ N:

(a) If ∆ is a partition of [a, b] ⊆ Rn, n ∈ N, a, b ∈ Rn, a < b, as in Def. D.2, then anotherpartition ∆′ of [a, b] given by partitions ∆′

k of [ak, bk], k ∈ 1, . . . , n, respectively,is called a refinement of ∆ if, and only if, each ∆′

k is a (1-dimensional) refinementof ∆k in the sense of [Phi16a, Def. 10.8(a)].

(b) If ∆ and ∆′ are partitions of [a, b] ⊆ Rn, n ∈ N, a, b ∈ Rn, a < b, then thesuperposition ∆+∆′ is given by the (1-dimensional) superpositions ∆k +∆′

k of the[ak, bk] in the sense of [Phi16a, Def. 10.8(b)]. Note that the superposition of ∆ and∆′ is always a common refinement of ∆ and ∆′.

Lemma D.9. Let n ∈ N, a, b ∈ Rn, a < b, I := [a, b], and suppose f : I −→ R isbounded with M := ‖f‖sup ∈ R+

0 . Let ∆′ be a partition of I and define

α :=n∑

k=1

#(ν(∆′

k) \ ak, bk)

(D.19)

i.e. α is the number of interior nodes that occur in the ∆′k. Then, for each partition ∆

of I, the following holds:

r(∆, f) ≤ r(∆ +∆′, f) ≤ r(∆, f) + 2αMφ(I)|∆|, (D.20a)

R(∆, f) ≥ R(∆ +∆′, f) ≥ R(∆, f)− 2αMφ(I)|∆|, (D.20b)

D RIEMANN INTEGRAL ON INTERVALS IN RN 170

where

φ(I) := max

|I|bk − ak

: k ∈ 1, . . . , n

(D.21)

is the maximal volume of the (n− 1)-dimensional faces of I.

Proof. We carry out the proof of (D.20a) – the proof of (D.20b) can be conductedcompletely analogous. Consider the case α = 1. Then there is a unique k0 ∈ 1, . . . , nsuch that there exists ξ ∈ ν(∆′

k0)\ak0 , bk0. If ξ ∈ ν(∆k0), then ∆+∆′ = ∆, and (D.20a)

is trivially true. If ξ /∈ ν(∆k0), then xk0,l−1 < ξ < xk0,l for a suitable l ∈ 1, . . . , Nk0.Recalling the notation from Def. D.2, we let

Pl(∆) :=(j1, . . . , jn) ∈ P (∆) : jk0 = l

, (D.22)

and define, for each (j1, . . . , jn) ∈ Pl(∆),

I ′j1,...,jn :=

k0−1∏

k=1

[xk,jk−1, xk,jk ]× [xk0,l−1, ξ]×n∏

k=k0+1

[xk,jk−1, xk,jk ], (D.23a)

I ′′j1,...,jn :=

k0−1∏

k=1

[xk,jk−1, xk,jk ]× [ξ, xk0,l]×n∏

k=k0+1

[xk,jk−1, xk,jk ], (D.23b)

and

m′p := inff(x) : x ∈ I ′p, m′′

p := inff(x) : x ∈ I ′′p for each p ∈ Pl(∆). (D.24)

Then we obtain

r(∆ +∆′, f)− r(∆, f) =∑

p∈Pl(∆)

(m′

p |I ′p|+m′′p |I ′′p | −mp |Ip|

)

=∑

p∈Pl(∆)

((m′

p −mp) |I ′p|+ (m′′p −mp) |I ′′p |

). (D.25)

Together with the observation

0 ≤ m′p −mp ≤ 2M, 0 ≤ m′′

p −mp ≤ 2M, (D.26)

(D.25) implies

0 ≤ r(∆ +∆′, f)− r(∆, f) ≤ 2M∑

p∈Pl(∆)

|Ip| = 2M (xk0,l − xk0,l−1)|I|

bk0 − ak0

≤ 2M |∆k0|φ(I) ≤ 2M |∆|φ(I). (D.27)

The general form of (D.20a) now follows by an induction on α.

Theorem D.10. Let n ∈ N, a, b ∈ Rn, a < b, I := [a, b], and let f : I −→ R bebounded.

D RIEMANN INTEGRAL ON INTERVALS IN RN 171

(a) Suppose ∆ and ∆′ are partitions of I such that ∆′ is a refinement of ∆. Then

r(∆, f) ≤ r(∆′, f), R(∆, f) ≥ R(∆′, f). (D.28)

(b) For arbitrary partitions ∆ and ∆′, the following holds:

r(∆, f) ≤ R(∆′, f). (D.29)

(c) J∗(f, I) ≤ J∗(f, I).

(d) For each sequence of partitions (∆k)k∈N of I such that limk→∞ |∆k| = 0, one has

limk→∞

r(∆k, f) = J∗(f, I), limk→∞

R(∆k, f) = J∗(f, I). (D.30)

In particular, if f ∈ R(I), then

limk→∞

r(∆k, f) = limk→∞

R(∆k, f) =

∫

I

f, (D.31a)

and if f ∈ R(I) and the ∆k are tagged, then also

limk→∞

ρ(∆k, f) =

∫

I

f. (D.31b)

(e) If there exists α ∈ R such that

α = limk→∞

ρ(∆k, f) (D.32)

for each sequence of tagged partitions (∆k)k∈N of I such that limk→∞ |∆k| = 0, thenf ∈ R(I) and α =

∫If .

Proof. (a): If ∆′ is a refinement of ∆, then ∆′ = ∆ + ∆′. Thus, (D.28) is immediatefrom (D.20).

(b): This also follows from (D.20):

r(∆, f)(D.20a)

≤ r(∆ +∆′, f)(D.11)

≤ R(∆ +∆′, f)(D.20b)

≤ R(∆′, f). (D.33)

(c): One just combines (D.12) with (b).

(d): Let (∆k)k∈N be a sequence of partitions of I such that limk→∞ |∆k| = 0, and let∆′ be an arbitrary partition of I with numbers α, M , and φ(I) defined as in Lem. D.9.Then, according to (D.20a):

r(∆k, f) ≤ r(∆k +∆′, f) ≤ r(∆k, f) + 2αMφ(I)|∆k| for each k ∈ N. (D.34)

From (b), we conclude the sequence(r(∆k, f)

)k∈N is bounded. Recall from [Phi16a,

Th. 7.27] that each bounded sequence (tk)k∈N in R has a smallest cluster point t∗ ∈ R,

D RIEMANN INTEGRAL ON INTERVALS IN RN 172

and a largest cluster point t∗ ∈ R. Moreover, by [Phi16a, Prop. 7.26], there exists atleast one subsequence converging to t∗ and at least one subsequence converging to t∗,and, in particular, the sequence converges if, and only if, t∗ = t∗ = limk→∞ tk. We canapply this to the present situation: Suppose (r(∆kl , f))l∈N is a converging subsequenceof (r(∆k, f))k∈N with

β := liml→∞

r(∆kl , f). (D.35)

First note β ≤ J∗(f, I) due to the definition of J∗(f, I). Moreover, (D.34) impliesliml→∞ r(∆kl + ∆′, f) = β. Since r(∆′, f) ≤ r(∆kl + ∆′, f) and ∆′ is arbitrary, weobtain J∗(f, I) ≤ β, i.e. J∗(f, I) = β. Thus, we have shown that every subsequence of(r(∆k, f))k∈N converges to β, showing

J∗(f, I) = β = limk→∞

r(∆k, f) (D.36)

as claimed. In the same manner, one conducts the proof of J∗(f, I) = limk→∞R(∆k, f).Then (D.31a) is immediate from the definition of Riemann integrability, and (D.31b)follows from (D.31a), since (D.11) implies

r(∆, f) ≤ ρ(∆, f) ≤ R(∆, f) (D.37)

for each tagged partition ∆ of I.

(e): Due to the definition of inf and sup,

∀∅6=A⊆I

∀ǫ>0

∃t∗∈A

f(t∗) < inff(x) : x ∈ A+ ǫ, (D.38a)

∀∅6=A⊆I

∀ǫ>0

∃t∗∈A

f(t∗) > supf(x) : x ∈ A − ǫ. (D.38b)

In consequence, for each partition ∆ of I and each ǫ > 0, there are tags τ∗ := (tp,∗)p∈P (∆)

and τ ∗ := (t∗p)p∈P (∆) such that

ρ(∆, τ∗, f) < r(∆, f) + ǫ ∧ ρ(∆, τ ∗, f) > R(∆, f)− ǫ. (D.39)

Now let (∆k)k∈N be a sequence of partitions of I such that limk→∞ |∆k| = 0. Accordingto the above, for each ∆k, there are tags τ k∗ := (tkp,∗)p∈P (∆k) and τ

k,∗ := (tk,∗p )p∈P (∆k) suchthat

∀k∈N

(ρ(∆k, τ k∗ , f) < r(∆k, f) +

1

k∧ ρ(∆k, τ k,∗, f) > R(∆k, f)− 1

k

). (D.40)

Thus,

J∗(f, I)(D.30)= lim

k→∞r(∆k, f)

(∗)= lim

k→∞ρ(∆k, τ k∗ , f) = α = lim

k→∞ρ(∆k, τ k,∗, f)

(∗∗)= lim

k→∞R(∆k, f)

(D.30)= J∗(f, I), (D.41)

where, at (∗) and (∗∗), we used (D.37), (D.40), and the Sandwich theorem [Phi16a, Th.7.16]. Since (D.41) establishes both f ∈ R(I) and α =

∫If , the proof is complete.

E WALLIS PRODUCT 173

Theorem D.11 (Riemann’s Integrability Criterion). Let I = [a, b] ⊆ Rn be an interval,a, b ∈ Rn, n ∈ N, a < b, and suppose f : I −→ R is bounded. Then f is Riemannintegrable over I if, and only if, for each ǫ > 0, there exists a partition ∆ of I such that

R(∆, f)− r(∆, f) < ǫ. (D.42)

Proof. Suppose, for each ǫ > 0, there exists a partition ∆ of I such that (D.42) issatisfied. Then

J∗(f, I)− J∗(f, I) ≤ R(∆, f)− r(∆, f) < ǫ, (D.43)

showing J∗(f, I) ≤ J∗(f, I). As the opposite inequality always holds, we have J∗(f, I) =J∗(f, I), i.e. f ∈ R(I) as claimed. Conversely, if f ∈ R(I) and (∆k)k∈N is a sequence ofpartitions of I with limk→∞ |∆k| = 0, then (D.31a) implies that, for each ǫ > 0, there isN ∈ N such that R(∆k, f)− r(∆k, f) < ǫ for each k > N .

E Wallis Product

Proposition E.1. For each n ∈ N0, consider the integral

I(n) :=

∫ π

0

(sin x)n dx . (E.1)

(a) The integrals of (E.1) satisfy the following recursion:

∀n≥2

I(n) =n− 1

nI(n− 2). (E.2)

(b) The integrals of (E.1) have the following product representations for each n ∈ N0:

I(2n) =

∫ π

0

(sin x)2n dx = πn∏

k=1

2k − 1

2k, (E.3a)

I(2n+ 1) =

∫ π

0

(sin x)2n+1 dx = 2n∏

k=1

2k

2k + 1. (E.3b)

(c) One has the following representation of π as an infinite product, known as the Wallisproduct:

π

2=

∞∏

k=1

(2k

2k − 1· 2k

2k + 1

)= lim

n→∞

n∏

k=1

(2k

2k − 1· 2k

2k + 1

)=

2

1· 23· 43· 45· 65· 67· · · .

(E.4)

E WALLIS PRODUCT 174

Proof. (a): We let n ≥ 2 and integrate by parts to obtain

I(n) =

∫ π

0

(sin x)n dx =[− (sin x)n−1 cos x

]π0+

∫ π

0

(n− 1) cos x (sin x)n−2 cos x dx

= 0 + (n− 1)

∫ π

0

(1− (sin x)2

)(sin x)n−2 dx

= (n− 1) I(n− 2)− (n− 1) I(n), (E.5)

implying n I(n) = (n− 1) I(n− 2) and (E.2).

(b): We use (a) to prove (E.3) via induction on n ∈ N0: For (E.3a), we have

I(0) = π = π∏

k∈∅

2k − 1

2k(E.6)

and, for each n ∈ N,

I(2n)(E.2)=

n− 1

nI(2n− 2)

ind.hyp.=

2n− 1

2nπ

n−1∏

k=1

2k − 1

2k= π

n∏

k=1

2k − 1

2k, (E.7)

as needed. Similarly, for (E.3b), we have

I(1) =

∫ π

0

sin x dx = [− cos x]π0 = −(−1)− (−1) = 2 = 2∏

k∈∅

2k

2k + 1(E.8)

and, for each n ∈ N,

I(2n+ 1)(E.2)=

2n+ 1− 1

2n+ 1I(2n− 1)

ind.hyp.=

2n

2n+ 12

n−1∏

k=1

2k

2k + 1= 2

n∏

k=1

2k

2k + 1, (E.9)

completing the induction and the proof of (b).

(c): For each x ∈ [0, 1], we have 0 ≤ sin x ≤ 1, implying

∀x∈[0,1]

∀n∈N0

1 ≥ (sin x)n ≥ (sin x)n+1 ≥ 0 (E.10)

and∀

n∈N0

I(n) ≥ I(n+ 1) ≥ 0. (E.11)

Thus,

∀n∈N

1 ≤ I(2n)

I(2n+ 1)≤ I(2n− 1)

I(2n+ 1)=

2n+ 1

2n, (E.12)

implying

1 = limn→∞

2n+ 1

2n= lim

n→∞I(2n)

I(2n+ 1)

(E.3)=

π

2limn→∞

n∏

k=1

(2k − 1

2k· 2k + 1

2k

), (E.13)

proving (c).

F IMPROPER RIEMANN INTEGRAL OF THE SINC FUNCTION 175

F Improper Riemann Integral of the Sinc Function

Definition F.1. The function

f : R −→ R, f(x) :=

sinxx

for x 6= 0,

1 for x = 0,(F.1)

is called the sinc function.

—

The sinc function is, clearly, continuous. We have seen in Ex. 2.58(a) that sinc is notLebesgue integrable. The goal of this section is to show that sinc is still improperlyRiemann integrable with ∫ ∞

0

sin x

xdx =

π

2. (F.2)

To this end, consider the function

g : R+ −→ R, g(t) :=

∫ ∞

0

e−tx sin x

xdx . (F.3)

For each t ∈ R+, the integrand and its derivative with respect to t are Lebesgue inte-grable, since they are dominated by the Lebesgue integrable function

x 7→

| sinx|x

for x < 1,

x e−tx for x ≥ 1.(F.4)

For each ǫ > 0 and each t ≥ ǫ the integrand in (F.3) is uniformly dominated by

x 7→

| sinx|x

for x < 1,

x e−ǫx for x ≥ 1.(F.5)

This shows that Cor. 2.24 applies on [ǫ,∞[, implying g to be differentiable (even on R+,since ǫ > 0 was arbitrary),

g′ : R+ −→ R, g′(t) := −∫ ∞

0

x e−tx sin x

xdx = −

∫ ∞

0

e−tx sin x dx . (F.6)

To compute the integral in (F.6) explicitly, we claim that an antiderivative of h(x) :=−e−tx sin x is given by

H(x) :=cos x+ t sin x

etx(1 + t2). (F.7)

Indeed,

H ′(x) =etx(t cosx− sin x)− tetx(cos x+ t sin x)

e2tx(1 + t2)

=− sin x− t2 sin x

e2tx(1 + t2)= −e−tx sin x = h(x). (F.8)

F IMPROPER RIEMANN INTEGRAL OF THE SINC FUNCTION 176

Thus, for each t ∈ R+,

g′(t) =

[cos x+ t sin x

etx(1 + t2)

]x=∞

x=0

= 0− 1

1 + t2= − 1

1 + t2. (F.9)

In consequence, both g and (− arctan) are antiderivatives of g′, implying

∃C∈R

∀t∈R+

g(t) = C − arctan(t). (F.10)

Using dominated convergence, we obtain

limt→∞

g(t) = limt→∞

∫ ∞

0

e−tx sin x

xdx

DCT= 0 (F.11)

andC = lim

t→∞

(g(t) + arctan(t)

)= 0 +

π

2=π

2, (F.12)

that means∀

t∈R+g(t) =

π

2− arctan(t). (F.13)

For each r > 0, we can apply dominated convergence once again to yield

∫ r

0

sin x

xdx

DCT= lim

t→0

∫ r

0

e−tx sin x

xdx . (F.14)

For each t ≥ 0 and each k ∈ N, let at,k :=∫ kπ

(k−1)πe−tx sinx

xdx . According to the Leibniz

criterion [Phi16a, Th. 7.85],

∀t>0

g(t) =∞∑

k=1

at,k, ∀j∈N

∣∣∣∣∣∞∑

k=j+1

at,k

∣∣∣∣∣ < |at,j+1| ≤ |a0,j+1|. (F.15)

and ∫ ∞

0

sin x

xdx =

∞∑

k=1

a0,k, ∀j∈N

∣∣∣∣∣∞∑

k=j+1

a0,k

∣∣∣∣∣ < |a0,j+1|. (F.16)

Thus, given ǫ > 0, we can choose R > 0, such that

∀r≥R

∀t>0

∣∣∣∣g(t)−∫ r

0

e−tx sin x

xdx

∣∣∣∣ =∣∣∣∣∫ ∞

r

e−tx sin x

xdx

∣∣∣∣ <ǫ

4(F.17)

and

∀r≥R

∣∣∣∣∫ ∞

0

sin x

xdx −

∫ r

0

sin x

xdx

∣∣∣∣ =∣∣∣∣∫ ∞

r

sin x

xdx

∣∣∣∣ <ǫ

4. (F.18)

Now let r ≥ R. Using (F.13) and (F.14), we choose δ > 0 such that

∀0<t<δ

π

2>

∫ ∞

0

e−tx sin x

xdx = g(t) >

π

2− ǫ

4(F.19)

G TOPOLOGICAL AND MEASURE-THEORETIC SUPPLEMENTS 177

and

∀0<t<δ

∣∣∣∣∫ r

0

sin x

xdx −

∫ r

0

e−tx sin x

xdx

∣∣∣∣ <ǫ

4. (F.20)

Altogether, we then obtain, for 0 < t < δ,∣∣∣∣π

2−∫ ∞

0

sin x

xdx

∣∣∣∣ ≤∣∣∣π2− g(t)

∣∣∣+∣∣∣∣g(t)−

∫ r

0

e−tx sin x

xdx

∣∣∣∣

+

∣∣∣∣∫ r

0

e−tx sin x

xdx −

∫ r

0

sin x

xdx

∣∣∣∣

+

∣∣∣∣∫ r

0

sin x

xdx −

∫ ∞

0

sin x

xdx

∣∣∣∣ < 4 · ǫ4= ǫ. (F.21)

As ǫ > 0 was arbitrary, this proves (F.2).

G Topological and Measure-Theoretic Supplements

G.1 Transitivity of Dense Subsets

Proposition G.1. Let (X, T ) be a topological space, B ⊆ A ⊆ X. If B is dense in Aand A is dense in X, then B is dense in X.

Proof. Let ∅ 6= O ∈ T . Since A is dense in X, O ∩ A 6= ∅. Since O ∩ A ∈ TA and B isdense in A, O ∩ A ∩B 6= ∅. Thus, O ∩ B 6= ∅, showing B to be dense in X.

G.2 Inverse Image and Trace for Semirings

Proposition G.2. Consider sets X, Y and a map f : X −→ Y .

(a) If S is a semiring on Y , then f−1(S) is a semiring on X.

(b) Consider B ⊆ X. If S is a semiring on X, then the trace S|B is a semiring on B.

Proof. (a): As ∅ = f−1(∅), ∅ ∈ f−1(S). If R, S ∈ S, then

f−1(R) ∩ f−1(S) = f−1(R ∩ S) ∈ f−1(S),

since R ∩ S ∈ S due to S being a semiring. Moreover, as S is a semiring, for R, S ∈ S,there are finitely many disjoint sets T1, . . . , Tn, n ∈ N, such that Ti ∈ S for eachi ∈ 1, . . . , n, and R \ S =

⋃n

i=1Ti. Then

f−1(R) \ f−1(S) = f−1(R \ S) = f−1

(⋃n

i=1Ti

)=⋃n

i=1f−1(Ti),

completing the proof that f−1(S) is a semiring.

(b): Due to (1.39), (b) follows immediately from (a).

G TOPOLOGICAL AND MEASURE-THEORETIC SUPPLEMENTS 178

G.3 Measurability with Respect to Completions

Proposition G.3. Let (X,A, µ), (X,B, ν) be measure spaces such that A ⊆ B ⊆ A,µ = ν A, ν = µB, where (X, A, µ) is the completion of (X,A, µ). If f : X −→ K isB-measurable, then there exists an A-measurable function g : X −→ K such that f = gµ-almost everywhere.

Proof. First, assume f ≥ 0 (i.e. f ∈ M+(B)) and choose a sequence (φk)k∈N in S+(B)such that φk ↑ f . Then

f = limN→∞

φN = limN→∞

(φ1 +

N∑

k=2

(φk − φk−1)

). (G.1)

Since φ1 ∈ S+(B) and each (φk−φk−1) ∈ S+(B), there exist sequences (αk)k∈N in R+ and(Bk)k∈N in B such that f =

∑∞k=1 αkχBk

. Since Bk ∈ B ⊆ A, there exist Ak, Nk ∈ A,Mk ⊆ Nk, such that Bk = Ak ∪Mk, µ(Nk) = 0. Define

g :=∞∑

k=1

αkχAk. (G.2)

Then g ∈ M+(A). We claim f = g µ-a.e.: Indeed, if x /∈ ⋃∞k=1(Bk \ Ak), then

f(x) = g(x) (if x /∈ ⋂∞k=1Bk, then f(x) = g(x) = 0; if x ∈ Bk implies x ∈ Ak for each

k ∈ N, then f(x) = g(x) =∑∞

k:x∈Bkαk). Since, for each k ∈ N, Bk \ Ak ⊆Mk ⊆ Nk,

µ

( ∞⋃

k=1

(Bk \ Ak)

)≤

∞∑

k=1

µ(Nk) = 0, (G.3)

we have shown f = g µ-a.e. For a general B-measurable f : X −→ K, we decompose finto (Re f)± and (Im f)± in the usual way, and apply the case f ≥ 0 to the functions(Re f)±, (Im f)±.

G.4 Interchanging Integrals with Uniform Limits

Proposition G.4. Let (X,A, µ) be a measure space with µ(X) <∞. Let f, fn : X −→K be measurable functions, n ∈ N, such that each fn is µ-integrable and f = limn→∞ fnuniformly µ-a.e. (i.e. there exists a µ-null set N ∈ A such that f = limn→∞ fn uniformlyon N c). Then f is µ-integrable with

limn→∞

∫

X

fn dµ =

∫

X

f dµ (G.4a)

and

limn→∞

∫

X

|fn − f | dµ = 0. (G.4b)

H POLAR COORDINATES 179

Proof. If µ(X) = 0, there is nothing to prove. Otherwise, given ǫ > 0, there existsN ∈ N such that, for each n > N , ‖fn − f‖∞ < ǫ/µ(X). Thus, for each n > N ,

∫

X

|fn − f | dµ ≤ ǫ

µ(X)· µ(X) = ǫ,

proving (G.4b). As

∀n∈N

∣∣∣∣∫

X

fn dµ −∫

X

f dµ

∣∣∣∣ ≤∫

X

|fn − f | dµ ,

(G.4a) is also proved. Since ‖f‖1 ≤ ‖f − fn‖1 + ‖fn‖ <∞, f is µ-integrable.

H Polar Coordinates

LetO := R+×]0, 2π[, U := R2 \ (x, 0) : x ≥ 0. (H.1)

We verify that the function defining polar coordinates, i.e.

φ : O −→ U, φ(r, ϕ) := (r cosϕ, r sinϕ), (H.2)

constitutes a C1-diffeomorphism:

We first show φ to be bijective with inverse map

φ−1 : U −→ O, φ−1(x, y) =(φ−11 (x, y), φ−1

2 (x, y)), (H.3a)

where, recalling the definition of arccot from [Phi16a, Def. and Rem. 8.27],

φ−11 (x, y) :=

√x2 + y2, (H.3b)

φ−12 (x, y) :=

arccot(x/y) for y > 0,

π for y = 0,

π + arccot(x/y) for y < 0.

(H.3c)

One verifies φ−1 φ = IdO:

∀(r,ϕ)∈O

(φ−1 φ)(r, ϕ) = φ−1(r cosϕ, r sinϕ) = (r, ϕ),

since

φ−11 (r cosϕ, r sinϕ) =

√r2(cos2 ϕ+ sin2 ϕ) = r,

φ−12 (r cosϕ, r sinϕ) =

arccot(cotϕ) = ϕ for 0 < ϕ < π,

π = ϕ for ϕ = π,

π + arccot(cotϕ) = π + ϕ− π = ϕ for π < ϕ < 2π;

REFERENCES 180

and φ φ−1 = IdU :

∀(x,y)∈U

(φ φ−1)(x, y) =(√

x2 + y2 cosφ−12 (x, y),

√x2 + y2 sinφ−1

2 (x, y))

= (x, y),

since

cosφ−12 (x, y) =

1√1+(cot arccot(x/y))−2

= 1√

1+ y2

x2

= x√x2+y2

for x > 0 (⇒ y 6= 0),

cos(π/2) = 0 = x√x2+y2

for x = 0, y > 0,

cos(3π/2) = 0 = x√x2+y2

for x = 0, y < 0,

−1√1+(cot arccot(x/y))−2

= x√x2+y2

for x < 0, y 6= 0,

cos π = −1 = x√x2+y2

for y = 0 (⇒ x < 0),

sinφ−12 (x, y) =

1√1+cot2 arccot(x/y)

= 1√

1+x2

y2

= y√x2+y2

for y > 0,

sin π = 0 = y√x2+y2

for y = 0,

−1√1+cot2 arccot(x/y)

= y√x2+y2

for y < 0.

Next, we note that φ has continuous first partials, where

∀(r,ϕ)∈O

Dφ(r, ϕ) =

(cosϕ −r sinϕsinϕ r cosϕ

), detDφ(r, ϕ) = r > 0. (H.4)

Thus, φ is a C1-diffeomorphism by Def. and Rem. 2.64.

References

[Alt06] Hans Wilhelm Alt. Lineare Funktionalanalysis, 5th ed. Springer-Verlag,Berlin, 2006 (German).

[Beh87] E. Behrends. Maß- und Integrationstheorie. Springer-Verlag, Berlin, 1987(German).

[Els07] Jurgen Elstrodt. Maß- und Integrationstheorie, 5th ed. GrundwissenMathematik, Springer-Verlag, Berlin, 2007 (German).

[Koe03] Max Koecher. Lineare Algebra und analytische Geometrie, 4th ed. Sprin-ger-Verlag, Berlin, 2003 (German), 1st corrected reprint.

[Kon64] H. Konig. Ein einfacher Beweis des Integralsatzes von Gauß. Jahrestagungder Deutschen Mathematischen Vereinigung. Vol. 66, 1964 (German).

[Kon04] Konrad Konigsberger. Analysis 2, 5th ed. Springer-Verlag, Berlin, 2004(German).

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[Oss09] E. Ossa. Topologie. Vieweg+Teubner, Wiesbaden, Germany, 2009 (German).

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[Phi16b] P. Philip. Analysis II: Topology and Differential Calculus of Several Variables.Lecture Notes, Ludwig-Maximilians-Universitat, Germany, 2016, available inPDF format athttp://www.math.lmu.de/~philip/publications/lectureNotes/philipPeter_Analysis2.pdf.

[RF10] Halsey Royden and Patrick Fitzpatrick. Real Analysis, 4th ed. PearsonEducation, Boston, USA, 2010.

[Rud87] W. Rudin. Real and Complex Analysis, 3rd ed. McGraw-Hill Book Company,New York, 1987.