real analysis ii (measure theory) notes

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Contents

Content to be covered iii

Chapter 1. Measure and Integration 1

1. σ- algebra and measure 1

2. Measurable Functions 5

3. Integration 9

Chapter 2. Signed measure and decompositions 23

4. Signed measure 23

5. Decompositions of signed measure space 26

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Content to be covered

Unit I

Measure space and different examples, finite, σ-finite, complete and saturated measures,

measurable functions and Lusin’s theorem and applications. Integration, general convergence

theorems.

Unit II

Signed measure, Hahn decomposition, Jordan decomposition. Lebesgue decomposition the-

orem , Radon-Nikodym theorem, Radon-Nikodym derivatives, Lebesgue Stiltjes integral.

Unit III

Cumulative distributions and properties, Lp-Spaces, Holder’s inequality, Minkowski inequal-

ity, Riesz-Fischer’s theorem, Riesz representation theorem, density in Lp-Spaces.

Unit IV

Caratheodory’s extension theorem, product measure, Fubini’s Theorem, Tonelli’s theorem,

regularity of Baire and Borel Measures.

Reference Books

(i) H.L.Royden, Real Analysis (3rd Edition) Mc. Millan, 1998.

(ii) G. de Berra, Introduction to Measure Theory, van-Nordstrand, 1974.

(iii) P.R. Halmos, Measure Theory, van-Nordstran, 1970.

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PADABHI

Real Analysis II Notes

Author:

Prakash A. Dabhi

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PADABHICHAPTER 1

Measure and Integration

1. σ- algebra and measure

Definition 1.1.1. Let X be a set. A subset A of the powerset P (X) of X is called a σ-

algebra if ∅ ∈ A , A is closed with respect to the formation of complement in X and A is

closed with respect to the formation of countable unions.

If A is a σ- algebra of subsets of X, then the pair (X,A ) is called a measurable space.

Let A be a σ- algebra of subsets of X. A subset A of X is called measurable if A ∈ A .

Note that {∅, X} and P (X) are the smallest and the largest σ- algebras of subsets of

X respectively.

Exercise 1.1.2. Show that the condition (3) in the definition 1.1.1 can be replaced by the

formation of countable intersections.

Definition 1.1.3. Let A be a σ- algebra of subsets of X. A map µ : A → [0,∞] is called

a measure if

(i) µ(∅) = 0,

(ii) µ is countably additive, i.e., if {En} is a sequence of pairwise disjoint measurable

subsets of X, then µ(⋃nEn) =

∑n µ(En)

If µ is a measure on a measurable space (X,A ), then the triplet (X,A , µ) is called a

measure space.

Examples 1.1.4.

(i) Let M be the σ- algebra of all measurable subsets of R, and let m be the Lebesgue

measure of R. Then (R,M ,m) is a measure space.

(ii) Let A be the collection of all measurable subsets of [0, 1] and let m be the Lebesgue

measure on [0, 1]. Then ([0, 1],A ,m) is a measure space.

(iii) Let B be the Borel σ algebra on R, and let m be the Lebesgue measure on R. Then

(R,B,m) is a measure space.

(iv) Let X be any set, and let A = {∅, X}. Let α > 0. Define µα : A → [0,∞] by

µα(∅) = 0 and µα(X) = α. Then (X,A , µα) is a measure space.

(v) Let X be a any set. Let ν : P (X) → [0,∞] be as follows. For a subset A of X put

ν(A) =∞ is A is an infinite set and put ν(A) to be the number of elements in A. Then

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2 1. MEASURE AND INTEGRATION

ν is a measure on X, called the counting measure (on X). The triplet (X,P (X), ν) is

a measure space.

(vi) Let X be an uncountable set. Let A = {E ⊂ X : either E or Ec is countable}. Then

A is a σ- algebra of subsets of X. Let α > 0 Define µα : →[0,∞] by µα(A) = 0 if A

is uncountable and µα(A) = α if A is countable. Then (X,A , µα) is a measure space.

(vii) Let X be a nonempty set, and let x ∈ X. Define δx : P (X) → [0,∞] by δx(A) = 1 if

x ∈ A and δx(A) = 0 if x /∈ A. Then δx is a measure on (X,P (X)), called the Dirac

measure concentrated at x. Then (X,P (X), δx) is a measure space.

(viii) Let (X,A , µ) be a measure space, and let X0 be a measurable subset of X. Let

A0 = {U ⊂ X : U ∈ A , U ⊂ X0} = {U ∩X0 : U ∈ A }. Then A0 is a σ- algebra of

subsets of X0. Define µ0 : A0 → [0,∞] by µ0(E) = µ(E), E ∈ A0. In fact, µ0 = µ|A0.

Then (X0,A0, µ0) is a measure space.

Lemma 1.1.5 (Monotonicity of a measure). Let (X,A , µ) be a measure space, and let E and

F be measurable subsets of X with F ⊂ E. Then µ(F ) ≤ µ(E). Furthermore, if µ(F ) <∞,

then µ(E − F ) = µ(E)− µ(F ).

Proof. Clearly, E = F ∪ (E − F ). As both E and F are measurable, E − F = E ∩ F c is

measurable. Since F and E − F are disjoint, µ(E) = µ(F ) + µ(E − F ) ≥ µ(F ).

Let µ(F ) < ∞. Since µ(E) = µ(F ) + µ(E − F ) and µ(F ) < ∞, we have µ(E − F ) =

µ(E)− µ(F ). �

Note that we cannot drop the condition that µ(F ) < ∞ in the above lemma. For

example, let E = F = R in (R,M ,m). Then m(E) = m(F ) =∞ and m(E−F ) = m(∅) = 0

but m(E)−m(F ) does not exist.

Lemma 1.1.6. Let (X,A , µ) be a measure space, and let {En} be a sequence of measurable

subsets of X. Then µ(⋃nEn) ≤

∑n µ(En).

Proof. Let F1 = E1 and for n > 1, let Fn = Fn − (⋃n−1k=1 Fk). Then each Fn is measurable,

Fn ∩ Fm = ∅ if n 6= m and⋃n Fn =

⋃nEn. Also note that Fn ⊂ En for all n. Therefore

µ(Fn) ≤ µ(En) for all n. Now, µ(⋃nEn) = µ(

⋃n Fn) =

∑n µ(Fn) ≤

∑n µ(En). �

Lemma 1.1.7. Let {En} be an increasing sequence of measurable subsets of a measure space

(X,A , µ). Then µ(⋃nEn) = limn µ(En) = supn µ(En).

Proof. Since {En} is increasing, µ(En) ≤ µ(En+1) for all n. Also, µ(En) ≤ µ(⋃nEn).

If µ(En0) = ∞ for some n0, then (µ(En) = ∞ for all n ≥ n0) clearly limn µ(En) = ∞and µ(

⋃nEn) = ∞. We are through in this case. Now assume that µ(En) < ∞ for all

n. Let F1 = E1 and for n > 1, let Fn = En − En−1. Then {Fn} is a sequence of pairwise

disjoint measurable subsets of X with⋃n Fn =

⋃nEn. Since µ(En) <∞ for all n, we have

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1. σ- algebra and measure 3

µ(Fn) = µ(En)− µ(En−1) for all n > 1 and µ(F1) = µ(E1). Now

µ(∪nEn) = µ(∪nFn) =∑n

µ(Fn) = limn

n∑k=1

µ(Fk)

= limn

[µ(E1) +n∑k=2

(µ(Ek)− µ(Ek−1))]

= limnµ(En).

Since the sequence {µ(En)} is increasing, limn µ(En) = supn µ(En). �

Corollary 1.1.8. Let {En} be a sequence of measurable subsets of a measure space (X,A , µ).

Then µ(⋃nEn) = limn µ(

⋃nk=1Ek)

Proof. For each n set Fn =⋃nk=1Ek. Then {Fn} is an increasing sequence of measurable

subsets of X and⋃n Fn =

⋃nEn. By above corollary we have

µ(⋃nEn) = µ(

⋃n Fn) = limn µ(Fn) = limn µ(

⋃nk=1Ek). �

Lemma 1.1.9. Let {En} be a decreasing sequence of measurable subsets of a measure space

(X,A , µ). If µ(E1) <∞, then µ(⋂nEn) = limn µ(En).

Proof. For each n, let Fn = E1 − En. Since {En} is a decreasing sequence, the sequence

{Fn} is an increasing sequence of measurable sets. As µ(E1) < ∞ (and hence µ(En) < ∞for all n), we have µ(Fn) = µ(E1)−µ(En). It is also clear that

⋃n Fn = E1− (

⋂nEn). Now,

µ(E1)− µ(⋂n

En) = µ(E1 − (⋂n

En)) = µ(⋃n

Fn)

= limnµ(Fn) = lim

n(µ(E1)− µ(Fn)) = µ(E1)− lim

nµ(En).

Hence µ(⋂nEn) = limn µ(En). �

We cannot drop the condition that µ(E1) < ∞ in the above lemma. For example

consider En = [n,∞) in (R,M ,m). Then {En} is a decreasing sequence of measurable

subsets of R. Note that m(En) =∞ for all n and so limnm(En) =∞ while⋂nEn = ∅ gives

m(⋂nEn) = 0.

Definition 1.1.10. Let (X,A , µ) be a measure space. The measure µ is called a finite

measure (or the measure space (X,A , µ) is called a finite measure space) if µ(X) <∞.

The measure µ is called a σ- finite measure (or the measure space (X,A , µ) is called a

σ- finite measure space) if X can be written has a countable union of measurable sets each

having finite measure, i.e., there is a sequence {En} of measurable subsets of X such that

µ(En) <∞ for all n and⋃nEn = X.

A subset E of a measure space (X,A , µ) is said to have σ- finite measure if it can be

written as a countable union of measurable subsets of X each having finite measure.

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Exercise 1.1.11.

(i) If (X,A , µ) is a finite measure space, then it is a σ- finite measure space. Give an

example to show that the converse is not true.

(ii) Any measurable subset of finite measure space has a finite measure.

(iii) If E is a measurable subset of σ- finite measure space, then E is of σ- finite measure.

(iv) If E1, . . . , En are sets of finite measure in a measure space, then their union is a set of

finite measure.

(v) Countable union of sets of σ- finite measures is of σ- finite measure.

Definition 1.1.12. A measure space (X,A , µ) (or the measure µ) is called complete if Acontains all subsets of sets of measure zero.

Example 1.1.13.

(i) The measure space (R,M ,m) is complete.

Let E ∈ M be such that m(E) = 0, and let F be a subset of E. Since F ⊂ E,

0 ≤ m∗(F ) ≤ m∗(E) = m(E) = 0, i.e., m∗(F ) = 0. Let A be any subset of R. Then

A ∩ F ⊂ F gives m∗(A ∩ F ) = 0. Now A ∩ F c ⊂ A gives m∗(A ∩ F c) ≤ m∗(A).

Therefore m∗(A) ≥ m∗(A ∩ F ) = m∗(A ∩ F c) + m∗(A ∩ F ). Hence F is measurable,

i.e., F ∈M .

(ii) The measure space (R,B,m) is not a complete measure space as the Cantor set C has

measure 0 and it contains a subset which is not a Borel set (Construct such a set!!!).

Theorem 1.1.14 (Completion of a measure space). Let (X,A , µ) be a measure space. Then

there is a complete measure space (X,A0, µ0) such that

(i) A ⊂ A0,

(ii) µ0(E) = µ(E) for every E ∈ A ,

(iii) If E ∈ A0, then E = A ∪ B for some A ∈ A and B ⊂ C for some C ∈ A with

µ(C) = 0.

Proof. Let A0 = {E ⊂ X : E = A ∪ B, A ∈ A , B ⊂ C for some C ∈ A with µ(C) = 0}.Clearly, A ⊂ A0 ( if E ∈ A , then E = E ∪∅ ∈ A0). Let E = a∪B ∈ A0. Then A ∈ A and

B ⊂ C for some C ∈ A with µ(C) = 0. Then Ec = Ac ∩ Bc = (Ac ∩ Cc) ∪ (Ac ∩ (C − B)).

Clearly, Ac ∩ Cc ∈ A , (Ac ∩ (C − B) ⊂ C. Therefore Ec ∈ A0. Let {En} be a countable

collection of elements of A0. Then En = An ∪ Bn, where An ∈ A and Bn ⊂ Cn for some

Cn ∈ A with µ(Cn) = 0. Now⋃nEn = (

⋃nAn)

⋃(⋃nBn). Clearly,

⋃nAn ∈ A and⋃

nBn ⊂⋃nCn,

⋃nCn ∈ A and 0 ≤ µ(

⋃nCn) ≤

∑n µ(Cn) = 0. Hence A0 is a σ- algebra

of subsets of X. Define µ0 on A0 as follows. If E = A∪B ∈ A0, then we put µ0(E) = µ(A).

Clearly, if E ∈ A , then µ0(E) = µ0(E ∪ ∅) = µ(E). By definition µ0(E) ≥ 0 for every

E ∈ A0. Let {En} be a sequence of pairwise disjoint elements of A0. Then En = An ∪ Bn,

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where An ∈ A , Bn ⊂ Cn for some Cn ∈ A with µ(Cn) = 0. Since En’s are pairwise disjoint,

An’s are pairwise disjoint. Now

µ0(⋃n

En) = µ0((⋃n

An)⋃

(⋃n

Bn)) = µ(⋃n

An) =∑n

µ(An) =∑n

µ0(En).

Therefore µ0 is a measure on (X,A0). It remains to show that µ0 is complete. For that let

E = A ∪ B ∈ A0 with µ0(E) = 0 = µ(A), and let F ⊂ E. Then F = ∅⋃

((C⋃A)⋂F ).

Obviously (C⋃A)⋂F ⊂ C

⋃A ∈ A and 0 ≤ µ(C

⋃A) ≤ µ(C) + µ(A) = 0. Hence

F ∈ A0. This finishes the proof. �

Definition 1.1.15. Let (X,A , µ) be a measure space. A subset E of X is said to be locally

measurable if E ∩ A ∈ A for every A ∈ A with µ(A) <∞.

A measure space (X,A , µ) is called saturated if every locally measurable subset of X

is measurable.

Note that every measurable set is locally measurable. The converse is not true (Example

???).

Lemma 1.1.16. Every σ- finite measure space is saturated.

Proof. Let (X,A , µ) be a σ- finite measure space. Then there is a sequence {En} of

measurable subsets of X such that⋃nEn = X and µ(En) <∞ for all n. Let E be a locally

measurable subset of X. Then clearly, E ∩ En ∈ A as µ(En) < ∞. Now E = E ∩ X =⋃n(E ∩ En). Therefore E is a countable union of measurable subsets of X and hence it is

measurable. This proves that (X,A , µ) is saturated. �

Since (R,M ,m) and (R,B,m) are σ- finite measure spaces, they are saturated.

2. Measurable Functions

Lemma 1.2.1. Let (X,A ) be a measurable space, and let f be an extended real valued

function on X. Then the following conditions are equivalent.

(i) {x ∈ X : f(x) > α} ∈ A for every α ∈ R.

(ii) {x ∈ X : f(x) ≥ α} ∈ A for every α ∈ R.

(iii) {x ∈ X : f(x) < α} ∈ A for every α ∈ R.

(iv) {x ∈ X : f(x) ≤ α} ∈ A for every α ∈ R.

All the above conditions imply that the set {x ∈ X : f(x) = α} ∈ A for every α ∈ R.

Proof. (i)⇒ (ii). Let α ∈ R. Then

{x ∈ X : f(x) ≥ α} =⋂n∈N

{x ∈ X : f(x) > α− 1

n}.

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DA

BH

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Since each set on the right side is measurable, it follows that {x ∈ X : f(x) ≥ α} is measur-

able.

(ii) ⇒ (iii) Let α ∈ R. Then {x ∈ X : f(x) < α} = {x ∈ X : f(x) ≥ α}c. Therefore

{x ∈ X : f(x) < α} is measurable.

(iii) ⇒ (iv) Let α ∈ R. Then {x ∈ X : f(x) ≤ α} =⋂n∈N{x ∈ X : f(x) < α + 1

n}.

Therefore {x ∈ X : f(x) ≤ α} is measurable.

(iv) ⇒ (i) Let α ∈ R. Then {x ∈ X : f(x) > α} = {x ∈ X : f(x) ≤ α}c. Therefore

{x ∈ X : f(x) > α} is measurable.

Since the all the above conditions are equivalent for any α ∈ R, the sets {x ∈ X : f(x) ≤α} and {x ∈ X : f(x) ≥ α} are measurable. Therefore their intersection {x ∈ X : f(x) = α}is measurable. �

Definition 1.2.2. Let (X,A ) be a measurable space, and let f : X → [−∞,∞]. Then f is

called measurable if {x ∈ X : f(x) > α} ∈ A for every α ∈ R.

Lemma 1.2.3. Let (X,A ) be a measurable space, and let E ⊂ X. Then E ∈ A if and only

if χE is measurable.

Proof. Assume that χE is measurable. Then the set {x ∈ X : χE(x) > 0} ∈ A . But

{x ∈ X : χE(x) > 0} = E. Therefore E is measurable.

Conversely, assume that E is measurable. Let α ∈ R. Then

{x ∈ X : χE(x) > α} =

∅ if α ≥ 1

E if 0 ≤ α < 1

X if α < 0

Hence χE is measurable. �

Theorem 1.2.4. Let (X,A ) be a measurable space, and let f and g be measurable functions

on X. Let c ∈ R. Then

(i) cf is measurable.

(ii) f + c is measurable.

(iii) f+, f− and |f | are measurable.

(iv) max{f, g} and min{f, g} are measurable.

Definition 1.2.5. Let (X,A , µ) be a measure space. A property P is said to hold almost

everywhere [µ] on X if there is a measurable subset E of X with µ(E) = 0 such that P holds

on X − E.

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Lemma 1.2.6. Let (X,A , µ) be a complete measure space, and let f be measurable. Let

g : X → [−∞,∞] be a map. If g = f a.e. [µ], then g is measurable.

Proof. Since f = g a.e. [µ], there is a measurable subset E of X with µ(E) = 0 such that

f = g on X−E. Let α ∈ R. Then {x ∈ X : g(x) > α} = ({x ∈ X : f(x) > α}∩(X−E))∪Ffor some subset F of E. Since µ is complete, F is measurable. Hence g is measurable. �

Theorem 1.2.7. Let (X,A , µ) be a measure space, and let f and g be measurable functions

on X which are finite a.e. [µ]. Then

(i) f ± g is measurable. (be very much careful in proving this.)

(ii) fg is measurable. (again be very much careful in proving this.)

(iii) f/g is measurable if g(x) 6= 0 for any x.

Lemma 1.2.8. Let (X,A ) be a measurable space, and let {fn} be a sequence of measurable

functions on X. Then supn fn, infn fn, lim supn fn, lim infn fn are measurable. In particular,

when fn → f (pointwise) on X, then f is measurable.

Proof. Let g = supn fn, and let h = infn fn. Let α ∈ R. Then {x ∈ X : g(x) ≤ α} =⋂n{x ∈ X : fn(x) ≤ α} and {x ∈ X : h(x) ≥ α} =

⋂n{x ∈ X : fn(x) ≥ α}. Therefore

both g and h are measurable. Let gk = supn≥k fn and hk = infn≥k fn. Then both gk and

hk are measurable for every k. Therefore lim supn fn = infk gk and lim infn fn = supk hk are

measurable. If a sequence {fn} converges to f , then f = lim supn fn = lim infn fn. Therefore

f is measurable. �

Definition 1.2.9. Let X be a topological space. Then the smallest σ- algebra of subsets of

X containing all open subsets of X is called the Borel σ- algebra on X. Any element of the

Borel σ- algebra is called a Borel set.

Question 1.2.10. Let A be a σ- algebra of subsets of X. Does there exist a topology on

X whose Borel σ- algebra is A ?

In particular, we know that the set M of all measurable subsets of R is a σ- algebra.

Does there exist a topology on R whose Borel σ- algebra is M ? (There is such a topology

find it or construct it).

Consider X = [−∞,∞]. The the standard topology on X is generated by a basis

{(a, b), (c,∞], [−∞, d) : a, b, c, d ∈ R, a < b}. Also note that any open subset of [−∞,∞]

can be written as a countable union of disjoint open sets of the form (a, b), (c,∞], [−∞, d).

Theorem 1.2.11. Let (X,A ) be a measurable space, and let f be an extended real valued

map. Then the following are equivalent.

(i) f is measurable.

(ii) {x ∈ X : f(x) > r} is measurable for every r ∈ Q.

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(iii) f−1(E) is measurable for every open subset E of [−∞,∞].

Proof. (i)⇒ (ii) is clear.

(ii)⇒ (i) Let a ∈ R. Then

f−1((a,∞]) = {x ∈ X : f(x) > a} =⋃

r∈Q,r>a

{x ∈ X : f(x) > r}.

Since each {x ∈ X : f(x) > r} is measurable and countable union of measurable sets is

measurable, {x ∈ X : f(x) > a} is measurable, i.e., f is measurable.

(iii) ⇒ (i) Assume that f−1(E) is measurable for every open subset E of [−∞,∞]. Let

α ∈ R. Then (α,∞] is an open subset of [−∞,∞]. Now f−1(α,∞] = {x ∈ X : f(x) > α}.Hence f is measurable.

(i)⇒ (iii) Assume that f is measurable. Let a ∈ R. Then f−1(a,∞] = {x ∈ X : f(x) > a}and f−1[−∞, a) = {x ∈ X : f(x) < a}. Since f is measurable, for any a ∈ R the sets

f−1(a,∞] and f−1[−∞, a) are measurable. Let a, b ∈ R, then f−1(a, b) = f−1[−∞, b) ∩f−1(a,∞], which is also measurable. Let E be an open subset of [−∞,∞], then it follows

from the structure theorem of open subsets of [−∞,∞] that E can be written as a countable

union of open intervals of the form [−∞, d), (a, b) and (c,∞], i.e., E =⋃nOn, where On

takes one of the form [−∞, d), (a, b) and (c,∞]. Now f−1(E) = f−1(⋃nOn) =

⋃n f−1(On).

Since each f−1(On) is measurable, it follows that f−1(E) is measurable. �

Definition 1.2.12. Let f be an extended real valued measurable function on a measurable

space (X,A ). Then for each α ∈ R the set Bα = {x ∈ X : f(x) < α} is measurable and it

satisfies Bα ⊂ Bα′ if α < α′. The sets Bα ’s are called the ordinate sets of f .

Theorem 1.2.13. Let (X,A ) be a measurable space, and let D be a dense subset of R.

Suppose that for each α ∈ D there is an associated Bα ∈ A such that Bα ⊂ Bα′ whenever

α < α′. Then there is a unique measurable function f on X such that f ≤ α on Bα and

f ≥ α on Bcα for every α ∈ D.

Proof. Define f : X → [−∞,∞] as f(x) = inf{α ∈ D : x ∈ Bα} if x ∈ Bα for some

α ∈ D and f(x) = ∞ if x is not in any Bα. Let x ∈ Bα. Then inf{β ∈ D : x ∈ Bβ} ≤ α,

i.e., f(x) ≤ α. Therefore f ≤ α on Bα. Let x ∈ Bcα. Suppose that f(x) = inf{β ∈ D :

x ∈ Bβ} < α. Then there is α′ ∈ D such that f(x) ≤ α′ < α and x ∈ Bα′ . Since α′ < α,

x ∈ Bα′ ⊂ Bα, which is a contradiction. Hence f ≥ α on Bcα. Now we prove that f is

measurable. Let λ ∈ R. Since D is dense in R, there is a sequence {αn} with αn < λ for

all n such that αn → λ as n → ∞. Let x ∈⋃nBαn . Then x ∈ Bαn for some n. Therefore

f(x) ≤ αn < λ, i.e., f(x) < λ. Let x ∈ X with f(x) = inf{α ∈ D : x ∈ Bα} < λ. Then

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3. Integration 9

there is β ∈ D such that f(x) ≤ β < λ and x ∈ Bβ. Since {αn} converges to λ, there is

n0 ∈ N such that β < αn0 < λ. Since β < αn0 and x ∈ Bβ, x ∈ Bαn0⊂⋃nBαn . Hence we

proved that {x ∈ X : f(x) < λ} =⋃nBαn . Since each Bαn is measurable, it follows that f

is measurable.

Let g be a measurable function on X such that g ≤ α on Bα and g ≥ α on Bcα. Let

x ∈ X. If x is not in any Bα, then g(x) ≥ α for every α ∈ D. Since D is dense in R,

g(x) =∞ = f(x). Let x ∈ Bα for some α. Then {α ∈ D : g(x) < α} ⊂ {α ∈ D : x ∈ Bα} ⊂{α ∈ D : g(x) ≤ α}. Therefore inf{α ∈ D : g(x) < α} ≥ inf{α ∈ D : x ∈ Bα} ≥ inf{α ∈D : g(x) ≤ α}, i.e., g(x) ≥ f(x) ≥ g(x). Hence f = g. �

Theorem 1.2.14. Let (X,A , µ) be a measure space. Suppose that for each α in a dense

set D of real numbers, there is assigned a set Bα ∈ A such that µ(Bα −Bβ) = 0 for α < β.

Then there is a measurable function f such that f ≤ α a.e. on Bα and f ≥ α a.e. on Bcα.

If g is any other function with this property, then g = f a.e.

Proof. (Verify the proof and write details)

Let C be a countable dense subset of D, let N =⋃α,β∈C,α<β(Bα−Bβ). Then µ(N) = 0.

Let B′α = Bα ∪ N . If α, β ∈ C with α < β, then B′α − B′β = (Bα − Bβ) − N = ∅. Thus

B′α ⊂ B′β. By above theorem there is a unique measurable function f such that f ≤ α on

B′α and f ≥ α on B′cα .

Let α ∈ D. Let {γn} be a sequence in C with γn > α and limn γn = α. Then

Bα−B′γn ⊂ Bα−Bγn . Therefore P =⋃n(Bα−B′γn) has measure 0. Let A =

⋂nB

′γn . Then

f ≤ infn γn ≤ α on A and Bα−A = Bα ∩ (⋂nB

′γn)c. Then f ≤ α a.e. on Bα and f ≥ α a.e.

on Bcα.

Let g be an extended real function with g ≤ γ on Bγ and g ≥ γ on Bcγ for each γ ∈ C.

Then g ≤ γ on B′γ and g ≥ γ on Bcγ except for a set of measure 0, say Qγ. Thus Q =

⋃γ∈C Qγ

is a set of measure 0 and f = g on X −Q, i.e., f = g a.e. [µ]. �

3. Integration

Definition 1.3.1. A function s : X → R is called a simple function if it assumes finitely

many distinct values.

Let s : X → R be a simple function, and let α1, α2, . . . , αn be distinct values assumed

by s. Let Ai = {x ∈ X : s(x) = αi}, 1 ≤ i ≤ n. Then Ai’s are pairwise disjoint and⋃ni=1Ai = X. Thus s admits a canonical representation s =

∑ni=1 αiχAi

.

Lemma 1.3.2. Let (X,A ) be a measurable space, and let s =∑n

i=1 αiχAibe a simple

function X. Then s is measurable if and only if each Ai is measurable.

Proof. Assume that s is measurable. Then for each i, the set {x ∈ X : s(x) = αi} is

measurable, i.e., Ai is measurable for each i.

PADABHI


Conversely, assume that each Ai is measurable. Then each χAiis measurable and hence∑n

i=1 αiχAi= s is measurable �

Definition 1.3.3. Let (X,A , µ) be a measure space, and let s =∑n

i=1 αiχAibe a non

negative measurable simple function on X. Then the Lebesgue integral of s over E ∈ A is

defined as ∫E

sdµ =n∑i=1

αiµ(Ai ∩ E).

By definition∫Xsdµ =

∑ni=1 αiµ(Ai ∩X) =

∑ni=1 αiµ(Ai).

Remarks 1.3.4. Let (X,A , µ) be a measure space, and let s =∑n

i=1 αiχAibe a non

negative measurable simple function on X.

(i) If E ∈ A , then∫E

sdµ =n∑i=1

αiµ(Ai ∩ E) ≥ 0 as αi ≥ 0 and µ(Ai ∩ E) ≥ 0 for each i.

(ii) If E,F ∈ A and E ⊂ F , then∫Esdµ ≤

∫Fsdµ.

SinceAi∩E ⊂ Ai∩F , we have µ(Ai∩E) ≤ µ(Ai∩F ). Now∫Esdµ =

∑ni=1 αµ(Ai∩E) ≤∑n

i=1 αµ(Ai ∩ F ) =∫Fsdµ.

(iii) If E1, E2, . . . , Em are pairwise disjoint measurable subsets of X, then∫∪mj=iEj

sdµ =∑mj=1

∫Ejsdµ.

Since E1, E2, . . . , Em are pairwise disjoint measurable subsets of X and Ai is measur-

able, µ((⋃mj=1Ej) ∩ Ai) =

∑mj=1 µ(Ej ∩ Ai). Now∫

⋃nj=1 Ej

sdµ =n∑i=1

αiµ((m⋃j=1

Ej) ∩ Ai) =n∑i=1

αi

(m∑j=1

µ(Ej ∩ Ai)

)

=m∑j=1

(n∑i=1

αiµ(Ej ∩ Ai)

)=

m∑j=1

∫Ej

sdµ.

(iv) Let E ∈ A . If µ(E) = 0 or s = 0 a.e. [µ] on E, then∫Esdµ = 0.

Assume that µ(E) = 0. Since E ∩ Ai ⊂ E, µ(E ∩ Ai) = 0 for all i. Hence∫Esdµ =∑n

i=1 αiµ(Ai ∩ E) = 0.

Now assume that s = 0 a.e. [µ] on E. Then there is a measurable subset F of E

with µ(F ) = 0 such that s = 0 on E − F . Now∫Esdµ =

∫Fsdµ +

∫E−F sdµ = 0 as

µ(F ) = 0 and s = 0 on E − F .

(v) If s ∈ A , then∫Esdµ =

∫XsχEdµ.

PA

DA

BH

I

3. Integration 11

Here s =∑n

i=1 αiχAi. Therefore sχE =

∑ni=1 αiχAi∩E.

Now∫XsχE =

∑ni=1 αiµ(Ai ∩ E) =

∫Esdµ.

(vi) If α ≥ 0 and E ∈ A , then∫Eαsdµ = α

∫Esdµ.

Here s =∑n

i=1 αiχAi. Therefore αs =

∑ni=1 ααiχAi

. So,∫E

αsdµ =n∑i=1

ααiµ(Ai ∩ E) = α

n∑i=1

αiµ(Ai ∩ E) = α

∫E

sdµ.

Lemma 1.3.5. Let s and t be non negative measurable simple functions on a measure space

(X,A , µ). Then∫X

(s+ t)dµ =∫Xsdµ+

∫Xtdµ.

Proof. Let s =∑n

i=1 αiχAiand t =

∑mj=1 βjχBj

be the canonical representations of s and

t respectively. Then Ai’s are pairwise disjoint measurable sets, Bj’s are pairwise disjoint

measurable sets and⋃ni=1Ai =

⋃mj=1Bj = X. For 1 ≤ i ≤ n and 1 ≤ j ≤ m, set Cij =

Ai ∩ Bj. Then each Cij is measurable and Cij’s are pairwise disjoint. Also⋃i,j Cij =⋃

i,j(Ai ∩ Bj) =⋃ni=1(Ai ∩ (

⋃mj=1Bj)) =

⋃ni=1Ai = X. If x ∈ X. Then x is in exactly one

Cij and on Cij, (s + t)(x) = αi + βj. Therefore s + t is a non negative measurable simple

function and s+ t =∑n

i=1

∑mj=1(αi + βj)χCij

is the canonical representation of s+ t. Now∫X

(s+ t)dµ =n∑i=1

m∑j=1

(αi + βj)µ(χCij)

=n∑i=1

m∑j=1

αiµ(Ai ∩Bj) +n∑i=1

m∑j=1

βjµ(Ai ∩Bj)

=n∑i=1

αi

m∑j=1

µ(Ai ∩Bj) +m∑j=1

βj

n∑i=1

µ(Ai ∩Bj)

=n∑i=1

αiµ(m⋃j=1

(Ai ∩Bj)) +m∑j=1

βjµ(n⋃i=1

(Ai ∩Bj))

=n∑i=1

αiµ(Ai ∩ (m⋃j=1

Bj)) +m∑j=1

βjµ((n⋃i=1

Ai) ∩Bj)

=n∑i=1

αiµ(Ai) +m∑j=1

βjµ(Bj)

=

∫X

sdµ+

∫X

tdµ.

This completes the proof �

PADABHI


Lemma 1.3.6. Let (X,A , µ) be a measure space, and let s be a non negative measurable

simple function on X. Define ϕ on A by

ϕ(E) =

∫E

sdµ (E ∈ A ).

Then ϕ is a measure on (X,A ).

Proof. Obviously, ϕ(∅) =∫∅ sdµ = 0 and ϕ(E) ≥ 0 for every E ∈ A . Let s =

∑ni=1 αiχAi

be the canonical representation of s. Let {En} be a sequence of pairwise disjoint measurable

subsets of X. Since En’s are pairwise disjoint and Ai’s are pairwise disjoint, Ai ∩ En are

pairwise disjoint. Now

ϕ(⋃j

Ej) =

∫⋃

j Ej

sdµ =n∑i=1

αiµ(Ai ∩ (⋃j

Ej))

=n∑i=1

αiµ(⋃j

(Ai ∩ Ej)) =n∑i=1

αi

(∑j

µ(Ai ∩ Ej)

)

=∑j

n∑i=1

αiµ(Ai ∩ Ej) =∑j

∫Ej

sdµ =∑j

ϕ(Ej).

Hence ϕ is a measure on (X,A ). �

Theorem 1.3.7 (Lusin’s Theorem). Let (X,A ) be a measurable space, and let f be a non

negative measurable function X. Then there is an increasing sequence {sn} of non negative

measurable simple functions on X converging to f (pointwise) on X. Further, if f is bounded,

then {sn} converges to f uniformly on X.

Proof. Let n ∈ N. For i = 1, 2, . . . , n2n, define Eni = {x ∈ x : i−12n≤ f(x) < i

2n} and

Fn = {x ∈ X : f(x) ≥ n}. Since f is measurable each Eni and Fn are measurable. Note

that Eni’s and Fn are pairwise disjoint and their union is X. For n ∈ N, define

sn =n2n∑i=1

i− 1

2nχEni

+ nχFn .

As each Eni and Fn are measurable, sn is a non negative measurable simple function.

First we prove that {sn} is increasing.For that let x ∈ X and n ∈ N.

Let f(x) ≥ n+ 1. Then f(x) > n. Therefore sn+1(x) = n+ 1 > n = sn(x).

Let n ≤ f(x) < n + 1. Then sn(x) = n. Since n ≤ f(x) < n + 1, there is i ∈{n2n+1, . . . , (n + 1)2n+1} such that x ∈ E(n+1)i. Therefore sn+1(x) = i−1

2n+1 for some i ∈{n2n+1, . . . , (n+ 1)2n+1}. Since i ≥ n2n+1, we have sn+1(x) = i−1

2n+1 ≥ n = sn(x).

PADABHI

3. Integration 13

Let f(x) < n. Then there is i ∈ {1, 2, . . . , n2n} such that i−12n≤ f(x) < i

2n, i.e.,

x ∈ Eni for some i ∈ {1, 2, . . . , n2n}. Therefore sn(x) = i−12n

. Since i−12n≤ f(x) < i

2n,

we have 2i−22n+1 ≤ f(x) < 2i

2n+1 . So, either 2i−22n+1 ≤ f(x) < 2i−1

2n+1 or 2i−12n+1 ≤ f(x) < 2i

2n+1 , i.e.,

x ∈ E(n+1)(2i−2) or x ∈ E(n+1)(2i−1). Therefore sn+1(x) = 2i−22n+1 or sn+1(x) = 2i−1

2n+1 . In any case,

sn+1(x) ≥ i−12n

= sn(x).

Hence the sequence {sn} is increasing.

Now prove that if x ∈ X, then sn(x)→ f(x) as n→∞.

For that let x ∈ X. If f(x) = ∞, then sn(x) = n for all n and clearly sn(x) = n →∞ = f(x) as n→∞.

Let f(x) < ∞. Then there is k = k(x) > 0 such that f(x) < k(x). Let n ∈ N be such

that n > k, i.e., f(x) < n. Then x ∈ Eni for some i ∈ {1, 2, , . . . , n2n}. Therefore sn(x) = i−12n

for some i ∈ {1, 2, , . . . , n2n}. Now sn(x) = i−12n≤ f(x) < i

2ngives |f(x) − sn(x)| =

f(x)− sn(x) ≤ i2n− i−1

2n= 1

2n→ 0 as n→∞. Therefore limn sn(x) = f(x) for every x ∈ X.

Now assume that f is bounded. Then there is k > 0 such that f(x) < k for every

x ∈ X. As we have done in last paragraph, we get |sn(x)− f(x)| < 12n

for all n ≥ k and for

all x ∈ X. Therefore supx∈X |sn(x) − f(x)| ≤ 12n

. Hence {sn} converges to f uniformly on

X. �

Example 1.3.8. Is it possible to approximate a non negative bounded measurable function

by a sequence of non negative measurable function vanishing outside a set of finite measure?

Consider the measure space (R,M ,m). Let f(x) = 1 for all x ∈ R. Let {sn} be a

sequence of non negative measurable simple functions on R such that sn = 0 on R− En for

some measurable En with m(En) < ∞. Since m(En) < ∞, R − En 6= ∅. Let x ∈ R − En.

Then sn(x) = 0. Now 1 = |f(x) − sn(x)| ≤ supy∈R |sn(y) − f(y)|. Therefore {sn} does not

converge to f uniformly on X.

Definition 1.3.9. Let (X,A , µ) be a measure space, and let f be a non negative measurable

function on X. Then the Lebesgue integral of f over E ∈ A (with respect to µ) is defined

by ∫E

fdµ = sup0≤s≤f

∫E

sdµ,

where s is a (non negative) measurable simple function.

Remarks 1.3.10. Let f and g be non negative measurable function on a measure space

(X,A , µ).

(i) If E ∈ A , then∫Efdµ ≥ 0.

If s is any non negative measurable simple function, then∫Esdµ ≥ 0. Therefore∫

Efdµ = sup0≤s≤f

∫Esdµ ≥ 0.

PADABHI


(ii) If E,F ∈ A and E ⊂ F , then∫Efdµ ≤

∫Efdµ.

If s is any non negative measurable simple function, then∫Esdµ ≥

∫Esdµ. Therefore

sup0≤s≤f∫Esdµ ≤ sup0≤s≤f

∫Esdµ, i.e.,

∫Efdµ ≤

∫Ffdµ.

(iii) Let E ∈ A . If µ(E) or f = 0 a.e. [µ] on E, then∫Efdµ = 0.

Let µ(E) = 0. Let s be a non negative measurable simple function with s ≤ f . Since

µ(E) = 0,∫Esdµ = 0. Therefore sup0≤s≤f

∫Esdµ = 0, i.e.,

∫Efdµ = 0.

Suppose that f = 0 a.e. [µ] on E. Then there is a measurable subset F of E with

µ(F ) = 0 such that f = 0 on E − F . Let s be any non negative measurable simple

function with s ≤ f . If x ∈ E−F , then 0 = f(x) ≥ s(x) ≥ 0, i.e., s(x) = 0. Therefore

s = 0 a.e. [µ] on E. Hence∫Esdµ = 0. This implies that sup0≤s≤f

∫Esdµ = 0, i.e.,∫

Efdµ = 0.

(iv) If E ∈ A and f ≤ g, then∫Efdµ ≤

∫Egdµ.

Let s be any non negative measurable simple function with s ≤ f . Then s ≤ g as

f ≤ g. Therefore∫Esdµ ≤

∫Egdµ. Therefore

∫Efdµ = sup0≤s≤f

∫Esdµ ≤

∫Egdµ.

(v) If α ≥ 0 and E ∈ A , then∫Eαfdµ = α

∫Efdµ.

If α = 0, then it is clear. Let α > 0. Let s be any non negative measurable simple func-

tion. Then 0 ≤ s ≤ αf if and only if 0 ≤ sα≤ f and s is a simple function if and only

if αs is a simple fucntion. Now α∫Efdµ = α sup0≤s≤f

∫Esdµ = sup0≤s≤f

∫Eαsdµ =

sup0≤αs≤αf∫Eαsdµ = sup0≤t≤αf

∫Etdµ =

∫Eαfdµ.

(vi) If E ∈ A , then∫XfχEdµ =

∫Efdµ.

Let s be a non negative measurable simple function with s ≤ fχE. Then s ≤ f on E.

Therefore∫Xsdµ =

∫Esdµ ≤

∫Efdµ. Since s is arbitrary

∫XfχEdµ ≤

∫Efdµ. Let s

be a non negative measurable simple function with s ≤ f on E. Then sχE ≤ fχE on

X. Therefore∫Esdµ =

∫XsχEdµ ≤

∫XfχEdµ. Again, since s is arbitrary, we have∫

Efdµ ≤

∫XfχEdµ. Hence

∫XfχEdµ =

∫Efdµ.

Theorem 1.3.11 (Monotone Convergence Theorem). Let (X,A , µ) be a measure space, and

let {fn} be an increasing sequence of non negative measurable functions on X converging to a

function f (pointwise) on X. Then∫Xfdµ = lim

n→∞

∫Xfndµ. (In other words

∫X

( limn→∞

fn)dµ =

limn→∞

∫Xfndµ.)

Proof. Since each fn is non negative and measurable, the limit function f is also non

negative and measurable and so∫Xfdµ exists. Since {fn} is an increasing sequence, f =

limn→∞ = supn fn. Therefore fn ≤ f for all n and hence∫Xfndµ ≤

∫Xfdµ. Since {fn} is

an increasing sequence,∫Xfndµ ≤

∫Xfn+1dµ for all n. Therefore {

∫Xfndµ} is an increasing

PADABHI

3. Integration 15

sequence of extended real numbers and hence limn→∞∫Xfndµ (it may be infinity). As∫

Xfndµ ≤

∫Xfdµ for all n,

limn→∞

∫X

fndµ ≤∫X

fdµ. (1.3.11.1)

Let 0 < c < 1. Let s be any non negative measurable simple function with s ≤ f .

Then clearly 0 ≤ cs(x) ≤ s(x) ≤ f(x) for every x ∈ X. For each n ∈ N, set En = {x ∈X : fn(x) ≥ cs(x)}. As both fn and cs are measurable, the set En is measurable. The

sequence {En} is increasing because the sequence {fn} is increasing. Clearly,⋃nEn ⊂ X.

Let x ∈ X. If s(x) = 0, then fn(x) ≥ cs(x) = 0 for all n, i.e., x ∈⋃nEn. Let s(x) > 0.

Then cs(x) < s(x) ≤ f(x). Since {fn(x)} converges to f(x), there is n0 ∈ N such that

cs(x) < fn0(x) ≤ f(x). Therefore x ∈ En0 ⊂⋃nEn. Hence

⋃nEn = X. Now on En,

cs ≤ fn. Therefore

c

∫En

sdµ =

∫En

csdµ ≤∫En

fndµ ≤∫

⋃n En

fndµ =

∫X

fndµ. (1.3.11.2)

Define ϕ on A by ϕ(E) =∫Esdµ, E ∈ A . Then ϕ is a measure on (X,A ). Since {En} is an

increasing sequence of measurable subsets of X, we have ϕ(X) = ϕ(⋃nEn) = limn→∞ ϕ(En),

i.e.,∫Xsdµ = limn→∞

∫Ensdµ. Taking limit n→∞ in equation (1.3.11.2) and applying this

equality we get c∫Xsdµ ≤ limn→∞

∫Xfndµ. Since the above is true for any non negative

measurable simple function, we get c∫Xfdµ ≤ limn→∞

∫Xfndµ. Since 0 < c < 1 is arbitrary,

taking c → 1, we get∫Xfdµ ≤ limn→∞

∫Xfndµ. The last inequality and the inequality in

equation (1.3.11.1) give the desired equality. �

Theorem 1.3.12. Let f and g be non negative measurable functions on a measure space

(X,A , µ). Then∫X

(f + g)dµ =∫Xfdµ+

∫Xgdµ.

Proof. By the Lusin’s theorem there exist increasing sequences {sn} and {tn} of non nega-

tive measurable simple functions converging to f and g respectively. Therefore by Monotone

Convergence Theorem∫Xfdµ = limn→∞

∫Xsndµ and

∫Xgdµ = limn→∞

∫Xtndµ. Note that

the {sn + tn} is an increasing sequence of non negative measurable functions converging

to f + g. Again the application of Monotone Convergence Theorem give∫X

(f + g)dµ =

limn→∞∫X

(sn + tn)dµ. Now∫X

(f + g)dµ = limn→∞

∫X

(sn + tn)dµ

= limn→∞

∫X

sndµ+ limn→∞

∫X

tndµ

PADABHI


=

∫X

fdµ+

∫X

gdµ.

Hence the proof. �

Corollary 1.3.13. Let f1, f2, . . . , fn be non negative measurable functions on a measure

space (X,A , µ). Then∫X

(∑n

i=1 fi)dµ =∑n

i=1

∫Xfidµ.

Proof. Use the Principle of Mathematical Induction. �

Theorem 1.3.14. Let {fn} be a sequence of non negative measurable functions on a measure

space (X,A , µ). Then∫X

(∑

n fn)dµ =∑

n

∫Xfndµ.

Proof. For each n ∈ N, let gn =∑n

k=1 fk. Then {gn} is an increasing sequence of non

negative measurable functions converging to∑

n fn. Therefore by Monotone Convergence

Theorem,∫X

(∑

n fn)dµ = limn

∫Xgndµ = limn

∫X

(∑n

k=1 fk)dµ = limn

∑nk=1

∫Xfkdµ =∑

n

∫Xfndµ. �

Theorem 1.3.15. Let f be a non negative measurable function on a measure space (X,A , µ).

Define ϕ on A by ϕ(E) =∫Efdµ, E ∈ A . Then ϕ is a measure on (X,A ).

Proof. Clearly, ϕ(∅) =∫∅ fdµ = 0 and ϕ(E) =

∫Efdµ ≥ 0 for every E ∈ A . Let {En}

be a sequence of pairwise disjoint measurable subsets of X. Observe that f =∑

n fχEn on⋃nEn. Also, we note that each fχEn is non negative and measurable. Therefore

ϕ(⋃n

En) =

∫⋃

En

fdµ =

∫⋃

En

(∑n

fχEn)dµ

=∑n

∫⋃

En

fχEndµ =∑n

∫En

fdµ

=∑n

ϕ(En).

Hence ϕ is a measure on (X,A ). �

Theorem 1.3.16 (Fatou’s Lemma). Let {fn} be a sequence of non negative measurable

functions on a measure space (X,A , µ). Then∫X

(lim infn

fn)dµ ≤ lim infn

∫Xfndµ.

Proof. For each n ∈ N, let gn = inf{fn, fn+1, . . .} = infk≥n fk. Then {gn} is a sequence of

non negative measurable functions and gn ≤ fn for all n. Therefore∫Xgndµ ≤

∫Xfndµ for

all n. Since {gn} is increasing, we have limn gn = supn gn = lim infn fn. It follows from the

Monotone Convergence Theorem that∫X

(lim infn

fn)dµ =

∫X

(limngn)dµ = lim

n

∫X

gndµ.

PADABHI

3. Integration 17

Since∫Xgndµ ≤

∫Xfndµ for all n, we get limn

∫Xgndµ = lim infn

∫Xgndµ ≤ lim infn

∫Xfndµ.

Hence∫X

(lim infn

fn)dµ ≤ lim infn

∫Xfndµ. �

Theorem 1.3.17 (Beppo Levi’s Theorem). Let {fn} be a sequence of non negative measur-

able functions on a measure space (X,A , µ) converging to f (pointwise) on X. If fn ≤ f

for all n, then∫Xfdµ = limn

∫Xfndµ.

Proof. Since {fn} converges to f and each fn is non negative and measurable, f is a non

negative measurable function on X. Here f = limn fn = lim infn fn. It follows from Fatous’

lemma that∫Xfdµ ≤ lim infn

∫Xfndµ. Since fn ≤ f for all n,

∫Xfndµ ≤

∫Xfdµ for all n.

Therefore lim supn∫xfndµ ≤

∫xfdµ. Now∫

X

fdµ ≤ lim infn

∫X

fndµ ≤ lim supn

∫X

fndµ ≤∫X

fdµ.

Therefore the sequence {∫Xfndµ} convergent and it converges to

∫Xfdµ, i.e., limn

∫Xfndµ =∫

Xfdµ. �

Example 1.3.18. Verify the Monotone Convergence Theorem, Fatou’s lemma and Beppo

Levi’s theorem for fn(x) = nx1+nx

on [1, 7].

Lemma 1.3.19. Let f be a non negative measurable function on a measure space (X,A , µ).

If∫Xfdµ = 0, then f = 0 a.e. [µ] on X.

Proof. Let E = {x ∈ X : f(x) 6= 0} = {x ∈ X : f(x) > 0}. Then E is a measurable

subset of X. Let En = {x ∈ X : f(x) > 1n} for n ∈ N. Then each En is measurable. One

can verify easily that E =⋃nEn. If µ(E) > 0, then µ(EN) > 0 for some N ∈ N. But

then 0 =∫Xfdµ ≥

∫EN

fdµ ≥ 1Nµ(EN) > 0, which is a contradiction. Hence µ(E) = 0, i.e.,

f = 0 a.e. [µ] on X. �

Definition 1.3.20. A non negative measurable function on a measure space (X,A , µ) is

called integrable if∫Xfdµ <∞.

Verify that f is integrable on (X,A , µ) if and only if∫Efdµ <∞ for every E ∈ A .

Now we define the integral of arbitrary measurable function (not necessarily non nega-

tive).

Definition 1.3.21. Let f be a measurable function on a measure space (X,A , µ). If at

least one of∫Xf+dµ and

∫Xf−dµ is finite, then we define the Lebesgue integral of f by∫

Xfdµ =

∫Xf+dµ −

∫Xf−dµ. The function f is called integrable if both

∫Xf+dµ and∫

Xf−dµ are finite.

Theorem 1.3.22. Let f be a measurable function on a measure space (X,A , µ). Then f is

integrable if and only if |f | is integrable.

PADABHI


Proof. Assume that f is integrable. Then both∫Xf+dµ and

∫Xf−dµ are finite. Now∫

X|f |dµ =

∫X

(f+ + f−)dµ =∫Xf+dµ +

∫X

+f−dµ < ∞. Therefore f is integrable. Con-

versely, assume that |f | is integrable. Then∫X|f |dµ =

∫Xf+dµ +

∫Xf−dµ < ∞. Since∫

Xf+dµ,

∫Xf−dµ ≤

∫Xf+dµ+

∫Xf−dµ, it follows that f is integrable. �

Lemma 1.3.23. Let f be a measurable function on a measure space (X,A , µ) such that∫Xfdµ exists. Then

∣∣∫Xfdµ

∣∣ ≤ ∫X|f |dµ.

Proof. If∫X|f |dµ =∞, then it is clear. Assume that

∫X|f |dµ <∞, i.e., |f | is integrable.

Then f is integrable. Now∣∣∫Xfdµ

∣∣ =∣∣∫Xf+dµ−

∫Xf−dµ

∣∣ ≤ ∣∣∫Xf+dµ

∣∣ +∣∣∫Xf−dµ

∣∣ =∫Xf+dµ+

∫Xf−dµ =

∫X

(f+ + f−)dµ =∫X|f |dµ. �

Lemma 1.3.24. Let f and g be integrable functions on a measure space (X,A , µ), and let

α ∈ R. Then

(i) f + g is integrable and∫X

(f + g)dµ =∫Xfdµ+

∫Xgdµ.

(ii) αf is integrable and∫Xαfdµ = α

∫Xfdµ.

Proof. Since f and g are measurable, f + g and αf are measurable. As both f and g are

integrable, both |f | and |g| are integrable.

(i) Now∫X|f + g|dµ ≤

∫X|f |dµ +

∫X|g|dµ < ∞. Therefore f + g is integrable. Let

h = f + g. Then h+ − h− = f+ − f− + g+ − g− gives h+ + f− + g− = h− + f+ = g+.

Therefore∫Xh+dµ +

∫Xf−dµ +

∫Xg−dµ =

∫Xh−dµ +

∫Xf+dµ +

∫Xg+dµ. As h, f and g

are integrable, all the numbers in above equation are finite. Therefore∫Xh+dµ−

∫Xh−dµ =∫

Xf+dµ−

∫Xf−dµ+

∫Xg+dµ−

∫Xg−dµ, i.e.,

∫X

(f + g)dµ =∫Xhdµ =

∫Xfdµ+

∫Xgdµ.

(ii) Since∫X|αf |dµ = |α|

∫X|f |dµ, the function αf is integrable. If α = 0, then clearly∫

Xαfdµ = α

∫Xfdµ.

Let α > 0. Then (αf)+ = αf+ and (αf)− = αf−. Now∫Xαfdµ =

∫X

(αf)+dµ −∫X

(αf)−dµ =∫Xαf+dµ−

∫Xαf−dµ = α(

∫Xf+dµ−

∫Xf−dµ) = α

∫Xfdµ.

Let α < 0. Then (αf)+ = (−α)f− and (αf)− = (−α)f+.∫Xαfdµ =

∫X

(αf)+dµ −∫X

(αf)−dµ =∫X

(−α)f−dµ−∫X

(−α)f+dµ = (−α)(∫Xf−dµ−

∫Xf+dµ) = α

∫Xfdµ. �

Corollary 1.3.25. If f1, f2, . . . , fn are integrable functions on a measure space (X,A , µ),

then∫X

(∑n

k=1 fk)dµ =∑n

k=1

∫Xfkdµ.

Proof. Use the Principle of Mathematical Induction. �

Remarks 1.3.26.

(i) If f and g are measurable functions on a measure space, f is integrable and |g| ≤ |f |,then g is integrable.

Here g is measurable and∫X|g|dµ ≤

∫X|f |dµ <∞. Therefore g is integrable.

PA

DA

BH

I

3. Integration 19

(ii) Let f and g be integrable over E ∈ A . If f ≤ g a.e. [µ] on E, then∫Efdµ ≤

∫Egdµ.

Since f and g are integrable over E, g − f is integrable over E and∫E

(g − f)dµ =∫Egdµ−

∫Efdµ. Since f ≤ g a.e. [µ] on E, there is a measurable subset F of E with

µ(F ) such that f ≤ g on E−F . Note that g−f is a non negative measurable function

on E. Therefore∫Egdµ−

∫Efdµ =

∫E

(g − f)dµ =∫F

(g − f)dµ+∫E−F (g − f)dµ ≥ 0

as µ(F ) = 0 and g − f ≥ 0 on E − F .

(iii) Let f be a measurable function on a measure space (X,A , µ), and let E ∈ A . If

µ(E) = 0 or f = 0 a.e. [µ] on E, then∫Efdµ = 0.

Since µ(E) = 0 and f+ and f− are non negative measurable functions, we get∫Ef+dµ =

∫Ef−dµ = 0. Therefore

∫Efdµ = 0.

If f = 0 a.e. [µ] on E, then f+ = 0 a.e. [µ] on E and f− = 0 a.e. [µ] on E. Therefore∫Ef+dµ =

∫Ef−dµ = 0 and hence

∫Efdµ = 0.

In particular, if f and g are integrable functions on X with f = g a.e. [µ] on X, then∫Xfdµ =

∫Xgdµ.

Theorem 1.3.27 (Lebesgue Dominated Convergence Theorem). Let {fn} be a sequence of

measurable functions on a measure space (X,A , µ) converging to a function f (pointwise)

on X. Let E ∈ A . If g is integrable over E and |fn| ≤ g on E for all n, then∫Efdµ =

limn

∫Efndµ.

Proof. Since {fn} converges to f and each fn is measurable, the function f is measurable.

As |fn| ≤ g on E for all n and fn → f as n→∞, we have |f | ≤ |g| on E. Since g is integrable

over E, each f is integrable over E and∫E|f |dµ ≤

∫Egdµ. Consider the sequences {g+ fn}

and {g − fn}. Then both are sequences of non negative measurable functions converging to

g + f and g − f respectively. Applying Fatou’s lemma we get∫E

limn

(g + fn)dµ =

∫E

lim infn

(g + fn)dµ ≤ lim infn

∫E

(g + fn)dµ,

it means∫E

gdµ+

∫E

fdµ ≤ lim infn

∫E

gdµ+ lim infn

∫E

fndµ =

∫E

gdµ+ lim infn

∫E

fndµ.

Therefore∫Efdµ ≤ lim infn

∫Efndµ. Applying Fatou’s lemma to the sequence {g − fn} we

obtain∫E

gdµ−∫E

fdµ ≤ lim infn

∫E

gdµ+ lim infn

(−∫E

fndµ) =

∫E

gdµ− lim supn

∫E

fndµ.

Therefore∫Efdµ ≥ lim supn

∫Efndµ. Hence

∫Efdµ ≤ lim infn

∫Efndµ ≤ lim supn

∫Efndµ ≤∫

Efdµ. It means that the sequence {

∫Efndµ} is convergent and it converges to

∫Efdµ, i.e.,

limn

∫Efndµ =

∫Efdµ. �

PADABHI


Corollary 1.3.28 (Bounded Convergence Theorem). Let {fn} be a sequence of measurable

functions on a measure space (X,A , µ) converging to a function f (pointwise) on X, and

let E ∈ A with µ(E) < ∞. If there is M > 0 such that |fn| ≤ M on E for all n, then∫Efdµ = limn

∫Efndµ.

Proof. Define g(x) = M for x ∈ E. Then g is integrable over E and it follows from the

last theorem that∫Efdµ = limn

∫Efndµ. �

Examples 1.3.29.

(i) Let µ1, µ2, . . . , µk be measures on (x,A ), and let α1, α2, . . . , αk be nonnegative real

numbers. Then α1µ1 + α2µ2 + · · ·+ αkµk is a measure on (X,A ).

Let µ = α1µ1 + α2µ2 + · · · + αkµk. Clearly µ(∅) = 0 and µ(E) ≥ 0 for every E ∈ A .

Let {En} be a sequence of pairwise disjoint measurable subsets of X. Then

µ(⋃n

En) = α1µ1(⋃n

En) + α2µ2(⋃n

En) + · · ·+ αkµk(⋃n

En)

=∑n

α1µ1(En) +∑n

α2µ2(En) + · · ·+∑n

αkµk(En)

=∑n

(α1µ1(En) + α2µ2(En) + · · ·+ αkµk(En)) (why?)

=∑n

µ(En)

Therefore µ is a measure.

(ii) Let µ and η be measures on a measurable space (X,A ). Is ν = max{µ, η} a measure

on (X,A )?

Consider the measure space (R,M ). Let m be the Lebesgue measure on R, and let

δ0 be the point mass measure at 0. Then ν([0, 1]) = 1, ν([0, 12]) = 1 and ν((1

2, 1]) = 1

2.

Hence ν([0, 12]) + ν((1

2, 1]) 6= ν([0, 1]). Therefore ν is not a measure on (R,M ).

(iii) Let (X,A ) be a measurable space, and let f : X → [−∞,∞] be a map. Then f is

measurable if and only if f−1({−∞}), f−1({∞}) are measurable and that f−1(E) is

measurable subset for every Borel subset E of R.

First assume that f−1({−∞}), f−1({∞}) are measurable and that f−1(E) is measur-

able subset for every Borel subset E of R. Let α ∈ R. Then {x ∈ X : f(x) > α} =

f−1((α,∞]) = f−1((α,∞))∪ f−1({∞}). As f−1({∞}) is measurable and f−1((α,∞))

is measurable (as (α,∞) is a Borel set), the set {x ∈ X : f(x) > α} is measurable.

Hence f is measurable.

Assume that f is measurable, then f−1({−∞}) =⋂n{x ∈ X : f(x) < −n} and

f−1({∞}) =⋂n{x ∈ X : f(x) > n} are measurable. Let a, b ∈ R. Then f−1((a,∞)) =

{x ∈ X : f(x) > a} − f−1({∞}), f−1((−∞, b)) = {x ∈ X : f(x) < b} − f−1({−∞})

PADABHI

3. Integration 21

and f−1((a, b)) = f−1((a,∞))∩f−1((−∞, b)). It follows that f−1((a,∞)), f−1((−∞, b))and f−1((a, b)) are measurable. Let C = {E ⊂ R : f−1(E)} is measurable. Then C is

a σ- algebra of subsets of R. As f−1((a,∞)), f−1((−∞, b)) and f−1((a, b)) are mea-

surable, (a,∞), (−∞, b), (a, b) ∈ C . Since any open subset of R is a countable union of

intervals of the form (a,∞), (−∞, b) and (a, b), it follows that C contains every open

subset of R. Since B is the smallest σ- algebra containing all open subsets of R, it

follows that B ⊂ C . Let E be a Borel set, i.e., E ∈ B ⊂ C . Then by definition of C ,

f−1(E) is measurable.

(iv) Let (X,A , µ) be a measure space which is not complete. If f = g a.e. [µ] on X and

f is measurable, show that g need not be measurable.

Since (X,A , µ) is not complete, there is a measurable subset E of X with µ(E) = 0

and E has a nonmeasurable subset say F . Define f = 0 and g = χF . Then f is

measurable and g is not measurable as F is not measurable. If x ∈ X − E, then

f(x) = g(x). Therefore f = g on X − E and µ(E) = 0. Hence f = g a.e. [µ] on X.

(v) Consider a measurable space (X,P (X)). Let η be a counting measure and let δx0 be

a Dirac measure at x0 ∈ X. Let f, g : X → [−∞,∞].

(a) Show that f = g a.e. [δx0 ] if and only if f(x0) = g(x0).

(b) Show that f = g a.e. [η] if and only if f(x) = g(x) for every x ∈ X.

(a) Assume that f = g a.e. [δx0 ]. Then there is a measurable subset E of X with

δx0(E) = 0 and f = g on X − E. Since δx0(E) = 0, x0 /∈ E. Hence f(x0) = g(x0).

Conversely, assume that f(x0) = g(x0). Then X − {x0} is a measurable subset of X

with δx0(X − {x0}) = 0. Therefore f = g a.e. [δx0 ].

(b) Assume that f = g a.e. [η]. Then there is a measurable subset E of X with

η(E) = 0 and f = g on X − E. Since η(E) = 0, E = ∅. Hence f(x) = g(x) for every

x ∈ X. Conversely, assume that f(x) = g(x) for every x ∈ X. Then clearly f = g a.e.

[η].

(vi) Let f and g be nonnegative integrable functions on a measure space (X,A , µ) with

g ≤ f . Show that f = g a.e. [µ] if and only of∫Xgdµ =

∫Xfdµ.

We may write f = (f−g)+g. Note that both f−g and g are nonnegative measurable

functions therefore∫Xfdµ =

∫X

(f − g)dµ+∫Xgdµ. We know that for a nonnegative

measurable function h, h = 0 a.e. [µ] if and only if∫Xhdµ = 0. Hence f = g a.e. [µ],

i.e., f − g = 0 a.e. [µ] if and only if∫X

(f − g)dµ = 0 if and only if∫Xfdµ =

∫Xgdµ.

(vii) Let f be a nonnegative measurable function on a measure space (X,A , µ), and let

X be a countable union of an increasing sequence {Xn} of measurable subsets of X.

Show that f integrable over X if and only if there is a nonnegative real number M

such that∫Xnfdµ ≤M for every n.

Assume that f is integrable, i.e.,∫Xfdµ <∞. Take M =

∫Xfdµ. Then M ≥ 0 as f

is nonnegative. Since f is nonnegative and Xn ⊂ X, we have∫Xnfdµ ≤

∫Xfdµ = M .

PADABHI


Conversely, assume that there is a nonnegative real number M such that∫Xnfdµ ≤M

for all n. We know that when f is a nonnegative measurable function, then ϕ(E) =∫Efdµ (E ∈ A ) is a measure. Since {Xn} is an increasing sequence of measurable

subsets of X with⋃nXn = X, we have∫

X

fdµ = ϕ(X) = ϕ(⋃n

Xn) = limn→∞

ϕ(Xn) = limn→∞

∫Xn

fdµ.

Since∫Xnfdµ ≤M for all n and limn→∞

∫Xnfdµ =

∫Xfdµ, we get

∫Xfdµ ≤M <∞.

Therefore f is integrable.

(viii) Let f be a measurable function on a measure space (X,A , µ). Then f is integrable

over X if and only if f is integrable over every measurable subset of X.

Assume that f is integrable over X, i.e.,∫Xf+dµ <∞ and

∫Xf−dµ <∞. Let E ∈ A .

Since both f+ and f− are nonnegative measurable function,∫Ef+dµ ≤

∫Xf+dµ <∞

and∫Ef−dµ ≤

∫Xf−dµ <∞. Therefore f is integrable over E.

Conversely, if f is integrable over every measurable subset of X, then in particular f

is integrable over X.

PADABHICHAPTER 2

Signed measure and decompositions

4. Signed measure

Definition 2.4.1. Let (X,A ) be a measurable space. A map ν : A → [−∞,∞] is called a

signed measure if

(i) ν assumes at most one of the values −∞ and ∞,

(ii) ν(∅) = 0,

(iii) ν is countably additive, i.e., if {En} is a sequence of pairwise disjoint measurable

subsets of X, then ν(⋃nEn) =

∑n ν(En).

If {En} is a sequence of pairwise disjoint measurable subsets of X and if ν(⋃nEn) is

finite, then the series∑

n ν(En) converges absolutely (as any rearrangement of the series is

convergent (because union can be taken in any order we like)), i.e.,∑

n |ν(En)| <∞.

Note that every measure on a measurable space is a signed measure and the converse

is not true?

Example 2.4.2. Consider the measurable space (R,M ). Let m the Lebesgue measure on R.

Let δ0 : M → [0,∞] be the point mass measure at 0, i.e., δ0(E) = 1 if 0 ∈ E and δ0(E) = 0 if

0 /∈ E. Take ν = δ0−m. Let E ∈M , then ν(E) = δ(E)−m(E) ≤ 1−m(E) ≤ 1. Therefore

ν cannot assume the value∞. Clearly, ν(∅) = 0. Let {En} be a sequence of pairwise disjoint

measurable subsets of R. Then ν(⋃nEn) = δ0(

⋃nEn)−m(

⋃nEn) =

∑n δ0(En)−

∑nm(En).

If 0 /∈ En for any n, then δ0(En) = 0 for all n. Therefore ν(⋃nEn) =

∑n δ0(En) −∑

nm(En) = −∑

nm(En) =∑

n(δ0(En)−m(En)) =∑

n ν(En). If 0 ∈⋃nEn, then 0 ∈ En

for exactly one n say N . Then ν(⋃nEn) =

∑n δ0(En)−

∑nm(En) = δ0(EN)−

∑nm(En) =

1−∑

nm(En) =∑

n(δ0(En)−m(En)) =∑

n ν(En). Therefore ν is countably additive. Hence

ν is a signed measure.

Definition 2.4.3. Let ν be a signed measure on a measurable space (X,A ).

(i) A subset E of X is called a positive set with respect to ν if E ∈ A and ν(F ) ≥ 0 for

every measurable subset F of E.

(ii) A subset E of X is called a negative set with respect to ν if E ∈ A and ν(F ) ≤ 0 for

every measurable subset F of E.

23

PADABHI

24 2. SIGNED MEASURE AND DECOMPOSITIONS

(iii) A subset E of X is called a null set with respect to ν if E ∈ A and ν(F ) = 0 for every

measurable subset F of E.

Remarks 2.4.4. Let ν be a signed measure on a measurable space (X,A ). Then

(i) Any measurable subset of a positive set (with respect to ν) is a positive set (with

respect to ν).

(ii) Any measurable subset of a negative set (with respect to ν) is a negative set (with

respect to ν).

(iii) Any measurable subset of a null set (with respect to ν) is a null set (with respect to

ν).

Obviously, if E is a positive set with respect to ν, then ν(E) ≥ 0. The converse need

not be true see the following.

Example 2.4.5. Consider the measurable space (R,M ). Let m the Lebesgue measure on

R. Let δ0 : M → [0,∞] be the point mass measure at 0, i.e., δ0(E) = 1 if 0 ∈ E and

δ0(E) = 0 if 0 /∈ E. Take ν = δ0 −m. Then ν is a signed measure.

(i) If E ∈M and 0 /∈ E, then E is a negative set.

Let 0 /∈ E. Let F be any measurable subset of E. Since 0 /∈ E, 0 /∈ F . Therefore

δ0(F ) = 0. Now ν(E) = δ0(E)−m(E) = 0−m(E) ≤ 0. Hence E is a negative set.

(ii) If E is a measurable subset of R and ν(E) ≥ 0, then E need not be a positive set.

Consider E = [0, 34] ∈M . Then ν(E) = δ0(E) −m(E) = 1 − 3

4> 0. Let F = [1

2, 34].

Then F is a measurable subset of E and ν(F ) = δ0(F )−m(F ) = 0− 14< 0. Therefore

E is not a positive set with respect to ν.

(iii) If E is a measurable subset of R and ν(E) ≤ 0, then E need not be a negative set.

Let E = [0, 2] ∈M . Then ν(E) = δ0(E) −m(E) = 1 − 2 < 0. Let F = [0, 12]. Then

F is a measurable subset of E and ν(F ) = δ0(F )−m(F ) = 1− 12> 0. Therefore E is

not a negative set with respect to ν.

(iv) If E is a measurable subset of R and ν(E) = 0, then E need not be a negative set.

Let E = [0, 1] ∈M . Then ν(E) = 0 and ν([0, 12]) > 0.

Lemma 2.4.6. Let ν be a signed measure on a measurable space (X,M ). Then

(i) countable union of positive sets is a positive set.

(ii) countable union of negative sets is a negative set.

(iii) countable union of null sets is a null set.

Proof. (i) Let {En} be a countable collection of positive sets. Then clearly E =⋃nEn

is measurable. Let F be a measurable subset of E. Let A1 = F ∩ E1 and for n > 1, set

An = (F ∩ En) − (⋃n−1i=1 (F ∩ Ei)). Then {An} is a pairwise disjoint measurable subsets of

X and F =⋃nAn. Also, An ⊂ En for all n. Since En is a positive set ν(An) ≥ 0 for all n.

PADABHI

4. Signed measure 25

Now ν(F ) = ν(⋃nAn) =

∑n ν(An) ≥ 0. Hence E is a positive set.

The proofs of (ii) and (iii) are similar. �

Lemma 2.4.7. Let ν be a signed measure on a measurable space (X,A ), and let A be a

positive set with respect to ν. Define νA on A by νA(E) = ν(A∩E) for every E ∈ A . Then

νA is a measure on (X,A ).

Proof. Clearly, νA(∅) = ν(A ∩ ∅) = 0. Let E ∈ A . Then A ∩ E is a measurable subset of

A. As A is a positive set νA(E) = ν(A∩E) ≥ 0. Let {En} be a sequence of pairwise disjoint

measurable subsets of X. Then {A ∩ En} is also a sequence of pairwise disjoint measurable

subsets of X. Now νA(⋃nEn) = ν(A ∩ (

⋃nEn)) = ν(

⋃n(A ∩ En)) =

∑n ν(A ∩ En) =∑

n νA(En). Hence νA is a measure on X. �

Lemma 2.4.8. Let ν be a signed measure on a measurable space (X,A ), and let B be a

positive set with respect to ν. Define νB on A by νB(E) = −ν(B ∩ E) for every E ∈ A .

Then νB is a measure on (X,A ).

Lemma 2.4.9. Let ν be a signed measure on a measurable space (X,A ), and let E ∈ Awith 0 < ν(E) <∞. Then E contains a positive set A with ν(A) > 0.

Proof. If E itself is a positive set, then we are done. If E is not a positive set, then E

has a subset which has negative ν- measure. Let n1 be the smallest positive integer such

that there is a measurable subset E1 of E such that ν(E) < − 1n1

. (i.e., if m is a positive

integer less than n1, then for any measurable subset K of E, ν(K) ≥ − 1m

). Note that

ν(E − E1) = ν(E) − ν(E1) ≥ ν(E) + 1n1> 0. If E − E1 is a positive set, we stop. If not,

then E − E1 has a measurable subset with negative ν- measure. Let n2 be the smallest

positive integer such that there is a measurable subset E2 of E−E1 such that ν(E2) < − 1n2

.

If E − (E1 ∪ E2) is a positive set, we stop. Otherwise continuing in this way, for each

k ∈ N, we get the smallest positive integer nk for which E − (E1 ∪ E2 ∪ · · · ∪ Ek−1) has

a measurable subset Ek for which ν(Ek) < − 1nk

. Note the Ek’s are pairwise disjoint. Let

A = E − (⋃k Ek). Then A is a measurable subset of E. Now E = A ∪ (

⋃k Ek) and so

ν(E) = ν(A) + ν((⋃k Ek)) = ν(A) +

∑k ν(Ek). Since ν(E) > 0 and ν(Ek) < 0 for all k,

it follows that ν(A) > 0. Since A ⊂ E and ν(E) < ∞, ν(A) < ∞. Therefore the series∑k ν(Ek) of negative real numbers is convergent and hence it converges absolutely. Now

0 <∑

k1nk≤∑

k |ν(Ek)| < ∞ gives 1nk→ 0 as n → ∞. Let F be a measurable subset of

A. Then F ⊂ E − (⋃k Ek) ⊂ E − (E1 ∪ E2 ∪ · · · ∪ Ek−1) for all k and E ∩ Ek = ∅. by the

minimal choice of nk we have ν(F ) ≥ − 1nk−1

for all k. Taking k → ∞ we get ν(F ) ≥ 0.

Hence A is a positive set. �

Lemma 2.4.10. Let ν be a signed measure on a measurable space (X,A ), and let E ∈ Awith −∞ < ν(E) < 0. Then E contains a negative set B with ν(B) < 0.

PADABHI


Proof. Take λ = −ν. Then λ is a signed measure on (X,A ). Replace ν by λ in the proof

of last lemma and get the conclusion. �

5. Decompositions of signed measure space

Theorem 2.5.1 (Hahn Decomposition). Let ν be a signed measure on a measurable space

(X,A ). Then X is a disjoint union of a positive set and a negative set, i.e, there exist a

positive set A and negative set B such that A ∪B = X and A ∩B = ∅.

Proof. We may assume that ν does not assume the value ∞ (otherwise work with −ν),

i.e., ν(E) ∈ [−∞,∞) for every E ∈ A . Let

λ = sup{ν(E) : E is a positive set with respect to ν}.

Then there is a sequence {An} of positive sets such that ν(An)→ λ as n→∞. Since each

An is a positive set ν(An) ≤ λ. Take A =⋃nAn. Then A is a positive set as countable union

of positive sets is a positive set. Therefore ν(A) ≤ λ. As A−An is a measurable subset of a

positive A, ν(A− An) ≥ 0 for all n. Let n ∈ N. Then ν(A) = ν(A− An) + ν(An) ≥ ν(An).

Therefore ν(A) ≥ limn ν(An) = λ. This implies that λ = ν(A) < ∞ as ν does not take the

value ∞.

Let B = Ac. We show that B is a negative set. If B is not a negative set, then B

contains a measurable set E with 0 < ν(E) < ∞. But then E has a subset, say F , which

is positive with ν(F ) > 0. Then clearly, A ∪ F is a positive set. As F ⊂ E ⊂ B = Ac,

F ∩A = ∅. Now λ ≥ ν(A ∪ F ) = ν(A) + ν(F ) = λ+ ν(F ) > λ, a contradiction. Hence B is

a negative set. �

Exercise 2.5.2. Give the proof of above theorem by reversing the roles of positive sets and

negative sets.

Hahn Decomposition of a signed measure space (X,A , ν) is not unique but it is unique

in the following sense.

Theorem 2.5.3 (Uniqueness of Hahn Decomposition). Let {A,B} and {A1, B1} be Hahn

decomposition of a signed measure space (X,A , ν). Then A∆A1 and B∆B1 are null sets.

Proof. We have A∆A1 = (A− A1) ∪ (A1 − A). Since A− A1 is a measurable subset of a

positive set A, it is a positive set. Also, A − A1 = A ∩ Ac1 ⊂ Ac1 = B1 and B1 is a negative

set. Therefore A−A1 is a negative set. Hence A−A1 is a null set. Similarly, it follows that

A1 − A is a null set. Therefore A∆A1 is a null set. Using similar arguments we have that

B∆B1 is a null set. �

Example 2.5.4. Consider the measurable space (R,M ). Let ν = δ0 −m, where δ0 is the

Dirac measure at 0 and m is the Lebesgue measure on R. Then ν is a signed measure on

(R,M ).

PADABHI


Let A be any measurable subset of R containing 0 such that m(A) = 0. Then A is a

positive set. As if F is a measurable subset of A, then m(F ) = 0 and ν(F ) = δ0(F )−m(F ) =

δ0(F ) ≥ 0. Let B = Ac. Then now we show that B is a negative set. Let F be a measurable

subset of B. Then F does not contain 0. Therefore ν(F ) = δ0(F ) −m(F ) = −m(F ) ≤ 0.

Therefore B is a negative set. Hence {A,B} is a Hahn decomposition of (R,M , ν).

In particular {Q,Qc} is a Hahn decomposition of (R,M , ν).

Definition 2.5.5. Let µ1 and µ2 be measures on a measurable space (X,A ). Then µ1 and

µ2 are called mutually singular, denoted by µ1 ⊥ µ2, if there exist disjoint measurable subsets

A and B of X with A ∪B = X such that µ1(B) = 0 and µ2(A) = 0.

Consider the measures δ0 and m on the measurable space (R,M ). Take A = Q and

B = Qc. Then A and B are disjoint measurable subsets of R with A ∪ B = R. Clearly,

δ0(B) = 0 and m(A) = 0. Therefore δ0 ⊥ m.

Theorem 2.5.6 (Jordan Decomposition). Let ν be a signed measure on a measurable space

(X,A ). Then there exist unique measures ν1 and ν2 such that ν = ν1 − ν2 and ν1 ⊥ ν2.

Proof. Let {A,B} be a Hahn decomposition of (X,A , ν), i.e, A is a positive set, B is a

negative set, A ∩ B = ∅ and A ∪ B = X. Define ν1 and ν2 on A by ν1(E) = ν(A ∩ E) and

ν2(E) = −ν(B ∩ E). Then ν1 and ν2 are measures on (X,A ). Now ν1(B) = ν(A ∩ B) =

ν(∅) = 0 and ν2(A) = ν(B ∩ A) = ν(∅) = 0. Therefore ν1 ⊥ ν2. Let E ∈ A . Then

ν(E) = ν(E ∩X) = ν(E ∩ (A ∪B)) = ν((E ∩ A) ∪ (E ∩B))

= ν(E ∩ A) + ν(E ∩B) = ν1(E)− ν2(E).

Therefore ν = ν1 − ν2.Uniqueness suppose that there exist measures ν ′1 and ν ′2 on (X,A ) such that ν ′1 ⊥ ν ′2

and ν = ν ′1−ν ′2. Then there exist two disjoint measurable subsets A1 and B1 with A1∪B1 =

X, ν ′1(B1) = 0 and ν ′2(A1) = 0.

First we show that {A1, B1} is Hahn decomposition of ν, i.e., A1 is a positive set and

B1 is a negative set. For that let E be a measurable subset of A1. Since ν ′2 is a measure,

ν ′2(A1) = 0 and E ⊂ A1, we have ν ′2(E) = 0. Also, since ν ′1 is a measure ν ′1(E) ≥ 0. Now

ν(E) = ν ′1(E)−ν ′2(E) = ν ′1(E) ≥ 0. Therefore A1 is a positive set with respect to ν. Similar

arguments shows that B1 is a negative set. Therefore {A1, B1} is a Hahn decomposition of

(X,A , ν). Let E ∈ A . Then

ν ′1(E) = ν ′1(E ∩X) = ν ′1(E ∩ (A1 ∪B1))

= ν ′1(E ∩ A1) + ν ′1(E ∩B1) (∵ 0 ≤ ν ′1(E ∩B1) ≤ ν ′1(B1) = 0)

= ν ′1(E ∩ A1)− ν ′2(E ∩ A1) (∵ 0 ≤ ν ′2(E ∩ A1) ≤ ν ′2(A1) = 0)

= ν(E ∩ A1) = ν((E ∩ A1) ∩X)

PADABHI


= ν((E ∩ A1) ∩ (A ∪B))

= ν(E ∩ A1 ∩ A) + ν(E ∩ A1 ∩B).

As E∩A1∩B is a measurable subset of a negative set B, ν(E∩A1∩B) ≤ 0. Since E∩A1∩Bis a measurable subset of a positive set A1, ν(E ∩ A1 ∩ B) ≥ 0. Hence ν(E ∩ A1 ∩ B) = 0.

Therefore ν ′1(E) = ν(E ∩ A1 ∩ A). Now ν1(E) = ν(E ∩ A) = ν((E ∩ A) ∩ (A1 ∪ B1)) =

ν(E ∩ A ∩ A1) + ν(E ∩ A ∩ B1) = ν(E ∩ A ∩ A1). Therefore ν1(E) = ν ′1(E), i.e., ν1 = ν ′1.

Since ν = ν1 − ν2 = ν ′1 − ν ′2 and ν1 = ν ′1, we have ν2 = ν ′2. �

Definition 2.5.7. Let ν be a signed measure on a measurable space (X,A ). Then there

exist two mutually singular measures ν+ and ν− on (X,A ) such that ν = ν+ − ν−. The

decomposition of ν as a difference of measures is called the Jordan decomposition of ν. The

measure ν+ is called the positive part or the positive variation of ν and the measure ν− is

called the negative part or the negative variation of ν.

Let ν = ν+ − ν− be the Jordan decomposition of a signed measure on a measurable

space (X,A ). Since ν takes at most one value from −∞ and∞, at least one of the measures

ν+ and ν− is a finite measure. If both the measures ν+ and ν− are finite measures, then ν

is called a finite signed measure.

If ν = ν+ − ν− is a signed measure on a measurable space (X,A ), then we define

|ν| = ν+ + ν−. Then clearly |ν| is a measure on (X,A ). The value |ν|(X) is called the total

variation of ν.

Note that if E is a positive set, then |ν|(E) = ν+(E). If E is a negative set, then

|ν|(E) = ν−(E). If E is a null set, then |ν|(E) = 0.

Exercise 2.5.8. Let f be an integrable function on a measure space (X,A , µ). Define ν on

A by ν(E) =∫Efdµ, E ∈ A . Find a Hahn decomposition and the Jordan decomposition

of ν.

Take A = {x ∈ X : f(x) ≥ 0} and B = {x ∈ X : f(x) < 0}. Then both A and B

are measurable, A ∪ B = X and A ∩ B = ∅. Note that f = f+ on any (measurable) subset

of A and f = f− on any (measurable) subset of B. Let E be a measurable subset of A.

Then ν(E) =∫Efdµ =

∫Ef+dµ ≥ 0. Therefore A is a positive set. It follows that B is a

negative set. Therefore {A,B} is a Hahn decomposition of (X,A , nu). Let E ∈ A , then

ν+(E) = ν(E ∩ A) =∫E∩A f

+dµ and ν−(E) = ν(E ∩ B) =∫E∩B f

−dµ. So ν = ν+ − ν− is

the Jordan decomposition ν on (X,A ).

Lemma 2.5.9. Let ν = ν+ − ν− be the Jordan decomposition of a signed measure ν on a

measurable space (X,A ). Then −ν− ≤ ν ≤ ν+ and |ν(E)| ≤ |ν|(E) for every E ∈ A .

Proof. Let E ∈ A . Then −ν−(E) ≤ ν+(E)− ν−(E) ≤ ν+(E). Therefore −ν− ≤ ν ≤ ν+.

Now |ν(E)| = |ν+(E)− ν−(E)| ≤ ν+(E) + ν−(E) = |ν|(E). �

PADABHI


Remark 2.5.10. If ν is a signed measure on (X,A ) and |ν| = 0, then ν = 0.

Let E ∈ A . Then 0 = |ν|(E) = ν+(E) +ν−(E). Therefore ν+(E) = ν−(E) = 0. Hence

ν(E) = ν+(E)− ν−(E) = 0 and so ν = 0.

We may have strict inequality in the inequality |ν(E)| ≤ |ν|(E). Consider the measur-

able space (R,M ). Let ν = δ0 − m. We know that {Q,Qc} is a Hahn decomposition of

(X,A ) with respect to ν. Therefore ν+ = δ0 and ν− = m. Let E = [−1, 1]. Then |ν(E)| = 1

and |ν|(E) = 3.

Definition 2.5.11. Let λ and µ be measures on a measurable space (X,A ). Then λ is

called absolutely continuous with respect to µ if E ∈ A and µ(E) = 0, then λ(E) = 0. If λ

is absolutely continuous with respect to µ, then we write λ� µ.

Example 2.5.12. Let f be a nonnegative integrable function on a measure space (X,A , µ).

Then the measure ϕ, defined by ϕ(E) =∫Efdµ (E ∈ A ), is absolutely continuous with

respect to µ.

Definition 2.5.13. Let ν and λ be signed measures on a measurable space (X,A ). Then

ν is said to be absolutely continuous with respect to λ if the measure |ν| is absolutely

continuous with respect to the measure |λ|. Likewise, ν and λ are called mutually singular

if the measures |ν| and |λ| are mutually singular.

Note that when µ is a measure on a measurable space (X,A ), then |µ| = µ.

Lemma 2.5.14. Let ν1 and ν2 be finite signed measures on a measurable space (X,A ), let

α, β ∈ R, and let µ be a measure on (X,A ). Then

(i) αν1 + βν2 is a finite signed measure.

(ii) |αν1| = |α||ν1|.(iii) |ν1 + ν2| ≤ |ν1|+ |ν2|.(iv) If ν1 ⊥ µ and ν2 ⊥ µ, then (αν1 + βν2) ⊥ µ.

(v) If ν1 � µ and ν2 � µ, then αν1 + βν2 � µ.

Proof. (i) Clearly, (αν1 + βν2)(∅) = 0 and αν1 + βν2 is countably additive as both the

signed measures are finite. Again since both the signed measures are finite, αν1 + βν2 is a

finite signed measure.

(ii) If α = 0, then there is nothing to prove. Let α > 0. Then (αν1)+ = αν+1 and (αν1)

− =

αν−1 . Now, |αν1| = (αν1)+ + (αν1)

− = αν+1 + αν−1 = α(ν+1 + ν−1 ) = |α||ν1|.Let α < 0. Then (αν1)

+ = −αν−1 and (αν1)− = −αν+1 . Now, |αν1| = (αν1)

++(αν1)− =

−αν−1 − αν+1 = −α(ν+1 + ν−1 ) = |α||ν1|.

PADABHI


(iii) Let {A,B} be a Hahn decomposition of ν1 + ν2. Let E ∈ A . Then

(ν1 + ν2)+ = (ν1 + ν2)(E ∩ A) = ν1(E ∩ A) + ν2(E ∩ A)

≤ ν+1 (E ∩ A) + ν+2 (E ∩ A)

≤ ν+1 (E) + ν+2 (E) = (ν+1 + ν+2 )(E),

i.e., (ν1 + ν2)+ ≤ ν+1 + ν+2 . Similarly, we get (ν1 + ν2)

− ≤ ν−1 + ν−2 . Now |ν1 + ν2| =

(ν1 + ν2)+ + (ν1 + ν2)

− ≤ ν+1 + ν+2 + ν−1 + ν−2 = |ν1|+ |ν2|.

(iv) Since ν1 ⊥ µ, there exist disjoint measurable set A1 and B1 with A1 ∪ B1 = X and

|ν1|(B1) = µ(A1) = 0. Also, as ν2 ⊥ µ there exist disjoint measurable set A2 and B2 with

A2 ∪ B2 = X and |ν2|(B2) = µ(A2) = 0. Take A = A1 ∪ A2 and B = B1 ∩ B2. Then both

A and B are measurable, A ∩ B = ∅, A ∪ B = X, µ(A) = 0 and 0 ≤ |αν1 + βν2|(B) ≤|α||ν1|(B) + |β||ν2|(B) ≤ |α||ν1|(B1) + |β||ν2|(B2) = 0. Hence (αν1 + βν2) ⊥ µ.

(v) Let E be a measurable subset of X with µ(E) = 0. Since ν1 � µ and ν2 � µ,

|ν1|(E) = |ν2|(E) = 0. Now 0 ≤ |αν1 + βν2|(E) ≤ |α||ν1|(E) + |β||ν2|(E) = 0. Hence

(αν1 + βν2)� µ. �

Exercise 2.5.15. Let ν be a signed measure and µ be a measure on a measurable space

(X,A ). If ν ⊥ µ and ν � µ, then ν = 0.

Since |ν| ⊥ µ. There exist disjoint measurable subsets A and B of X such that A∩B =

X and |ν|(B) = µ(A) = 0. Since |ν| � µ and µ(A) = 0, |ν|(A) = 0. Let E ∈ A . Then

|ν|(E) = |ν|(E ∩ (A ∪ B)) = |ν|((E ∩ A) ∪ (E ∩ B)) = |ν|(E ∩ A) + |ν|(E ∩ B) = 0. Hence

|ν| = 0 and so ν = 0.

We know that if f is a nonnegative measurable function on a measure space (X,A , µ),

then ϕ(E) =∫Efdµ (E ∈ A ) is a measure on (X,A ) and ϕ � µ. The converse of this is

known as the Radon Nikodym Theorem.

Theorem 2.5.16 (Radon Nikodym Theorem). Let ν and µ be σ- finite measures on a

measurable space (X,A ), and let ν � µ. Then there is a unique (a.e. [µ]) nonnegative

measurable function f such that

ν(E) =

∫E

fdµ (E ∈ A ).

Proof. Assume first that µ is finite, i.e., µ(X) < ∞. Observe that ν − αµ is a signed

measure for each α ∈ Q. Let {Aα, Bα} be a Hahn decomposition of ν − αµ, i.e., Aα is a

positive set of ν − αµ and Bα is a negative set of ν − αµ. For α = 0, take A0 = X and

B = ∅. Let α < β. Since Bα − Bβ ⊂ Bα and Bα is a negative set of ν − αµ, Bα − Bβ is a

PADABHI


negative set of ν − αµ. Similarly, as Bα−Bβ ⊂ Bcβ = Aβ and Aβ is a positive set of ν − βµ,

Bα −Bβ is a positive set of ν − βµ. Now 0 ≤ (ν − βµ)(Bα −Bβ) ≤ (ν − αµ)(Bα −Bβ) ≤ 0,

i.e., (ν − βµ)(Bα − Bβ) = (ν − αµ)(Bα − Bβ) = 0. This gives (β − α)µ(Bα − Bβ) = 0,

i.e., µ(Bα − Bβ) = 0. Hence for each α ∈ Q, there assigned a measurable set Bα such that

µ(Bα−Bβ) = 0 if α < β. Therefore there is a measurable function f such that f ≤ α on Bα

a.e. [µ] and f ≥ α on Bcα = Aα a.e. [µ]. Since A0 = X, f ≥ 0 a.e. [µ]. So, we may assume

that f is a nonnegative.

Fix N ∈ N. Let E ∈ A . For k ∈ {0, 1, . . .}, set Ek = E ∩ (B k+1N− B k

N) and

E∞ = E − (⋃∞k=0B k

N). Then E = E∞ ∪ (

⋃∞k=0Ek). Let i < j. Then Ei ∩ Ej = E ∩

(B i+1N∩ Bc

iN

) ∩ (B j+1N∩ Bc

jN

) ⊂ (B i+1N∩ Bc

iN

). Since µ(B i+1N∩ Bc

iN

) = 0, we get µ(Ei ∩ Ej) =

0. As ν � µ, ν(Ei ∩ Ej) = 0. Also Ei ∩ E∞ = ∅. As µ and ν are measures, we get

ν(E) = ν(E∞) +∑∞

k=0 ν(En) and µ(E) = µ(E∞) +∑∞

k=0 µ(En). Since µ is a finite measure,∑n µ(En) = µ(E)− µ(E∞) and hence the series

∑n µ(En) is convergent.

Let x ∈ Ek = E ∩ (B k+1N−B k

N). Then k

N≤ f(x) ≤ k+1

N. Therefore

k

Nµ(Ek) ≤

∫Ek

fdµ ≤ k + 1

Nµ(Ek) (2.5.16.1)

Note that (ν − kNµ)(Ek) ≥ 0 and (ν − k+1

Nµ)(Ek) ≤ 0. Therefore

k

Nµ(Ek) ≤ ν(Ek) ≤

k + 1

Nµ(Ek) (2.5.16.2)

It follows from the equations (2.5.16.1) and (2.5.16.2) that if k ∈ {0, 1, . . .}, then

ν(Ek)−1

Nµ(Ek) ≤

k

Nµ(Ek) ≤

∫Ek

fdµ ≤ ν(Ek) +1

Nµ(Ek). (2.5.16.3)

If µ(E∞) = 0, then ν(E∞) = 0 as ν � µ. Also, as µ(E∞) = 0,∫E∞

fdµ = 0. So, in this

case ν(E∞) =∫E∞

fdµ.

Let µ(E∞) > 0. If x ∈ E∞ = E − (⋃k B k

N), i.e., x ∈ B k

Nfor all k. Therefore

f(x) ≥ kN

for all k, i.e., f(x) =∞. So, we have∫E∞

fdµ =∞. Now, E∞ = E − (⋃k B k

N) =

E ∩ (⋃k B k

N)c = E ∩ (

⋂k B

ckN

) ⊂ BckN

= A kN

, i.e., E∞ ⊂ A kN

for all k. Since A kN

is a positive

set of ν − kNµ, it follows that ν(E∞) ≥ k

Nµ(E∞) for all k and hence ν(E∞) =∞. So, in this

case too ∞ = ν(E∞) =∫E∞

fdµ. It follows from the equation (2.5.16.3) that

ν(E∞) +∑k

ν(Ek)−1

N(µ(E∞) +

∑k

µ(Ek)) ≤∑k

∫Ek

fdµ+

∫E∞

fdµ

≤ ν(E∞) +∑k

ν(Ek) +1

N(µ(E∞) +

∑k

µ(Ek)).

PADABHI


Therefore

ν(E)− 1

Nµ(E) ≤

∫E

fdµ ≤ ν(E) +1

Nµ(E).

Since µ is a finite measure and N ∈ N is arbitrary, it follows that ν(E) =∫Efdµ.

Now let µ be σ- finite. Then there is a sequence {Xn} of pairwise disjoint measurable

subsets of X such that⋃nXn = X and µ(Xn) <∞ for all n. Since µ is a finite measure on

the σ- algebra of all measurable subsets of Xn and v � µ, by above there is a nonnegative

measurable function fn on Xn such that ν(En) =∫Enfdµ for every measurable subset En of

Xn. Define f on X by f =∑

n fnχXn . Then f is a nonnegative measurable function of X.

Let E be a measurable subsets of X. Then

ν(E) = ν(E ∩ (⋃n

Xn)) =∑n

ν(E ∩Xn) =∑n

∫E∩Xn

fndµ

=∑n

∫E

fnχXndµ =

∫E

(∑n

fnχXn)dµ =

∫E

fdµ.

Let g be a nonnegative measurable function on X such that ν(E) =∫Egdµ (E ∈ A ).

Since ν is σ finite measure, there is a sequence {Yn} of measurable subsets of X such that⋃n Yn = X and ν(Yn) <∞ for all n. Fix n. Now

∫Yngdµ =

∫Ynfdµ = ν(Yn) <∞. Therefore

f and g are integrable on Yn and hence they are integrable on any measurable subset of Yn.

It follows that∫F

(f − g)dµ = 0 for every measurable subset F of Yn and hence f = g a.e.

[µ] on Yn. Let En be a measurable subset of Yn with µ(En) = 0 such that f = g on Yn−En.

But then f = g on X − (⋃nEn). Since µ(

⋃nEn) = 0, it follows that f = g a.e. [µ] on

X. �

Corollary 2.5.17. Let (X,A , µ) be a σ- finite measure space, and let ν be a finite signed

measure on (X,A ) that is absolutely continuous with respect to µ. Then there is an integrable

function f on X (with respect µ) such that ν(E) =∫Efdµ for every E ∈ A .

Proof. Here ν � µ, i.e., ν+ + ν− = |ν| � µ. Therefore both the measures ν+ and ν− are

absolutely continuous with respect to µ. Note that both ν+ and ν− are finite measures (hence

σ- finite measures) as ν is a finite signed measure. Therefore by Radon Nikodym theorem

there exist nonnegative measurable functions f1 and f2 on X such that ν+(E) =∫Ef1dµ

and ν−(E) =∫Ef2dµ for every E ∈ A . Since both ν+ and ν− are finite measures, both f1

and f2 are integrable (with respect to µ). Therefore f = f1 − f2 is integrable (with respect

to µ). Let E ∈ A . Then

ν(E) = ν+(E)− ν−(E) =

∫E

f1dµ−∫E

f2dµ

PADABHI


=

∫E

(f1 − f2)dµ =

∫E

fdµ.

This completes the proof. �

Definition 2.5.18. Let ν and µ be σ- finite measures on a measurable space (X,A ), and

let ν be absolutely continuous with respect to µ. Then there is a nonnegative function f on

X such that ν(E) =∫Efdµ for every E ∈ A . Then function f is called the Radon Nikodym

derivative of ν with respect to µ and it is denoted by [ dνdµ

].

Thus if ν and µ are measures on a measurable space (X,A ) and ν � µ, then ν(E) =∫E

[ dνdµ

]dµ for every E ∈ A .

Examples 2.5.19.

(i) Let ν and µ be σ- finite measures on a measurable space (X,A ), and let ν � µ. If f is

a nonnegative measurable function on X, then∫Efdν =

∫Ef [ dν

dµ]dµ for every E ∈ A .

First note that ν(E) =∫E

[ dνdµ

]dµ for every E ∈ A . Let s =∑n

i=1 αiχAibe a nonnega-

tive measurable function on X. Let E ∈ A . Then∫E

sdν =n∑i=1

αiν(Ai ∩ E) =n∑i=1

αi

∫Ai∩E

[dν

dµ

]dµ

=n∑i=1

∫E

αi

[dν

dµ

]χAi

dµ

=

∫E

(n∑i=1

αiχAi

)[dν

dµ

]dµ

=

∫E

s

[dν

dµ

]dµ.

By Lusin’s theorem there is an increasing sequence {sn} of nonnegative measurable

simple functions converging to f (pointwise) on X. Then {sn[ dνdµ

]} is an increasing

sequence of nonnegative measurable functions on X converging to f [ dνdµ

]. It follows

from the Monotone Convergence theorem that∫E

fdν = limn

∫E

sndν = limn

∫E

sn

[dν

dµ

]dµ =

∫E

f

[dν

dµ

]dµ.

(ii) Let ν1, ν2 and µ be σ- finite measures on a measurable space (X,A ) and that both ν1and ν2 are absolutely continuous with respect to µ. Then [d(ν1+ν2)

dµ] = [dν1

dµ] + [dν2

dµ].

PADABHI


Since both ν1 and ν2 are absolutely continuous with respect to µ, ν1 + ν2 is absolutely

continuous with respect µ. Let E ∈ A . Then (ν1 + ν2)(E) = ν1(E) + ν2(E) gives∫E

[d(ν1+ν2)dµ

]dµ =∫E

[dν1dµ

]dµ +∫E

[dν2dµ

]dµ =∫E

([dν1dµ

] + [dν2dµ

])dµ. Because the Radon

Nikodym derivative of ν1 + ν2 with respect to µ is unique, we have [d(ν1+ν2)dµ

] = [dν1dµ

] +

[dν2dµ

].

(iii) (Chain rule) Let ν, µ and λ be σ- finite measures on a measurable space (X,A ), and

let ν � µ� λ. Then [ dνdλ

] = [ dνdµ

][dµdλ

].

Let E be a measurable subset of X. Because [ dνdµ

] is a nonnegative measurable function,

it follows from (i) that∫E

[ dνdµ

]dµ =∫E

[ dνdµ

][dµdλ

]dλ. now∫E

[dν

dλ

]dλ = ν(E) =

∫E

[dν

dµ

]dµ =

∫E

[dν

dµ

] [dµ

dλ

]dλ.

Because the Radon Nikodym derivative is unique, it follows that [ dνdλ

] = [ dνdµ

][dµdλ

].

(iv) If ν is a σ- finite measure on (X,A ), then [dνdν

] = 1.

Clearly ν � ν. Let E be a measurable subset of X. Then∫E

1dν = ν(E) =∫E

[dνdν

]dν.

Therefore [dνdν

] = 1.

(v) Let ν and µ be σ- finite measures on a measurable space (X,A ), and that ν and µ

are absolutely continuous with respect to each other. Then [ dνdµ

][dµdν

] = 1.

We have ν � µ� ν. It follows from (iii) and (iv) that 1 = [dνdν

] = [ dνdµ

][dµdν

].

Theorem 2.5.20 (Lebesgue Decomposition Theorem). Let ν and µ be σ- finite measures

on a measurable space (X,A ). Then there exists a unique pair of measures ν0 and ν1 such

that ν0 ⊥ µ, ν1 � µ and ν = ν1 + ν2.

Proof. Take λ = ν + µ. Then λ is a measure on (X,A ) and both ν and µ are absolutely

continuous with respect to λ. Forst we show that λ is a σ- finite measure. As both ν

and µ are σ- finite measures, there exist countable collection {Xn} and {X ′m} of pairwise

disjoint measurable subsets of X such that⋃nXn =

⋃mX

′m = X, ν(Xn) < ∞ for all n

and µ(X ′m) < ∞ for all m. Take Xnm = Xn ∩ X ′m. Then Xnm is a measurable subset

of X. Also note that⋃n,mXnm = X and Xnm’s are pairwise disjoint. Now λ(Xnm) =

ν(Xnm) + µ(Xnm) = ν(Xn ∩X ′m) + µ(Xn ∩X ′m) ≤ ν(Xn) + µ(X ′m) <∞. Therefore λ is a σ-

finite measure on (X,A ). By Radon Nikodym theorem there exist nonnegative measurable

function f and g on X such that µ(E) =∫Efdλ and ν(E) =

∫Egdλ for every E ∈ A . Take

A = {x ∈ X : f(x) > 0} and B = {x ∈ X : f(x) = 0}. Then both A and B are measurable

subsets of X with A ∪B = X and A ∩B = ∅. Define ν0 and ν1 on A by ν0(E) = ν(E ∩B)

and ν1(E) = ν(E ∩ A). Then both ν0 and ν1 are measures on (X,A ). Let E ∈ A . Then

ν(E) = ν(E ∩ (A ∪B)) = ν((E ∩A) ∪ (E ∩B)) = ν(E ∩B) + ν(E ∩A) = ν0(E) + ν1(E) =

(ν0 + ν1)(E). Therefore ν = ν0 + ν1.

PADABHI


Now ν0(A) = ν(A ∩B) = ν(∅) = 0. Also, µ(B) =∫Bfdλ =

∫B

0dλ = 0. Hence ν0 ⊥ µ.

Let E ∈ A with µ(E) = 0. Then µ(E ∩ A) =∫E∩A fdλ = 0. Since f is a nonnegative

measurable function, f = 0 a.e. [λ] on E ∩ A, i.e., λ{x ∈ E ∩ A : f(x) > 0} = 0. But

E ∩ A = {x ∈ E ∩ A : f(x) > 0}. Therefore λ(E ∩ A) = 0. Now 0 = λ(E ∩ A) =

µ(E ∩ A) + ν(E ∩ A) = ν(E ∩ A), i.e, ν1(E) = ν(E ∩ A) = 0. Hence ν1 � µ.

Now we show that ν = ν0 + ν1 is a unique decomposition of ν with ν0 ⊥ µ and ν1 � µ.

Suppose that there exist measure ν ′0 and ν ′1 with ν = ν ′0 + ν ′1, ν′0 ⊥ µ and ν ′1 � µ. As ν

is a σ- finite measure, all the measures ν0, ν′0, ν1 and ν ′1 are σ- finite measures. As in first

paragraph, there is a countable collection {Yn} of pairwise disjoint measurable subsets of X

such that X =⋃n Yn, νi(Yn) < ∞ and ν ′i(Yn) < ∞ for all n and i = 1, 2. Fix n. Then ν0,

ν1, ν′0 and ν ′1 are finite measures on Yn. Also note that both ν0 ⊥ µ, ν ′0 ⊥ µ and both ν1

and ν ′1 are absolutely continuous with respect to µ on Yn. Now ν = ν0 + ν1 = ν ′0 + ν ′1 gives

ν0 − ν ′0 = ν1 − ν ′1 on Yn. But (ν0 − ν ′0) ⊥ µ and ν0 − ν ′0 = ν1 − ν ′1 � µ on Yn. Hence ν0 = ν ′0and ν1 = ν ′1 on any measurable subset of Yn. Let E be a measurable subset of X. Then

ν0(E) = ν0(E ∩X) = ν0(E ∩ (⋃n

Yn)) = ν0(⋃n

(E ∩ Yn))

=∑n

ν0(E ∩ Yn) =∑n

ν ′0(E ∩ Yn) = ν ′0(⋃n

(E ∩ Yn))

= ν ′0(E ∩ (⋃n

Yn)) = ν ′0(E ∩X) = ν ′0(E).

This proves that ν0 = ν ′0. Similarly, ν1 = ν ′1. �

Remark 2.5.21. If we take A = {x ∈ X : g(x) > 0} and B = {x ∈ X : g(x) = 0}. Then

we get a decomposition of µ as µ = µ0 + µ1 such that µ0 ⊥ ν and µ1 � ν.

real analysis ii (measure theory) notes

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