[problem books in mathematics] exercises in analysis || measures and topology

Chapter 4

Measures and Topology

4.1 Introduction

4.1.1 Borel and Baire σ-Algebras

In the previous chapter, we considered measures defined on abstractσ-algebras of sets. However, in most cases the underlying measurespace has a natural topological structure. When we combine the mea-sure theoretic and topological structures, we get a richer theory whichwe outline in this chapter.

Definition 4.1Let (X, τ) be a topological space. The Borel σ-algebra of X, denotedby B(X), is the σ-algebra generated by τ (i.e., by the open sets, B(X) =σ(τ)).

Definition 4.2Let (X, τ) be a topological space and let

Cc(X)def=

{f : X −→ R : f is τ -continuous and has compact support

}.

The Baire σ-algebra of X, denoted by Ba(X), is the smallestσ-algebra of subsets of X, which makes each function in Cc(X)measurable.

Remark 4.3Therefore Ba(X) is the σ-algebra generated by the sets

{x ∈ X :

f(x) � λ}with f ∈ Cc(X), λ ∈ R. Evidently {f � λ} can be replaced

by {f � λ} or {f > λ} or {f < λ}.

L. Gasinski and N.S. Papageorgiou, Exercises in Analysis: Part 1,Problem Books in Mathematics, DOI 10.1007/978-3-319-06176-4 4,© Springer International Publishing Switzerland 2014

633

634 Chapter 4. Measures and Topology

The Baire σ-algebra is most interesting when X is locally compact.

Theorem 4.4If X is a locally compact topological space (see Definition 2.92),then Ba(X) is the σ-algebra generated by the compact Gδ-sets.

The next theorem compares the two σ-algebras B(X) and Ba(X).

Theorem 4.5If X is a topological space, then(a) Ba(X) ⊆ B(X).(b) When X is locally compact, C is compact, U is open and C ⊆ U ,then there exist V,K ∈ Ba(X) such that V is σ-compact open and Kis compact Gδ such that C ⊆ V ⊆ K ⊆ U .(c) When X is locally compact separable and metrizable, then B(X) =Ba(X).

We consider the Borel and Baire σ-algebras for products of topo-logical spaces.

Theorem 4.6If X and Y are topological spaces, then(a) B(X)⊗B(Y ) ⊆ B(X × Y ).(b) When X and Y are both second countable (see Definition 2.24),then B(X × Y ) = B(X)⊗ B(Y ).(c) When X and Y are both second countable and locally compact,then Ba(X × Y ) = Ba(X)⊗ Ba(Y ).

Definition 4.7Let X be a topological space and let f : X −→ R be a function.(a) We say that f is Borel measurable (or a Borel function) if itis measurable when X is endowed with the Borel σ-algebra B(X) (onR as always we consider the Borel σ-algebra).(b) We say that f is Baire measurable (or a Baire function) ifit is measurable when X is endowed with the Baire σ-algebra Ba(X).

Remark 4.8Evidently the elements of Cc(X) are Baire functions and every Bairefunction is Borel.

4.1. Introduction 635

4.1.2 Regular and Radon Measures

Definition 4.9Let X be a topological space, let Σ be a σ-algebra and let μ be a measuredefined on Σ.(a) We say that μ is outer regular if for all A ∈ Σ, we have

μ(A) = inf{μ(U) : U ∈ Σ, U is open and A ⊆ U

}.

(b) We say that μ is inner regular if for all A ∈ Σ, we have

μ(A) = sup{μ(C) : C ∈ Σ, C is closed and C ⊆ A

}.

(c) We say that μ is regular if it is both outer and inner regular.(d) We say that μ is inner regular with respect to compact setsif for all A ∈ Σ, we have

μ(A) = sup{μ(K) : K ∈ Σ, K is compact and K ⊆ A

}.

(e) We say that μ is Radon (or tight), if μ(K) < +∞ for all compactsets K ⊆ X, μ is regular and inner regular with respect to compact sets.

Remark 4.10If Σ contains B(X), then in the above definitions the requirements thatU ∈ Σ, C ∈ Σ and K ∈ Σ are of course redundant. The notions ofregularity and Radoness will be applied usually on σ-algebras Σ thatcontain at least Ba(X). Finally, if μ is a signed measure, then μ isregular (respectively, Radon) if and only if |μ| is regular (respectively,Radon).

Theorem 4.11If X is a metrizable space,then every finite Borel measure (i.e., a measure defined on B(X)) isregular.

In Polish spaces, we can improve this theorem.

Theorem 4.12If X is a Polish space (see Definition 2.150),then every finite Borel measure is Radon.


Definition 4.13Let X be a topological space and let μ : B(X) −→ R

∗ = R ∪ {±∞} bea signed measure. The support of μ, denoted by suppμ, is a closedset in X such that(a) |μ|((suppμ)c) = 0; and(b) if U is an open set such that U ∩ suppμ = 0, then

|μ|(U ∩ suppμ) > 0.

Remark 4.14The support of μ need not exist. If it exists, it is unique. According tothe above definition, the support of μ is the smallest closed set whosecomplement has |μ|-measure zero.

Theorem 4.15If X is a topological space and μ : B(X) −→ [0,+∞] is a measure,then if either X is second countable (see Definition 2.24) or μ is Radon,then suppμ exists.

Definition 4.16For any set X and x ∈ X, δx denotes the Dirac measure at x, whichis the probability measure on 2X having all its mass at x, that is

δx(A)def=

{1 if x ∈ A,0 if x ∈ A.

Proposition 4.17If X is a topological space, then δx is a Radon measure.

Concerning image measures, we have the following result.

Theorem 4.18If X and Y are topological spaces, X is compact, f : X −→ Y is acontinuous function, μ is a finite tight Borel measure on X and μf−1

is its image measure by f on Y (i.e., (μf−1)(A) = μ(f−1(A)

)for all

A ∈ B(Y )),then μf−1 is a finite Radon measure on B(Y ).

4.1.3 Riesz Representation Theorem for ContinuousFunctions

Definition 4.19Let X be a locally compact topological space (see Definition 2.92).


We introduce the following vector spaces of continuous functionsf : X −→ R:

• Cc(X) is the space of continuous functions f : X −→ R which havecompact supports.

• C0(X) is the space of continuous functions f : X −→ R which vanishat infinity, i.e., for every ε > 0, there exists a compact set K ⊆ Xsuch that

∣∣f(x)∣∣ < ε for x ∈ K.

• Cb(X)is the space of bounded continuous functions f : X −→ R.

Clearly we have

Cc(X) ⊆ C0(X) ⊆ Cb(X).

Moreover, if X is compact, then

Cc(X) = C0(X) = Cb(X).

If X is not compact, then each of the inclusions is strict. On Cb(X)we introduce the supremum norm ‖ · ‖Cb

, defined by

‖f‖Cb

def= sup

x∈X

∣∣f(x)

∣∣.

This norm is restricted on C0(X) and Cc(X). Note that on them thesupremum in the above definition is a maximum. We denote theserestrictions by ‖ · ‖C0 and ‖ · ‖Cc , respectively.

Proposition 4.20If X is a locally compact topological space,

then(Cb(X), ‖ · ‖Cb

)is complete and C0(X) is a closed subset of it.

Finally Cc(X) is dense in C0(X).

For locally compact spaces X, we can introduce their Alexandrovone-point compactification X∗ (see Remark 2.97 and Theorem 2.98).We have the following result.

Proposition 4.21If X is a locally compact topological space and X∗ is its Alexandrov

one-point compactification (see Remark 2.97), Y ={f ∈ C(X∗) :

f(∞) = 0}and for every f ∈ Y , f = f |X ,

then f �−→ f is a linear isometry of Y onto C0(X).


Definition 4.22Let X be a locally compact topological space. By Mb(X) we denotethe space of signed Radon (tight) Borel measures μ (i.e., |μ| is Radon;see Remark 4.10) and bounded (i.e., |μ|(X) < +∞). We equip Mb(X)with the norm ‖μ‖ = |μ|(X) and then

(Mb(X), ‖·‖) becomes a Banach

space. The elements of Mb(X) are also known as bounded signedRadon measures.

The following theorem relates the Banach space Mb(X) with thedual of C0(X) (i.e., the space of continuous linear functionals onC0(X), denoted by C0(X)∗; see Chap. 5).

Theorem 4.23 (Riesz Representation Theorem)For every continuous linear functional l on C0(X), there exists aunique μ ∈ Mb(X) such that

l(f) =

∫

X

f dμ ∀ f ∈ C0(X).

Moreover, we have ‖l‖∗ = sup{∣∣l(f)

∣∣ : ‖f‖∞ � 1

}= ‖μ‖ = |μ|(X).

Remark 4.24Using the terminology of Chap. 5, the above theorem says thatC0(X)∗ = Mb(X) (i.e., the two Banach spaces C0(X)∗ and Mb(X)are isometrically isomorphic). If l is positive (i.e., l(f) � 0 for allf � 0), then μ � 0.

4.1.4 Space of Probability Measures: ProhorovTheorem

Definition 4.25Let X be a metric space and let M+

1 (X) consist of all probabilitymeasures on B(X). For every f ∈ Cb(X) consider the functionalθf : M

+1 (X) −→ R, defined by

θf (μ) =

∫

X

f(x) dμ.

Then the weak topology on M+1 (X) is defined to be the topology

w({θf}f∈Cb(X)

)(see Definition 2.62).


In the next theorem, we state several equivalent useful definitionsof this topology.

Theorem 4.26 (Portmanteau Theorem)Let X be a metric space and let {μα}α∈J ⊆ M1

+(X) be a net. Thefollowing statements are equivalent:(a) μα

w−→ μ;(b) for every f ∈ UCb(X) where

UCb(X)def=

{f : X −→ R : f is bounded and uniformly continuous

},

we have ∫

Ω

f dμα −→∫

Ω

f dμ;

(c) for every closed set C ⊆ X, we have lim supα∈J

μα(C) � μ(C);

(d) for every open set U ⊆ X, we have μ(U) � lim infα∈J

μα(U);

(e) for every A ∈ B(X) such that μ(∂A) = 0, we have limα∈J

μα(A) =

μ(A).

It is interesting to know when this weak topology on M+1 (X) is

metrizable, in which case we can use sequences. The next theoremprovides conditions for the metrizability of the weak topology.

Theorem 4.27The space

(M+

1 (X), w)is separable metrizable if and only if X is a

separable metric space.

In the next theorem, we indicate a concrete countable dense subsetsof

(M+

1 (X), w).

Theorem 4.28If X is a separable metric space,then the set of all convex combinations of Dirac measures (see Defini-tion 4.16) is dense in

(M+

1 (X), w).


(M+

1 (X), w)is compact metrizable if and only if X is a

compact metric space.



(M+

1 (X), w)is Polish if and only if X is a Polish space.

In the case of Polish spaces, we can have a nice description of thecompact subsets of

(M+

1 (X), w).

Theorem 4.31 (Prohorov Theorem)If X is a Polish space and C ⊆ M+

1 (X),

then Cwis w-compact if and only if C is uniformly tight (i.e., for every

ε > 0, we can find a compact set Kε ⊆ X such that μ(Kε) � 1− ε forall μ ∈ C).

Theorem 4.32If X is a separable metric space and {μn}n�1 ⊆ M+

1 (X) is a sequence,

then μnw−→ μ if and only if

limn→+∞ sup

f∈C

∣∣∫

X

f dμn −∫

Ω

f dμ∣∣ = 0

for every equicontinuous and uniformly bounded family C ⊆ Cb(X)(see Definition 1.83).

Proposition 4.33If X and Y are locally compact, second countable topological spaces(see Definitions 2.92 and 2.24) and μ and ν are Radon measures onX and Y , respectively,then μ× ν is a Radon measure on X × Y .

4.1.5 Polish, Souslin and Borel Spaces

In Definitions 2.150 and 2.156 we introduced the notions of Polish andSouslin spaces, respectively. In the sequel, we will relate these notionsto the Borel sets. We start with an interesting property of Souslinspaces known as the “separation property”.

Theorem 4.34 (Separation Property)If X is a topological space and {An}n�1 is a sequence of pairwisedisjoint Souslin subspaces of X,then there exists a sequence {Cn}n�1 of pairwise disjoint Borel subsetsof X such that An ⊆ Cn for all n � 1.


Corollary 4.35If X is a topological space, X =

⋃

n�1An, with {An}n�1 being pairwise

disjoint Souslin subsets of X,then An ∈ B(X) for every n � 1.In particular, if two complementary subsets of X are Souslin, then bothare Borel sets.

Corollary 4.36If τ1 and τ2 are two comparable Souslin topologies on a set X,then B(Xτ1) = B(Xτ2).

Proposition 4.37If X and Y are two topological spaces, f : X −→ Y is a function and

Gr f ={(x, y) ∈ X × Y : y = f(x)

}is a Souslin subspace of X × Y ,

then f is a Borel function (i.e., the inverse image of every Borel setof Y is a Borel set of X).

Corollary 4.38If X and Y are Souslin spaces,then f : X −→ Y is a Borel function if and only if Gr f ⊆ X × Y isSouslin if and only if Gr f ⊆ X × Y is Borel. Also direct and inverseimages by f of Souslin sets are Souslin sets.

Definition 4.39Let (X,Σ) and (Y,Y) be measurable spaces. A bijection functionf : X −→ Y is an isomorphism if f is (Σ,Y)-measurable and f−1

is (Y,Σ)-measurable. Then we say that spaces X and Y are isomor-phic. If E ⊆ X and F ⊆ Y , then we say that E and F are iso-morphic if

(E,B(X) ∩ E

)and

(F,B(Y ) ∩ F

)are isomorphic. If X

and Y are separable metrizable and Σ = B(X), Y = B(Y ) (the Borelσ-algebras), then we use the term Borel isomorphic.

Definition 4.40A topological space X is said to be a Borel space if there exists aPolish space Y and A ∈ B(Y ) such that X is homeomorphic to A.The empty set is by definition a Borel space.

Proposition 4.41Every Borel space is metrizable and separable.


Theorem 4.42Let X and Y be Borel spaces. Then X and Y are Borel isomorphic ifand only if they have the same cardinality.

Corollary 4.43Every uncountable Borel space is Borel isomorphic to every other un-countable Borel space. In particular, every uncountable Borel space isisomorphic to [0, 1] and has the cardinality of the continuum.

Theorem 4.44 (Kuratowski Theorem)If X is a Borel space, Y is a separable metrizable space andf : X −→ Y is an injective Borel function,then f(X) ∈ B(Y ) and f−1 is a Borel set. In particular, if Y is a Borelspace, then X and f(X) are Borel isomorphic (see Definition 4.39).

Definition 4.45Let (X,Σ) be a measurable space, let μ be a probability measure on Xand let Σμ be the μ-completion of Σ (see Definition 3.23). Then

Σdef=

⋂{Σμ : μ is a probability measure on Σ

}

is called the universal completion of Σ.

Proposition 4.46

(a) Every probability measure μ on Σ can be extended uniquely to aprobability measure μ on Σ and the function μ �−→ μ is a bijectionfrom the set of probability measures on Σ onto the set of probabilitymeasures on Σ.(b) If (X,Σ) and (Y,Y) are measurable spaces and f : X −→ Y is a(Σ,Y)-measurable function, then f is also

(Σ, Y)

-measurable.

Definition 4.47Let X be a topological space and let B(X) be its Borel σ-algebra. Theuniversal completion of B(X) is denoted by Bu(X) and it is called theuniversal σ-algebra of X. If X and Y are two topological spacesand f : X −→ Y is a function, then we say that f is universallymeasurable if and only if f is

(Bu(X),Bu(Y ))-measurable.

Proposition 4.48If X is a Borel space,then every Souslin subset of X (see Definition 2.156) is universallymeasurable.


4.1.6 Measurable Multifunctions: Selection Theorems

Next we will present a few basic things about measurable multifunc-tions, culminating to the main selection theorem. So, let (Ω,Σ) be ameasurable space and let (X, d

X) be a separable metric space. Addi-

tional hypotheses will be introduced as needed.

Definition 4.49Let F : Ω −→ 2X be a multifunction.(a) We say that F is measurable if for every open set U ⊆ X, wehave F−(U) =

{ω ∈ Ω : F (ω) ∩ U = ∅} ∈ Σ.

(b) We say that F is graph measurable if Gr f ={(ω, x) ∈ Ω×X :

x ∈ F (ω)} ∈ Σ⊗ B(X).

Remark 4.50It is customary to say that the domain of F , denoted by domF , isthe set

domFdef=

{ω ∈ Ω : F (ω) = ∅}.

It is clear from Definition 4.49(a), that domF ∈ Σ and so, whendealing with measurable multifunctions, it is not a loss generality toassume that their domain is all of Ω.

In what follows for a topological space X, we use the followingnotation:

Pf

(X) def

={A ⊆ X : A is nonempty and closed

}

Pk

(X) def

={A ⊆ X : A is nonempty and compact

}

Pf (X)def= Pf

(X) ∪ {∅}.

Moreover, if X is a normed space, then

Pfc

(X) def

={A ∈ Pf

(X): A is convex

}

Pkc

(X) def

={A ∈ Pk

(X): A is convex

}

Pwkc

(X) def

={A ⊆ X : A is nonempty, w-compact and convex

}.

The next theorem summarizes some important properties andequivalent definitions of measurable multifunctions.


Theorem 4.51Let (Ω,Σ) be a measurable space and let (X, d

X) be a separable metric

space. Consider a multifunction F : Ω −→ Pf (X) and the followingproperties:(1) For every A ∈ B(X), we have

F−(A) ={ω ∈ Ω : F (ω) ∩A = ∅} ∈ Σ.

(2) For every closed set C ⊆ X, we have

F−(C) ={ω ∈ Ω : F (ω) ∩ C = ∅} ∈ Σ.

(3) F is measurable (see Definition 4.49(a)).(4) For every x ∈ X, the R+-valued function

ω �−→ dX

(x, F (ω)

)= inf

{dX(x, u) : u ∈ F (ω)

}

is Σ-measurable (where R+ = [0,+∞]).(5) F is graph measurable (see Definition 4.49(b)).

Then we have the following relations between these properties:(a) (1) =⇒ (2) =⇒ (3) ⇐⇒ (4) =⇒ (5).(b) If X is σ-compact, then (2) ⇐⇒ (3).(c) If Σ = Σ (see Definition 4.45) and X is complete (i.e., a Polishspace), then all five properties are equivalent.

Remark 4.52If (Ω,Σ, μ) is a complete σ-finite measure space, then Σ = Σ.

In Theorem 2.170, we stated an important selection theorem, thecelebrated Michael selection theorem. There the emphasis was topo-logical and so the selector produced was continuous. Here the empha-sis is measure theoretic and so we are looking for measurable selec-tors, i.e., a Σ-measurable single valued function f : Ω −→ X such thatf(ω) ∈ F (ω) for all ω ∈ Ω.

Theorem 4.53 (Kuratowski–Ryll Nardzewski Selection Theorem)If (Ω,Σ) is measurable space, X is a Polish space and

F : Ω −→ Pf

(X)is a measurable multifunction,

then there exists a Σ-measurable function f : Ω −→ X such thatf(ω) ∈ F (ω) for all ω ∈ Ω.


Remark 4.54In fact the above selection theorem remains true, if we drop the com-pleteness hypothesis on X (i.e., X is only separable metrizable) andinstead assume that for every ω ∈ Ω, the set F (ω) is a complete subsetof X.

Theorem 4.53 can be strengthened as follows.

Theorem 4.55If (Ω,Σ) is a measurable space, X is a Polish space and

F : Ω −→ Pf

(X)is a multifunction,

then the following two statements are equivalent:(a) F is measurable (see Definition 4.49(a)).(b) There exists a sequence {fn}n�1 of Σ-measurable functionsfn : Ω −→ X, n � 1 such that for all ω ∈ Ω, we have

fn(ω) ∈ F (ω) ∀ ω ∈ Ω, n � 1

andF (ω) =

{fn(ω)

}n�1

∀ ω ∈ Ω.

This theorem leads to the following stronger version ofTheorem 4.51.

Theorem 4.56Let (Ω,Σ) be a measurable space, let X be a separable metric space andlet F : Ω −→ Pf

(X)be a multifunction. We consider the following

properties:(1) For every set A ∈ B(X), we have F−(A) ∈ Σ.(2) For every closed set C ⊆ X, we have F−(C) ∈ Σ.(3) F is measurable (see Definition 4.49(a)).(4) For every x ∈ X, the function ω �−→ dist

(x, F (ω)

)is

Σ-measurable.(5) There exists a sequence {fn : Ω −→ X}n�1 of Σ-measurable func-tions such that

fn(ω) ∈ F (ω) ∀ ω ∈ Ω, n � 1

andF (ω) =

{fn(ω)

}n�1

∀ ω ∈ Ω.

(6) GrF ∈ Σ ⊗ B(X) (i.e., F is graph measurable; seeDefinition 4.49(b)).


Then we have the following relations between these properties:(a) (1) =⇒ (2) =⇒ (3) ⇐⇒ (4) =⇒ (6).(b) If X is complete (i.e., a Polish space), then (4) ⇐⇒ (5).(c) If X is σ-compact, then (2) ⇐⇒ (3).(d) If Σ = Σ (see Definition 4.45) and X is complete (i.e., a Polishspace), then all six properties are equivalent.

In the second measurable selection theorem the hypotheses on themeasurable space are stronger, but the conditions on the range spaceare weakened considerably.

Theorem 4.57 (Yankov–von Neumann–Aumann Selection Theorem)If (Ω,Σ) is a complete measurable space (i.e., Σ = Σ), X is a Souslin

space and F : Ω −→ 2X \ {∅} is a multifunction such that GrF ∈Σ⊗ B(X),then there exists a Σ-measurable function f : Ω −→ X such that f(ω) ∈F (ω) for all ω ∈ Ω.

Remark 4.58If instead we assume that (Ω,Σ, μ) is a σ-finite measurable spaceand the other hypotheses remain unchanged, then we can have aΣ-measurable function f : Ω −→ X such that f(ω) ∈ F (ω) μ-almosteverywhere on Ω.

Again we can improve Theorem 4.57 and have a whole sequence ofselectors which are dense in F .

Theorem 4.59If (Ω,Σ) is a complete measurable space (i.e., Σ = Σ), X is a Souslin

space and F : Ω −→ 2X \ {∅} is a multifunction such that GrF ∈Σ⊗ B(X),then there exists a sequence of Σ-measurable functions fn : Ω −→ X,n � 1 such that

fn(ω) ∈ F (ω) ∀ ω ∈ Ω, n � 1

andF (ω) ⊆ {

fn(ω)}n�1

∀ ω ∈ Ω.


Remark 4.60Again, if instead we assume that (Ω,Σ, μ) is a σ-finite measure spaceand the rest are the same, then we can find a sequence of Σ-measurablefunctions fn : Ω −→ X, n � 1 such that

fn(ω) ∈ F (ω) μ-almost everywhere on Ω, for all n � 1

andF (ω) ⊆ {

fn(ω)}n�1

μ-almost everywhere on Ω.

4.1.7 Projection Theorems

The projection of a Borel set in R2 on a coordinate axis need not

be Borel. In fact this observation was the starting point for Souslinto develop his theory of Souslin (or analytic) sets. This raises twoimportant questions:(a) When can we guarantee that the projection of a Borel set is stillBorel?(b) More generally, can we characterize the projection of a measurableset?

Proposition 4.61If X and Y are Polish spaces, C ⊆ B(X × Y ) = B(X) ⊗ B(Y ) and

for every x ∈ X, we have C(x) ={y ∈ Y : (x, y) ∈ C

}is σ-compact,

then projXC ∈ B(X).

There is another such projection result, for which we need to in-troduce the following class of spaces.

Definition 4.62A topological space X is said to be of class σMK if X =

⋃

n�1Kn,

where Kn is metrizable and compact for all n � 1.

Remark 4.63Evidently a separable metrizable locally compact space is of classσMK . But a σMK space need not be metrizable. Anticipating somematerial from the theory of Banach spaces (see Chap. 5), we can seethat, if X∗ is the topological dual of a separable Banach space, then

X∗ is a σMK -space, since X∗ =⋃

n�1nB

∗1, with B

∗1

def=

{x∗ ∈ X∗ :

‖x∗‖∗ � 1}.


Proposition 4.64If X is a Borel space, Y is a space of class σMK, C ∈ B(X ×Y ) and

for all x ∈ X, the set C(x) ={y ∈ Y : (x, y) ∈ C

}is closed,

then projXC ∈ B(X).

Now we are ready to answer the second question and characterizethe projection of a measurable sets.

Theorem 4.65 (Yankov–von Neumann–Aumann ProjectionTheorem)If (Ω,Σ) is a measurable space, X is a Souslin space and C ∈Σ⊗ B(X),then proj

ΩC ∈ Σ.

Finally, concerning the Borel σ-algebra of a Souslin space X, wehave the following interesting property.

Proposition 4.66If X is a Souslin space, then B(X) is separable (see Defini-tion 3.14(b)).

4.1.8 Dual of Lp(Ω) for 1 � p � ∞In Chap. 3 we introduced the Banach spaces Lp(Ω), 1 � p � +∞. ByLp(Ω)∗ we denote the dual of the Banach space Lp(Ω), namely thelinear space of all continuous, linear functionals ξ : Lp(Ω) −→ R (seealso Chap. 5). Endowed with the norm

‖ξ‖∗ = sup{∣∣ξ(u)

∣∣ : ‖u‖p � 1},

the space Lp(Ω)∗ becomes a Banach space.In the next theorem, we give a very convenient characterization of

the dual space Lp(Ω)∗ for p ∈ [1,+∞).

Theorem 4.67 (Riesz Representation Theorem)If (Ω,Σ, μ) is a measure space and 1 � p < +∞,

then ξ ∈ Lp(Ω)∗ if and only if there exists a unique v ∈ Lp′(Ω) (where1p +

1p′ = 1) such that

ξ(u) =

∫

Ω

uv dμ ∀ u ∈ Lp(Ω) and ‖ξ‖∗ = ‖v‖p′ .


Therefore, the Banach space v ∈ Lp′(Ω) (where 1p + 1

p′ = 1) andLp(Ω)∗ are isometrically isomorphic.

Remark 4.68The function Lp(Ω)∗ � ξ �−→ v ∈ Lp′(Ω) introduced above is a linearisometry which is surjective and permits the identification of Lp(Ω)∗

with Lp′(Ω) (where 1p + 1

p′ = 1). In the sequel we will always use thisidentification, i.e., that

Lp(Ω)∗ = Lp′(Ω) ∀ p ∈ [1,+∞),

with 1p +

1p′ = 1. We also say that p and p′ are conjugate exponents.

Note that, if p = 1, then p′ = +∞ and so L1(Ω)∗ = L∞(Ω). Then thedual of L∞(Ω) contains L1(Ω) and the inclusion is strict.

To describe the elements of L∞(Ω)∗, we need the following defini-tion.

Definition 4.69Let (Ω,Σ, μ) be a σ-finite measure space.(a) A function ξ ∈ L∞(Ω)∗ is said to be absolutely continuous ifthere exists u ∈ L1(Ω) such that

ξ(v) =

∫

Ω

vu dμ ∀ v ∈ L∞(Ω).

(b) A function ξ ∈ L∞(Ω)∗ is said to be singular if there exists adecreasing sequence {Cn}n�1 ⊆ Σ such that μ(Cn) ↘ 0 and

ξ(v) = ξ(χ

Cnv) ∀ n � 1, v ∈ L∞(Ω).

The next theorem characterizes the dual of L∞(Ω).

Theorem 4.70 (Yosida–Hewitt Theorem)Every ξ ∈ L∞(Ω)∗ admits a unique decomposition ξ = ξa + ξs, with ξaabsolutely continuous and ξs singular. Moreover, ‖ξ‖∗ = ‖ξa‖∗+‖ξs‖∗.

4.1.9 Sequences of Measures: Weak Convergencein Lp(Ω)

Next we turn our attention to sequences of measures. We start by gen-eralizing the notion of uniform integrability of a sequence of functions(see Definition 3.124), to sequences of signed measures.


Definition 4.71Let (Ω,Σ, ν) be a measure space.(a) If Y ⊆ 2Ω and μ is a set function on Y, we say that μ is Vitalicontinuous if for every decreasing sequence {Cn}n�1 ⊆ Y such that⋂

n�1Cn = ∅, we have μ(Cn) −→ 0.

(b) A sequence {μk}k�1 of Vitali continuous set functions μk on Yis said to be Vitali equicontinuous if for every decreasing sequence{Cn}n�1 ⊆ Y such that

⋂

n�1Cn = ∅, we have that for every ε > 0, we

can find n0 = n0(ε) � 1 such that

∣∣μk(Cn)∣∣ � ε ∀ k � 1, n � n0.

(c) If Y = Σ, then a sequence {μk}k�1 of set functions on Σ isuniformly ν-absolutely continuous provided that for a given ε > 0,we can find δ = δ(ε) > 0 such that for all C ∈ Σ with ν(C) < δ, wehave ∣

∣μk(C)∣∣ < ε ∀ k � 1.

Now, for a given measurable space (Ω,Σ), by M(Σ) we denote thelinear space of bounded signed measures on Σ. For μ ∈ M(Σ), we set

‖μ‖ = variation of μ = |μ|(Ω)

(the total variation norm on M(Ω); see Definition 3.148). Then(M(Σ), ‖ · ‖) is a Banach space (see Proposition 3.149). Also let

‖μ‖∞ = sup{∣∣μ(C)

∣∣ : C ∈ Σ

}.

This is another norm on M(Σ) equivalent to ‖ · ‖, since

‖μ‖∞ � ‖μ‖ � 4‖μ‖∞ ∀ μ ∈ M(Σ).

Recall that, if μ ∈ M(Σ), then |μ| ∈ M(Σ).

Theorem 4.72 (Nikodym Theorem)If (Ω,Σ) is a measurable space, {μn}n�1 ⊆ M(Σ) and lim

n→+∞μn(C) =

μ(C) for C ∈ Σ,then μ ∈ M(Σ) and {μn}n�1 ⊆ M(Σ) is Vitali equicontinuous (seeDefinition 4.71(b)).


The next theorem is a slight improvement of the Vitali–Hahn–Sakstheorem (see Theorem 3.151) and can be proved using the Nikodymtheorem (see Theorem 4.72).

Theorem 4.73If (Ω,Σ) is a measurable space, ν is a nonnegative element in M(Σ)and {μn}n�1 ⊆ M(Σ) is a sequence such that(a) for all A ∈ Σ, μn(A) −→ μ(A); and(b) μn � ν for all n � 1,then the sequence {μn}n�1 is uniformly ν-absolutely continuous, μ ∈M(Σ) and μ � ν.

From Theorem 4.67, we know that L1(Ω)∗ = L∞(Ω). So, we canspeak about weak convergence of sequences in L∞(Ω).

Definition 4.74Let (Ω,Σ, μ) be a measure space. We say that the sequence {un}n�1 ⊆L1(Ω) is weakly convergent to u ∈ L1(Ω) (denoted by un

w−→ u inL1(Ω)), if for every h ∈ L∞(Ω), we have

〈un, h〉 =

∫

Ω

unhdμ −→∫

Ω

uhdμ = 〈u, h〉 .

Similarly, we say that the sequence {un}n�1 ⊆ Lp(Ω) is weakly conver-

gent to u ∈ Lp(Ω), with 1 < p < +∞ (denoted by unw−→ u in Lp(Ω)),

if for every h ∈ Lp′(Ω) (where 1p +

1p′ = 1), we have

∫

Ω

unhdμ −→∫

Ω

uhdμ.

The next theorem characterizes weak convergent sequences inL1(Ω).

Theorem 4.75 (Dunford–Pettis Theorem)Let (Ω,Σ, μ) be a finite measure space. A sequence {un}n�1 ⊆ L1(Ω)admits a weakly convergent subsequence (converging to some u ∈L1(Ω)) if and only if the sequence {un}n�1 ⊆ L1(Ω) is uniformly in-tegrable (see Definition 3.124).


By introducing topological structure on the space Ω, we can havethe following alternative version of Dunford–Pettis Theorem (see alsoTheorem 3.125).

Theorem 4.76If Ω is a locally compact topological space, (Ω,Σ, μ) is a measure spacewith a tight measure μ (see Definition 4.9(e)) and {un}n�1 ⊆ L1(Ω)is a sequence,then un

w−→ u ∈ L1(Ω) if and only if(a) for every ε > 0 there exists δ > 0 such that

μ(A) < δ =⇒∫

A

|un| dμ < ε ∀ n � 1;

(b) for every ε > 0, there exists a compact set K ⊆ X such that

∫

Kc

|un| dμ < ε ∀ n � 1;

(c) the sequence {un}n�1 is L1-bounded, i.e., there exists M > 0 suchthat ‖un‖1 � M for all n � 1.

4.1.10 Covering Theorems

Now we turn out attention to the relation between Lebesgue integra-tion and differentiation of functions, leading to generalizations of thefundamental theorem of calculus. A first result in this direction wasthe Radon–Nikodym theorem (see Theorem 3.152).

We start with some covering results.

Definition 4.77

(a) A collection F of closed balls in RN is a cover of A ⊆ R

N ifA ⊆ ⋃

B∈FB.

(b) A cover F is a Vitali or fine cover of A ⊆ RN if

inf{diamB : x ∈ B, B ∈ F}

= 0 ∀ x ∈ A.


Remark 4.78We could define Vitali covers using cubes instead of balls. The condi-tion for a cover to be a Vitali cover is equivalent to saying that “forevery ε > 0 and every x ∈ A there exists B ∈ F such that x ∈ B anddiamB < ε”.

Theorem 4.79 (Vitali Covering Theorem)If A ⊆ R

N is a Lebesgue measurable set with finite Lebesgue measureand F is a Vitali cover of A,then for every ε > 0, we can find a sequence {Bn}n�1 ⊆ F of pairwisedisjoint balls such that

λN(A \

⋃

n�1

Bn

)= 0 and λN

( ⋃

n�1

Bn

)� λN (A) + ε

(here by λN we denote the N -dimensional Lebesgue measure).

Remark 4.80The above theorem does not claim that

⋃

n�1Bn covers A. The covering

is only in a measure theoretic sense. However, as a by-product of theproof, we obtain that

A ⊆⋃

B∈FB ⊆

⋃

n�1

Bn,

where Bn denotes the concentric closed ball with radius 5 times theradius of Bn. Theorem 4.79 remains valid if balls are replaced by cubes.Its proof relies on the structure of the Lebesgue measures λN and soit cannot be extended if λN is replaced by a general tight measure onRN (a Radon measure on R

N ). The next covering theorem is moresuitable for arbitrary Radon measures on R

N , since it does not requireenlargement of the balls (to pass from B to B).

Theorem 4.81 (Besicovitch Covering Theorem)If F is a collection of closed balls in R

N such that sup{diamB : B ∈

F}< +∞, and A is the set of centres of balls in F ,

then there exist F1, . . . ,Fm(N) ⊆ F such that each Fk (k =1, . . . ,m(N)) is a countable family of disjoint balls in F , A ⊆m(N)⋃

k=1

⋃

B∈Fk

B and the number m(N) of these subcollections depends only

on the dimension N of the space.


Remark 4.82Again the above theorem remains true if the balls are replaced bycubes.

The next result is a consequence of Theorem 4.81 and can be viewedas a measure theoretic reformulation of it. It says that for any givenopen set, we can fill it up with a countable collection of disjoint ballsin such a way that the remainder set has μ-measure zero.

Proposition 4.83If μ is a Borel measure on R

N , F is a collection of closed balls in

RN , A is the set of centres of these balls, μ(A) < +∞ and for each

a ∈ A, we have inf{r : Br(a) ∈ F}

= 0 (where Br(a)def=

{x ∈ R

N :‖x− a‖ � r

}),

then for every open set U ⊆ RN , there exists a countable subcollection

F0 ⊆ F of disjoint balls such that⋃

B∈F0

⊆ U and μ((A ∩ U) \ (

⋃

B∈F0

B))

= 0.

4.1.11 Lebesgue Differentiation Theorem

Using the covering theorems presented above, we can study the differ-entiation of Radon measures on R

N (recall that a Borel measure μ onRN is Radon if it is finite on compact sets, outer regular and for all

A ∈ B(RN ), we have μ(A) = sup{μ(K) : K ⊆ A, F is compact

}).

Definition 4.84Let μ, ν be two Radon measures on R

N . For every x ∈ RN , we define

Dμν(x)def=

⎧⎨

⎩lim supn→+∞

ν(Br(x))μ(Br(x))

if μ(Br(x)

)> 0 for all r > 0,

+∞ if μ(Br(x)

)= 0 for some r > 0,

Dμν(x)def=

⎧⎨

⎩lim infn→+∞

ν(Br(x))μ(Br(x))

if μ(Br(x)

)> 0 for all r > 0,

+∞ if μ(Br(x)

)= 0 for some r > 0,

If Dμν(x) = Dμν(x) < +∞, then we say that ν is differentiablewith respect to μ at x and we write

Dμν(x) = Dμν(x) = Dμν(x).

We call Dμν(x) the derivative of ν with respect to μ at x.


Theorem 4.85If μ and ν are Radon measures on R

N ,

then Dμν exists and is finite μ-almost everywhere on RN , it is μ-

measurable and for every Borel set A ⊆ RN , we have

ν(A) = νs(A) +

∫

A

Dμν dμ,

with νs ⊥ μ and Dμνs(x) = 0 μ-almost everywhere on RN . If ν � μ,

then νs = 0 and

ν(A) =

∫

A

Dμν dμ.

As a consequence of Theorem 4.85, we have the following theorem.

Theorem 4.86 (Lebesgue Differentiation Theorem on RN)

If μ is a Radon measure and u ∈ L1(RN ;μ), then

limr→0+

1μ(Br(x))

∫

Br(x)

u dμ = u(x) for μ-almost all x ∈ RN .

Corollary 4.87If μ is a Radon measure, 1 � p < +∞ and u ∈ Lp

loc

(RN ;μ

), then

limr→0+

1μ(Br(x))

∫

Br(x)

∣∣u− u(x)

∣∣p dμ = 0 for μ-almost all x ∈ R

N .

In fact, if μ = λN (the Lebesgue measure on RN ), then we have a

stronger version of the above corollary.

Proposition 4.88If 1 � p < +∞ and u ∈ Lp

loc

(RN), then

limB↘{x}

1λN (B)

∫

Br(x)

∣∣u− u(x)

∣∣p dλN = 0 for λN -almost all x ∈ R

N

(here the limit is taken over all closed balls B such that x ∈ B anddiamB −→ 0; we emphasize that balls need not be centred at x).


Definition 4.89Let μ be a Radon measure on R

N and let u : RN −→ R be aμ-measurable function. The set of points x0 ∈ R

N for which

limr→0

1μ(Br(x))

∫

Br(x0)

∣∣u− u(x0)∣∣ dμ = 0,

is the Lebesgue set of u. The elements of this set are called theLebesgue points of u.

Also the set of points x0 ∈ RN for which

limr→0

1μ(Br(x))

∫

Br(x0)

u dμ = u(x0),

is the differentiability set of u.

Remark 4.90According to Corollary 4.87, if u ∈ Lp

loc

(RN ;μ

)(1 � p < +∞), then

μ-almost all points x ∈ RN belong to the Lebesgue set of u. Moreover,

the Lebesgue set of u is a subset of the differentiability set of u andthe inclusion can be strict.

Another interesting consequence of Corollary 4.87 is the followingresult.

Proposition 4.91If A ⊆ R

N is a Lebesgue measurable set, then

limr→0+

λN (Br(x)∩A)

λN (Br(x))= 1 for λN -almost all x ∈ A and

limr→0+

λN (Br(x)∩A)

λN (Br(x))= 0 for λN -almost all x ∈ R

N \ A.

Definition 4.92Let A ⊆ R

N . A point x ∈ RN is a point of density for A if

limr→0+

λN (Br(x)∩A)

λN (Br(x))= 1.

Remark 4.93According to Proposition 4.91, λN -almost all points of A are densitypoints for A. We can think of the set of points of density for A as akind of a measure theoretic interior for the set A.


We can generalize the Lebesgue differentiation theorem (seeTheorem 4.86), by using the notion of a regular family.

Definition 4.94Let μ be a Radon measure on R

N . For a given x ∈ RN , a family Fx of

μ-measurable subsets of RN is said to be regular at x if the followingholds:(a) for every ε > 0, there exists C ∈ Fx such that diamC � ε;(b) there exists a constant c � 1 such that for every C ∈ Fx, we haveμ(Bx) � cμ(C), where Bx is the smallest closed ball with centre at x,which contains C.

Remark 4.95In the above definition property (a) says that the sets in Fx shrink tox, without requiring that x belongs in any of the sets of Fx. Property(b) says that each set in Fx is comparable to a ball centred at x. Ifμ = λN , then regular families at the origin (and by translation to anyother point in R

N) are the balls, cubes, ellipsoids or regular polygonscentred at the origin. On the other hand, as we already mentioned,the sets of the regular family need not contain the origin, for exampleconsider the collection of annuli

S� ={x ∈ R

N : 12 < ‖x‖ <

},

with > 0.

Proposition 4.96If μ is a Radon measure on R

N , u ∈ L1loc

(RN ;μ

), x ∈ R

N is aLebesgue point of u (see Definition 4.89) and Fx is a regular family atx, then

limdiamC → 0

C ∈ Fx

1μ(C)

∫

C

∣∣f − f(x)

∣∣ dμ = 0.

In particular, we have

limdiamC → 0

C ∈ Fx

1μ(C)

∫

C

f dμ = f(x).


Remark 4.97In the particular case N = 1 (functions of one variable), Proposi-tion 4.96 implies that for every u ∈ L1

loc(R), we have

limh→0

1h

x+h∫

x

f dλ = f(x) for λ-almost all x ∈ R,

with λ being the Lebesgue measure on R.

4.1.12 Bounded Variation and Absolutely ContinuousFunctions

Next we turn our attention to functions of bounded variation of onevariable. Since, as we will see, every function of bounded variation canbe written as the difference of two increasing functions, we start witha few basic facts about monotone functions. By monotone (respec-tively, strictly monotone), we mean increasing or decreasing (respec-tively, strictly increasing or strictly decreasing). A monotone functionneed not be continuous. Nevertheless, the discontinuity set of such afunction has a precise description.

Theorem 4.98If T ⊆ R is an interval and u : T −→ R is a monotone function,then u has at most countably many jump discontinuity points.Conversely, for a given countable set D, we can find a monotone func-tion u : R −→ R with discontinuity set D.

Remark 4.99If D = Q, then we see that there is a monotone function u : R −→ R

which is continuous on the irrationals and discontinuous on the ratio-nals.

Concerning the differentiability properties of a monotone function,we have the following theorem.

Theorem 4.100 (Lebesgue Theorem on Monotone Functions)If T ⊆ R is an interval and u : T −→ R is a monotone function,then u is differentiable almost everywhere on T .

This theorem is sharp.


Theorem 4.101If D is a Lebesgue-null set,then there exists a continuous monotone function which is not differ-entiable on D.

The next theorem gives some consequences of the Lebesgue theo-rem on monotone functions (see Theorem 4.100).

Theorem 4.102If T ⊆ R is an interval and u : T −→ R is a monotone function,then u′ is Lebesgue measurable and for every [0, b] ⊆ T , we have

b∫

a

∣∣u′(t)∣∣ dt �

∣∣u(b)− u(a)∣∣,

i.e., u ∈ L1loc(T ). Moreover, if u is bounded the u′ ∈ L1(T ) and for

mu = supT , ml = inf T , we have∫

T

∣∣u′(t)∣∣ dt �

∣∣ limt→m+

u

u(t)− limt→m−

l

u(t)∣∣ � sup

Tu− inf

Tu.

Remark 4.103The Cantor function (see Remark 3.83) is a continuous increasingfunction such that u′ = 0 λ-almost everywhere on [0, 1] (λ being theLebesgue measure on R). Actually, we can find a continuous strictlyincreasing function u : [0, 1] −→ R such that u′ = 0 λ-almost every-where on [0, 1].

For the differentiability of series of monotone functions, we havethe following theorem.

Theorem 4.104 (Fubini Theorem on Series of Monotone Functions)If T ⊆ R is an interval and

{un : T −→ R

}n�1

is a sequence of

increasing functions such that∑

n�1un(t) converges pointwise to u(t)

for all t ∈ T ,then the convergence is uniform on compact subsets of T , u is λ-almosteverywhere differentiable and u′(t) =

∑

n�1u′n(t) for almost all t ∈ T .

The set of monotone functions is not a vector space (clearly thedifference of two monotone functions need not be monotone). It is only


a cone. In order to characterize the smallest vector space of functionsu : T −→ R which contains the cone of monotone functions, we areled to the notion of functions of bounded variation. In what follows,for a given interval T ⊆ R, by a partition of T , we mean a finite setP = {tk}nk=0 ⊆ T such that t0 < t1 < . . . < tn.

Definition 4.105Let T ⊆ R be an interval and let u : T −→ R be a function. The totalvariation of u on T , denoted by Var u, is defined by

Var udef= sup

p

{ n∑

k=1

∣∣u(tk)− u(tk−1)

∣∣ : P = {uk}nk=0

},

with P = {uk}nk=0 being a partition of T . We say that u is of boundedvariation if Var u < +∞. We denote the space of all functionsu : T −→ R of bounded variation, by BV (T ).

Remark 4.106It is easy to see that if T contains one of its endpoints a = inf Tor b = supT , then we can always consider partitions with t0 = a(if a ∈ T ) and tn = b (if b ∈ T ). We will say that u : T −→ R is locallyof bounded variation if u ∈ BV [a, b] for all intervals [a, b] ⊆ T .The space of all functions u : T −→ R locally of bounded variation isdenoted by BVloc(T ). In some occasions, in order to emphasize thedependence on the interval T , we write Var Tu. Moreover, if U ⊆ R

is open, then we know that U =⋃

n�1Tn, where {Tn}n�1 are pairwise

disjoint open intervals. Then for a function u : U −→ R, we define

VarUvdef=

∑

n�1

VarTnu

and we say that u is of bounded variation on U if VarUu < +∞.The corresponding space is denoted by BV (U). Finally, if we havefunctions u : T −→ R

N , we can still consider the total variation of uand consequently speak of functions of bounded variation, if in Defi-nition 4.105, we replace | · | by ‖ · ‖ (the norm of RN ). Then the spaceare denoted by BV (T ;RN ) and BVloc(T ;R

N ).

A first natural question is whether a monotone function is ofbounded variation.


Proposition 4.107If T ⊆ R

N is an interval and u : T −→ R is a monotone function,then for every interval I ⊆ T , we have

VarIu = supI

u− infIu.

Consequently u ∈ BVloc(T ). Moreover, u ∈ BV (T ) if and only if u isbounded.

In the study of functions of bounded variation, the following func-tion is important.

Definition 4.108Let T ⊆ R be an interval, t0 ∈ T and u ∈ BVloc(T ). The indefinitevariation of u is the function

V (t)def=

{Var [t0,t]u if t0 � t,

−Var [t,t0]u if t < t0.

Sometimes, in order to emphasize the dependence on t0 (on (t0, u)),we write Vt0 (or Vt0,u). If inf T ∈ T , then we define

V∞(t)def= Var T∩(−∞,t]u ∀ t ∈ T.

Proposition 4.109If T ⊆ R is an interval and u ∈ BVloc(T ),then for all t, s ∈ T with s < t, we have

∣∣u(t)− u(s)∣∣ � V (t)− V (s) = Var [s,t]u

and so the functions V and V + u are increasing.

As a consequence of this proposition, we have the followingtheorem.

Theorem 4.110If T ⊆ R is an interval,then the smallest vector space containing all monotone functions(respectively, all bounded monotone functions) is the space BVloc(T )(respectively, the space BV (T )).


Remark 4.111So, according to this theorem every u ∈ BVloc(T ) (respectively, everyu ∈ BV (T )) can be written as the difference of two increasing functions(respectively, of two bounded increasing functions).

Proposition 4.112If T ⊆ R is an interval and u ∈ BVloc(T ),then u has at most countably many jump discontinuities, u is differ-entiable λ-almost everywhere and for every [a, b] ⊆ T , we have

b∫

a

∣∣u′(t)∣∣ dt � Var [a,b]u.

Moreover, if u ∈ BV (T ), then u is bounded, u′ ∈ L1(T ) and

∫

T

|u′| dt �∫

T

|V ′| dt � supT

V − infT

V = Var u.

Now, let us look closer the spaces BVloc(T ) and BV (T ). First weexamine compositions with functions of bounded variation.

Theorem 4.113If T ⊆ R is an interval and f : R −→ R,then f◦u ∈ BVloc(T ) (respectively, f◦u ∈ BV (T )) for all u ∈ BVloc(T )(respectively, all u ∈ BV (T )) if and only if f is locally Lipschitz.

Remark 4.114The result is also true for the spaces BVloc(T ;R

N ), BV (T ;RN ) andf : RN −→ R.

So far we know that BV (T ) is a vector space. Is it possible toequip BV (T ) with a norm and make it a normed space or even bettera Banach space? Note that

Var (λu) = |λ|Var u ∀ λ ∈ R

andVar (u+ v) � Var u+Var v ∀ u, v ∈ BV (T ).

However, Var u = 0 does not imply that u = 0, only that u is constant.So, a possible norm on BV (T ) should involve more than Var u.


Proposition 4.115If T ⊆ R is an interval,

then for any fixed t0 ∈ T , u �−→ ‖u‖ =∣∣u(t0)

∣∣ + Var u is a norm onBV (T ). Moreover,

(BV (T ), ‖ · ‖) is a Banach space.

The completeness part in the above proposition is based on thefollowing theorem.

Theorem 4.116 (Helly Selection Theorem; Helly First Theorem)If T ⊆ R is an interval and C ⊆ BV (T ) is an infinite set such that

there exist t0 ∈ T and M > 0 for which, we have∣∣u(t0)

∣∣ � M andVaru � M for all u ∈ C,then there exists a sequence {un}n�1 ⊆ C and u ∈ BV (T ) such thatun(t) −→ u(t) for all t ∈ T .

Remark 4.117The Banach space

(BV (T ), ‖ · ‖) is not separable.

Before moving to absolutely continuous functions, we will charac-terize those continuous functions which are of bounded variation. Firsta definition.

Definition 4.118Let X and Y be two nonempty sets, let u : X −→ Y be a function andlet A ⊆ X. For every y ∈ Y , let

Su(y;A)def=

{x ∈ A : u(x) = y

}.

The function Nu(·;A) : Y −→ N0 ∪ {+∞}, defined by

Nu(y;A)def=

{cardSu(y;A) if Su(y;A) is finite,+∞ if Su(y;A) is infinite

is called the Banach indicatrix of u on A. If A = X, then we dropthe dependence on A and we write Nu(·).Theorem 4.119If T ⊆ R is an interval and u : T −→ R is continuous,then Nu is Borel measurable and

+∞∫

−∞Nu(y) dy = Var u.

Therefore u ∈ BV (T ) if and only if Nu ∈ L1(R).


The problem of reconstructing a function from its derivative (i.e., afundamental theorem of calculus for Lebesgue integral) leads to thenotion of absolutely continuous functions. Continuous monotonefunctions (which are λ-almost everywhere differentiable; see Theo-rem 4.100) in general cannot be recovered from the integral of itsderivative. The Cantor function u : [0, 1] −→ R is continuous increas-ing and u′(t) = 0 for λ-almost all t ∈ [0, 1] (see Remark 3.83). However,we have

1∫

0

u′(t) dt = 0 < u(1) − u(0) = 1.

So, we need to restrict ourselves to a subclass of BV (T ).

Definition 4.120Let T ⊆ R be an interval. A function u : T −→ R is said to be abso-lutely continuous on T , if for every ε > 0 there exists a δ > 0 suchthat for every finite family

{(sk, tk)

}n

k=1of disjoint subintervals of T ,

we have

n∑

k=1

(tk − sk) � δ =⇒n∑

k=1

∣∣u(tk)− u(sk)∣∣ � ε.

The space of all absolutely continuous functions u : T −→ R is denotedby AC(T ). A function u : T −→ R is locally absolutely continuousif it is absolutely continuous on every interval [a, b] ⊆ T . The spaceof all locally absolutely continuous functions u : T −→ R is denotedby ACloc(T ). If U ⊆ R is an open set, we can still define the notionof absolute continuity, provided that [sk, tk] ⊆ U for all k = 1, . . . , n.We denote the space of such functions by AC(U). Finally, we candefine the notion of absolute continuity for functions u : T −→ R

N ,if in the above definition we replace | · | by ‖ · ‖ the norm of R

N .Then for the corresponding spaces, we use the notions AC(T ;RN ) andACloc(T ;R

N ).

Remark 4.121It is clear from the above that an absolutely continuous functionu : T −→ R is uniformly continuous (just take n = 1 in the defini-tion). The converse is not true.

The next result gives a relation between spaces BVloc(T ) andACloc(T ).


Proposition 4.122If T ⊆ R is an interval,then ACloc(T ) ⊆ BVloc(T ) and AC(T ) ⊆ BV (T ).In particular, if u ∈ ACloc(T ) (respectively, u ∈ AC(T )), then u′ existsλ-almost everywhere on T and u′ ∈ L1

loc(T ) (respectively, u ∈ L1(T )).

Even if u ∈ BV (T ) ∩ C(T ), u need not be absolutely continuous(the Cantor function is such an example). What is missing is the socalled Lusin N-property .

Definition 4.123Let T ⊆ R be an interval and let u : T −→ R be a function. We saythat u satisfies the Lusin N-property if u maps Lebesgue-null subsetsof T onto Lebesgue-null subsets of R.

Theorem 4.124 (Banach–Zaretsky Theorem)If T ⊆ R is an interval, then u ∈ ACloc(T ) if and only if(a) u is continuous; and(b) u is differentiable λ-almost everywhere on T with u′ ∈ L1

loc(T );and(c) u has the Lusin N -property.

Corollary 4.125If T ⊆ R is an interval and u : T −→ R is a continuous function suchthat(i) u′ exists λ-almost everywhere and u′ ∈ Lp(T ) for some1 � p < +∞; and(ii) u satisfies the Lusin N -property,then u ∈ AC(T ).

Corollary 4.126If T ⊆ R is an interval, then u ∈ ACloc(T ) if and only if(a) u is continuous; and(b) u ∈ BVloc(T ); and(c) u has the Lusin N -property.

Now we can say that the functions in ACloc(T ) are precisely thefunctions which can be reproduced by their derivatives, i.e., the fun-damental theorem of the Lebesgue calculus holds.


Theorem 4.127If T ⊆ R is an interval, then u ∈ ACloc(T ) if and only if(a) u is continuous; and(b) u is λ-almost everywhere differentiable on T with u′ ∈ L1

loc(T );and(c) for all t, t0 ∈ T , we have u(t) = u(t0) +

∫ tt0u′(s) ds.

Corollary 4.128If T ⊆ R is an interval and u : T −→ R is everywhere differentiablewith u ∈ L1

loc(T ),

then for all t, t0 ∈ T , we have u(t) = u(t0) +∫ tt0u′(s) ds.

Theorem 4.129If T ⊆ R is an interval, u ∈ BVloc(T ) (respectively, u ∈ BV (T )) and[a, b] ⊆ T ,then u ∈ AC

([a, b]

)(respectively, u ∈ AC(T )) if and only if

b∫

a

|u′| dt = Var [a,b]u (respectively,

∫

T

|u′| dt = Var u).

Next we examine the validity of the chain rule and the change ofvariable formula for absolutely continuous functions.

Theorem 4.130 (Chain Rule)If T ⊆ R is an interval, u ∈ BVloc(T ) and f : R −→ R is locallyLipschitz,then f ◦ u : T −→ R is differentiable λ-almost everywhere on T and

(f ◦ u)′(t) = f∗(u(t))u′(t) for λ-almost all t ∈ T,

where f∗ : R −→ R is any Borel function such that f∗ = f ′ λ-almosteverywhere on R (recall that f being locally Lipschitz is differentiableλ-almost everywhere on R).

Remark 4.131If f : RN −→ R is locally Lipschitz (N � 2) and u ∈ ACloc(T ;R

N ),then f ◦ u ∈ ACloc(T ), but the chain rule of Theorem 4.130 may fail.For the chain rule which is formulated in Theorem 4.130, we needadditional conditions on the set

Sfdef=

{x ∈ R

N : f is not differentiable at x}.

In Theorem 4.160 below we will give these conditions.


Theorem 4.132 (Change of Variables)If u : [a, b] −→ [c, d] is absolutely continuous and f : [c, d] −→ R isbounded measurable, then

u(b)∫

u(a)

f(y) dy =

b∫

a

f(u(t)

)u′(t) dt.

Theorem 4.133 (Area Formula)If T ⊆ R is an interval, f : T −→ [0,+∞] is a Borel measurablefunction and u : T −→ R is differentiable λ-almost everywhere on Tand has the Lusin N -property, then

∫

R

∑

x∈u−1(y)

f(x) dy =

∫

T

f(t)∣∣u′(t)

∣∣ dt.

Remark 4.134By definition

∑

x∈u−1(u)

f(x) = sup{∑

x∈Ff(x) : F ⊆ u−1(y) is finite

}.

Theorem 4.135 (Integration by Parts Formula)If T ⊆ R is an interval and u, v ∈ ACloc(T ),then for all t, t0 ∈ T , we have

u(t)v(y) − u(t0)v(t0) =

t∫

t0

uv′ ds +t∫

t0

u′v ds.

Definition 4.136

(a) Let T ⊆ R be an interval. A nonconstant function u : T −→ R

is said to be singular if it is differentiable λ-almost everywhere andu′(t) = 0 for λ-almost all t ∈ T .(b) Let T ⊆ R be an interval. An increasing function u : T −→ R isa saltus function, if u(t) =

∑

n∈Aun(t), where A ⊆ N and

un(t) =

⎧⎨

⎩

0 if t < rn,sn if t = rn,sn + tn if rn < t,


for some sets {rn}n∈A ⊆ T and {sn}n∈A, {tn}n∈A ⊆ [0,+∞] with sn+tn > 0 for all n ∈ A.

Remark 4.137The Cantor function is singular. Also, if u : [a, b] −→ R is increasingand for every t ∈ (a, b), we have

u+(t) = lims→t+

u(s) and u−(t) = lims→t−

u(s),

then

us(t) =∑

s ∈ [a, b]s < t

[u+(t)− u−(t)

]+ u(t)− u−(t)

is a saltus function.

Theorem 4.138If T ⊆ R is an interval and u ∈ BVloc(T ),then u = uac + ucs + us, where uac ∈ ACloc(T ), ucs is continuoussingular and us is a saltus function. Moreover

Var u = Var uac +Var ucs +Var us.

Next we extend the notion of bounded variation to functions ofseveral variables.

Definition 4.139Let Ω ⊆ R

N be an open set and let u ∈ L1loc(Ω). We say that u is of

bounded variation if

sup{∫

Ω

udiv hdz : h ∈ C1c

(Ω;RN

),∥∥h(z)

∥∥ � 1for allz ∈ Ω

}< +∞.

We denote the space of functions of bounded variation by BV (Ω). Wesay that u ∈ L1

loc(Ω) is of local bounded variation if for all openU ⊆ Ω with U compact and U ⊆ Ω, we have u ∈ BV (U). We de-note the space of functions which are locally of bounded variation byBVloc(Ω). A Lebesgue measurable set A ⊆ R

N has a finite perime-ter in Ω, if χA ∈ BV (Ω). We say that the Lebesgue set A ⊆ R

N haslocally finite perimeter in Ω, if χA ∈ BVloc(Ω).


Remark 4.140So, according to the above definition u ∈ L1

loc(Ω) is of bounded vari-ation, if the distributional partial derivatives ∂u

∂zi, i ∈ {1, . . . , N} are

bounded signed Radon measures (see Definition 4.22), i.e., for everyi ∈ {1, . . . , N}, there exists a bounded signed measure μi : B(Ω) −→ R

(B(Ω) being the Borel σ-algebra of Ω) such that∫

Ω

u ∂h∂zi

dz = −∫

Ω

hdμi ∀ i ∈ {1, . . . , N} , h ∈ C∞c (Ω).

The measures μi are the distributional (or weak) derivatives of u withrespect to zi, i ∈ {1, . . . , N} and we denote them by Diu. Therefore,if u ∈ BV (Ω), then Diu ∈ Mb(Ω) for all i ∈ {1, . . . , N} and Du =(D1u, . . . ,DNu

) ∈ Mb

(Ω;RN

)(Du being the gradient of u). So, we

can define the total variation |Du| of the vector measure Du by

|Du|(A) def= sup {

n∑

k=1

∥∥Du(Ck)

∥∥ : {Ck}nk=1 disjoint Borel

partition of A}.We know that |Du|(·) is a finite Radon measure. By Theorem 4.23,we have

|Du|(Ω) = ‖Du‖Mb(Ω;RN )

= sup{ n∑

i=1

∫

Ω

hid(Diu) : h ∈ C0

(Ω;RN

), ‖h‖∞ � 1

}

< +∞.

This leads to the following definition.

Definition 4.141Let Ω ⊆ R

N be an open set and let u ∈ L1loc(Ω). The variation of u

in Ω is defined by

V (u; Ω)def= sup

{ N∑

i=1

∫

Ω

∂hi∂zi

u dz : h ∈ C∞c (Ω;RN ), ‖h‖∞ � 1

}.

Proposition 4.142If Ω ⊆ R

N is an open set,then BV (Ω) equipped with the norm ‖u‖ = ‖u‖1+|Du|(Ω) is a Banachspace.


Based on Remark 4.140, we have the following structure theoremfor BVloc(Ω).

Theorem 4.143If Ω ⊆ R

N is an open set and u ∈ BVloc(Ω),

then there exists μ ∈ Mb(Ω) and a μ-measurable function ξ : Ω −→ RN

such that ‖ξ(z)‖ � 1 for μ-almost all z ∈ Ω and

∫

Ω

udiv hdz = −∫

Ω

(h, ξ)RN

dμ ∀ h ∈ C∞c (Ω;RN ),

the measure μ is denoted by |Du|.

Remark 4.144If A ⊆ R

N is of locally finite perimeter in Ω and u = χA , then u ∈BVloc(Ω) (see Definition 4.139) and then by Theorem 4.143, we have

∫

A

div hdz = −∫

Ω

(h, ξ)RN

d|Dχa | ∀ h ∈ C∞c (Ω;RN )

(see also Remark 4.148).

From Remark 4.140, we have that |Du|(·) is the variational measureof u and |∂A|(·) is the perimeter measure of A, with |∂A|(Ω) being theperimeter of A in Ω.

Theorem 4.145If Ω ⊆ R

N is an open set and u ∈ BV (Ω),then there exists a sequence of functions {un}n�1 ⊆ BV (Ω) ∩ C∞(Ω)such that un −→ u in L1(Ω) and |Dun|(Ω) −→ |Du|(Ω).

Remark 4.146Note that for u ∈ BV (Ω) ∩ C∞(Ω), we have

|Du|(Ω) =

∫

Ω

‖Du‖ dz.

Definition 4.147Let A ⊆ R

N be a set with locally finite perimeter in RN . The reduced

boundary of A, denoted by ∂∗A ⊆ RN , is the set of all z ∈ R

N suchthat:


(a) |DχA |(Br(z)

)> 0; and

(b) the limit νA(z) = − limr→0+

DχA(Br(z))

|DχA|(Br(z))

exists in RN and

‖νA(z)‖ = 1.

Remark 4.148The minus sign in the definition of νA indicates that the normal vectoris directed outward from the set A, i.e., in the direction opposite tothe gradient of χA . Note that for all x ∈ R

N

n(x) = limr→0+

DχA(Br(x))

|Dχa |(Br(x)),

when this limit exists, is the Radon–Nikodym derivative of DχAwith

respect to |DχA| and so for all C ∈ B(RN ), we have

DχA(C) =

∫

C

n d|DχA|,

hence∫

A

div hdz =

∫

RN

(h, n)RN

d|DχA| ∀ h ∈ C∞

c

(RN ;RN

)

and thus∥∥n(z)

∥∥ = 1 |DχA|-almost everywhere and n = −ξ in Theo-

rem 4.143.Because |Dχ

A|(RN \ ∂∗A

)= 0, we see that νA(·) is a multidi-

mensional analogue of the Radon–Nikodym derivative of DχA(·) withrespect to |DχA |(·).

Next we generate the notion of boundary for a Lebesgue measur-able set.

Definition 4.149

(a) Let A ⊆ RN be a Lebesgue measurable set. A unit vector n is

a measure-theoretic outer normal of A at z, if the following twoconditions hold:

limr→0+

1rN

λN(Br(z) ∩

{x ∈ R

N : (x− z, n)RN

< 0, x ∈ A})

= 0

and

limr→0+

1rN

λN(Br(z) ∩

{x ∈ R

N : (x− z, n)RN

> 0, x ∈ A})

= 0.


We denote by n(x,A), the measure-theoretic outer normal of A atx ∈ R

N .(b) Let A ⊆ R

N be a Lebesgue measurable set. The measure-theoretic boundary of A is the set

∂∗Adef=

{x ∈ R

N : n(x,A) exists}.

Remark 4.150If A ⊆ R

N is a set with locally finite perimeter, then ∂∗A ⊆ ∂∗A. Notethat, if for a unit vector n ∈ R

N , we introduce the hyperplane

H(z)def=

{x ∈ R

N : (x− z, n)RN

= 0}

and the corresponding two half-spaces

H+(z)def=

{x ∈ R

N : (x− z, n)RN

> 0},

H−(z)def=

{x ∈ R

N : (x− z, n)RN

< 0},

then n is a measure theoretic normal to A at z, if the following twoconditions hold:

limr→0+

λN (Br(z)∩A∩H+(z))rN

= 0 and limr→0+

λN ((Br(z)\A)∩H−(z))rN

= 0.

4.1.13 Hausdorff Measures: Change of Variables

We will use the notions introduced in Definitions 4.147 and 4.149, toproduce multidimensional generalizations of the fundamental theoremof calculus. But first we need to say a few basic things about Hausdorffmeasures.

For s ∈ [0,+∞), let

ωsdef= π

s2

Γ( s2+1) , where Γ(t) =

∞∫

0

st−1e−s ds

is the Euler Γ-function. This constant coincides with the Lebesguemeasure of the unit ball of RN if s = N .


Definition 4.151Let s ∈ [0,+∞) and A ⊆ R

N . The s-dimensional Hausdorffmeasure of A is defined by

Hs(A)def= lim

δ→0Hs

δ (A),

where Hsδ is defined by

Hsδ (A)

def=

ωs

2sinf

{∑

k�1

(diamCk)s : A ⊆

⋃

k�1

Ck, diamCk < δ}.

Remark 4.152In the above definition the covering sets are arbitrary. However, with-out any loss of generality, we may assume that they are open convexor closed convex. Also, we can replace the condition diamCk < δ, bythe condition diamCk � δ without changing the value of Hs

δ (A). Ifs = 0, then H0 is the counting measure.

Theorem 4.153For s ∈ [0,+∞), Hs is a Borel measure.

Remark 4.154However, Hs is not a Radon measure, since for s ∈ (0, N), RN is notσ-finite with respect to Hs.

The next theorem summarizes some basic properties of theHausdorff measure.

Theorem 4.155

(a) HN = λN .(b) Hs = 0 on R

N for all s > N .(c) For all s > 0, all A ⊆ R

N and all λ > 0, we have Hs(λA) =λsHs(A).(d) For all s � 0, all A ⊆ R

N and every affine isometryL : RN −→ R

N , we have Hs(L(A)

)= Hs(A).

(e) If s � 0, A ⊆ RN and ξ : A −→ R

k is Lipschitz continuous withLipschitz constant η > 0, then Hs

(ξ(A)

)� ηsHs(A).

(f) If s > s′ > 0 and A ⊆ RN , then

[Hs(A) > 0 =⇒ Hs′(A) = +∞]


and also [Hs′(A) < +∞ =⇒ Hs(A) = 0

].

Definition 4.156The Hausdorff dimension of a set A ⊆ R

N is defined by

H-dim(A)def= inf

{s � 0 : Hs(A) = 0

}= sup

{s � 0 : Hs(A) > 0

}.

Remark 4.157Note that H- dim may be any number in [0,+∞], not necessary aninteger. Moreover, if d = H-dim(A), then Hs(A) = 0 for all s > d andHs(A) = +∞ for all s ∈ (0, d).

Proposition 4.158If A,C ⊆ R

N and dist(A,C) = inf(‖a− c‖ : a ∈ A, c ∈ C

)> 0,

then Hs(A ∪ C) = Hs(A) +Hs(A) for all s � 0.

Definition 4.159

(a) A set A ⊆ RN is said to be H1-rectifiable if there exists a

sequence of Lipschitz continuous functions un : R −→ RN such that

H1(A \ ⋃

n�1un(R)

)= 0.

(b) A set A ⊆ RN is said to be purely H1-rectifiable if

H1(A ∩ u(R)

)= 0,

for every Lipschitz continuous u : R −→ RN .

Using these notions we can have the multidimensional version ofthe chain rule (see Theorem 4.130).

Theorem 4.160If f : RN −→ R is a locally Lipschitz function and T ⊆ R is aninterval,then f ◦ u ∈ ACloc(T ) for every u ∈ ACloc(T ;R

N ).Moreover, if the set Sf =

{x ∈ R

N : f is not differentiable at x}is

purely H1-rectifiable, then

(f ◦ u)′(x) =

N∑

i=1

∂f∂zi

(u(x)

)u′i(x) for almost all x ∈ T,

with ∂f∂zi

(u(x)) u′i(x) being zero whenever u′i(x) = 0.


Using Hausdorff measures, we can extend Theorem 4.133, to func-tions of several variables.

Theorem 4.161 (Area Formula)If N � k and f : RN −→ R

k is a Lipschitz continuous function,

then for every Lebesgue measurable set A ⊆ RN , we have

∫

A

Jf(z) dz =

∫

Rk

H0(A ∩ f−1(y)

)dHN (y),

with Jf(z) = [[detDf(z)]], where for linear L : RN −→ Rk (with

N � k), we write L = O ◦ S, with S : Rk −→ Rk symmetric and

O : RN −→ Rk orthogonal and [[L]] = |detS|.

Theorem 4.162 (Change of Variable Formula)If N � k and f : RN −→ R

k is a Lipschitz continuous function and

u ∈ L1(RN ),then ∫

RN

u(z)Jf(z) dz =

∫

Rk

( ∑

z∈u−1(y)

u(z))dHn(y).

Theorem 4.163 (Isodiametric Inequality)

For all sets A ⊆ RN , we have λN (A) � ωN

(diamA

2

)N.

Finally, the area formula (see Theorem 4.161) has a dual version.

Theorem 4.164 (Coarea Formula)If N � k and f : RN −→ R

k is a Lipschitz continuous function,

then for every Lebesgue measurable set A ⊆ RN , we have

∫

A

Jf(z) dz =

∫

Rk

HN−k(A ∩ f−1(y)

)dy.

4.1.14 Caratheodory Functions

We conclude this chapter with three theorems that are very usefulboth in theory and applications. We start with a definition.

Definition 4.165Let (Ω,Σ) be a measurable space and let X and Y be two topologicalspaces. We say that f : Ω × X −→ Y is a Caratheodory func-tion if


(a) for every x ∈ X, the function ω �−→ f(ω, x) is(Σ,B(Y )

)-

measurable;(b) for every ω ∈ Ω, the function x �−→ f(ω, x) is continuous.

Theorem 4.166If (Ω,Σ) is a measurable space, X is a separable metric space, Y is ametric space and f : Ω×X −→ Y is a Caratheodory function,then f is jointly measurable.

The next theorem is a parametric version of the Lusin theorem (seeTheorem 3.77).

Theorem 4.167 (Scorza–Dragoni Theorem)If T and X are two Polish spaces, Y is a separable metric space, μ is afinite tight Borel measure on T and f : T ×X −→ Y is a Caratheodoryfunction,then for every ε > 0 we can find a compact subset Tε ⊆ T such thatμ(T \ Tε) < ε and f |

Tε×Xis continuous.

If Y = R ∪ {+∞}, then we can extend Theorem 4.167 as follows.

Theorem 4.168If T and X are Polish spaces, μ is a finite tight Borel measure onT and f : T ×X −→ R ∪ {+∞} is a Borel function such that for allf ∈ T , f(t, ·) is lower semicontinuous,then for every ε > 0, we can find a compact set Tε ⊆ T such thatμ(T \ Tε) < ε and f |

Tε×Xis lower semicontinuous.

4.2. Problems 677

4.2 Problems

Problem 4.1 ��

Suppose that A ⊆ [0, 1] is a Lebesgue measurable set and λ(A) > 0 (λbeing the Lebesgue measure on R). Show that we can find x, u ∈ A,x = u such that x− u ∈ Q.

Problem 4.2 ��

Show that the Borel σ-algebra of a topological space X is the smallestfamily of subsets of X which contains all open sets, all closed sets andwhich is closed under countable intersections and countable disjointunions.

Problem 4.3 �

Let X be a metrizable set. Show that the Borel σ-algebra B(X) is thesmallest family of subsets of X which includes the open sets and whichis closed under countable intersections and under countable disjointunions.

Problem 4.4 ��

Suppose that X is a locally compact topological space and let X isa subbasis for the topology of X (see Definition 2.19(a)). Show thatBa(X) ⊆ σ(X ) ⊆ B(X).

Problem 4.5 �

Suppose that X is a topological space,(X, B(X), μ

)is a complete

measure space (B(X) is the μ-completion of the Borel σ-algebra withrespect to μ; see Definition 3.23) and u : X −→ R is a function such

that: there exists a partition {Cn}n�1 ⊆ B(X) of X for which C0 isμ-null and for all n � 1, u|

Cnis lower semicontinuous (respectively,

upper semicontinuous, continuous). Show that u is B(X)-measurable.

Problem 4.6 ��

Suppose that X is a locally compact topological space and U ⊆ X isa σ-compact open set. Show that U ∈ Ba(X).

Problem 4.7 ��

Suppose that X is a locally compact, σ-compact topological spaceand X∗ is the Aleksandrov one-point compactification of X (see Re-mark 2.97). Show that B(X) = X ∩ B(X∗).


Problem 4.8 ��

Suppose that (Ω,Σ) is a measurable space, X is a topological space andG ∈ Σ⊗B(X). Show that there exists Σ0 ⊆ Σ, a countably generatedsub-σ-algebra of Σ (see Definition 3.14(a)) such that G ∈ Σ0 ⊗ B(X).

Problem 4.9 �

Let X be a locally compact topological space. Show that the openBaire sets form a basis for the topology of X.

Problem 4.10 ��

Let X be a topological space and let C ⊆ X be a nonempty closed setendowed with the subspace topology. Show that in general it is nottrue that B(C) ⊆ B(X).

Problem 4.11 ��

Let C ∈ B([0, 1]) with λ(C) > 0 (λ being the Lebesgue measure).Show that C has the cardinality of the continuum (do not use thecontinuum hypothesis).

Problem 4.12 ��

Suppose that X is a separable metric space, μ is a Borel measure onX and u : X −→ R+ = [0,+∞] is a Borel function. Show that

∫

X

u dμ =

+∞∫

0

μ({

x ∈ X : u(x) � η})

dη.

Problem 4.13 ��

Let λ be the Lebesgue measure on R. Suppose that A ⊆ R is aLebesgue measurable set and assume that λ

(A∩ (a, b)

)� 1

2(b− a) forall a, b ∈ R such that a < b. Show that λ(A) = 0.

Problem 4.14 �

Suppose that X is a metrizable space, B(X) is its Borel σ-algebra andμ, ν are two finite measures on B(X). Show the following:(a) If μ and ν coincides on the open sets of X, then μ = ν.(b) If X is σ-compact and μ, ν coincide on compact sets, then μ = ν.

4.2. Problems 679

Problem 4.15 ��

Suppose that X is a metrizable space and μ : B(X) −→ R+ = [0,+∞]is a σ-finite measure. Show that every set A ∈ B(X) with finite μ-measure is inner regular , i.e., for every ε > 0 there exists a closedset Cε ⊆ A such that μ(A \ Cε) < ε.

Problem 4.16 ��

Suppose that X is a topological space and μ : B(X) −→ R+ = [0,+∞]is a measure which is outer regular (see Definition 4.9), finite on com-pact sets and for every open set U ⊆ X, we have μ(U) = sup

{μ(K) :

K ⊆ U, K is compact}. Show that for every A ∈ B(X) which is

σ-finite for μ, we have μ(A) = sup{μ(D) : D ⊆ A, D is compact

}.

Problem 4.17 ��

Suppose that X is a locally compact topological space and let μ, ν betwo Radon measures on B(X) (see Definition 4.9(e)). Show that ν � μ(i.e., for all A ∈ B(X), ν(A) � μ(A)) if and only if

∫

X

u dν �∫

X

u dμ ∀ u ∈ Cc(X), u � 0.

Problem 4.18 ��

Suppose that X is a topological space, B(X) is its Borel σ-algebra andμ : B(X) −→ R+ = [0,+∞] is σ-bounded measure, i.e., there exists acountable open cover {Un}n�1 of X with μ(Un) < +∞ for all n � 1.Show the following:(a) Every compact set in X has finite μ-measure.(b) Suppose that X is a σ-compact metric space. Then every set A ∈B(X) with μ(A) < +∞ is inner regular with respect to compactsets, i.e., μ(A) = sup

{μ(K) : K ⊆ A, K is compact

}.

Problem 4.19 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ : B(X) −→ R+ = [0,+∞] is a σ-finite measurewhich is outer regular, finite on all compact sets and inner regular withrespect to compact sets on all open sets (i.e., for any open set U ⊆ X,we have μ(U) = sup

{μ(K) : K ⊆ U, K is compact

}). For every

A ∈ B(X), show that the following statements hold:(a) For a given ε > 0, we can find an open set U and a closed set Csuch that C ⊆ A ⊆ U and μ(U \ C) < ε.


(b) There exist an Gδ-set D and an Fσ-set E such that E ⊆ A ⊆ Dand μ(D \E) = 0.

Problem 4.20 �

Suppose that X is a topological space, B(X) is its Borel σ-algebra andμ : B(X) −→ R+ = [0,+∞) is an additive set function. Show that(a) μ is outer regular if and only if it is inner regular (see Defini-tion 4.9);(b) if μ(A) = sup


}for all A ∈ B(X),

then μ is regular.

Problem 4.21 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ : B(X) −→ R+ = [0,+∞] is a measure. Supposethat u ∈ Cc(X), u � 0 and assume that

∫

X

u dμ = 0. Show that

u|suppμ = 0.

Problem 4.22 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ is a signed Radon measure on B(X). Show that

∫

X

u dμ = 0 ∀ u ∈ Cc(X), u|supp μ = 0.

Problem 4.23 ��

Suppose that X is a topological space, B(X) is its Borel σ-algebraand μ : B(X) −→ R+ = [0,+∞)] is an additive set function, which isouter regular and inner regular with respect to the compact sets (i.e.,μ(A) = sup


}for all A ∈ B(X)). Show

that μ is a measure.

Problem 4.24 ��

Let X be a metrizable space. Show that a finite Borel measure μ isRadon if and only if for each ε > 0, there exists a compact set K suchthat μ(X)− ε < μ(K).

Problem 4.25 ��

Suppose that X is a topological space and μ is a Borel measure onX. Suppose that X is second countable or alternatively that μ is

4.2. Problems 681

outer regular and also inner regular with respect to compact sets (seeDefinition 4.9). Show that μ has a support (necessarily unique; seeRemark 4.14).

Problem 4.26 �

Suppose that (X,Σ) is a measurable space and μ, ν are two finitemeasures on Σ. Is it true that the smallest measure on Σ not less thanμ or ν is max

{μ(A), ν(A)

}, A ∈ Σ? Justify your answer.

Problem 4.27 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ and m are two Radon measures on B(X) withμ being σ-finite. Show that σ = min{μ,m} (the largest measure lessthan or equal to μ and m) is also a Radon measure on B(X).

Problem 4.28 ��

Show that the N -Lebesgue measure λN is Radon.

Problem 4.29 ��

Suppose that X is a locally compact topological space, B(X) is theBorel σ-algebra and μ is a Radon measure on B(X). Show that Cc(X)is dense in Lp(X,μ) (1 � p < +∞).

Problem 4.30 �

Find a bounded measurable function f : R −→ R for which there is nosequence {fn}n�1 ⊆ C(R) such that ‖fn − f‖∞ −→ 0.

Problem 4.31 �

On B(R) we introduce the set function μ0, defined by

μ0(A)def= card

{x : x ∈ Q ∩A

}.

Show that μ0 is a measure which is σ-finite but it is not regular.

Problem 4.32 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ : B(X) −→ [0,+∞) is a finite additive set func-tion with the following property

μ(A) = sup{μ(K) : K ⊆ A, K is compact

} ∀ A ∈ B(X).

Show that μ is a measure (i.e., μ is countably additive).


Problem 4.33 �

Consider the space

Vdef=

{u ∈ C∞

c (R) :

∫

R

u dx = 0}.

Is V dense in L1(R)? Justify your answer.

Problem 4.34 ��

Let (X, dX) be a metric space. We say that two sets A,C ⊆ X are

metrically separated if inf{dX(a, c) : a ∈ A, c ∈ C

}> 0, i.e., the

two sets are a positive distance apart. We say that an outer measureμ is additive on metrically separated sets, if μ(A ∪C) = μ(A) +μ(C), whenever A and C are metrically separated. Show that an outermeasure which is additive on metrically separated sets is in fact a Borelmeasure, i.e., every Borel set of X is μ-measurable.

Problem 4.35 �

Suppose that X is a locally compact metric space and {μn}n�1 is asequence of Radon measures such that

∫

X

f dμn −→∫

X

f dμ ∀ f ∈ C0(X)

for some Radon measure μ. Show that:(a) for all compact sets K ⊆ X, we have lim sup

n→+∞μn(K) � μ(K);

(b) for all open sets U ⊆ X, we have μ(U) � lim infn→+∞ μn(U).

Problem 4.36 ��

Suppose that X is a metric space and μ is a finite Borel measure onX. We say that μ is uniformly distributed if

0 < μ(Br(x)

)= μ

(Br(y)

) ∀ x, y ∈ X, r > 0.

Suppose that μ and ν are two uniformly distributed finite Borel mea-sures on X. Show that μ = cν for some c > 0.

Problem 4.37 ��

Suppose that {fn}n�1 ⊆ L1(0, 1) is a sequence such that

(a)∣∣fn(x)

∣∣ � h(x) almost everywhere on [0, 1], with h ∈ L1(0, 1); and

4.2. Problems 683

(b) we have

1∫

0

fng dx −→ 0 ∀ g ∈ C([0, 1]

).

Show that for every Borel set A ⊆ [0, 1], we have

∫

A

fn dx −→ 0.

Problem 4.38 ��

Suppose that X is a locally compact topological space and μ and νare two tight measures on B(X). Suppose that

∫

X

f dμ =

∫

X

f dν ∀ f ∈ Cc(X).

Show that μ = ν.

Problem 4.39 ��

Suppose that (X,Σ) is a measurable space, Y is a metric space andf : X −→ Y is a function. Show that the following two statements areequivalent:(a) f is Σ-measurable;(b) for every continuous function ϕ : Y −→ R, the function ϕ ◦ f isΣ-measurable.

Problem 4.40 �

Let (Ω,Σ) be a measurable space, let X be a topological space andlet f : Ω −→ X be a function. Show that the following statements areequivalent:(a) f is

(Σ,Ba(X)

)-measurable (see Definition 4.2).

(b) For every continuous function ϕ : X −→ R, ϕ◦f is Σ-measurable.


Problem 4.41 ��

Let A ⊆ R be a Lebesgue measurable set with λ(A) < +∞ (λ being theLebesgue measure on R) and consider the function f : R −→ [0,+∞),defined by

f(x)def= λ

((A+ x) ∩A

).

Show that:(a) f is continuous; and(b) lim

x→+∞ f(x) = 0. (Compare with Problem 3.44.)

Problem 4.42 ��

Suppose that A ⊆ R is a Lebesgue measurable set with positiveLebesgue measure and let us set

ϕ( )def=

∫

R

χA(t )χA(t) dt.

Show that ϕ is continuous at = 1.

Problem 4.43 ��

Find a decreasing sequence of measures on B(RN ) (N � 1) such thatthe limit set function is not a measure.

Problem 4.44 ��

Suppose that X is a compact metric space and μ is a finite Borelmeasure on X such that for every x ∈ X, we have μ

({x}) = 0. Showthat for a given ε > 0, we can find δ = δ(ε) > 0 such that for allA ∈ B(X) with diamA < δ and we have μ(A) < ε.

Problem 4.45 ��

Suppose that X is a compact topological space and μ : B(X) −→[0,+∞) is an additive set function such that

μ(A) = infU is openA ⊆ U

μ(U) = supK is compact

K ⊆ A

μ(K) ∀ A ∈ B(X).

Show that μ is σ-additive.

Problem 4.46 ��

Suppose that X is a compact metric space and μ is a finite nonatomicBorel measure on X. Show that there is a basis {Un}n�1 for the metrictopology of X such that μ(∂Un) = 0 for all n � 1.

4.2. Problems 685

Problem 4.47 ��

Suppose that X is a metric space and μ is a Borel measure on X. Wesay that μ is doubling if μ is finite on bounded sets and there existsc > 0 such that

μ(B2r(x)

)� cμ

(Br(x)

) ∀ x ∈ X, r > 0.

Suppose that μ is a doubling Borel measure on X. Show that thereexist t, c > 0 such that

μ(Br(x))μ(BR(y)) � c

(rR

)t ∀ x, y ∈ X, R � r > 0, x ∈ BR(y).

Problem 4.48 �

Suppose thatX is a metric space, μ is a Borel measure onX, A ∈ B(X)and μ(A) < +∞. Show that for a given ε > 0, we can find a closedset C ⊆ A such that μ(A \ C) � ε.

Problem 4.49 ��

Let A ⊆ RN be a Lebesgue measurable set such that λN (A) > 0 (λN

being the Lebesgue measure on RN ). Show that cardA = c. (Compare

Problem 4.11).

Problem 4.50 �

Suppose that {fn}n�1 ⊆ C([0, 1]

)is a sequence of functions such that

fn(x) −→ f(x) for almost all x ∈ [0, 1] and let ϑ ∈ (0, 1). Show thatthere exists a compact set K ⊆ [0, 1] such that λ(K) > ϑ and f |

Kis

continuous (λ being the Lebesgue measure on R).

Problem 4.51 �

Show that in R, a compact set with positive Lebesgue measure mayhave empty interior.

Problem 4.52 ��

Let f : [0, 1] −→ R be a Lebesgue measurable function. Show that wecan find a sequence of continuous functions hn : [0, 1] −→ R such that

hnλ−→ f (λ being the Lebesgue measure on R).


Problem 4.53 �

Let X and Y be two topological spaces such that

ΔYdef=

{(y, y) : y ∈ Y

} ∈ B(Y )⊗ B(Y )

and let f : X −→ Y be a Borel function. Show that

Gr f ∈ B(X)⊗ B(Y ).

Problem 4.54 �

Let (Ω,Σ) and (Y,Y) be two measurable spaces and assume that (Y,Y)is countably separated (i.e., there exists a sequence {Cn}n�1 ⊆ Ysuch that for any u, v ∈ Y , we can find n � 1 such that χCn

(u) =χCn

(v)). Suppose that f : Ω −→ Y is measurable. Show that Gr f ∈Σ⊗ Y.

Problem 4.55 �

Let (Ω,Σ, μ) be a σ-finite measure space, let f ∈ L1(Ω) and supposethat ∫

A

f(ω) dμ � λ ∀ A ∈ Σ, μ(A) < +∞

for some λ ∈ R. Show that∫

Ω

f(ω) dμ � λ.

Problem 4.56 �

Let C ⊆ Lp(0, 1) (with 1 < p < +∞) be a bounded set. Show that Cis uniformly integrable.

Problem 4.57 �

Suppose that (Ω,Σ) is a measurable space, X is a perfectly normalspace and f : Ω −→ X is a function. Show that f is

(Σ,B(X)

)-

measurable if and only if for every continuous function ϕ : X −→ R ,the function ϕ ◦ f is Σ-measurable.

Problem 4.58 ��

Let (Ω,Σ) be a measurable space and let X be a separable metricspace. Suppose that f : Ω −→ X is Σ-measurable. Show that thereexists a sequence of measurable functions fn : Ω −→ X with at mostcountable values such that fn ⇒ f in X.

4.2. Problems 687

Problem 4.59 ��

Let f, g : (0, 1) −→ [0,+∞) be two measurable functions. We saythat f and g are equimeasurable if for all ϑ > 0, we haveλ({f > ϑ}) = λ

({g > ϑ}) (λ being the Lebesgue measure on R). Showthe following:(a) If f and g are equimeasurable, then

1∫

0

f dx =

1∫

0

g dx.

(b) If ξ : [0,+∞) −→ [0,+∞) is a Borel function and f, g are equimea-surable, then ξ ◦ f and ξ ◦ g are equimeasurable too.

Problem 4.60 ��

Let f : (0, 1) −→ [0,+∞) be a measurable function. Show thefollowing:(a) There exists a unique function f∗ : (0, 1) −→ [0,+∞) which isdecreasing, right continuous and the functions f and f∗ are equimea-surable (see Problem 4.59).(b) For every Lebesgue measurable set A ⊆ (0, 1), we have

∫

A

f dx �λ(A)∫

0

f∗ dx

and if ξ : (0, 1) −→ [0,+∞) is decreasing, then

1∫

0

fξ dx �1∫

0

f∗ξ dx.

Problem 4.61 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra, μ is a Radon measure on B(X) and u ∈ L1(X,μ). Weset

ν(A) =

∫

A

u dμ ∀ A ∈ B(X).

Show that ν ∈ Mb(X) (see Definition 4.22).


Problem 4.62 �

Let X be a topological space and let μ be a Radon measure on X.Show that, if ν is a measure on X and ν � μ, then ν is Radon too.

Problem 4.63 �

Suppose that X is a locally compact topological space, μ ∈ Mb(X)and u ∈ Cc(X) are such that |u(x)| � ξ for all x ∈ suppμ. Show that

∣∣∫

Ω

u dμ∣∣ � ξ‖μ‖.

Problem 4.64 ��

Suppose that X is a locally compact topological space, B(X) is itsBorel σ-algebra and μ : B(X) −→ [−∞,+∞] is a signed measure withcompact support. Show that μ is finite.

Problem 4.65 ��

Let X be a locally compact topological space in which every open setis σ-compact (for example when X is strongly Lindeloff, in particularwhen X is second countable; see Proposition 2.164). Show that everyBorel measure which is finite on compact sets is Radon.

Problem 4.66 ��

Suppose that X is a locally compact topological space and {μn}n�1 isa sequence of Radon measures defined on the Borel σ-algebra B(X).Suppose that

∫

X

f dμn −→∫

X

f dμ ∀ f ∈ Cc(X),

for some Radon measure μ defined on the Borel σ-algebra B(X). Showthat μ(X) � lim inf

n→+∞ μn(X).

Problem 4.67 ��

Suppose that X is a locally compact topological space and {μn}n�1 is asequence of Borel measures on X. Suppose that ξ = sup

n�1μn(X) < +∞

and ∫

X

f dμn −→∫

X

f dμ ∀ f ∈ Cc(X).

4.2. Problems 689

Show that μ(X) < +∞ and

∫

X

hdμn −→∫

X

hdμ ∀ h ∈ C0(X).

Problem 4.68 ��

Let X be a compact topological space. Show that a continuous linearfunctional ϕ on C(X) is positive if and only if ‖ϕ‖∗ = ϕ(1) (1 is theconstant function on X identically equal to 1).

Problem 4.69 ��

Suppose that X is an uncountable locally compact topological space.Show that C0(X)∗ = Mb(X) is not separable.

Problem 4.70 ��

Suppose that X and Y are two compact metric spaces and f : X −→ Yis a continuous surjection. Show that for every Radon measure ϑ onY , we can find a Radon measure μ on X such that μf−1 = ϑ (seeTheorem 4.18).

Problem 4.71 ��

Suppose thatX is a locally compact topological space and ϕ ∈ C0(X)∗,ϕ � 0. Show that the following two statements are equivalent:(a) ϕ

(max{f, h}) = max

{ϕ(f), ϕ(h)

}for all f, h ∈ C0(X);

(b) ϕ = cδx for some c � 0 and some x ∈ X.

Problem 4.72 ��

Suppose that X is a locally compact topological space, μ is a Radonmeasure on X and Y is a family of lower semicontinuous functions fromX into R+ = [0,+∞] which is upward directed (i.e., if h1, h2 ∈ Y, thenthere exists h ∈ Y such that h1 � h and h2 � h). Let f = sup

{h :

h ∈ Y}. Show that

∫

X

f dμ = suph∈Y

∫

X

hdμ.

Problem 4.73 ��

Suppose that X is a locally compact topological space, μ is a Radonmeasure on X and f : X −→ R+ = [0,+∞] is a Borel function. Showthat:


(a) we have∫

X

f dμ = inf{∫

X

hdμ : h is lower semicontinuous on X, h � f}.

(b) we have∫

X

f dμ = sup{∫

X

η dμ : η is upper semicontinuous on X, 0 � η � f}

(recall that η is upper semicontinuous if and only if −η is lower semi-continuous).

Problem 4.74 �

Suppose that X is a locally compact topological space and μ is asigned Borel measure on X satisfying that for every A ∈ B(X) andevery ε > 0, we can find a compact set K ⊆ X and an open set U ⊆ Xsuch that K ⊆ A ⊆ U and

∣∣μ(B)

∣∣ < ε ∀ B ∈ B(X), B ⊆ U \K.

Show that μ ∈ Mb(X) (see Definition 4.22). (Compare with Prob-lem 4.19.)

Problem 4.75 ��

Suppose that X and Y are two separable metric spaces andξ : M+

1 (X) ⊗M+1 (Y ) −→ M+

1 (X × Y ) is the function defined by

ξ(m,μ)def= m× μ,

where m× μ is the product of the two measures m and μ. Show thatξ is continuous when M+

1 (X), M+1 (Y ) and M+

1 (X × Y ) are furnishedwith their respective weak topologies (see Definition 4.25).

Problem 4.76 ��

Suppose that X is a metric space and f : X −→ R∗ = R ∪

{±∞} is a lower semicontinuous, bounded below (respectively, up-per semicontinuous, bounded above) function. Show that the functionξ :

(M+

1 (X), w) −→ R

∗ (see Definition 4.25), defined by

ξ(μ) =

∫

X

f dμ

is lower semicontinuous (respectively, upper semicontinuous).

4.2. Problems 691

Problem 4.77 ��

Suppose that X is a metric space, U ⊆ X is an open set, C ⊆ X is aclosed set and η ∈ R. Show that:(a) The set D =

{μ ∈ M+

1 (X) : μ(U) > η}is weakly open.

(b) The set E ={μ ∈ M+

1 (X) : μ(C) � η}is weakly closed.

Problem 4.78 ��

Suppose that X is a metric space, B(X) is its Borel σ-algebra andfor each A ∈ B(X) we consider the function ξA : M+

1 (X) −→ [0, 1],defined by

ξA(μ)def= μ(A).

We furnishM+1 (X) with the weak topology (see Definition 4.25). Show

that ξA is Borel measurable.

Problem 4.79 �

Let X be a metric space and consider the space Cb(X) fur-nished with the sup-norm and the space M+

1 (X) furnished withthe weak topology (see Definition 4.25). Show that the functionγ : Cb(X)×M+

1 (X) −→ R, defined by

γ(u, μ)def=

∫

X

u dμ

is jointly continuous.

Problem 4.80 ��

Let X be a metrizable space. Show that X is homeomorphic to asequentially closed subspace of M+

1 (X) with the weak topology (seeDefinition 4.25).

Problem 4.81 ��

Suppose that X and Y are two metrizable spaces and f : X −→ Y isa homeomorphism into Y . Let ϑ : M+

1 (X) −→ M+1 (Y ) be defined by

ϑ(μ)(A)def= μ

(f−1(A)

) ∀ μ ∈ M+1 (X), A ∈ B(Y ).

Show that ϑ is a homeomorphism too.

Problem 4.82 ��

Let X be a locally compact metric space. A set A ⊆ X is said to bebounded if A ⊆ K for some compact set K ⊆ X. Let B(X) be the


Borel σ-algebra of X, let {μn}n�1 be a sequence of Radon measuresof X. We say that the sequence {μn}n�1 converges vaguely to a

Radon measure μ (denoted by μnv−→ μ), if

∫

X

f dμn −→∫

X

f dμ ∀ f ∈ Cc(X).

Show that the following two statements are equivalent:(a) μn

v−→ μ;(b) μn(A) −→ μ(A) for all bounded set A ∈ B(X) with μ(∂A) = 0.

Problem 4.83 ��

LetX be a Polish space and let C ⊆ M+1 (X). Show that C is uniformly

tight if and only if there exists a function ϕ : X −→ [0,+∞] such thatfor every λ > 0, the set {ϕ � λ} is compact in X (i.e., ϕ is inf-compact) and

supμ∈C

∫

X

ϕ(x) dμ < +∞.

Problem 4.84 ��

Suppose that {μn}n�1 is a sequence of probability measures defined

on B(RN ) and

∫

RN

f(x) dμn −→∫

RN

f(x) dμ ∀ f ∈ C∞c

(RN)

for some probability measure μ defined on B(RN ).Show that the set

{μn : n � 1

} ∪ {μ} is uniformly tight and

μnw−→ μ in M+

1 (RN ).

Problem 4.85 ��

Suppose that X and Y are two separable metric spaces and{ϑn : X −→ Y

}n�1

is a sequence of Borel functions such that ϑn −→ ϑuniformly on compact sets and ϑ is continuous. Let the sequence{μn}n�1 ⊆ M+

1 (X) be uniformly tight and μnw−→ μ. Show that

μnϑ−1n

w−→ μϑ−1 in M+1 (Y ).

4.2. Problems 693

Problem 4.86 �

Suppose that X1,X2 and Y are three separable metric spaces,ϑ1 : Y −→ X1 and ϑ2 : Y −→ X2 are two continuous functions suchthat ϑ = (ϑ1, ϑ2) : Y −→ X1 × X2 is proper (i.e., the inverse imageof a compact set is compact) and E ⊆ M+

1 (Y ) is such that the setsE1 = Eϑ−1

1 ⊆ M+1 (X1) and E2 = Eϑ−1

2 ⊆ M+1 (X2) are uniformly

tight. Show that E is uniformly tight.

Problem 4.87 ��

Suppose that X is a metric space, μ, ν ∈ M+1 (X) and

∫

X

f dμ =

∫

X

f dν ∀ f ∈ UCb(X).

Show that μ = ν (see also Problem 4.38).

Problem 4.88 ��

Suppose that X and Y are two separable metric spaces, m : X ×B(Y ) −→ [0, 1] is a function such that the function x �−→ m(x, ·)is continuous from X into M+

1 (Y ) furnished with the weak topology(see Definition 4.25) and f ∈ Cb(X × Y ). Show that the function

x �−→ h(x) =

∫

Y

f(x, y)m(x, dy)

is continuous.

Problem 4.89 �

Suppose that X is a separable metric space and C ⊆ X is a nonemptyclosed set. Let

M+1 (C)

def=

{μ ∈ M+

1 (X) : suppμ ⊆ C}.

Show that the set M+1 (C) is closed in M+

1 (X) when the latter is fur-nished with the weak topology (see Definition 4.25).

Problem 4.90 ��

Let X be a separable metric group. For any two measures μ, ν ∈M+

1 (X), the convolution μ � ν is defined to be the set function

(μ � ν)(A)def=

∫

X

μ(Ax−1) dν(x) ∀ A ∈ B(X).


Show that � : M+1 (X) × M+

1 (X) −→ M+1 (X) and it is continuous

(M+1 (X) is furnished with the weak topology; see Definition 4.25).

Problem 4.91 ��

Suppose that X is a separable metric space and Y ⊆ 2X is such thatσ(Y) = B(X) and Y is closed under finite intersections (for exampleY can be the family of closed subsets of X). For every A ∈ B(X), letηA : M+

1 (X) −→ [0, 1] be the function, defined by

ηA(μ)def= μ(A) ∀ μ ∈ M+

1 (X).

Show thatB(M+

1 (X))

= σ( ⋃

A∈Yη−1A

(B(R)))

(M+1 (X) is furnished with the weak topology; see Definition 4.25).

Problem 4.92 ��

Suppose that (Ω,Σ) is a measurable space, X is a Polish space andF : Ω −→ 2X \ {∅} is a graph measurable multifunction. Let

S(ω)def=

{μ ∈ M+

1 (X) : μ(F (ω)

)= 1

} ∀ ω ∈ Ω.

Show that S : Ω −→ 2M+1 (X) \ {∅} is graph measurable (M+

1 (X) isfurnished with the weak topology; see Definition 4.25).

Problem 4.93 ��

Suppose that X and Y are separable metric spaces, Y ⊆ B(Y ) is afamily such that Y is closed under finite intersections and σ(Y) = B(X)and ξ : X −→ M+

1 (X) is a function (spaces M+1 (X) and M+

1 (Y ) arefurnished with the weak topologies; see Definition 4.25). Show thatξ is Borel measurable if and only if for every A ∈ Y, the functionϑA : X −→ [0, 1], defined by

ϑA(x)def= ξ(x)(A)

is Borel measurable.

Problem 4.94 ��

Suppose that X and Y are separable metric spaces, ξ : X −→ M+1 (Y )

is a continuous function (M+1 (Y ) is furnished with the weak topology;

see Definition 4.25). Show that for every function f : X ×Y −→ R∗ =

4.2. Problems 695

R ∪ {+∞}, which is lower semicontinuous and bounded below, thefunction

X � x �−→∫

Y

f(x, y)ξ(x, dy)

is lower semicontinuous and bounded below.

Problem 4.95 ��

Let (X,Σ) be a separable measurable space. Show that there exists asubspace A of {0, 1}N such that (X,Σ) and

(A,B(A)) are isomorphic.

Problem 4.96 ��

Let X and Y be two topological spaces and let f : X −→ Y be afunction such that Gr f is a Souslin subset of X × Y . Show that f isBorel function.

Problem 4.97 ��

Let X be a Borel space (see Definition 4.40). Show that M+1 (X)

equipped with the weak topology (see Definition 4.25) is a Borel spacetoo.

Problem 4.98 ��

Suppose that X is a Polish space and {μn}n�1 ⊆ M+1 (X) is a sequence

such that μnw−→ μ for some μ ∈ M+

1 (X). Suppose that f : X −→ R isa continuous function, g : X −→ (−∞,+∞] is a lower semicontinuousfunction and

limk→+∞

supn�1

∫

{|f |�k}

|f | dμn = 0 and limk→+∞

supn�1

∫

{g−�k}

g− dμn = 0.

Show that

limn→+∞

∫

X

f dμn =

∫

X

f dμ and lim infn→+∞

∫

X

g dμn �∫

X

g dμ > −∞.

Problem 4.99 ��

Suppose that X is a Polish space and {μn}n�1 ⊆ M+1 (X) is a sequence

such that μnw−→ μ for some μ ∈ M+

1 (X). Suppose that f ∈ C(X) ∩L1(X,μn) for all n � 1, f � 0 and

lim supn→+∞

∫

X

f dμn �∫

X

f dμ < +∞.


Show that

limk→+∞

supn�1

∫

{f�k}f dμn = 0.

Problem 4.100 �

Suppose thatX and Y are two Polish spaces, A ∈ B(X), f : X −→ Y isa Borel measurable, injective function, C = f(A) ⊆ Y and C ∈ B(Y ).Show that f−1 : C −→ X is Borel measurable.

Problem 4.101 ��

Let (X,Σ) be a measurable space such that there exists a Polish spaceZ and a Souslin (analytic) set A ⊆ Z such that the measurable spaces(X,Σ) and

(A,B(A)) are isomorphic (see Definition 4.39). Let Y

be a Polish space and let f : X −→ Y be a Σ-measurable function.Show that for every C ∈ Σ, the set f(C) is Souslin (analytic; seeRemark 2.157).


Suppose that X is a Borel space (see Definition 4.40) and f : X −→R∗ = R ∪ {±∞} is a Borel measurable function. We defined the

function γf : M+1 (X) −→ R

∗, by

γf (μ)def=

∫

X

f dμ.

Show that this function is Borel measurable when M+1 (X) is furnished

with the weak topology (see Definition 4.25).


Let f : RN −→ RM be a Borel measurable map and let D ∈ B(RN ).

Show that f(D) is a Souslin set.


Let X be a Souslin space which is completely regular (see Prob-lem 2.42). Show that X is perfectly normal.


Let X be a topological space and A ∈ Ba(X). Show that there existsa sequence {fk}k�1 ⊆ C(X) and a set B ∈ B(R∞) such that

(�) A ={x ∈ X :

{fk(x) : k � 1

} ∈ B}.

4.2. Problems 697

In addition every set of the form (�) is Baire and we can take {fk}k�1 ⊆Cb(X).

Problem 4.106 �

Let X be a topological space and let μ be a Baire measure on X. Showthat for every B ∈ Ba(X) and every ε > 0, there exists a continuousfunction ξ : X −→ [0, 1] such that

∣∣∫

X

ξ dμ− μ(B)∣∣ < ε.


Let (Ω,Σ) be a measurable space and let (X, dX) and (Y, d

Y) be two

metric spaces with X being separable. Suppose that D is a countable,dense subset of X and let f : Ω×X −→ Y be a map such that:(i) for every x ∈ D, the function ω �−→ f(ω, x) is measurable;(ii) for every ω ∈ Ω, the function x �−→ f(ω, x) is continuous.

Show that f is jointly measurable.

Problem 4.108 �

Suppose that Ω is a set, A is a nonempty subset of 2Ω, X is a metricspace and G : Ω −→ 2X \ {∅} is a multifunction such that

G−(C)def=

{ω ∈ Ω : G(ω) ∩ C = ∅} ∈ A ∀ C ⊆ X, C closed.

Suppose that A is closed under countable unions. Show that for everyopen set U ⊆ X, we have G−(U) ∈ A.


Suppose that (Ω,Σ) is a measurable space, (X, dX) is a metric space

and{fn : Ω −→ X

}n�1

is a sequence of(Σ,B(X)

)-measurable func-

tions such that fn(ω) −→ f(ω) for all ω ∈ Ω. Show that f is(Σ,B(X)

)-measurable.

Problem 4.110 �

Suppose that (X,Σ, μ) is a σ-finite measure space, X is a separableBanach space and let F : Ω −→ 2X \ {∅} is a graph measurable multi-function. For 1 � p � +∞, let

SpF

def=

{f ∈ Lp(Ω;X) : f(ω) ∈ F (ω) μ-almost everywhere on Ω

}.


Show that SpF = ∅ if and only if inf

x∈F (ω)‖x‖ = m(ω) � h(ω) μ-almost

everywhere, with some h ∈ Lp(Ω).


Suppose that (Ω,Σ) is a complete measurable space (i.e., Σ = Σ; seeDefinition 4.45), X is a Souslin space, ξ : Ω ×X −→ R

∗ = R ∪ {±∞}is a jointly measurable function and F : Ω −→ 2X \ {∅} is a graphmeasurable multifunction. Let us set

m(ω)def= inf

x∈F (ω)ξ(ω, x) ∀ ω ∈ Ω.

Show that the function m : Ω −→ R = R ∪ {+∞} is Σ-measurable.

Problem 4.112 �

Let X and Y be two Souslin spaces and let f : X × Y −→ R be aBorel measurable and bounded below function (respectively, boundedabove). We set

m(x)def= inf

y∈Yf(x, y) (respectively, M(x) = sup

y∈Yf(x, y)).

Show that m (respectively, M) is measurable with respect to everyBorel measure on X.

Problem 4.113 �

Suppose that (Ω,Σ) is a measurable space, X is a Polish space, ξ : Ω×X −→ R

∗ = R ∪ {±∞} is a jointly measurable function such thatfor all ω ∈ Ω, the function x �−→ ξ(ω, x) is upper semicontinuous andF : Ω −→ Pf

(X)is a measurable multifunction. Let us set

m(ω)def= inf

x∈F (ω)ξ(ω, x) ∀ ω ∈ Ω,

Show that the function m : Ω −→ R = R ∪ {+∞} is Σ-measurable.


Suppose that (Ω,Σ, μ) is a σ-finite measure space, X is a separableBanach space, F : Ω −→ 2X \{∅} is a graph measurable multifunction,ξ : Ω×X −→ R

∗ = R ∪ {±∞} is a jointly measurable function,

Iξ(u)def=

∫

Ω

ξ(ω, u(ω)

)dμ

4.2. Problems 699

is well defined (maybe +∞ or −∞) for all u ∈ SpF (1 � p � +∞;

see Problem 4.110) and there exists u0 ∈ SpF such that Iξ(u0) > −∞.

Show that

supu∈Sp

F

Iξ(u) =

∫

Ω

supx∈F (ω)

ξ(ω, x) dμ.

Problem 4.115 �

Suppose that (Ω,Σ, μ) is a σ-finite measure space, X is a separableBanach space and F : Ω −→ 2X \ {∅} is a graph measurable multi-function (see Definition 4.49). Show that the set Sp

F ⊆ Lp(Ω;X) (seeProblem 4.110) is bounded if and only if the function ω �−→ ∣

∣F (ω)∣∣

belongs to Lp(Ω) (1 � p � +∞).

Problem 4.116 �

Suppose that (Ω,Σ) is a measurable space, X is a σ-compact metricspace and F : Ω −→ Pf

(X)is a multifunction such that for every

compact set K ⊆ X, we have

F−(K)def=

{ω ∈ Ω : F (ω) ∩K = ∅} ∈ Σ.

Show that F is measurable (see Definition 4.49(a)).


Suppose that (Ω,Σ) is a complete measurable space, X is a Polishspace, Y is a σ-compact Polish space and F : Ω −→ Pf (X × Y ) isa multifunction. Consider the multifunction G : Ω × X −→ Pf (Y ),defined by

G(ω, x)def=

{y ∈ Y : (x, y) ∈ F (ω)

}.

Show that F is measurable if and only if G is measurable.


Suppose that (Ω,Σ) is a measurable space, X is a Polish space andu : X −→ Ω is a function such that(a) for every ω ∈ Ω, we have u−1(ω) ∈ Pf

(X); and

(b) for every open set V ⊆ X, we have u(V ) ∈ Σ.

Show that there exists a Σ-measurable function f : Ω −→ X such thatu(f(ω)

)= ω for all ω ∈ Ω.



Suppose that (Ω,Σ) is a measurable space, X is a Polish space, Yis a metric space, g : Ω × X −→ Y is a Caratheodory function,U : Ω −→ Pf

(X)is a measurable multifunction, we set

G(ω)def= g

(ω,U(ω)

) ∀ ω ∈ Ω.

Show that for all open sets V ⊆ Y , we have G−(V ) ∈ Σ.

Problem 4.120 �

Suppose (Ω,Σ) is a complete measurable space, T is a Souslin space,X is a metric space, g : Ω× T −→ X is a jointly measurable function,U : Ω −→ 2T \ {∅} is a multifunction such that GrU ∈ Σ ⊗ B(T ) andh : Ω −→ X is a Σ-measurable function such that h(ω) ∈ g

(ω,U(ω)

)

for all ω ∈ Ω. Show that there exists a(Σ,B(T ))-measurable function

u : Ω −→ T such that u(ω) ∈ U(ω) and h(ω) = g(ω, u(ω)

)for all

ω ∈ Ω.


Suppose that (Ω,Σ) is a measurable space, X is a Polish space andF : Ω −→ 2X \ {∅} is a multifunction such that for every x ∈ Xand every r > 0, we have F−(Br(x)

) ∈ Σ (here Br(x) ={x′ ∈

X : d(x′, x) � r}

with d being the metric of X). Show that F ismeasurable.


Suppose that (Ω,Σ) is a measurable space, X is a separable metricspace, Y is a metric space, f : Ω×X −→ Y is a Caratheodory functionand U ⊆ Y is an open set and for every ω ∈ Ω there exists x ∈ X suchthat f(ω, x) ∈ U . Show that the multifunction F : Ω −→ 2X \ {∅},defined by F (ω)

def=

{x ∈ X : f(ω, x) ∈ U

}is graph measurable.


Suppose that (Ω,Σ) is a measurable space, X is a separable metricspace, Y is a topological space and f : Ω×X −→ Y is a Caratheodoryfunction. Let U ⊆ Y be an open set and let us set

F (ω)def=

{x ∈ X : f(ω, u) ∈ U

}.

Show that for all closed sets C ⊆ X, we have that

F−(C)def=

{ω ∈ Ω : F (ω) ∩ C = ∅} ∈ Σ.

4.2. Problems 701


Let (Ω,Σ) be a measurable space and let X be a Polish space. Supposethat f : Ω×X −→ R is a Caratheodory function. Let

E(ω)def= epi f(ω, ·) = {

(x, λ) ∈ X × R : f(ω, x) � λ}

for all ω ∈ Ω. Show that for every open set U ⊆ X × R, we have

E−(U)def=

{ω ∈ Ω : E(ω) ∩ U = ∅} ∈ Σ.


Let (Ω,Σ, μ) be a σ-finite measure space and let X be a separableBanach space. Suppose that F : Ω −→ 2X \ {∅} is a graph mea-surable multifunction with open values. Suppose that S1

F = ∅ (seeProblem 4.110) and let

∫

Ω

F (ω) dμ ={∫

Ω

f(ω) dμ : f ∈ S1F

}.

Show that the set∫

Ω

F (ω) dμ is open in X.


Let (Ω,Σ, μ) be a finite measure space. Suppose that {fn}n�1 ⊆ Lp(Ω)(with 1 < p < +∞), ‖fn‖p � M for all n � 1 with some M > 0.Assume also that for all A ∈ Σ, we have

∫

A

fn dμ −→∫

A

f dμ,

for some f ∈ Lp(Ω) such that ‖f‖p � M . Show that fnw−→ f in

Lp(Ω) (see Definition 4.74).


Let (Ω,Σ, μ) be a σ-finite measure space and let f : Ω −→ R be aΣ-measurable function such that fg ∈ L1(Ω) for all g ∈ Lp(Ω) (with1 � p < +∞). Show that f ∈ Lp′(Ω) (where 1

p + 1p′ = 1).


Suppose that (X,Σ, μ) is a measure space and {fn}n�1 ⊆ L1(X) is asequence such that the limit

limn→+∞

∫

A

fn dμ exists and is finite for all A ∈ Σ.


Show that for every ε > 0, we can find δ = δ(ε) > 0 such that

if μ(A) < δ, then

∫

A

|fn| dμ < ε for all n � 1.


Suppose that (X,Σ, μ) is a finite measure space and {fn}n�1 ⊆ L1(X)is a sequence such that

(i) for all A ∈ Σ, the limit limn→+∞

∫

A

fn dμ exists and is finite; and

(ii) supn�1

‖fn‖1 = M < +∞.

Show that there exists f ∈ L1(X) such that fnw−→ f in L1(X).


Let (X,Σ, μ) be a measure space and let {μn}n�1 ⊆ M(Σ) be a Vitaliequicontinuous sequence (see Definition 4.71) such that μn � μ forall n � 1. Show that the sequence {μn}n�1 is uniformly μ-absolutelycontinuous.


Suppose that (X,Σ, μ) is a finite measure space and {μn}n�1 ⊆ M(Σ)is a uniformly μ-absolutely continuous sequence (see Definition 4.71).Show that the sequence {μn}n�1 is Vitali equicontinuous.


Find a sequence of functions {fn}n�1 ⊆ Lp(a, b) (with 1 � p � +∞)

such that fnw−→ 0 but fn −→ 0 in Lp(a, b) (see Definition 4.74).

Problem 4.133 �

Let (Ω,Σ, μ) be a finite measure space and let {un}n�1 ⊆ L1(Ω) bea uniformly integrable sequence such that un(ω) −→ u(ω) μ-almosteverywhere in Ω. Show that un

w−→ u in L1(Ω).


Let f be a real valued function defined in a neighbourhood of a pointx0 ∈ R. If there exists a set D ⊆ R such that

dist(x0,D) = 0 and limx → x0

x ∈ D

f(x) = f(x0),

4.2. Problems 703

then we say that f is an approximately continuous function atx0. If f is approximately continuous at all points of its domain, wesay that f is an approximately continuous function . Show thata measurable, almost everywhere finite function is approximately con-tinuous at almost every point.


Suppose that f ∈ L1(0, 1) and for every x ∈ (0, 1) and every ε > 0, wecan find an open interval Ix ⊆ (0, 1) such that

x ∈ Ix, λ(Ix) < ε, and

∫

Ix

f dλ = 0.

Show that for every open interval I ⊆ (0, 1), we have∫I f dλ = 0.

Problem 4.136 �

Let λ be the Lebesgue measure on R. Does there exist a Lebesguemeasurable set A ⊆ R such that λ(A ∩ I) = 1

2λ(I) for every boundedinterval I? Justify your answer.


Let f ∈ L1(0, 1) and let Lf be the Lebesgue set of f (see Defini-tion 4.89). Show that Lf = S, where

Sdef=

{x ∈ [0, 1] : d

dx

x∫

0

∣∣f(t)− r∣∣ dt =

∣∣f(x)− r∣∣ for some r ∈ Q

}.


For any set B ⊆ R, any x ∈ R (not necessarily in B), we introduce thefollowing two quantities:

lim supλ(I)→0

λ(B∩I)λ(I) is the upper density of B at x

and

lim infλ(I)→0

λ(B∩I)λ(I) is the lower density of B at x,

where lim sup and lim inf are taken over all intervals I such that x ∈ I(here λ denotes the Lebesgue measure on R). If these two quantities


are equal, then their common value is the density of B at thepoint x (cf. Definition 4.92).

Let A,C ⊆ R be two nonempty sets which are metrically separated(see Problem 4.34). Show that the density of each of sets A and C atalmost all points of the other set is zero.


Show that the set of points at which the density of a Lebesgue mea-surable set A ⊆ R exists but it is not equal to 0 or 1, is of measurezero and of first category.(see Problem 4.138).

Problem 4.140 �

Suppose that A ⊆ [0, 1] is a Lebesgue measurable set and that thereexists ϑ > 0 such that λ

(A∩ [a, b]

)� ϑ(b− a) for all 0 � a � b � 1 (λ

being the Lebesgue measure on R). Show that λ(A) = 1. (Comparewith Problem 4.13.)

Problem 4.141 �

Let μ be a finite Borel measure on [0, 1] which is absolutely continuouswith respect to the Lebesgue measure λ on R. Show that

limh→0

μ([0, 1] ∩ (x− h, x+ h))

λ([0, 1] ∩ (x− h, x+ h))exists for λ-almost all x ∈ [0, 1].

Problem 4.142 �

Does there exist a Lebesgue measurable set A ⊆ [0, 1] such that

λ(A ∩ [a, b]

)=

b− a

2∀ 0 � a < b � 1

(λ being the Lebesgue measure on R).

Problem 4.143 �

Let f : R −→ R be a Lipschitz continuous function and letg : [a, b] −→ R be an absolutely continuous function. Show thatf ◦ g : [a, b] −→ R is absolutely continuous.

Problem 4.144 �

Let A ⊆ [0, 1] be a Lebesgue measurable set and consider the followingsequence

fn(x) = n

1n∫

0

χA(x+ t) dt ∀ x ∈ R, n � 1.

4.2. Problems 705

Show that:(a) 0 � fn � 1 for all n � 1;(b) fn ∈ AC

([0, 1]

)for all n � 1;

(c) fn −→ χAalmost everywhere on [0, 1];

(d) fn −→ χAin L1(R).


Suppose that A ⊆ R is a Lebesgue measurable set and qA = A forall q ∈ Q \ {0} (here qA =

{qx : x ∈ A

}). Show that A or R \ A is

Lebesgue-null.


Let D ⊆ [0, 1] be a Lebesgue-null set. Find an increasing and abso-lutely continuous function f : [0, 1] −→ R such that f ′(x) = +∞ forall x ∈ D.


Show that the function f : [0, 1] −→ R, defined by

f(x)def=

{0 if x = 0,x2 cos 1

x2 if 0 < x � 1

is everywhere differentiable but it is not of bounded variation.

Problem 4.148 �

Let f : [a, b] −→ R be an absolutely continuous function. Show that fis constant if and only if f ′(x) = 0 for almost all x ∈ [a, b].


Let f : R −→ R be a Lipschitz continuous function and let D ⊆ R bea Lebesgue-null set. Show that f ′(x) = 0 for almost all x ∈ f−1(D).


Assume that {un : [a, b] −→ R}n�1 is a sequence of functions such thatun(x) −→ u(x) for all x ∈ [a, b]. Show that Var u � lim inf

n→+∞ Varun.


Let f : [a, b] −→ R be a differentiable function. Show that f is abso-lutely continuous if and only if f is of bounded variation.


Problem 4.152 �

Suppose that u : [a, b] −→ R is a continuous function and ϑ < Varu(Var u can be 0 or +∞). Show that there exists δ > 0 such that

n∑

k=1

∣∣u(xk)− u(xk−1)∣∣ > ϑ,

for every partition a = x0 < x1 < . . . < xn = b such thatmax1�k�n |xk − xk−1| < δ.

Problem 4.153 �

Let u ∈ BV([a, b]

) ∩ C([a, b]

). Show that the set

Adef=

{y ∈ R : cardu−1({y} = +∞)

}

is Lebesgue-null.


Let {un : [0, 1] −→ R}n�1 be a sequence of absolutely continuous func-tions such that un(t) −→ u(t) for all t ∈ [0, 1] and {u′n}n�1 ⊆ L1(0, 1)

is uniformly integrable. Show that u ∈ AC([0, 1]

).

Problem 4.155 �

Suppose that f : [a, b] −→ R is absolutely continuous and p � 1. Showthat |f |p : [a, b] −→ R is absolutely continuous too.


Let A,C ⊆ RN be two Borel sets. Show that:

(a) If A ⊆ C, then H- dimA � H- dimC.(b) H- dim(A ∪ C) = max

{H- dimA, H- dimC

}.


Let γ : [a, b] −→ RN be a continuous curve. Show that

∥∥γ(b)− γ(a)∥∥ � H1

(γ[a, b]

).


Let γ : [a, b] −→ RN be a Lipschitz injective curve (a simple curve).

Show that H1(γ([a, b]

))= Var γ < +∞.

4.2. Problems 707


Suppose that C ⊆ RN is a connected compact set and 0 < r <

12diamC. Show that H1

(C ∩Br(x)

)� r for all x ∈ C.


Let (λN )∗ denote the Lebesgue outer measure on RN . Show that there

exists c > 0 such that cHN (A) � (λN )∗(A) � HN (A) for all A ⊆ RN .


Let A ⊆ RN be a set of positive outer measure and let f : RN −→ R

M

be a function. Let Gr f |A def=

{(x, f(x)

): x ∈ A

}. Show that

H- dimGr f |A � N and if f is Lipschitz, then H- dimGr f |A = N .


Show that for every connected set A ⊆ RN , we have H1(A) � diamA.


Show that, if A ⊆ RN has H- dimA < 1, then A is totally disconnected

(i.e., all connected components are singletons).


Suppose that (Ω,Σ) is a measurable space, X is a compact met-ric space, Y is a separable metric space and f : Ω × X −→ Y is aCaratheodory function. Show that the function f : Ω −→ C(X;Y ),defined by

f(ω)(·) def= f(ω, ·) ∀ ω ∈ Ω

is measurable when C(X;Y ) is furnished with the uniform con-vergence topology (i.e., the metric topology induced by the metricd(h, g) = max

x∈XdY

(h(x), g(x)

)for all h, g ∈ C(X;Y ) and with d

Ybeing

the metric on Y ; this topology is the c-topology; see Definition 2.174and Theorem 2.183).

Problem 4.165 �

Suppose that (Ω,Σ), X, Y and C(X;Y ) are as Problem 4.164 andthat f : Ω −→ C(X;Y ) is a Σ-measurable function. Show that thefunction

Ω×X � (ω, x) �−→ f(ω, x) = f(ω)(x) ∈ Y

is Caratheodory.


Problem 4.166 �

Let X be a locally compact topological space. For every f ∈ Cc(X),we consider the linear functional ϕf : Mb(X) −→ R, defined by

ϕf (μ)def= μ(f) =

∫

X

f dμ ∀ μ ∈ Mb(X).

Then we consider the weak (initial) topology on Mb(X) introduced bythis family of functions (see Definition 2.62). This topology is knownamong probabilities and measures theorists as vague topology (seealso Problem 4.82) and among functional analysts as w∗-topology (seeDefinition 5.63). We will denote it by w∗. Suppose that E ⊆ Mb(X).Show that the following statements are equivalent:

(a) Ew∗

is w∗-compact;(b) for every compact set K ⊆ X, there exists ηK > 0 such that|μ|(K) � ηK for all μ ∈ E.


Suppose that X is a locally compact topological space, ϑ > 0 and

consider the set Edef=

{μ ∈ Mb(X) : ‖μ‖ � ϑ

}. Show that E is

w∗-compact.

Is the set Ddef=

{μ ∈ Mb(X) : ‖μ‖ = ϑ

}w∗-closed? Justify your

answer.Finally show that if X is compact, then

D+def=

{μ ∈ Mb(X) : μ � 0, ‖μ‖ = ϑ

}

is w∗-compact (for the definition of the w∗-topology see Prob-lem 4.166).


Let X be a locally compact topological space. Show that the functionσ : X −→ M+

b (X), defined by σ(x) = δx is a homeomorphism of Xinto M+

b (X) (M+b (X) is endowed with the relative w∗-topology).

Moreover, show that if X is not compact, then for every ε > 0 andf ∈ Cc(X), there is a compact set K ⊆ X such that

∣∣δx(f)

∣∣ < ε for all

x ∈ K.


Let X be a locally compact topological space. Show that M+b (X) with

the relative w∗-topology (see Problem 4.166) is separable metrizable ifand only if X is second countable.

4.2. Problems 709


Suppose that X is a separable metric space and {μn}n�1 ⊆ M+1 (X) is

such that μnw∗−→ μ (see Problem 4.166). Show that

suppμ ⊆ lim infn→+∞ suppμn

(see Problem 1.171).


Suppose that X is a locally compact separable metric space and

{μn}n�1 ⊆ Mb(X) is a sequence such that μnw∗−→ μ in Mb(X) for

some μ ∈ Mb(X) (see Problem 4.166). Show that for every lowersemicontinuous function h : X −→ [0,+∞), we have

∫

X

hdμ � lim infn→+∞

∫

X

hdμn.


Suppose that {μn}n�1 ⊆ Mb(R) is a sequence of measures, μ ∈ Mb(R),

ϕn(x) = μn

((−∞, x]

)for n � 1 and ϕ(x) = μ

((−∞, x]

). Suppose that

‖μn‖ � M for all n � 1 and for all x, continuity points of ϕ, we have

ϕn(x) −→ ϕ(x). Show that μnw∗−→ μ (see Problem 4.166).


4.3 Solutions

Solution of Problem 4.1Let {qn}n�1 be the rational points of [−1, 1]. Let

An = A+ qn ={x+ qn : x ∈ A

}.

From the translation invariance of the Lebesgue measure, we have

λ(An) = λ(A) > 0

and clearly An ⊆ [−1, 2] for all n � 1. Then⋃

n�1An ⊆ [−1, 2].Suppose that the sets An are pairwise disjoint, i.e., if n = m, thenAn ∩ Am = ∅. Then, by σ-additivity, we have

∑

n�1

λ(An) = λ( ⋃

n�1

An

)� λ

([−1, 2]

)= 3,

but

λ( ⋃

n�1

An

)=

∑

n�1

λ(An) = λ(A)∑

n�1

1 = +∞

(since λ(A) > 0), a contradiction. This means that we can findm,n � 1 such that Am ∩An = ∅. Let a ∈ Am ∩An. Then

a = x+ qm = u+ qn with x, u ∈ A, x = u,

so x− u = qn − qm ∈ Q.

Solution of Problem 4.2Let B(X) be the Borel σ-algebra of X and let Y be the family ofsubsets of X which contains all open sets, all closed sets and whichis closed under countable intersections and countable disjoint unions.Evidently Y ⊆ B(X). Let

F ={A ∈ Y : Ac ∈ Y}.

Then F ⊆ Y ⊆ B(X) and F contains all open and all closed setsin X. We will show that F is a σ-algebra and then we will haveF = Y = B(X). We do this in steps.

4.3. Solutions 711

(a) If A,C ∈ F , then A \ C ∈ F .

By definition Ac, Cc ∈ F . Since Y is closed under finite inter-sections, we have that A \ C = A ∩ Cc ∈ Y. Also Y is closed undercountable disjoint unions. Hence

(A \ C)c = (A ∩Cc)c = Ac ∪C = Ac ∪ (A ∩ C) ∈ Y.Therefore A \ C ∈ F .

(b) F is closed under finite unions.

Let A,C ∈ F . Then A,C,Ac, Cc ∈ F and so(A ∪C)c = Ac ∩ Cc ∈ Y. From (a), we have that

A \ C ∈ F , C \A ∈ F and A ∩ C ∈ F .

Hence the disjoint union

(A \ C) ∪ (C \ A) ∪ (A ∩ C) = A ∪ C ∈ F ,

which proves that A ∪ C ∈ F and by induction, we prove that F isclosed under finite unions.

(c) F is an algebra of sets.

By part (b) and since X, ∅ ∈ F , it remains to show that F isclosed under intersections. Let A,C ∈ F . Then A ∩ C ∈ Y and(A ∩ C)c = Ac ∪ Cc ∈ F (see (b) and recall that Ac, Cc ∈ F). HenceA ∩ C ∈ F and by induction we show that F is closed under finiteintersections, hence F is an algebra.

(d) F is a σ-algebra of sets.

Let {An}n�1 ⊆ F . Inductively we define

C1 = A1, Cn = An \ (n−1⋃

k=1

Ak

) ∀ n > 1.

Then from (c), we have

n−1⋃

k=1

Ak ∈ F ∀ n > 1


and so from (a), it follows that

Cn = An \ (n−1⋃

k=1

Ak

) ∈ F .

Therefore {Cn}n�1 is a sequence of disjoint F-sets and so

⋃

n�1

An =⋃

n�1

Cn ∈ F .

On the other hand

( ⋃

n�1

An

)c=

⋂

n�1

Acn ∈ F

and so it follows that⋃

n�1An ∈ F and this proves that F is a σ-algebra.

Therefore we conclude that F = Y = B(X).

Solution of Problem 4.3From Problem 1.79, we know that every closed set is Gδ (i.e., countableintersection of open sets). Hence every family of subsets of X thatincludes the open sets and is closed under countable intersections mustcontain also closed sets. Then use Problem 4.2 to conclude that thefamily of sets coincides with the Borel σ-algebra of X.

Solution of Problem 4.4Recall that the finite intersections of the elements of X form a basisof the topology of X. So, σ(X ) contains a basis for the topology of X.Let K ⊆ X be a compact Gδ-set. Then K =

⋂

n�1Un, with Un being

open sets. Since X is locally compact (see Definition 2.92), for everyn � 1, we can find an open set Vn such that

4.3. Solutions 713

K ⊆ Vn ⊆ Un and Vn is a finite union of basic open sets.

So Vn ∈ σ(X ) and thus

K =⋂

n�1

Vn ∈ σ(X ).

Invoking Theorem 4.4, we conclude that Ba(X) ⊆ σ(X) ⊆ B(X).

Solution of Problem 4.5We will do the solution for the lower semicontinuous case, the othertwo cases can be treated similarly. For every λ ∈ R and every n � 1,we have

f |−1Cn

((λ,+∞)

)is open in Cn,

hence it has the form Cn ∩ Un with Un ⊆ X being an open set. Thismeans that

f |−1Cn

((λ,+∞)

) ∈ B(X).

In addition, since μ is complete, μ(C0) = 0 and f |−1C0

((λ,+∞)

) ⊆ C0,we infer that

f |−1C0

((λ,+∞)

) ∈ B(X)

and it is μ-null. Finally, we have

f−1((λ,+∞)

)=

⋃

n�0

f |−1Cn

((λ,+∞)

) ∈ B(X),

so f is B(X)-measurable.

Solution of Problem 4.6We have

U =⋃

n�1

Kn,

with Kn ⊆ U being compact for all n � 1. Since X is locally compact,by Proposition 2.94, for all n � 1, we can find an open set Vn ⊆ Xsuch that

V n is compact and Kn ⊆ Vn ⊆ V n ⊆ U.


Invoking Problem 2.103, for all n � 1, we can find a continuous func-tion gn : X −→ [0, 1] such that

gn|Kn= 1 and gn|V c

n= 0

(hence gn ∈ Cc(X) for all n � 1). Let

Kn ={x ∈ X : gn(x) � 1

2

} ∀ n � 1.

Evidently the sets Kn are compact and

Kn =⋂

m�3

{x ∈ X : gn(x) >

12 − 1

m

} ∀ n � 1,

hence they are Gδ-sets and Kn ⊆ Kn ⊆ U for all n � 1. So, U =⋃

n�1Kn, which expresses U as a countable union of compact Gδ-sets

and this by Theorem 4.4 implies that U ∈ Ba(X).

Solution of Problem 4.7Note that every compact set in X is also compact in X∗. Because inthis case the compact sets generate the Borel σ-algebra of X, we seethat

B(X) ⊆ X ∩ B(X∗).

To show that the opposite inclusion also holds, it suffices to show thatfor every open U ⊆ X∗, the set U\{∞} ∈ B(X). Since by hypothesisXis σ-compact, by Proposition 2.100, we can find an increasing sequence{Kn}n�1 of compact sets on X such that

X =⋃

n�1

Kn and Kn ⊆ intKn+1 ∀ n � 1.

Then the set U ∩ Kn+1 is open in X∗ and relatively compact in X,hence it belongs in B(X). Note that U \ {∞} =

⋃n�1(U ∩ intKn+1)

and so U \ {∞} ∈ B(X). This proves that B(X) = X ∩ B(X∗).

4.3. Solutions 715

Solution of Problem 4.8Let Y be the family of such sets G ∈ Σ⊗ B(X) for which there existsΣ0 ⊆ Σ, a countably generated sub-σ-algebra of Σ such that G ∈Σ0 ⊗B(X). Evidently Y = ∅ and it is closed under complementation.Also, if {Gn}n�1 ⊆ Y, then Gn ∈ Σ0n ⊗ B(X) with Σ0n being acountably generated sub-σ-algebra of Σ (see Definition 3.14). Hence

⋃

n�1

Gn ∈ σ( ⋃

n�1

(Σ0n ⊗ B(X)

)) ⊆ σ( ⋃

n�1

Σ0n

)⊗ B(X)

andΣ0 = σ

( ⋃

n�1

Σ0n

)

is a countable generated sub-σ-algebra of Σ. Therefore Y is a σ-algebracontaining all cylinders, therefore Y = Σ⊗ B(X).

Solution of Problem 4.9Let x ∈ X and U ∈ N (x). Then from the local compactness of X andProblem 2.103, we know that we can find f ∈ Cc(X) such that

f(x) = 1 and f(y) = 0 ∀ y ∈ U c.

Let us set

V ={f > 1

2

}.

This is an open Baire set and x ∈ V ⊆ U . Therefore the open Bairesets form a basis for the topology of X (see Definition 4.2).

Solution of Problem 4.10Let X = R

2 furnished with the topology which has as basis all setsof the form [a, b) × [c, d) with a, b, c, d ∈ R. Clearly X endowedwith this topology is a topological space (i.e., it is Hausdorff). AlsoB(X) = B(R2) where R

2 is the plane equipped with the usual Eu-clidean topology. Let C be the straight line on the plane passing fromthe origin and with slope −1, i.e.,

C ={(x, y) ∈ R

2 : x = −y}.


Then C ⊆ X is closed. For any x ∈ R, the set [x, x+1)× [−x,−x+1)is open and intersects C at precisely one point (x,−x) ∈ C. Henceevery point of C is open and so B(C) = 2C , hence B(C) is not a subsetof B(X).

Solution of Problem 4.11We know that λ is regular (see Theorem 4.11). So, without any lossof generality, we may assume that C is closed. Let A = (0, 1) \ C andlet us set

ϑ(t) = λ([0, t] ∩ C

) ∀ t ∈ [0, 1].

Then ϑ : [0, 1] −→ [0, λ(C)

]is a continuous surjection and is constant

in any connected component of A, hence ϑ(A) is at most countable.Since ϑ(C) contains

[0, λ(C)

] \ ϑ(A), it follows that C has cardinalityat least that of the continuum. Therefore the cardinality of C equalsthat of the continuum.

Solution of Problem 4.12Let

G ={(x, η) ∈ X × R+ : u(x) � η

}.

Then, we have

+∞∫

0

μ({

x ∈ X : u(x) � η})

dη =

+∞∫

0

μ({

x ∈ X : (x, η) ∈ G})

dη

=

∫

X

λ({

η � 0 : (x, η) ∈ G})

dμ(x)

=

∫

X

λ([0, u(x)

])dμ(x) =

∫

X

u dμ

(λ being the Lebesgue measure).

4.3. Solutions 717

Solution of Problem 4.13Suppose that λ(A) > 0. Then we can find an integer n such that

λ(A ∩ (n, n+ 1)

)> 0.

By Theorem 4.11, for a given ε ∈ (0, λ

(A ∩ (n, n + 1)

)), we can find

an open set U ⊆ (n, n+ 1) such that

A∩ (n, n+1) ⊆ U ⊆ (n, n+1) and λ(U) < λ(A∩ (n, n+1)

)+ ε.

We know that

U =⋃

k�1

(ak, bk),

where{(ak, bk)

}k�1

is a sequence of disjoint intervals. Then we have

A ∩ (n, n+ 1) =⋃

k�1

(A ∩ (ak, bk)

),

so, using the σ-additivity of λ and the assumption, we have

λ(A ∩ (n, n+ 1)

)=

∑

k�1

λ(A ∩ (ak, bk)

)�

∑

k�1

12(bk − ak)

= 12λ(U) < 1

2

[λ(A ∩ (n, n+ 1)

)+ ε

]

and thus

λ(A ∩ (n, n+ 1)

)< ε < λ

(A ∩ (n, n+ 1)

),

a contradiction.

Solution of Problem 4.14

(a) From Theorem 4.11, we know that μ and ν are regular (see Def-inition 4.9). So, using regularity of μ and ν, for every A ∈ B(X), wehave

μ(A) = inf{μ(U) : U ⊆ X is open, A ⊆ U

}

= inf{ν(U) : U ⊆ X is open, A ⊆ U

}

= ν(A)


(b) Since X is σ-compact (see Definition 2.99), every closed set is theunion of an increasing sequence of compact sets. Then exploiting theregularity of μ and ν, for every A ⊆ B(X), we have

μ(A) = sup{μ(K) : K ⊆ X is compact, K ⊆ A

}

= sup{ν(K) : K ⊆ X is compact, K ⊆ A

}

= ν(A).

Solution of Problem 4.15Since by hypothesis μ is σ-finite, we can find an increasing sequence{En}n�1 of Borel sets with

μ(En) < +∞ ∀ n � 1 and X =⋃

n�1

En.

Let μn : B(X) −→ R+ = [0,+∞) be the finite measure defined by

μn(A) = μ(A ∩ En) ∀ A ∈ B(X).

Theorem 4.11 implies that μn is regular (see Definition 4.9) and so wecan find a closed set Cn ⊆ A such that

μn(A \ Cn) < 1n .

Let us set

Cn =n⋃

k=1

Ck.

Evidently {Cn}n�1 is an increasing sequence of closed sets such that

μn(A \ Cn) � μn(A \ Cn) < 1n .

Moreover, for every m � n, we have μm � μn (recall that the sequence{En}n�1 is increasing) and so

μm(A \ Cn) < 1n ∀ m � n.

Henceμm

(A \

⋃

n�1

Cn

)= 0 ∀ m � 1

4.3. Solutions 719

and so finally

μ(A \

⋃

n�1

Cn

)= lim

m→+∞μm

(A \

⋃

n�1

Cn

)= 0.

If μ(A) < +∞, then μ( ⋃

n�1Cn

)< +∞ and so, since μ(Cn) ↗ μ(A) as

n → +∞, we have

μ(A \ Cn) = μ(A)− μ(Cn) −→ 0,

which proves the inner regularity of the set A.

Solution of Problem 4.16First assume that μ(A) < +∞. Due to the outer regularity of μ (seeDefinition 4.9), for a given ε > 0, we can find an open set U suchthat A ⊆ U and μ(U) < μ(A) + ε. Then by hypothesis there exists acompact setK ⊆ U such that μ(U) < μ(K)+ε. Note that μ(U\A) < εand pick an open set V such that U \ A ⊆ V and μ(V ) < ε (outerregularity of μ). Let C = K \ V . Then C is compact, C ⊆ A and

μ(C) = μ(K)− μ(K ∩ V ) > μ(U)− ε− μ(V ) � μ(A)− 2ε,

so

μ(A) = sup{μ(D) : D ⊆ A, D is compact

}.

Now assume that μ(A) = +∞. Then A =⋃

n�1An, with an increasing

sequence {An}n�1 ⊆ B(X) such that μ(An) < +∞ for all n � 1. Wehave μ(An) −→ +∞ and so for every k ∈ N, we can find n0 � 1such that μ(An0) > k. Then from the first part of the solution, wecan find a compact set K ⊆ An0 such that μ(K) > k. Therefore+∞ = μ(A) = sup

{μ(D) : D ⊆ A, D is compact

}.



“=⇒”: Let

s =

n∑

k=1

akχAkwith ak � 0, Ak ∈ B(X)

be a positive simple function. Then∫

X

s dν �∫

X

s dμ.

For a given u ∈ Cc(X), we can find a sequence of simple functions{sn}n�1 such that 0 � sn � u and

sn(x) ↗ u(x) ∀ x ∈ X.

Then by the Lebesgue monotone convergence theorem (see Theo-rem 3.92), we have

∫

X

u dν = limn→+∞

∫

X

sn dν � limn→+∞

∫

X

sn dμ =

∫

X

u dμ.

“⇐=”: Let K ⊆ X be a compact set. For a given ε > 0, we can findan open set U such that K ⊆ U and

μ(U) < μ(K) + ε

(due to the regularity of μ; see Definition 4.9). Since X is locallycompact (see Definition 2.92), invoking Proposition 2.94, we can findan open set V ⊆ X such that V is compact and K ⊆ V ⊆ V ⊆ U .Hence, applying Problem 2.103 for the pair {K,V }, we can find acontinuous function u : X −→ [0, 1] such that

u|K

= 1 and suppu ⊆ V.

Then u ∈ Cc(X) and χK� u � χ

V. Hence

ν(K) =

∫

X

χK dν �∫

X

u dν �∫

X

u dμ �∫

X

χV dμ

= μ(V ) � μ(U) < μ(K) + ε.

Since ε > 0 was arbitrary, we let ε ↘ 0, to conclude that

ν(K) � μ(K) ∀ K ⊆ X, K is compact.

4.3. Solutions 721

Since μ, ν are Radon measures, from the above inequality, we concludethat ν � μ.


(a) Let K ⊆ X be a compact set. Then because μ is σ-bounded, we

can find an open cover {Vn}n�1 of K such that μ(Vn) < +∞ for alln � 1. Due to the compactness of K, we can find N � 1 such that

K ⊆N⋃

n=1

Vn.

Hence

μ(K) � μ( N⋃

n=1

Vn

)�

N∑

n=1

μ(Vn) < +∞.

(b) By Problem 4.15, for a given ε > 0, we can find a closed set C ⊆ A,with μ(A \ C) < ε

2 . Because X is σ-compact (see Definition 2.99), wecan find an increasing sequence of compact sets {Kn}n�1 such that

X =⋃

n�1

Kn.

Then

C =⋃

n�1

(C ∩Kn),

with C ∩Kn being compact and μ(C ∩Kn) ↗ μ(C) as n → +∞. So,we can find n0 � 1 such that

μ(C \ (C ∩Kn0)

)< ε

2 .

Let Kε = C ∩Kn0 ⊆ X be a compact set. Then

μ(A \Kε) = μ(A \ C) + μ(C \Kε) < ε2 + ε

2 = ε.

Thus, A is inner regular with respect to compact sets.



(a) Since μ is σ-finite, we can write

A =⋃

n�1

An,

with μ(An) < +∞ for all n � 1. Due to the outer regularity of μ (seeDefinition 4.9), we can find open sets Un such that

An ⊆ Un and μ(Un) < μ(An) +ε

2n+1 .

LetU =

⋃

n�1

Un.

Then the set U is open, A ⊆ U and

μ(U \ A) �∑

n�1

μ(Un \ An) < ε2 .

Similarly, working this time with Ac ∈ B(X), we produce an open setV such that

Ac ⊆ V and μ(V \ Ac) < ε2 .

Let C = V c. Then C is closed, C ⊆ A ⊆ U and

μ(U \C) = μ(U \A)+μ(A\C) = μ(U \A)+μ(V \Ac) < ε2+

ε2 = ε.

(b) From part (a), for every n � 1, we can find a closed set Cn andan open set Un such that

Cn ⊆ A ⊆ Un and μ(Un \ Cn) < 1n ∀ n � 1.

ThenE =

⋃

n�1

Cn is an Fδ-set

andD =

⋂

n�1

Un is a Gσ-set

andE ⊆ A ⊆ D and μ(D \ E) < 1

n ∀ n � 1,

4.3. Solutions 723

hence μ(D \E) = 0.


(a) “=⇒”: Suppose that μ is an outer regular measure. Then for all

A ∈ B(X), we have

μ(A) = inf{μ(U) : A ⊆ U, U is open

}.

Since μ is finite, we have

μ(X)− μ(A) = μ(Ac) = inf{μ(U) : Ac ⊆ U, U is open

}

= μ(X)− sup{μ(C) : C ⊆ A, C is closed

}.

So,μ(A) = sup

{μ(C) : C ⊆ A, C is closed

}

and thus μ is inner regular.

“⇐=”: The proof of this implication is similar.

(b) Let A ∈ B(X). Since every compact set is closed, we have

μ(A) = sup{μ(K) : K ⊆ A, K is compact

}

� sup{μ(C) : C ⊆ A, C is closed

}

� μ(A),

so μ is inner regular. Hence by part (a), μ is outer regular too, henceregular.

Solution of Problem 4.21Let x ∈ X be such that u(x) > 0. Then we can find U ∈ N (x) withU being compact (since X is locally compact) such that

u(z) � ξ > 0 ∀ z ∈ U.

Let h ∈ Cc(X), h � 0 be such that supph ⊆ U . Then

h(y) � ‖h‖∞ξ u(y) ∀ y ∈ X,


so ∫

X

hdμ � ‖h‖∞ξ

∫

X

u dμ = 0

and thus ∫

X

hdμ = 0.

This means that x ∈ suppμ (see Definition 4.13) and so u|supp μ = 0.

Solution of Problem 4.22Suppose that u ∈ Cc(X), K = suppu, E = suppμ and u|

E= 0 (see

Definition 4.13). For a given ε > 0, let

Uε ={x ∈ X :

∣∣u(x)

∣∣ < ε

}.

Then Uε ⊆ X is an open set and E ⊆ Uε. Then Ec is an open setcontaining the compact set U c

ε . Invoking Problem 2.103, we can finda continuous function g : X −→ [0, 1] such that

g|Ucε

= 1 and supp g ⊆ Ec.

Then supp (ug) ∩ E = ∅ and so by hypothesis∫

X

ug dμ = 0.

Also,u = ug on K ∩ U c

ε and |ug| � |u| in X.

Therefore, from the definition of Uε, we have∣∣u(x)− (ug)(x)

∣∣ � 2ε ∀ x ∈ X.

Since supp (u− ug) ⊆ K, we have

∣∣∫

X

(u− ug) dμ∣∣ � 2Mε,

for some M = M(K) > 0 and so

∣∣∫

X

u dμ∣∣ � 2Mε.

4.3. Solutions 725

Let ε ↘ 0, to conclude that

∫

X

u dμ = 0.

Solution of Problem 4.23By Proposition 3.20, it suffices to show that, if {An}n�1 ⊆ B(X) is adecreasing sequence such that

⋂

n�1

An = ∅,

then μ(An) ↘ 0. We argue indirectly. So, suppose that

0 < β = limn→+∞μ(An).

Let ε ∈ (0, 12β

). Then for each n � 1, we can find a compact set

Kn ⊆ An such that

μ(An \Kn) < ε2n .

We claim that the sequence {Kn}n�1 has the finite intersection prop-erty. Indeed, note that

μ(Am \

m⋂

n=1

Kn

)<

m∑

n=1

ε2n = ε,

so, from the choice of ε > 0, we have

μ( m⋂

n=1

Kn

)> 0,


thusm⋂

n=1

Kn = ∅

and finally, from Theorem 2.81, we have

⋂

n�1

Kn = ∅.

But then⋂

n�1

An = ∅

(as Kn ⊆ An for all n � 1), a contradiction.


“=⇒”: This is immediate from Definition 4.9(a).

“⇐=”: From Theorem 4.11, we know that μ is regular (see Defini-tion 4.9). Therefore to show that μ is Radon, it suffices to show that,for every closed set C ⊆ X, we have

μ(C) = sup{μ(K) : K ⊆ C, K is compact

}.

Arguing by contradiction, suppose that C ⊆ X is a closed set suchthat there exists ε ∈ (

0, μ(X))for which, we have

sup{μ(K) : K ⊆ C, K is compact

}� μ(C)− ε.

If D ⊆ X is compact, then D ∩ C ⊆ C is also compact and so

μ(D ∩C) � μ(C)− ε.

Then

μ(D) = μ(D∩C)+μ(D∩Cc) � μ(C)−ε+μ(Cc) = μ(X)−ε.

4.3. Solutions 727

Since D ⊆ X was arbitrary compact set, we have

sup{μ(D) : D ⊆ X, D is compact

}� μ(X) − ε,

a contradiction to our hypothesis.

Solution of Problem 4.25First assume that X is second countable (see Definition 2.24). Let

D =⋃

U is openμ(U) = 0

U and C = Dc.

From Proposition 2.164(a), we know that X is strongly Lindeloff.Hence because D is open, we can find a sequence of open sets {Un}n�1

with

μ(Un) = 0 ∀ n � 1

such that

D =⋃

n�1

Un.

It follows that μ(D) = 0. Let V ⊆ X be an open set such thatV ∩ C = ∅. We claim that μ(V ∩ C) > 0. Indeed, if μ(V ∩ C) = 0,then

μ(V ) = μ(V ∩ C) + μ(V ∩D) = 0

and so V ⊆ D, a contradiction. Therefore according to Definition 4.13,we have that C = suppμ.

Next assume that μ is outer regular and also inner regular withrespect to compact sets (see Definition 4.9). Let D be the open setintroduced in the previous part of the solution. Let K ⊆ D be acompact set. Then we can find open sets V1, . . . , Vn, with μ(Vk) = 0for all k ∈ {1, . . . , n} such that

K ⊆n⋃

k=1

Vk.


Hence μ(K) = 0. Exploiting our hypothesis on μ, we have

μ(D) = sup{μ(K) : K ⊆ D, K is compact

}= 0

and so, if C = Dc, then as above, we conclude that C = suppμ.

Solution of Problem 4.26No. Consider X = {0, 1} and Σ = 2X . Let μ = δ0 and ν = δ1. Thenϑ = max{μ, ν} satisfies

ϑ(A) =

{1 if A ∈ Σ, A = ∅,0 if A = ∅

and in fact this is not a measure. In his case the smallest measure notless than μ or ν is μ+ ν and

(μ+ ν)(X) = 2.

Solution of Problem 4.27We know that σ is a measure on B(X) and in view of the fact thatσ � μ, we have that σ is σ-finite. Also σ � μ (see Definition 3.150).

Let A ∈ B(X) be such that μ(A) < +∞. We consider the Borelσ-algebra B(A) of A. We know that

B(A) = A ∩ B(X).

The Radon–Nikodym theorem (see Theorem 3.152) implies that thereexists a unique function u ∈ L1(A,B(A), μ), u � 0 such that

σ(A ∩C) =

∫

A∩Cu dμ ∀ C ∈ B(X).

Because μ is a Radon measure, for a given ε > 0, we can find a compactset K ⊆ A such that

0 � σ(A)− σ(K) =

∫

A

u dμ−∫

K

u dμ =

∫

A\Ku dμ � ε,

4.3. Solutions 729

soσ(A) = sup

{σ(K) : K ⊆ A, K is compact

}.

Since σ is clearly finite on compact sets, to show that it is Radon, itremains to show that it is outer regular (see Definition 4.9). So, letD ∈ B(X). We know that

σ(D) = inf{μ(E) +m(D \ E) : E ∈ B(X), E ⊆ D

}.

Letη = inf

{σ(U) : U ⊇ D, U is open

}

and ε > 0. For a given E ∈ B(X), E ⊆ D, we choose open sets U andV such that

E ⊆ U, μ(U) � μ(E)+ε, D\E ⊆ V and m(V ) � m(D\E)+ε.

Then, we have

σ(D) � η � σ(U ∪ V ) � σ(U) + σ(V ) � μ(U) +m(V )

� μ(E) +m(D \E) + 2ε,

so σ(D) � η � σ(D) + 2ε. Let ε ↘ 0, to conclude that σ(D) = η andso σ is Radon.

Solution of Problem 4.28Let A ∈ B(RN ). Let

Bn(0) ={x ∈ R

N : ‖x‖ � n}.

Then Bn(0) is a compact metric space and so by Theorem 4.12,λN |

Bn(0)is Radon. Therefore

λN(A ∩Bn(0)

)= sup

{λN (K) : K ⊆ A ∩Bn(0), K is compact

}.

Since A ∩Bn(0) ↗ A, we infer that λN(A ∩Bn(0)

) ↗ λN (A) and so

λN (A) = sup{λN (K) : K ⊆ A, K is compact

}.

If λN (A) = +∞, then obviously, we have

λN (A) = inf{λ(U) : A ⊆ U, U is open

}.


Hence, we assume that λN (A) < +∞. For a given ε > 0 and sinceλN |

Bn(0)is Radon and

Bn(0) ={x ∈ R

N : ‖x‖ < n}

is open, we have

λN(A ∩Bn(0)

)= inf{λN

(U ∩Bn(0)

): A ∩Bn(0) ⊆ U ∩Bn(0),

U is open}= inf{λN

(U ∩Bn(0)

): A ∩Bn(0) ⊆ U ∩Bn(0),

U is open}= inf

{λN (U) : A ∩Bn(0) ⊆ U, U is open

}.

So, for every n � 1, we can find open set Un ⊆ RN such that

λN(Un \ (A ∩Bn(0)

))� ε

2n .

Let

U =⋃

n�1

Un ⊆ RN .

Then U is open, A ⊆ U and

U \A =⋃

n�1

Un \⋃

n�1

(A ∩Bn(0)

) ⊆⋃

n�1

(Un \ (A ∩Bn(0)

)),

so

0 � λN (U)−λN (A) = λN (U \A) �∑

n�1

λN(Un\

(A∩Bn(0)

))� ε,

thus λN (A) = inf{λN (U) : A ⊆ U, U is open

}and so λN is Radon.

Solution of Problem 4.29Recall that simple functions are dense in Lp(X,μ) (see Proposi-tion 3.110). So, it suffices to show that for any A ∈ B(X) withμ(A) < +∞, the characteristic function χ

Acan be approximated in

the Lp-norm by functions in Cc(X). Because μ is Radon, for a given

4.3. Solutions 731

ε > 0, we can find a compact set K ⊆ A and an open set U ⊇ A suchthat

μ(U \K) � ε

(see Problem 4.19). As before, because X is locally compact (seeDefinition 2.92), we can find a function h ∈ Cc(X) such that

h|K

= 1, 0 � h � 1 and supph ⊆ U

(see Problem 2.103). Then

∥∥χ

A− h

∥∥pp

=

∫

X

∣∣χ

A− h

∣∣p dμ �

∫

X

χpU\K dμ = μ(U \K) � ε,

so Cc(X) is dense in Lp(X,μ) for all 1 � p < +∞.

Solution of Problem 4.30Let a, b ∈ R with a < b and let

f(x) = χ(a,b)

(x) ∀ x ∈ R.

Suppose that f can be approximated in the L∞-norm by continuousfunctions. Let ε ∈ (

0, 12)and let g ∈ C(R) be such that

‖f − g‖∞ < ε.

Then ∣∣χ

(a,b)(x)− g(x)

∣∣ < ε almost everywhere on R.

For every δ > 0, there are points y ∈ (a, a+ δ) and z ∈ (a− δ, a) suchthat ∣

∣1− g(y)∣∣ < ε and

∣∣g(z)

∣∣ < ε.

Hence

lim supy→a+

g(y) � 1− ε and lim infy→a−

g(y) � ε,

a contradiction to the continuity of g.


Solution of Problem 4.31Clearly μ0 is a measure. Also, since

R =( ⋃

q∈Q{q}) ∪ (R \Q),

we see that μ0 is σ-finite. For every nonempty open set U ⊆ R,card (U ∩ Q) = +∞ and for every x ∈ R, we have μ0({x}) < +∞.However

inf{μ0(U) : U is open and x ∈ U

}= +∞

and so μ0 is not regular.

Solution of Problem 4.32Suppose that μ is not a measure. Then by Theorem 3.19 and Propo-sition 3.20, we can find a decreasing sequence {An}n�1 ⊆ B(X) suchthat

⋂

n�1

An = ∅ and limn→+∞μ(An) = inf

n�1μ(An) > 0.

By hypothesis, for every n � 1, we can find a compact set Kn ⊆ An

such that

μ(An) � μ(Kn) +1

2n+1 infm�1

μ(Am).

So, we have

μ(An \

n⋂

m=1

Kn

)�

n∑

m=1

μ(Am \Km) � 12 infn�1

μ(Am),

thus

μ( n⋂

m=1

Kn

) = 0, hencen⋂

m=1

Km = ∅.

Let

Kn =

n⋂

m=1

Kn.

4.3. Solutions 733

Note that {Kn}n�1 is a decreasing sequence of nonempty, compact setscontained in the compact set K1. Therefore

⋂

m�1

Km = ∅

(see Theorem 2.81), which contradicts the fact that

⋂

n�1

An = ∅.

Solution of Problem 4.33No. Consider the function

f = χ[0,1]

∈ L1(R).

Suppose that g ∈ L1(R) satisfies

‖f − g‖1 < 12 .

We have

1−∫

R

g dx =

∫

R

(f − g) dx �∫

R

|f − g| dx < 12 ,

so ∫

R

g dx > 12

and so g ∈ V .

Solution of Problem 4.34Evidently, it suffices to show that closed sets are μ-measurable. So, letC ⊆ X be a closed set and let

E0 ={x ∈ X : dist(x,C) � 1

},

En ={x ∈ X : 1

2n � dist(x,C) < 12n−1

} ∀ n � 1.


Due to the subadditivity of μ, it suffices to show that

μ(D) � μ(D ∩ C) + μ(D ∩ Cc) ∀ D ⊆ X.

Clearly we may assume that μ(D) < +∞. Note that, if |n −m| � 2,then En and Em are metrically separated and so, we have

μ(D) � μ(D ∩ ( n⋃

m=0

E2m

))=

n∑

m=0

μ(D ∩ E2m) ∀ n � 1.

The same inequality is also true with 2m replaced by 2m + 1. So, itfollows that ∑

m�0

μ(D ∩Em) < +∞.

Note that the sets C andn⋃

m=0Em (n � 0) are metrically separated.

Hence

μ(D) � μ((D ∩ C) ∪ (

D ∩ ( n⋃

m=0

En

)))

= μ(D ∩ C) + μ(D ∩ ( n⋃

m=0

En

))

� μ(D ∩ C) + μ(D ∩Cc)−∑

k>n

μ(D ∩Ek)

(from the subadditivity of μ). Passing to the limit as n → +∞, weconclude that

μ(D) � μ(D ∩ C) + μ(D ∩ Cc),

i.e., C is μ-measurable.


(a) Let K ⊆ X be a compact set. By the regularity of μ and the localcompactness of the space X, for a given ε > 0, we can find an openset V ⊆ X such that

K ⊆ V, V is compact and μ(V ) � μ(K) + ε.

4.3. Solutions 735

By Problem 2.103, we can find f ∈ Cc(X) such that

0 � f � 1, f |K

= 1 and f |V c = 0.

Then

μ(K) � μ(V )− ε �∫

X

f dμ− ε = limn→+∞

∫

X

f dμn − ε

� lim supn→+∞

μn(K)− ε.

Since ε > 0 was arbitrary, we let ε ↘ 0 to conclude that

lim supn→+∞

μn(K) � μ(K).

(b) The proof of this part is similar to that of part (a) using this timethe approximation of μ(u) by the μ-measure of compact subsets of U(see Definition 4.9).


g(r) = μ(Br(x)

)and h(r) = ν

(Br(x)

) ∀ x ∈ X, r > 0.

Also let U ⊆ X be a nonempty open set. Then

limr↘0

ν(U∩Br(x))h(r) = 1 ∀ x ∈ U

and using the Fatou lemma (see Theorem 3.95) and the Fubini theorem(see Theorem 3.115), we have

μ(U) =

∫

U

limr↘0

ν(U∩Br(x))h(r) dμ � lim inf

r↘0

1h(r)

∫

U

ν(U ∩Br(x)

)dμ

= lim infr↘0

1h(r)

∫

U

μ(Br(x)

)dν =

(lim infr↘0

g(r)h(r)

)ν(U).


If in the above argument we interchange μ and ν, we obtain

ν(U) �(lim infr↘0

h(r)g(r)

)μ(U).

It follows that the limit c = limr↘

g(r)h(r) exists and so μ(U) = cν(U) for

all open sets U ⊆ X. Then Theorem 4.11 implies that μ = cν.

Solution of Problem 4.37For a given ε > 0, we can find δ = δ(ε) > 0 such that

∫

D

hdx < ε ∀ D ∈ B([0, 1]), λ(D) < δ

(λ being the Lebesgue measure on R). Theorem 4.12 and Problem 4.19imply that we can find a compact set K ⊆ A and an open set U withA ⊆ U such that λ(U \K) < δ. Also, we can find a continuous functiong : [0, 1] −→ R such that

0 � g � 1, g|K

= 1, g|Uc = 0.

Then, using also hypotheses (a) and (b) and recalling that λ(U \K) <δ, we have

lim supn→+∞

∣∣∫

A

fn dx∣∣ = lim sup

n→+∞

∣∣

1∫

0

fnχAdz

∣∣

� lim supn→+∞

∣∣

1∫

0

fng dz∣∣+ lim sup

n→+∞

∣∣

1∫

0

fn(χA − g) dx∣∣

� lim supn→+∞

∣∣1∫

0

fng dx∣∣+

1∫

0

fχU\K dx � ε.

4.3. Solutions 737

Since ε > 0 was arbitrary, we let ε ↘ 0, to conclude that

limn→+∞

∫

A

fn dx = 0.

Solution of Problem 4.38Let K ⊆ X be a compact set. The tightness of μ and ν (see Defini-tion 4.9) implies that we can find two sequences of relatively compactopen sets {Un}n�1 and {Vn}n�1 such that

μ(Un), ν(Vn) < +∞, K ⊆ Un, K ⊆ Vn ∀ n � 1

andμ(K) = lim

n→+∞μ(Un) and ν(K) = limn→+∞ ν(Vn).

Let us set

Wn =

n⋂

k=1

(Uk ∩ Vk) ∀ n � 1.

Then {Wn}n�1 is a decreasing sequence of open sets in X with

μ(K) = limn→+∞μ(Wn), ν(K) = lim

n→+∞ ν(Wn).

By Problem 2.103, we can find fn ∈ Cc(X) such that

fn|K = 1, fn|Wcn

= 0 and 0 � fn � 1 ∀ n � 1.

Let hn = min{f1, . . . , fn

}. We have that {hn}n�1 is a decreasing

sequence in Cc(X) and hn ↘ χK. Then using the Lebesgue monotone

convergence theorem (see Theorem 3.92) and the hypothesis of theproblem, we have

μ(K)=

∫

X

χKdμ= lim

n→+∞

∫

X

hn dμ= limn→+∞

∫

X

hn dν=

∫

X

χKdν=ν(K).


So, we have proved that for every compact set K ⊆ X, we haveμ(K) = ν(K). Since μ and ν are tight, we conclude that μ = ν(see Definition 4.9(e)).


“(a) =⇒ (b)”: This is immediate since ϕ is Borel measurable.

“(b) =⇒ (a)”: Let U be an open subset of Y . Let hU ∈ C(Y ) bedefined by

hU (y) = dist(y, U c) ∀ y ∈ Y,

if U = Y and

hU (y) = 1 ∀ y ∈ Y,

if U = Y . We have

U ={y ∈ Y : hU (y) > 0

}.

Note that f−1(U) ={x ∈ X : (hU ◦ f)(x) > 0

} ∈ Σ. Since U is anarbitrary open subset of Y and open sets generate the Borel σ-algebraof Y , we conclude that f is Σ-measurable.


“(a) =⇒ (b)”: Obvious (since the composition of measurable mapsis measurable).

“(b) =⇒ (a)”: Let X ={A ∈ Ba(X) : f−1(A) ∈ Σ

}. Evidently X

is a σ-algebra and since ϕ ◦ f is Σ-measurable, it contains all sets ofthe form ϕ−1

([λ,+∞)

). Therefore, by Remark 4.3, X = Ba(X) and

so f is(Σ,Ba(X)

)-measurable.

4.3. Solutions 739

Solution of Problem 4.41First we show the following Claim.

Claim. limx→0

λ((A+ x) ∩A

)= λ(A).

If

A =

n⋃

k=1

(ak, bk),

where{(ak, bk)

}n

k=1are pairwise disjoint open bounded intervals, then

λ(A) =n∑

k=1

(bk − ak)= limx→0

n∑

k=1

(min {bk, bk + x}−max {ak, ak+x} )

= limx→0

n∑

k=1

λ((ak + x, bk + x) ∩ (ak, bk)

)

= limx→0

λ( n⋃

k=1

(ak+x, bk+x) ∩ (ak, bk))� lim inf

x→0λ((A+ x) ∩A

)

� lim supx→0

λ((A+ x) ∩A

)� λ(A),

so

limx→0

λ((A+ x) ∩A

)= λ(A).

Now suppose that the set A is compact. For a given ε > 0, we canfind an open set U ⊆ R such that

A ⊆ U, λ(U \ A) < ε and U =n⋃

k=1

(ak, bk),

with{(ak, bk)

}n

k=1mutually disjoint. Let V = U \ A. This is open

too. Then

λ(A) � λ(U) = limx→0

λ((U + x) ∩ U

)

= limx→0

λ(((A+x) ∩A

)∪((A+x)∩V )∪((V+x)∩A)∪((V+x) ∩ V))

� lim infx→0

λ((A+ x) ∩A

)+ 3λ(V ) � lim inf

x→0λ((A+ x) ∩A

)+ 3ε

� lim supx→0

λ((A+ x) ∩A

)+ 3ε � λ(A) + 3ε.


Since ε > 0 is arbitrary, we let ε ↘ 0, to conclude that

λ(A) = limx→0

λ((A+ x) ∩A

),

where A is compact.Finally for the general case, we can find an increasing sequence

{An}n�1 such that

An is compact, An ⊆ A for all n � 1, limn→+∞λ(An) = λ(A)

(since λ is a Radon measure on R; see Problem 4.28). Then

λ(A) = limn→+∞λ(An) = lim

n→+∞ limx→0

λ((An + x) ∩An

)

� lim infx→0

λ((A+ x) ∩A

)� lim sup

x→0λ((A+ x) ∩A

)� λ(A),

soλ(A) = lim

x→0λ((A+ x) ∩A

).

This proves the Claim.

Using this Claim, we can prove the two statements of theproblem.(a) For all y, x ∈ R, we have

∣∣f(y)− f(x)

∣∣ =

∣∣λ((A+ y) ∩A

)− λ((A+ x) ∩A

)∣∣.

Note that

λ((A+y)∩A

)= λ

(((A+y)\(A+x)

)∩A)+λ

((A+y)∩ (A+x)∩A

)

and

λ((A+x)∩A

)= λ

(((A+x)\(A+y)

)∩A)+λ

((A+y)∩(A+x)∩A

).

So, subtracting the above two quantities, we obtain∣∣f(y)− f(x)

∣∣ =∣∣λ(((A+ y) \ (A+ x)

) ∩A)

−λ(((A+ x) \ (A+ y)

) ∩A)∣∣

� λ((A+ y) \ (A+ x)

)+ λ

((A+ x) \ (A+ y)

)

= λ((A+ (y − x)

) \ A)+ λ((A+ (x− y)

) \ A)

= λ(A)− λ((A+ (y − x)

) ∩A)+ λ(A)− λ

((A+ (x− y)

) ∩A).

Using the Claim, we get that∣∣f(y)− f(x)

∣∣ −→ 0 as y → x.

4.3. Solutions 741

(b) If A is a compact set, then we can find R0 > 0 such that for anyx � R0, we have (A+ x) ∩A = ∅ and so

f(x) = 0 ∀ x � R0.

For the general case, as before exploiting the fact that λ is a Radonmeasure, we can find an increasing sequence {An}n�1 such that

An is compact, An ⊆ A for all n � 1, limn→+∞λ(An) = λ(A)

Then we have

limx→+∞λ

((A+ x) ∩A

)

= limx→+∞

[λ((A+ x) ∩A

)−λ((An + x) ∩An

)+λ

((An + x) ∩An

)]

= limn→+∞ lim

x→+∞[λ(((A+x)∩A)\((An+x)∩An

))+λ

((An+x) ∩An

)]

� limn→+∞

[λ((A+ x) \ (An + x)

)+ λ(A \ An)

]= 0.

Solution of Problem 4.42Let B ⊆ R be a bounded open set of the form

B =

m⋃

k=1

(ak, bk)

for some m � 1 and the intervals (ak, bk) being pairwise disjoint. Then

m⋃

k=1

(( ak, bk) ∩ (ak, bk)

) ⊆ B ∩B ∀ > 0.

If λ denotes the Lebesgue measure on R, then

λ(B) =m∑

k=1

(bk − ak)

= lim�→1

m∑

k=1

λ(( ak, bk) ∩ (ak, bk)

)

� lim inf�→1

λ( B ∩B)

� λ(B).


Let K ⊆ R be a compact set. Then we can find a decreasing sequence{Bn}n�1 of bounded open sets of the above form such that

K =⋂

n�1

Bn.

Then from the subadditivity of λ, we have

λ( Bn ∩Bn) � λ( (Bn \K)

)+ λ(Bn \K) + λ( K ∩K)

and so

λ(K) = limn→+∞λ(Bn) = lim

n→+∞ lim�→1

λ( Bn ∩Bn)

� lim infn→+∞ lim inf

�→1

(λ( (Bn \K)

)+ λ(Bn \K) + λ( K ∩K)

)

= lim infn→+∞

(lim inf�→1

λ( K ∩K) + 2λ(Bn \K))

= lim inf�→1

λ( K ∩K) � lim sup�→1

λ( K ∩K) � λ(K).

Since λ is Radon (see Problem 4.28), we can find an increasing sequence{Kn}n�1 of compact sets such that

⋃

n�1

Kn ⊆ A and λ(Kn) ↗ λ(A).

So, we have

λ(A) = limn→+∞λ(Kn) = lim

n→+∞ lim�→1

λ( Kn ∩Kn)

� lim inf�→1

λ( A ∩A) � lim sup�→1

λ( A ∩A) � λ(A).

We have

∣∣ϕ( ) − ϕ(1)

∣∣ =

∣∣+∞∫

−∞

∣∣χ 1

�A(t)χ

A(t)− χ

A(t)

∣∣ dt

∣∣

= λ(A) − λ(1nA ∩A

),

4.3. Solutions 743

soϕ( ) −→ ϕ(1) as → 1.

This proves the continuity of ϕ at = 1.

Solution of Problem 4.43Let N = 1 and let λ be the Lebesgue measure on R. For each n � 1and each A ∈ B(R), let

λn(A) = λ(A ∩ (n,+∞)

).

Evidently {λn}n�1 is a decreasing family of Borel measures. Let

μ(A) = limn→+∞λn(A) ∀ A ∈ B(R).

The set function μ is not a measure, since

μ([0,+∞)

)= lim

n→+∞λn

([0,+∞)

)= +∞

and it is not equal to

∑

k�1

μ([k − 1, k)

)=

∑

k�1

limn→+∞λn

([k − 1, k)

)= 0.

hence μ is not countably additive, thus μ is not measure.

Solution of Problem 4.44Note that X is a Polish space and so Theorem 4.12 implies that μ is aRadon measure. Since by hypothesis μ

({x}) = 0 so, for each x ∈ X,we can find δ(x) > 0 such that

μ(Bδ(x)(x)

)< ε.

Due to the compactness of X, we can find a finite set {xk}nk=1 ⊆ Xsuch that

X =n⋃

k=1

Bδ(xk)(xk).


Let us set δ = 12 lim1�k�n

δ(xk) > 0. Let A ∈ B(X) be such that

diamA < δ. We can find k ∈ {1, . . . , n} such that

A ∩B 12δ(xk)

= ∅,

so A ⊆ Bδ(xk)(xk) and thus μ(A) < ε.

Solution of Problem 4.45Let {Ak}k�1 ⊆ B(X) be a sequence of disjoint Borel sets and let usset

A =⋃

k�1

Ak ∈ B(X).

Then for a given ε > 0, we can find a compact set Kε ⊆ A such that

μ(A) � μ(Kε) + ε.

Also, for every k � 1, we can find an open set Uk ⊆ X such that

Ak ⊆ Uk and μ(Uk) � μ(Ak) +ε2k

∀ k � 1.

For every integer n � 1, we have

n∑

k=1

μ(Ak) = μ( n⋃

k=1

Ak

)� μ(A)

(due to the additivity of μ), so passing to the limit as n → +∞, wehave ∑

k�1

μ(Ak) � μ(A).

Since Kε is compact and Kε ⊆ A, we can find n1 � 1 such that

Kε ⊆n1⋃

k=1

Uk

and so

μ(A) � μ(Kε)+ε �n1∑

k=1

μ(Uk)+ε �∑

k�1

μ(Uk)+ε �∑

k�1

μ(Ak)+2ε.

4.3. Solutions 745

Because ε > 0 is arbitrary, we let ε ↘ 0, to obtain

μ(A) �∑

k�1

μ(Ak),

so finally

μ(A) =∑

k�1

μ(Ak),

i.e., μ is σ-additive.

Solution of Problem 4.46If x ∈ X and r1 = r2, then

∂Br1(x) ∩ ∂Br2(x) = ∅.

Therefore at most countably many of the sets{∂Br(x)

}r>0

have posi-

tive measure. So, for every x ∈ X, we can find a sequence{rn(x)

}n�1

such that

0 < rn(x) < 1n and μ

(∂Brn(x)(x)

)= 0 ∀ n � 1.

Since X is compact, for every n � 1, we can find a finite set{xk}Nn

k=1 ⊆ X such that

X =

Nn⋃

k=1

Brn(xk)(xk).

If U ⊆ X is an open set and x ∈ U , then we can find r > 0 such thatBr(x) ⊆ U . Let n � 1 be such that r > 2

n . We have x ∈ Brn(xk)(xk)for some k ∈ {1, . . . , Nn}. If y ∈ Brn(xk)(xk), then

d(x, y) � diamBrn(xk)(xk) < 2n < r

and so we have

x ∈ Brn(xk)(xk) ⊆ Br(x) ⊆ U,


thusB =

{Brn(xk)(xk)

}1�k�Nn,n�1

= {Un}n�1

is a basis for the topology and μ(∂Un) = 0 for all n � 1.

Solution of Problem 4.47Let x, y ∈ X, R � r > 0 and x ∈ BR(y). For every integer k � 0, weset

Rk = 2kr and m = min{k � 0 : BRk

(x) ⊇ BR(y)}.

Then from the doubling property of μ, after m-iterations, we have

μ(BR(y)

)� μ

(BRk

(x))

� cmμ(Br(x)

),

soμ(Br(x))μ(BR(y)) � 1

cm .

But note that BR(y) ⊆ BRm−1(x), hence Rm−1 � 2R (recall thatx ∈ BR(y)) and m � log2

(4Rr

). So, setting c = 1

c2and t = − log2 c,

we conclude thatμ(Br(x))μ(BR(y)) � c

(rR

)t.

Solution of Problem 4.48Let μA : B(X) −→ [0,+∞) be the finite Borel measure, defined by

μA(C) = μ(A ∩ C) ∀ C ∈ B(X).

Invoking Theorem 4.11, we have that μA is regular and so we can finda closed set C ⊆ A such that

μA(A) � μA(C) + ε.

Thenμ(A) � μ(A ∩ C) + ε

4.3. Solutions 747

and so

μ(A)− μ(A ∩ C) = μ(A \ C) � ε.

Solution of Problem 4.49Since λN is a Radon measure (see Problem 4.28), we can find a compactset K ⊆ A such that λN (K) > 0. According to the Cantor–Bendixsontheorem (see Problem 1.16 and Remark 2.28), we have that K = P∪Nwith P ⊆ R

N being a perfect set (see Definition 1.13) and N ⊆ RN

being countable set. Moreover, P = ∅ or otherwise K = N andso λ(K) = 0, a contradiction. Now recall that for a perfect set P ,cardP = c (see Problem 1.31). Hence

c � cardA � cardP = c

so cardA = c.

Solution of Problem 4.50By the Egorov theorem (see Theorem 3.76), for a given ε ∈ (0, 1− ϑ),we can find a Lebesgue measurable set A ⊆ [0, 1] such that

λ(A) > 1− ε and fn ⇒ f on A.

Hence the function f |Ais continuous and because λ is regular, we can

find a compact set K ⊆ A such that

λ(K) > 1− ε > ϑ.

Solution of Problem 4.51Let ε > 0 and let {qn}n�1 be an enumeration of the rational numbersin [0, 1]. Let us set

K = [0, 1] \⋃

n�1

(qn − ε

2·2n , qn + ε2·2n

).


ThenK is compact and intK = ∅. Moreover, if λ denotes the Lebesguemeasure on R, then

λ(K) � 1−∑

n�1

ε2·2n � 1− ε

since λ([0, 1] \K)

� 1− (1− ε) = ε.


hn =

{f(t) if |f(t)| � n,n · sgn f(t) if |f(t)| > n,

where

sgn f(t) =

{1 if f(t) > 0,−1 if f(t) < 0

∀ n � 1.

Evidently

hn(t) −→ f(t) ∀ t ∈ [0, 1]

and so hnλ−→ f (see Proposition 3.130). According to Proposi-

tion 3.110, we can find simple functions converging uniformly to hnfor all n � 1. Therefore, if suffices to consider a simple function

f(t) =m∑

k=1

akχAk(t),

with Ak disjoint measurable subsets of [0, 1]. Using Problem 4.28, wemay assume that each set Ak is compact. For every i � 1, we can finddisjoint sets Uk such that Ak ⊆ Uk and

λ( m⋃

k=1

(Uk \ Ak))

< 1i .

4.3. Solutions 749

Let ξ = max1�k�n

|ak|. By Problem 2.103 (see also the Urysohn lemma;

Theorem 2.136), we can find a continuous function fi : [0, 1] −→ [−c, c]such that

fi = f on

m⋃

k=1

Ak and fi = 0 on [0, 1] \m⋃

k=1

Uk.

So, λ({

t ∈ [0, 1] : fi(t) = f(t)})

< 1i , which means that fi

λ−→ f .

Solution of Problem 4.53Let h : X × Y −→ Y × Y be the function, defined by

h(x, y) =(f(x), y

).

Evidently h is(B(X)⊗ B(Y ), B(Y )⊗ B(Y )

)-measurable. So

Gr f ={(x, y) ∈ X × Y : f(x) = y

}= h−1(ΔY ) ∈ B(X)⊗B(Y ).

Solution of Problem 4.54Let {Cn}n�1 ⊆ Y be a separating sequence for Y . Let

Δ ={(y, y) : y ∈ Y

}

be the diagonal of Y × Y . Then

(Y × Y ) \Δ =( ⋃

n�1

Cn × (Y \ Cn)) ∪ ( ⋃

n�1

(Y \ Cn)× Cn

),

so Δ ∈ Σ⊗ Y.


The function ξ : Ω× Y −→ Y × Y , defined by

ξ(ω, u) =(f(ω), u

)

is measurable. Finally, note that Gr f = ξ−1(Δ).

Solution of Problem 4.55By recalling that f = f+ − f−, we see that we may assume withoutany loss of generality, that f � 0. Let {Ωn}n�1 ⊆ Σ be such thatΩn ↗ Ω and μ(Ωn) < +∞ for all n � 1. Then, let us set fn = χ

Ωnf .

We see that fn ↗ f and so by the Lebesgue monotone convergencetheorem (see Theorem 3.92), we have

∫

Ω

fn dμ ↗∫

Ω

f dμ.

But ∫

Ω

fn dμ =

∫

Ωn

f dμ � λ,

hence ∫

Ω

f(ω) dμ � λ.

Solution of Problem 4.56Let M > 0 be such that

‖u‖p � M ∀ u ∈ C.

For every Lebesgue measurable set A ⊆ [0, 1], using the Holderinequality (see Theorem 3.103), we have

∫

A

|u| dμ � ‖u‖p‖χA‖p′ � Mλ(A)

1p′ ∀ u ∈ C,

4.3. Solutions 751

where λ denotes the Lebesgue measure on R. For a given ε > 0, let

δ =(

εM

)p′> 0. Hence, if λ(A) � δ, then

∫

A

|u| dμ � ε ∀ u ∈ C,

so the set C is uniformly integrable.

Solution of Problem 4.57“=⇒”: For every open set U ⊆ R

N , the set ϕ−1(U) ⊆ X is open (sinceϕ is continuous). Hence (ϕ◦f)−1(U) ∈ Σ and so ϕ◦f is Σ-measurable.“⇐=”: Let C ⊆ X be a closed set. Since X is perfectly normal, forsome continuous function ϕ : X −→ R, we have C = ϕ−1(0). Then

f−1(C) = (ϕ ◦ f)−1(0) ∈ Σ

and so f is(Σ,B(X)

)-measurable.

Solution of Problem 4.58For every n � 1, we cover X by a finite or countable family of balls{Bn

k }k�1 of diameter less than or equal to 1n . From this cover, we

generate a cover of X consisting of disjoint Borel sets Cn,k, k � 1 ofdiameter less than or equal to 1

n , namely

Cn,k = Bnk \

n−1⋃

i=1

Bni ∀ k � 1.

From every Cn,k we choose ξk and set

fn(x) = ξk ∀ x ∈ f−1(Cn,k).

ThendX

(fn(x), f(x)

)� 1

n ∀ x ∈ X,


hencefn ⇒ f in X.


(a) From Problem 3.149, we know that

1∫

0

f dx =

+∞∫

0

λ({f > ϑ}) dϑ and

1∫

0

g dx =

+∞∫

0

λ({g > ϑ}) dϑ.

So, if f and g are equimeasurable, then we conclude that

1∫

0

f dx =

1∫

0

g dx.

(b) Let

Y ={A ⊆ [0,+∞) : f−1(A), g−1(A) ⊆ (0, 1) are Lebesgue

measurable and λ(f−1(A)

)= λ

(g−1(A)

)}.

It is easy to see that Y is a σ-algebra. If f and g are equimeasurable,then Y contains all half-lines of the form (ϑ,+∞), ϑ > 0 and so itcontains all Borel sets of [0,+∞). Then

λ(f−1

(ξ−1(ϑ,+∞)

))= λ

(f−1

(ξ−1(ϑ,+∞)

)) ∀ ϑ > 0,

so ξ ◦ f and ξ ◦ g are equimeasurable too.


(a) We start by observing that if η : (0,+∞) −→ [0,+∞) is decreas-ing, right continuous and

limϑ→+∞

η(ϑ) = 0,

then for every r > 0, the set {η > r} is a bounded open interval of theform

(0, η∗(r)

), with η∗ : (0,+∞) −→ [0,+∞) also decreasing, right

4.3. Solutions 753

continuous and vanishing at infinity (it is the inverse of η and it isdefined by

η∗(r) = inf{ϑ ∈ [0,+∞) : η(ϑ) � r

},

hence (η∗)∗ = η).Let

η(ϑ) = λ({f > ϑ}) ∀ ϑ > 0

(the distribution function of f). Then η is decreasing, right con-tinuous and

η(ϑ) −→ 0 as ϑ → +∞.

Let h : (0, 1) −→ [0,+∞) be a decreasing, right continuous function.Let h : (0,+∞) −→ [0,+∞) be the extension by zero of h. If f and hare equimeasurable, then

η(ϑ) = λ({f > ϑ}) = λ

({h > ϑ}) = λ({h > ϑ}) = h∗(ϑ),

so h = η∗.

(b) We have

∫

A

f dx =

1∫

0

χAf dx =

1∫

0

(χAf)∗ dr

(see Problem 4.59(a)). For every ϑ > 0, we have

λ({χ

Af > ϑ}) � λ(A)

and so(χ

Af)∗(ϑ) = 0 ∀ ϑ � λ(A),

hence(χ

Af)∗ � f∗.

It follows that

∫

A

f dx =

λ(A)∫

0

(χAf)∗ dr �

λ(A)∫

0

f∗ dr.

In particular, if t ∈ (0, 1), the

t∫

0

f dx �t∫

0

f∗ dx


and so, we conclude that

1∫

0

fξ dx �1∫

0

f∗ξ dx.

Solution of Problem 4.61Since u = u+−u−, without any loss of generality, we may assume thatu � 0. Consider the set

E ={x ∈ X : u(x) > 0

}.

For every n � 1, let

En ={x ∈ X : u(x) � 1

n

}.

Then En is μ-measurable and since

μ(En)n �

∫

En

u dμ �∫

X

u dμ < +∞,

we haveμ(En) < +∞ ∀ n � 1.

SinceE =

⋃

n�1

En,

we see that E is σ-finite for μ. For every D ∈ B(X), we see that

ν(D) = ν(D ∩ E).

So, for a given ε > 0, we can find n � 1 such that

ν(D)− ν(D ∩ En) < ε2 .

Since ν � μ (see Definition 3.150) and μ is a Radon measure, we canfind a compact set K ⊆ D ∩En such that

ν(D ∩ En)− ν(K) =

∫

D∩En

u dμ−∫

K

u dμ < ε2 ,

4.3. Solutions 755

so

ν(D)−ν(K) =(ν(D)−ν(D∩En)

)+(ν(D∩En)−ν(K)

)< ε

2+ε2 = ε

and thus ν is inner regular with respect to compact sets.Also, since ν � μ and μ is a Radon measure (hence outer regular;

see Definition 4.9), for every n � 1, we can find an open set Un suchthat

D ∩ En ⊆ Un and ν(Un)− ν(D ∩ En) < ε2n .

Then

U =⋃

n�1

Un is an open set

and D ⊆ U . Since

U \D ⊆⋃

n�1

(Un \ (D ∩ En)

),

we have

0 � ν(U)−ν(D) = ν(U \D) �∑

n�1

ν(Un\(D∩En)

)�

∑

n�1

ε2n = ε

and so ν is outer regular.Finally note that ν is finite (since u ∈ L1(X,μ)). Hence, we con-

clude that ν ∈ Mb(X).

Solution of Problem 4.62By the Radon–Nikodym theorem (see Definition 3.150 and Theo-rem 3.152), we can find f ∈ L1(X,μ)+ such that

ν(A) =

∫

A

f(x) dμ(x) ∀ A ∈ Σ.

Recall that the Lebesgue integral is absolutely continuous. So, for agiven ε > 0, we can find δ = δ(ε) > 0 such that

μ(C) � δ =⇒ ν(C) =

∫

C

f dμ � ε.


From this and the fact that μ is a Radon measure, we see that ν isregular and

ν(A) = sup{ν(K) : K ⊆ A, K is compact

}

(inner regular with respect to compact sets). Therefore ν is a Radonmeasure too.

Solution of Problem 4.63Let g : X −→ [0, 1] be a continuous function such that

g|supp u = 1 and supp g is compact

(see Problem 2.103 and recall that X is locally compact). We have

∣∣u(x)

∣∣ � ξg(x) ∀ x ∈ suppμ

(see Definition 4.13), so

∫

X

|u| d|μ| � ξ

∫

X

g d|μ| � ξ‖μ‖

(since suppμ = supp |μ|) and thus

∣∣∫

X

u dμ∣∣ �

∫

X

|u| d|μ| � ξ‖μ‖.

Solution of Problem 4.64Since |μ| is a measure of the same support as μ (see Definition 4.13),without any loss of generality we may assume that μ � 0. As in thesolution of Problem 4.21, exploiting the local compactness of X, wecan find h ∈ Cc(X), 0 � h � 1 with h|supp μ = 1 (see Definition 4.13).Then, for all u ∈ Cc(X), we have

∣∣u(x)

∣∣ � ‖u‖∞h(x) ∀ x ∈ suppμ,

4.3. Solutions 757

so ∫

X

|u| dμ � ‖u‖∞∫

X

hdμ

and thus μ is finite (since u ∈ Cc(X) was arbitrary; see Theo-rem 4.23).

Solution of Problem 4.65Let μ be a Borel measure on X which is finite on compact sets. ThenCc(X) ⊆ L1(X,μ) and l : Cc(X) −→ R, defined by

l(u) =

∫

X

u dμ ∀ u ∈ Cc(X)

is a positive linear functional, so by Theorem 4.23, we can find aRadon measure m corresponding to l. If U ⊆ X is an open set, thenby hypothesis

U =⋃

n�1

Kn

with Kn ⊆ X being compact for all n � 1. Exploiting the localcompactness of X and using Problem 2.103, we can find a functionh1 ∈ Cc(X) such that

h1|K1= 1, 0 � h1 � 1 and supph1 ⊆ U.

Then by induction, for every n � 2, we can find hn ∈ Cc(X) such that

hn∣∣

n⋃

i=1Ki

= hn∣∣

n⋃

i=1supphi

= 1, 0 � hn � 1 and supphn ⊆ U.

Evidently hn ↗ χU

and so by the Lebesgue monotone convergencetheorem (see Theorem 3.92), we have

μ(U) = limn→+∞

∫

X

hn dμ = limn→+∞

∫

X

hn dm = m(U).


By Problem 4.19(a), if A ∈ B(X) and ε > 0, we can find an open setW ⊇ A and a closed set D ⊆ A such that m(W \D) < ε. But the setW \D is open. So, from the previous part of the proof, we have

μ(W \D) = m(W \D) < ε,

so μ(W ) � μ(A) + ε and thus μ is outer regular (see Definition 4.9).Also, we have μ(A) � μ(D) + ε and D is σ-compact (since X is σ-compact; see Definition 2.99). So, we can find a compact set K ⊆ Asuch that

μ(A) � μ(K) + 2ε

and so μ is inner regular with respect to compact sets, i.e., μ isRadon.

Solution of Problem 4.66For every f ∈ Cc(X) with 0 � f � 1, we have

∫

X

f dμn � μn(X) ∀ n � 1.

Hence ∫

X

f dμ � lim infn→+∞ μn(X)

and so

μ(X) = sup{∫

X

f dμ : f ∈ Cc(X), 0 � f � 1}

� lim infn→+∞ μn(X)

(see Theorem 4.23).

Solution of Problem 4.67By Problem 4.66, we have μ(X) � lim inf

n→+∞ μn(X) and so μ(X) < +∞.

Recall that the embedding Cc(X) ⊆ C0(X) is dense. So, for a givenε > 0 and h ∈ C0(X), we can find fε ∈ Cc(X) such that

‖fε − h‖∞ � ε.

4.3. Solutions 759

We have

∣∣∫

X

fε dμn −∫

X

hdμn

∣∣ � ‖fε − h‖∞μn(X) � εξ ∀ n � 1

and∣∣∫

X

fε dμ−∫

X

hdμ∣∣ � εξ.

So, by the triangle inequality, we have

∣∣∫

X

hdμn −∫

X

hdμ∣∣ � 2εξ +

∣∣∫

X

fε dμn −∫

X

fε dμ∣∣ ∀ n � 1.

But, by hypothesis∫

X

fε dμn −→∫

X

fε dμ.

So, passing to the limit as n → +∞, we have

lim supn→+∞

∣∣∫

X

hdμn −∫

X

hdμ∣∣ � 2εξ.

Since ε > 0 was arbitrary, we let ε ↘ 0, to conclude that∫

X

hdμn −→∫

X

hdμ ∀ h ∈ C0(X).


“=⇒”: Suppose that ϕ is a continuous linear functional on C(X).Note that

{u ∈ C(X) : ‖u‖∞ � 1

}=

{u ∈ C(X) :

∣∣u(x)

∣∣ � 1 for all x ∈ X

}.

Therefore, if ϕ is positive, then by the Riesz representation theorem(see Theorem 4.23; note that in our case C(X) = C0(X)), we have

‖ϕ‖∗ = sup{ϕ(u) : u ∈ C(X), ‖u‖∞ � 1

}

= sup{ϕ(u) : u ∈ C(X),

∣∣u(x)

∣∣ � 1for all x ∈ X

}= ϕ(1).


“⇐=”: Suppose that ‖ϕ‖∗ = ϕ(1). Let u ∈ C(X), u � 0, u = 0and let us set v = u

‖u‖∞ . Then ‖1 − v‖∞ � 1 and so ϕ(1) − ϕ(v) =

ϕ(1 − v) � ‖ϕ‖∗ = ϕ(1), hence ϕ(v) � 0 and thus

ϕ(u) = ‖u‖∞ϕ(v) � 0 ∀ u ∈ C(X), u � 0,

hence ϕ is positive.

Solution of Problem 4.69For each x ∈ X, we define the positive continuous linear functionalϕx : C(X) −→ R, by

ϕx(u) = u(x) ∀ u ∈ C(X).

Then, for all x, y ∈ X, x = y, we have ‖ϕx − ϕy‖∗ = 2. Since by hy-pothesis X is uncountable, the set {ϕx}x∈X ⊆ C0(X)∗ is uncountabletoo. Then

{B1(ϕx)

}x∈X is an uncountable family of mutually disjoint

open balls. This implies that no countable subset of C0(X)∗ can bedense in it. Therefore C0(X)∗ = Mb(X) is not separable.

Solution of Problem 4.70We consider the functional ξ : C0(X) −→ R, defined by

ξ(u) = ϑ(Y )max{u(x) : x ∈ X

} ∀ u ∈ C0(X).

It is easy to see that ξ(u+ v) � ξ(u) + ξ(v) and

ξ(λu) = λξ(u) ∀ u, v ∈ C0(X), λ � 0.

The surjectivity of f allows us to introduce the following linear sub-space of C0(X),

V ={v ◦ f : v ∈ C0(Y )

}.

We consider the linear functional l : V −→ R, defined by

l(v ◦ f) =

∫

Y

v dϑ.

4.3. Solutions 761

Then

l(u ◦ f) � ϑ(Y )max{v(y) : y ∈ Y

}

= ϑ(Y )max{(v ◦ f)(x) : x ∈ X

}= ξ

(v ◦ f)

(since f is surjective). Then the Hahn–Banach theorem (see Theo-rem 5.24) implies that we can find a linear functional l : C0(X) −→ R

such that l|V= l and l(u) � ξ(u) for all u ∈ C0(X). Note that l(u) � 0

if u � 0 and so

l(u) = −l(−u) ∀ u ∈ C0(X), u � 0.

Clearly l is continuous. Therefore l is a positive, continuous, linearfunctional on C0(X) and so by Theorem 4.23, we can fine μ ∈ Mb(X),μ � 0 which is necessarily Radon, since X is compact (see Theo-rem 4.12) such that

l(u) =

∫

X

u dμ ∀ u ∈ C0(X).

Since l extends l, we have

∫

Y

v dϑ =

∫

X

(v ◦ f) dμ =

∫

Y

v d(μ ◦ f−1) ∀ v ∈ C0(Y )

and so ϑ = μf−1 by the uniqueness of the measure in Theorem 4.23.


“(b) =⇒ (a)”: Clearly, if ϕ = cδx for some c � 0 and x ∈ X, thedesired lattice property holds.

“(a) =⇒ (b)”: Suppose that ϕ ∈ C0(X)∗, ϕ � 0 has the describedlattice property. By Theorem 4.23, we can find μ ∈ Mb(X), μ � 0such that

ϕ(f) =

∫

X

f dμ ∀ f ∈ C0(X).


If x, y ∈ suppμ (see Definition 4.13) and x = y, then we can findf, h ∈ Cc(X) such that

f � 0, h � 0, min{f, h} = 0 and f(x) = h(y) = 1.

So, we have

ϕ(max{f, h}) = ϕ(f + h) = ϕ(f) + ϕ(h) > max

{ϕ(f), ϕ(h)

},

a contradiction. Therefore suppμ is a singleton, i.e., suppμ = {x}.Let c = μ

({x}) > 0. Then, for every f ∈ C0(X), we have

ϕ(f) =

∫

X

f dμ = f(x)μ({x}) = cf(x),

so ϕ = cδx.

Solution of Problem 4.72Let ξ : X −→ R+ = [0,+∞] be a lower semicontinuous function, letx ∈ X be such that ξ(x) > 0 and choose ϑ ∈ (

0, ξ(x)). Then the

set {ξ > ϑ} is nonempty open and so by Problem 2.103, we can findg ∈ Cc(X) such that

g(x) = ϑ and 0 � g � ϑχU � ξ.

Hence, if ξ(x) > 0, then

ξ(x) = sup{g(x) : g ∈ Cc(X), 0 � g � ξ

}.

Clearly the same is true if ξ(x) = 0. Therefore

ξ = sup{g : g ∈ Cc(X), 0 � g � ξ

}.

From Proposition 2.55(a) we know that f is lower semicontinuous.Also let X∗ be the Alexandrov one-point compactification of X (seeRemark 2.97). Since C(X∗) is separable, Y is directed and usingProposition 4.21 and the last equation, we can find an increasing se-quence {gn}n�1 ⊆ Cc(X) such that

0 � gn � hn ∀ n � 1

4.3. Solutions 763

for some hn ∈ Y and gn ↗ f . Then from the Lebesgue monotoneconvergence theorem (see Theorem 3.92), we have

∫

X

f dμ = limn→+∞

∫

X

gn dμ = supn�1

∫

X

gn dμ.

But note that

supn�1

∫

X

gn dμ � suph∈Y

∫

X

hdμ �∫

X

f dμ.

Therefore, we conclude that

∫

X

f dμ = suph∈Y

∫

X

hdμ.

Remark. Therefore, if X is a locally compact topological space, μ isa Radon measure on X and f : X −→ R+ = [0,+∞] is lower semicon-tinuous, then

∫

X

f dμ = sup{∫

X

g dμ : g ∈ Cc(X), 0 � g � f}.

Solution of Problem 4.73Let {sn}n�1 be a sequence of nonnegative simple functions such that

sn(x) ↗ f(x) ∀ x ∈ X.

We have

f = s1 +∑

n�2

(sn − sn−1)

and in the series each term is a nonnegative simple function. So, wewrite that

f =∑

k�1

ϑkχAk,

with ϑk > 0 and Ak ∈ B(X) for all k � 1.


(a) Since μ is a Radon measure, for a given ε > 0, we can find anopen set Uk ⊇ Ak such that

μ(Uk) � μ(Ak) +ε

2kϑk.

Let us seth =

∑

k�1

ϑkχUk.

Recall that the characteristic function of an open set is lower semicon-tinuous. So, h is lower semicontinuous and clearly

f � h and

∫

X

hdμ �∫

X

f dμ+ ε.

Since ε > 0 was arbitrary, we conclude that∫

X

f dμ = inf{∫

X

hdμ : h is lower semicontinuous on X, h � f}.

(b) Let

σ <

∫

X

f dμ.

Then for N � 1 large enough, we will have

N∑

k=1

ϑkμ(Ak) > σ.

Since μ is a Radon measure, for all k � 0, we can find a compact setKk ⊆ Ak such that

N∑

k=1

ϑkμ(Ak) > σ.

Recall that the characteristic function of a closed set is upper semi-continuous. Therefore

η =

N∑

k=1

ϑkχKk� 0

is upper semicontinuous,

0 � η � f and

∫

X

η dμ > σ.

4.3. Solutions 765

Since σ <∫X f dμ was arbitrary, we conclude that

∫

X

f dμ=sup{ ∫

X

η dμ : η is upper semicontinuous on X, 0 � η � f}.

Solution of Problem 4.74Let A ∈ B(X) and let ε > 0 be given. We can find a compact setK ⊆ X and an open set U ⊆ X such that

∣∣μ(B)

∣∣ < ε ∀ B ∈ B(X), B ⊆ U \K.

Then we have

0 � μ+(A)−μ+(K)=μ+(A\K)= sup{μ(B) : B ∈ B(X), B ⊆ A\K}

and

0 � μ+(U)−μ+(A) = μ+(U\A) = sup{μ(B) : B ∈ B(X), B ⊆ U\A}

(see Theorem 3.146). Therefore

μ+(A)− μ+(K) < ε and μ+(U)− μ+(A) < ε.

This implies that μ+ is a finite Radon measure on X. Similarly forμ−. So, we conclude that μ = μ+ − μ− ∈ Mb(X).

Solution of Problem 4.75Due to the separability of X and Y , we can treat X and Y as subsets ofthe Hilbert cubeH = [0, 1]N (see Theorem 2.154(b)). Let d be a metricon H×H consistent with its topology and let g ∈ UCb(X × Y ) (hereUCb(X × Y ) denotes the space of bounded d-uniformly continuousfunctions on X × Y with values in R). Then g admits a uniformlycontinuous extension g ∈ UC(H×H) = C(H×H). From the Stone–Weierstrass theorem, we know that for a given ε > 0, we can findfunctions fk, hk ∈ C(H) = UC(H), for k � 1 such that

supx,y∈H

∣∣

m∑

k=1

fk(x)hk(y)− g(x, y)∣∣ � ε.


If mnw−→ m in M+

1 (X) and μnw−→ μ in M+

1 (Y ) and fk = f |X ,hk = h|Y , then

lim supn→+∞

∣∣∫

X×Y

g d(mn × μn)−∫

g d(m× μ)∣∣

� lim supn→+∞

∣∣∫

X×Y

(g −

m∑

k=1

fkhk)d(mn × μn)

∣∣

+m∑

k=1

limn→+∞

∣∣∫

X

fk dmn

∫

Y

hk dμn −∫

X

fk dm

∫

Y

hk dμ∣∣

+ limn→+∞

∣∣∫

X×Y

( m∑

k=1

fkhk − g)d(m× μ)

∣∣

� 2ε.

Let ε ↘ 0 and use the Portmanteau theorem (see Theorem 4.26), toconclude that

ξ(mn, μn)w−→ ξ(m,μ),

so ξ is continuous.

Solution of Problem 4.76We do the solution for f being lower semicontinuous and boundedbelow. The other case can be done similarly. To show the weak lowersemicontinuity of ξ, we need to show that for every η ∈ R, the set

Iη ={μ ∈ M+

1 (X) : ξ(μ) � η}

is weakly closed. So, let {μα}α∈I ⊆ Iη be a net and assume that

μαw−→ μ in M+

1 (X).

From Problem 2.20, we know that there exists a sequence {fn}n�1 ⊆Cb(X) such that

fn(x) ↗ f(x) ∀ x ∈ X.

We have∫

X

fn dμα �∫

X

f dμα = ξ(μα) � η ∀ α ∈ I, n � 1.

4.3. Solutions 767

From the Portmanteau theorem (see Theorem 4.26), we have

limα∈I

∫

X

fn dμα =

∫

X

fn dμ � η ∀ n � 1

and by the Lebesgue monotone convergence theorem (see Theo-rem 3.92), we have

∫

X

fn dμ ↗∫

X

f dμ = ξ(μ) � η,

thus μ ∈ Iη and so Iη is weakly closed, i.e., ξ is lower semicontinuous.

Solution of Problem 4.77(a) Let f = χ

U: X −→ R be the characteristic function of U . Since

U is open, χUis lower semicontinuous. So, by Problem 4.76 above, the

function

ξ(μ) =

∫

X

χUdμ

is weakly lower semicontinuous. So, the set

D ={μ ∈ M+

1 (X) : μ(U) > η}

={μ ∈ M+

1 (X) : ξ(μ) > η}

is open.

(b) In this case χC is upper semicontinuous and so the function

ξ(μ) =

∫

X

χCdμ

is upper semicontinuous (see Problem 4.76). Therefore, the set

E ={μ ∈ M+

1 (X) : μ(C) � η}

={μ ∈ M+

1 (X) : ξ(μ) � η}

is closed.


Solution of Problem 4.78Let τ denote the topology of X and let

X ={A ∈ B(X) : ξA is Borel

}.

From Problem 4.77, we see that X ⊇ τ . In particular X ∈ X . Also, ifA,C ∈ X and A ⊆ C, then ξC\A = ξC − ξA, which shows that ξC\A isBorel and so C \ A ∈ X . Finally, if {An}n�1 ⊆ X is a sequence suchthat An ↗ A, then ξAn ↗ ξA and so ξA is Borel, hence A ∈ X . So, weinfer that X is a λ-class (Dynkin class; see Definition 3.7(a)). Becauseτ ⊆ X and τ is closed under finite intersections, we apply Theorem 3.9and obtain X = σ(τ) = B(X).

Solution of Problem 4.79Let {uα}α∈I ⊆ Cb(X) and {μα}α∈I ⊆ M+

1 (X) be nets such that

uα‖·‖∞−→ u and μα

w−→ μ.

Then, we have

∣∣γ(uα, μα)− γ(u, μ)∣∣ =

∣∣∫

X

uα dμα −∫

X

u dμ∣∣

�∣∣∫

X

(uα − u) dμα

∣∣+

∣∣∫

X

u dμα −∫

X

u dμ∣∣

� ‖uα − u‖∞ +∣∣∫

X

u dμα −∫

A

u dμ∣∣ −→ 0

(by the Portmanteau theorem; see Theorem 4.26). Thus γ is jointlymeasurable.

Solution of Problem 4.80Let ξ : X −→ M+

1 (X) be defined by ξ(x) = δx. Evidently ξ is injective.Also, if {xα}α∈I is a net such that xα −→ x in X, then for all u ∈Cb(X), we have

∫

X

u dδxα = u(xα) −→ u(x) =

∫

X

u dδx,

4.3. Solutions 769

so, using the Portmanteau theorem (see Theorem 4.26), we have

ξ(xα) = δxα

w−→ δx = ξ(x)

and thus ξ is continuous.Conversely, let

ξ(xα) = δxα

w−→ δx = ξ(x) in M+1 (X).

Suppose that

ξ−1(ξ(x0)

)= xα −→ x = ξ−1

(ξ(x)

)in X.

Then we can find U ∈ N (x) and a subnet {xβ}β of {xα}α∈I such that

xβ ∈ X \ U ∀ β.

Recall that a metric space is completely regular (see Problem 2.42 andTheorem 1.43). So, we can find a continuous function h : X −→ [0, 1]such that

h(x) = 0 and h|X\U = 1,

so ∫

X

hdδxβ= h(xβ) = 1 −→ 0 = h(x) =

∫

X

hdδx,

a contradiction. This proves that ξ−1 is continuous too and so ξ is ahomeomorphism into M+

1 (X).Finally, we show that ξ(X) is sequentially closed in M+

1 (X). So,let {xn}n�1 ⊆ X be a sequence and assume that

δxn

w−→ μ in M+1 (X).

If {xn}n�1 does not have a convergent subsequence, then the set D ={xn : n � 1

}is a closed subset of X and so is every C ⊆ D. By the

Portmanteau theorem (see Theorem 4.26), we have

lim supn→+∞

δxn(C) � μ(C).

Hence, for every infinite set A ⊆ D, we have μ(A) = 1, a contradictionto the fact that μ is a measure. So, we can find a subsequence

{xnk

}k�1

of a sequence {xn}n�1 such that

xnk−→ x in X,


soξ(xnk

) = δxnk

w−→ δx = ξ(x) in M+1 (X).

Thus μ = δx and ξ(X) is sequentially closed in M+1 (X).

Solution of Problem 4.81Without any loss of generality, we assume that f(X) = Y . We start byshowing that ϑ is injective. So, let μ,m ∈ M+

1 (X), μ = m. By Prob-lem 4.14(a), we can find an open set U ⊆ X such that μ(U) = m(U).Since f is a homeomorphism, we have that the set V = f(U) ⊆ Y isopen and U = f−1(V ). Then

ϑ(μ)(V ) = μ(f−1(V )

)= μ(U) = m(U) = m

(f−1(V )

)= ϑ(m)(V )

and so ϑ is injective.Next we show the continuity of ϑ. So, assume that


1 (X)

and let V ⊆ Y be an open set. Then the set f−1(V ) = U ⊆ X is openand so by the Portmanteau theorem (see Theorem 4.26), we have

lim infα

ϑ(μα)(V ) = lim infα

μα(U) � μ(U) = ϑ(μ)(V ).

Since V ⊆ Y was an arbitrary open set, Theorem 4.26 implies that

ϑ(μα)w−→ ϑ(μ) in M+

1 (Y )

and this proves the continuity of ϑ.Finally we show that ϑ−1 is continuous. To this end, suppose that

ϑ(μα)w−→ ϑ(μ) in M+

1 (Y ).

Let U ⊆ X be an open set. Then f(U) = V ⊆ Y is an open set too.We have

lim infα

μα(U) = lim infα

μα

(f−1(V )

)= lim inf

αϑ(μα)(V )

� ϑ(μ)(V ) = μ(f−1(V )

)= μ(U)

(see Theorem 4.26). Thus


1 (X)

4.3. Solutions 771

(see Theorem 4.26). Therefore ϑ−1 is continuous and so ϑ is a home-omorphism.


“(a) =⇒ (b)”: Let A ∈ B(X) be a bounded set with μ(∂A) = 0 andlet U ⊆ X be a bounded and open set such that A ⊆ U . For a giveninteger k � 1, we define

Dk ={x ∈ U : dist(x,A) < 1

k

}.

By Problem 2.103, we can find a continuous function fk : X −→ [0, 1]such that

fk∣∣A

= 1 and fk∣∣Dc

k

= 0.

Since the set U is bounded, fk ∈ Cc(X). We have

∣∣μn(A)− μ(A)

∣∣ �

∣∣∫

X

(χ

A− fk

)dμn

∣∣+

∣∣∫

X

fk dμn −∫

X

fk dμ∣∣

+∣∣∫

X

(fk − χ

A

)dμ

∣∣.

Note that (fk − χA

)∣∣A

= 0 and(fk − χA

)∣∣Dc

k

= 0.

Hence, we have

∣∣∫

X

(fk − χ

A

)dμ

∣∣ � μ(Dk \A) � μ

(Dk \ intA

).

Similarly, we obtain that

∣∣∫

X

(χ

A− fk

)dμn

∣∣ � μn

(Dk \ intA

).

Moreover, since by hypothesis (statement (a)), μv−→ μ and f ∈

Cc(X), we have∣∣∫

X

fk dμn −∫

X

f dμ∣∣ −→ 0.


The set Dk \ intA is closed and so the function χDk\intA is upper

semicontinuous. Then according to Problem 2.20(c), we can find asequence {hkm}m�1 ⊆ C(X) with supphkm ⊆ U such that hkm ↘χ

Dk\intA as m → +∞. Since hkm ∈ Cc(X), we have

μn

(Dk \ intA

)=

∫

X

χDk\intA(x) dμn

�∫

X

hkm(x) dμn −→∫

X

hkm dμ as n → +∞,

so

lim supn→+∞

μn

(Dk \ intA

)�

∫

X

hkm(x) dμ.

Because supphkm ⊆ U and μ(U) < +∞ (recall that μ is a Radonmeasure and the set U is compact; see Definition 4.9(e)), we can letm → +∞ and then by the Lebesgue monotone convergence theorem(see Theorem 3.92), we have

lim supn→+∞

μn

(Dk \ intA

)� μ

(Dk \ intA

).

Therefore, finally we have

lim supn→+∞

∣∣μn(A) − μ(A)∣∣ � 2μ

(Dk \ intA

).

Sending k → +∞, we have

Dk \ intA ↘ A \ intA = ∂A.

But by hypothesis μ(∂A) = 0. Therefore, we conclude thatμn(A) −→ μ(A).

“(b) =⇒ (a)”: Let f ∈ Cc(X). By Proposition 2.94, we can find twobounded, open sets V and W in X such that

K ⊆ V ⊆ V ⊆ W,

where K is compact and supp f ⊆ K. We have

V =⋂

δ>0

Dδ,

4.3. Solutions 773

whereDδ =

{x ∈ W : dist(x, V ) < δ

}.

Evidently each Dδ is open and

∂Dδ ={x ∈ W : dist(x, V ) = δ

}.

Hence, the sets {Dδ}δ>0 have disjoint boundaries and becauseμ(W ) < +∞, we have μ(Dδ) = 0 for some δ > 0. Therefore, withoutany loss of generality, we can have μ(∂V ) = 0.

If A is a Borel subset of V , the closure of A relative to V is

given by AV

= A ∩ V . Hence, the boundary of A relative to V is∂V A = (∂A) ∩ V . Suppose that μ(∂V A) = 0. Then μ

((∂A) ∩ V

)= 0.

Also ∂A ∩ V c ⊆ A ∩ V c ⊆ V \ V = ∂V . Hence μ(∂A ∩ V c) = 0.Therefore μ(∂A) = 0. If μ′

n = μn|V and μ′ = μ|V , then

μ′n

w−→ μ′ in M+1 (V )

(see the Portmanteau theorem; Theorem 4.26). Hence∫

V

f dμn −→∫

V

f dμ,

so ∫

X

f dμn −→∫

X

f dμ

(recall that supp f ⊆ K ⊆ V ) and thus μnv−→ μ (since f ∈ Cc(X)

was arbitrary).


“=⇒”: Suppose that C ⊆ M+1 (X) is uniformly tight (see Theo-

rem 4.31). Let {εn}n�0 ⊆ R+ be a sequence such that

∑

n�0

εn < +∞.

Let Kn = Kεn be an increasing sequence of compact subsets of Xsuch that

μ(Kn) � 1− 1n ∀ n � 1, μ ∈ C.


Letϕ(x) = inf

{n � 0 : x ∈ Kn

}=

∑

n�0

χX\Kn

(x).

Then clearly

supμ∈C

∫

X

ϕ(x) dμ < +∞

and ϕ has compact sublevel sets.

“⇐=”: If

supμ∈C

∫

X

ϕ(x) dμ < +∞,

then by Markov inequality (see Proposition 3.90), the definition ofuniform tightness is satisfied by the compact sublevel sets of ϕ.

Solution of Problem 4.84Let g ∈ C∞

c

(RN)be such that

0 � g � 1, g(x) =

{1 if ‖x‖ � 1

2 ,0 if ‖x‖ � 1.

Letgk(x) = g

(1kx

) ∀ k � 1.

Then by hypothesis, we have

lim infn→+∞ μn

(Bk(0)

)� lim

n→+∞

∫

RN

gk(x) dμn =

∫

RN

gk(x) dμ.

Also, from the Lebesgue dominated convergence theorem (see Theo-rem 3.94), we have

limk→+∞

∫

RN

gk(x) dμ = 1.

So, for a given ε > 0, we can find n0 � 1 and k � 1 large enough suchthat

μn

(Bk(0)

)� 1− ε ∀ n � n0.

4.3. Solutions 775

Then from this and Theorem 4.12, we conclude that the set{μn :

n � 1} ∪ {μ} is uniformly tight. Then the Prohorov theorem (see

Theorem 4.31) implies that we can find a subsequence{μnk

}k�1

of

{μn}n�1, such that

μnk

w−→ μ.

Hence ∫

X

f dμ =

∫

X

f dμ ∀ f ∈ C∞c

(RN),

so μ = μ.But from Theorem 4.30, we know that

(M+

1 (RN ), w)is a Polish

space. So, by the Urysohn criterion (see Problem 1.3), we concludethat the original sequence {μn}n�1 converges weakly to μ.

Solution of Problem 4.85Let f ∈ Cb(Y ). By adding a constant if necessary, we may assumethat f � 0. If K ⊆ X is compact, then

f ◦ ϑn ⇒ f ◦ ϑ on K

and

limn→+∞

∫

K

((f ◦ ϑn)− (f ◦ ϑ)) dμn = 0

(since K is compact and the sequence{μn(K)

}n�1

is bounded), so

lim infn→+∞

∫

X

(f ◦ ϑn) dμn � lim infn→+∞

∫

K

(f ◦ ϑn) dμn

= limn→+∞

∫

K

((f ◦ ϑn)− (f ◦ ϑ)) dμn

+ lim infn→+∞

∫

K

(f ◦ ϑ) dμn

= lim infn→+∞

∫

K

(f ◦ ϑ) dμn


�(− sup f

)supn�1

μn(X \K)

+ lim infn→+∞

∫

X

(f ◦ ϑ) dμn

=(− sup f

)supn�1

μn(X \K)

+

∫

X

(f ◦ ϑ) dμ

(since f ◦ ϑ ∈ C(K)). Since {μn}n�1 is uniformly tight (see Theo-rem 4.31), we can find an increasing sequence {Km}m�1 of compactsets in X such that

limm→+∞ sup

n�1μn(X \Km) = 0.

So, using in the above inequality K = Km and letting m → +∞, weconclude that

lim infn→+∞

∫

Y

f d(μnϑ−1n ) �

∫

Y

f d(μϑ−1).

replacing f by −f , we also have

lim supn→+∞

∫

Y

f d(μnϑ−1n ) �

∫

Y

f d(μϑ−1),

so

limn→+∞

∫

Y

f d(μnϑ−1n ) =

∫

Y

f d(μϑ−1)

and thus μnϑ−1n

w−→ μϑ−1 in M+1 (Y ).

Solution of Problem 4.86Let μ ∈ M+

1 (Y ) (see Definition 4.25) and let

μ1 = μϑ−11 and μ2 = μϑ−1

2 .

By hypothesis, for every ε > 0, we can find compact sets K1 ⊆ X1 andK2 ⊆ X2 such that

μ1(X1 \K1) � ε2 and μ2(X2 \K2) � ε

2 ∀ μ ∈ E.

4.3. Solutions 777

It follows that

μ(Y \ ϑ−1

1 (K1))

� ε2 and μ

(Y \ ϑ−1

2 (K2))

� ε2

and so

μ(Y \ (ϑ−1

1 (K1) ∩ ϑ−12 (K2)

))�μ

(Y \ ϑ−1

1 (K1))+ μ

(Y \ ϑ−1

2 (K2))� ε

∀ μ ∈ E.

But note that

ϑ−11 (K1) ∩ ϑ−1

2 (K2) = ϑ−1(K1 ×K2)

is compact, since by hypothesis ϑ is proper and K1 × K2 is com-pact. Therefore, we conclude that E ⊆ M+

1 (Y ) is uniformly tight (seeTheorem 4.31).

Solution of Problem 4.87Let C ⊆ X be a nonempty closed set and let

Un ={x ∈ X : dist(x,C) < 1

n

} ∀ n � 1.

Then Un is open for every n � 1 and

C =⋂

n�1

Un.

Then sets C and U cn are disjoint closed and

inf{d(x, y) : x ∈ C, y ∈ U c

n

}� 1

n > 0.

So, if we set

fn(x) =dist(x,U c

n)

dist(x,U cn) + dist(x,C)

∀ x ∈ X,

then

fn ∈ UCb(X), 0 � fn � 1, fn|Ucn

= 0, fn|C = 1.

We have

μ(C) �∫

X

fn dμ =

∫

X

fn dν =

∫

Un

fn dν � ν(Un).


Letting n → +∞, we have μ(C) � ν(C). Reversing the roles of μ andν in the above argument, we infer that

μ(C) = ν(C) ∀ C ⊆ X, C is closed.

But μ and ν are regular (see Theorem 4.11). So, we conclude thatμ = ν.

Solution of Problem 4.88Let u : X −→ M+

1 (X × Y ) be defined by

u(x) = δx ×m(x, ·).

Since both functions X � x �−→ δx ∈ M+1 (X) and X � x �−→

m(x, ·) ∈ M+1 (Y ) are continuous, we see that u is continuous. Also,

let ηf : M+1 (X × Y ) −→ R be defined by

ηf (ν) =

∫

X×Y

f dν ∀ ν ∈ M+1 (X × Y ).

From the definition of the weak topology this function is continuoustoo. Finally note that h = ηf ◦ u, so h is continuous.

Solution of Problem 4.89Because of Theorem 4.27, we can work with sequences. So, let{μn}n�1 ⊆ M+

1 (C) be a sequence such that

μnw−→ μ in M+

1 (X).

Let U = Cc. Then U ⊆ X is an open set and by the Portmanteautheorem (see Theorem 4.26), we have

0 = lim infn→+∞ μn(U) � μ(U),

4.3. Solutions 779

i.e., μ(U) = 0. Thus suppμ ⊆ C (see Definition 4.13) and so μ ∈M+

1 (C).

Solution of Problem 4.90That μ � ν ∈ M+

1 (X) follows at once from the Fubini theorem (seeTheorem 3.115) and we also have

(μ � ν)(A) =

∫

X

ν(x−1A) dμ(x).

Because of Theorem 4.27, it suffices to show the sequential continuity.Let {μn}n�1 , {νn}n�1 ⊆ M+

1 (X) be two sequences and assume that

μnw−→ μ, νn

w−→ ν in M+1 (X).

Then for every f ∈ Cb(X), we have∫

X

f d(μ � ν) =

∫

X

f(xy) dμ(x) dν(y) =

∫

X

f(xy) d(μ× ν)(x, y).

Since f ∈ C(X), we have that the function (x, y) �−→ f(xy) belongsin C(X ×X). From Problem 4.75, we have

∫

X

f(xy) d(μn × νn) −→∫

X

f(xy) d(μ× ν),

so ∫

X

f d(μn � νn) −→∫

X

f d(μ � ν)

and thusμn � νn

w−→ μ � ν in M+1 (X),

i.e., � is continuous.

Solution of Problem 4.91Let F be the smallest σ-algebra on M+

1 (X) that makes all functions{ηA}A∈F measurable. Let

D ={A ∈ B(X) : ηA is Borel measurable

}.


It is easy to see that D is a λ-class (a Dynkin class; see Defini-tion 3.7(b)) and contains all closed (or open) sets in X (see the Port-manteau theorem; Theorem 4.26). By the Dynkin theorem (see Theo-rem 3.9), we have that D = B(X). Therefore F ⊆ B(M+

1 (X)).

The function

M+1 (X) � μ �−→ ϑf (μ) =

∫

X

f dμ ∈ R

is F-measurable, when f = χA for all A ∈ B(X). It follows that ϑf isF-measurable, when f is Borel simple. Let f ∈ Cb(X). We can findan increasing sequence of simple functions {sn}n�1 such that

−M � sn � f ∀ n � 1

for some M > 0 and sn ↗ f . By the Lebesgue monotone conver-gence theorem (see Theorem 3.92), we have ϑsn ↗ ϑf , hence ϑf isF-measurable. Hence for every ε > 0, μ ∈ M+

1 (X) and f ∈ Cb(X),the subbasic open set

Uε(μ; f) ={ν ∈ M+

1 (X) :∣∣∫

X

f dν −∫

X

f dμ∣∣ < ε

}

is F-measurable. It follows that F = B(M+1 (X)

).

Solution of Problem 4.92Let C ∈ Σ⊗B(X) and let ηC : Ω×M+

1 (X) −→ [0, 1] be the function,defined by

ηC(ω, μ) = (δω × μ)(C) ∀ (ω, μ) ∈ Ω×M+1 (X).

Let

Y ={C ∈ Σ⊗ B(X) : ηC is Σ⊗ B(X)-measurable

}.

If C = A×B with A ∈ Σ, B ∈ B(X), then

ηC(ω, x) = χA(ω)μ(B)

4.3. Solutions 781

and so by Problem 4.91, we have that ηC is Σ ⊗ B(X)-measurable,hence {

A×B : A ∈ Σ, B ∈ B(X)} ⊆ Y.

Note that Y is an algebra and a monotone class (see Definition 3.10).Therefore invoking Theorem 3.12, we conclude that Y = Σ⊗ B(X).

Now note that

GrS={(ω, μ) ∈ Ω×M+

1 (X) : μ(F (ω)

)= 1

}=η−1

GrF

({1}) ∈ Σ⊗B(X).

Solution of Problem 4.93“=⇒”: Let A ∈ Y and let ηA : M+

1 (Y ) −→ [0, 1] be defined by

ηA(μ) = μ(A) ∀ μ ∈ M+1 (Y ).

We see that ϑA = ηA◦ξ. By hypothesis we know ξ is Borel measurable,while from Problem 4.91, we have that ηA is Borel measurable too.Therefore, it follows that ϑA = ηA ◦ ξ is Borel measurable.“⇐=”: By Problem 4.91, we have

ξ−1(B(M+

1 (Y )))

= ξ−1(σ( ⋃

A∈Yη−1A

(B(R))))

= σ( ⋃

A∈Yξ−1

(η−1A

(B(R)))) = σ( ⋃

A∈Yϑ−1A

(B(R))) ⊆ B(X)

(recall that ϑA = ηA ◦ ξ).

Solution of Problem 4.94According to Problem 2.20(a), there exists a sequence {fn}n�1 ⊆Cb(X × Y ) such that fn ↗ f . Let

ϑn(x) =

∫

Y

fn(x, y)ξ(x, dy) ∀ x ∈ X.


From Problem 4.88, we know that ϑn ∈ Cb(X) and from the Lebesguemonotone convergence theorem (see Theorem 3.92), we have ϑn ↗ ϑand this by Problem 2.20(a) implies that ϑ is lower semicontinuousand of course bounded below.

Solution of Problem 4.95Since (X,Σ) is separable, Σ is countably generated (see Defini-tion 3.14). Let {Cn}n�1 be a sequence of generators of Σ. We claimthat the sequence {Cn}n�1 separates points of X. Suppose that forsome x, x′ ∈ X, x = x′, we have

χCn(x) = χCn

(x′) ∀ n � 1.

We define Σ0 ={C ∈ 2X : χ

C(x) = χ

C(x′)

}. Evidently Σ0 is a

σ-algebra and by hypothesis {Cn}n�1 ⊆ Σ0 and so Σ ⊆ Σ0, whichcontradicts the separability of (X,Σ) (see Definition 3.14(b)). Letf : X −→ {0, 1}N be defined by f(x) =

{χ

Cn(x)

}n�1

. From the sepa-

rating property of {Cn}n�1 we see that f is an injection and of coursef is measurable. Let A = f(X). We need to show that f−1 : A −→ Xis measurable. To this end, we need to show that if C ∈ Σ, thenf(C) ∈ B(A). Let

Σ1 ={C ∈ 2X : f(C) ∈ B(X)

}.

Clearly Σ1 is a σ-algebra and {Cn}n�1 ⊆ Σ1, because

f(Cn) ={ {λk}k�1 ∈ {0, 1}N : λn = 1

} ∩A.

Therefore Σ ⊆ Σ1 and so f−1 is measurable, hence f is an isomorphism(see Definition 4.39).

Solution of Problem 4.96Let A ∈ B(Y ) and let proj

X(respectively, proj

Y) be the projection on

X (respectively, on Y ). We have

f−1(A) = projX

(proj−1

Y(A) ∩Gr f

).

4.3. Solutions 783

Note that the set proj−1Y

(A)∩Gr f is a Borel subset of Gr f and so it isa Souslin subset of Gr f (see Definition 2.156). Therefore f−1(A) ⊆ Xis a Souslin set. Similarly, we show that the set f−1(Ac) ⊆ X is aSouslin set. Since f−1(A) and f−1(Ac) form a Souslin partition of X,from Corollary 4.35, we infer that they are Borel sets. Hence f is aBorel function.

Solution of Problem 4.97Since X is a Borel space, we can find a Polish space Y (see Defini-tion 2.150) and a homeomorphism f : X −→ Y such that Z = f(X) ∈B(Y ). By Problem 4.81, the function ϑ : M+

1 (X) −→ M+1 (Y ), defined

by

ϑ(μ)(A) = μ(f−1(A)

) ∀ A ∈ B(X)

is a homeomorphism too. Then

ϑ(M+

1 (X))

={m ∈ M+

1 (Y ) : m(Z) = 1}

= ξ−1Z ({1}),

where ξZ : M+1 (Y ) −→ [0, 1] is the function, defined by

ξZ(m) = m(Z) ∀ m ∈ M+1 (Y ).

From Problem 4.78, we know that ξZ is Borel. Hence

ξ−1Z ({1}) = ϑ

(M+

1 (X)) ⊆ M+

1 (Y ) is Borel.

Since ϑ is a homeomorphism, we conclude thatM+1 (X) is a Borel space

too.

Solution of Problem 4.98Let gm = max{g,−m} for m � 1. Then gm is still lower semicontinu-ous, it is bounded below and gm � g for all m � 1. Since μn

w−→ μ,by Problem 4.76, we have

lim infn→+∞

∫

X

gm dμn �∫

X

gm dμ ∀ m � 1.


Note that

supn�1

( ∫

X

gm dμn−∫

X

g dμn

)� sup

n�1

∫

{g−�m}

g− dμn −→ 0 asm → +∞.

Recalling that gm � g for all m � 1, we conclude that

∫

X

g dμ � lim infn→+∞

∫

X

g dμn.

Applying this result to g = f and g = −f , we obtain that

limn→+∞

∫

X

f dμn =

∫

X

f dμ.

Solution of Problem 4.99For every k � 1, let

fk = min{f, k} and Ck = {f � k}.

Clearly fk is continuous and bounded and Ck is closed. We have

lim supn→+∞

∫

{f�k}f dμn = lim sup

n→+∞

( ∫

X

(f − fk) dμn + kμn(Ck))

�∫

X

(f − fk) dμ + kμ(Ck) =

∫

Ck

f dμ

[see Problem 4.98 and the Portmanteau theorem (Theorem 4.26)]. Forany ε > 0, we can find k � 1 large enough such that

∫

Ck

f dμ < ε.

4.3. Solutions 785

Solution of Problem 4.100Note that Gr f−1 = u(Gr f), where u : X × Y −→ Y × X is thehomeomorphism, defined by

u(x, y) = (y, x).

Hence Gr f−1 ∈ B(Y × X) = B(Y ) ⊗ B(X) if and only if Gr f ∈B(X × Y ) = B(X)⊗B(Y ) (see Theorem 4.6). Then the result followsfrom Corollary 4.38.

Solution of Problem 4.101Let h : A −→ X be the isomorphism between the measurable spaces(A,B(A)) and (X,Σ). Let D ∈ Σ. Then h−1(D) ∈ B(A) and so thereexists a set E ∈ B(Z) such that

h−1(D) = E ∩A,

which is a Souslin (analytic) set in Z (see Definition 2.156). Then

f(D) = (f ◦ h)(E ∩A)

and the latter is a Souslin set (see Corollary 4.38).

Solution of Problem 4.102First suppose that f is a characteristic functions, i.e., f = χ

Afor some

A ∈ B(X). Then

γf (μ) = γχA(μ) =

∫

X

χAdμ = μ(A),

so the function μ �−→ γf (μ) is Borel measurable by Problem 4.91.The linearity of the integral implies the measurability of γf , when

f is a simple function. Finally let f be a general Borel measurablefunction. Then according to Theorem 3.68, we can find a sequence{sn}n�1 of simple functions such that

0 � |s1| � . . . � |sn| � . . . � |f |


and

sn(x) −→ f(x) μ-almost everywhere on X.

Then by the Lebesgue monotone convergence theorem (see Theo-rem 3.92), we have

γsn(μ) =

∫

X

sn dμ −→∫

X

f dμ = γf (μ),

so the function μ �−→ γf (μ) is Borel measurable.

Solution of Problem 4.103From Corollary 4.38, we know that

Gr f ∈ B(RN ×RM)

= B(RN)⊗ B(RM).

Note that f(D) = projRM

Gr f . The projection on RM is a continuous

map and the continuous image of Souslin sets in RN×R

M (in particularof Borel sets in R

N × RM ) are still Souslin. Therefore f(D) ⊆ R

M isa Souslin set (see Definition 2.156).

Solution of Problem 4.104Let U ⊆ X be an nonempty open set. The complete regularity ofX implies that for every x ∈ U , we can find a continuous functionfx : X −→ [0, 1] such that

fx(x) = 1 and fx|Uc = 0.

Let

Ux ={u ∈ X : fx(u) > 0

}.

Then {Ux}x∈U is an open cover of X. Recall that a Souslin spaceis strongly Lindeloff (see Definitions 2.156 and 2.163 and Corol-lary 2.165). So, we can find a countable subcover

{Uxn

}n�1

of U .Let

f =∑

n�1

12n fxn.

4.3. Solutions 787

Evidently f is continuous and f |Uc = 0. For every u ∈ U , we have

u ∈ Uxn for some n � 1 and so fxn(u) > 0. Hence

U ={u ∈ X : f(u) > 0

}

and this proves that X is perfectly normal (see Definition 2.137).

Solution of Problem 4.105First we show that every set of the form

(�){x ∈ X :

{fk(x) : k � 1

} ∈ B}.

is Baire. Note that this is true if B is closed. Indeed, since R∞

is a Polish space, B has the form B = η−1(0) for some continuousfunction η : R∞ −→ R and the function x �−→ η

({fn(x) : n � 1

})is

continuous. Let us fix a sequence {fn}n�1 and let

D0 ={B ∈ B(R∞) :

{x ∈ X : {fn(x) : n � 1} ∈ B} ∈ Ba(X)

}.

It is easy to check that D0 is a σ-algebra. Hence it contains B(R∞)and so D0 = B(R∞). On the other hand, the class X of all sets Arepresentable in the form (�) with fk ∈ Cb(X) contains all sets of theform {f > 0} with f ∈ C(X). In addition, it is easy to see that X isa σ-algebra. Hence X = Ba(X).

Solution of Problem 4.106By Problem 4.105, we may assume that X = R

∞. Since R∞ is a Polishspace, we can find a closed set C ⊆ B and an open set U ⊇ B suchthat |μ|(U \C) < ε

2 . Let ξ : X −→ [0, 1] be a continuous function suchthat

ξ∣∣C

= 1 and ξ∣∣Uc

= 0.

Clearly ξ is the required function.


Solution of Problem 4.107Let C ⊆ X be a closed set. Then for (ω, x) ∈ Ω × X, we have thatf(ω, x) ∈ C if and only if for every n � 1, there exists x′ ∈ D suchthat

dX(x, x′) � 1

n and dY

(f(ω, x), f(ω, x′)

)� 1

n .

Therefore, we have

f−1(C) =⋂

n�1

⋃

x′∈D

{ω ∈ Ω : dist

Y(f(ω, x′), C) � 1

n

}

×{x ∈ X : d

X(x, x′) � 1

n

},

so f−1(C) ∈ Σ× B(X) and thus f is jointly measurable.

Solution of Problem 4.108Let U be an open subset of X. Since X is a metric space, we havethat U is an Fσ-set (see Problem 1.79). Hence

U =⋃

n�1

Cn,

with Cn closed in X. Then

G−(U) =⋃

n�1

G−(Cn).

But by hypothesis

G−(Cn) ∈ A ∀ n � 1

and A is closed under countable unions. Therefore, we conclude thatG−(U) ∈ A.

Solution of Problem 4.109For any open set U ⊆ X and any integer k � 1, we set

Uk ={x ∈ U : dist

X(x,U c) > 1

k

}.

4.3. Solutions 789

Since U is open, we have U =⋃

k�1

Uk =⋃

k�1

Uk. Then, for every ω ∈ Ω,

we have the following implications:

f(ω) ∈ U =⇒ ∃k � 1 : limn→+∞ fn(ω) ∈ Uk

=⇒ ∃k � 1 ∃N � 1 ∀n � N : fn(ω) ∈ Uk

=⇒ ∃k � 1 : f(ω) ∈ Uk

=⇒ f(ω) ∈ U.

Therefore, it follows that f−1(U) =⋃

k�1

lim infn→+∞ f−1

n (Uk), so f−1(U) ∈ Σ

and thus f is Σ-measurable.


“=⇒”: Immediate from the definition of SpF .

“⇐=”: According to Theorem 4.59, we can find{fn : Ω −→ X

}n�1

,a sequence of Σ-measurable selections of F such that

F (ω) ⊆ {fn(ω)

}n�1

μ-almost everywhere on Ω.

Thenm(ω) = inf

n�1

∥∥fn(ω)

∥∥ for μ-almost all ω ∈ Ω,

so m ∈ Lp(Ω)+.Let ε : Ω −→ R+\{0} be a Σ-measurable function such that ε ∈ Lp(Ω).We introduce the multifunction

Sε(ω) ={x ∈ F (ω) : ‖x‖ � m(ω) + ε(ω)

}.

Evidently GrSε ∈ Σ ⊗ B(X) and so we can apply the Yankov–vonNeumann–Aumann selection theorem (see Theorem 4.57; see also Re-mark 4.58) and obtain a Σ-measurable function f : Ω −→ X such thatf(ω) ∈ Sε(ω) μ-almost everywhere on Omega. Then f ∈ Sp

F and soSpF = ∅.


Solution of Problem 4.111It suffices to show that for every ϑ ∈ R, the set

{ω ∈ Ω : m(ω) < ϑ

}

belongs in the σ-algebra Σ. Note that

{ω ∈ Ω : m(ω) < ϑ

}= proj

Ω

{(ω, x) ∈ GrF : ξ(ω, x) < ϑ

}.

But the joint measurability of ξ and the graph measurability of Fimply that

{(ω, x) ∈ GrF : ξ(ω, x) < ϑ

} ∈ Σ⊗ B(X).

Then invoking the Yankov–von Neumann–Aumann projection theorem(see Theorem 4.65), we infer that

projΩ

{(ω, x) ∈ GrF : ξ(ω, x) < ϑ

} ∈ Σ

and so for every ϑ ∈ R, we have

{ω ∈ Ω : m(ω) < ϑ

} ∈ Σ,

which proves that m is Σ-measurable.

Solution of Problem 4.112Let λ ∈ R and let

Lλ ={x ∈ X : m(x) < λ

}.

ThenLλ = proj

X

{(x, y) ∈ X × Y : f(x, y) < λ

}.

Invoking the Yankov–von Neumann–Aumann projection theorem (seeTheorem 4.65), we conclude thatm is measurable with respect to everyBorel measure on X. Similarly for M .

Solution of Problem 4.113From Theorem 4.55, we know that there is a sequence {fn}n�1 ofΣ-measurable selections of F , i.e., for all n � 1, fn : Ω −→ X isΣ-measurable and fn(ω) ∈ F (ω) for all ω ∈ Ω) such that

F (ω) ={fn(ω)

}n�1

∀ ω ∈ Ω.

4.3. Solutions 791

The upper semicontinuity of ξ(ω, ·) for all ω ∈ Ω implies that

m(ω) = infn�1

ξ(ω, fn(ω)

).

But since ξ is jointly measurable, for every n � 1, the function ω �−→ξ(ω, fn(ω)

)is Σ-measurable. Hence the function m is Σ-measurable

too.

Solution of Problem 4.114From Problem 4.111, we know that the function

ω �−→ η(ω) = supx∈F (ω)

ξ(ω, x)

is Σ-measurable and so the integral∫Ω η(ω) dμ is well defined (possibly

+∞).For every u ∈ Sp

F , we have

ξ(ω, u(ω)

)� η(ω) μ-almost everywhere

and so

supu∈Sp

F

Iξ(u) �∫

Ω

η dμ.

In particular, we have

−∞ < Iξ(u0) �∫

Ω

η dμ.

If Iξ(u0) = +∞, then there is nothing to prove. So, we may assumethat Iξ(u0) < +∞. This means that ξ

(·, u0(·)) ∈ L1(Ω). Let

β <

∫

Ω

η dμ.

If we can find u ∈ SpF such that

Iξ(u) > β,


then we have the desired equality of the problem. To this end let{Ωn}n�1 ⊆ Σ be an increasing sequence such that

Ω =⋃

n�1

Ωn and μ(Ωn) < +∞ ∀ n � 1.

Also, let δ ∈ L1(Ω) be such that

δ(ω) > 0 ∀ ω ∈ Ω.

We setAn = Ωn ∩ {

ω ∈ Ω : ξ(ω, u0(ω)

)� n

} ∈ Σ

and

ηn(ω) =

⎧⎪⎨

⎪⎩

η(ω)− δ(ω)n if ω ∈ An and η(ω) � n,

n− δ(ω)n if ω ∈ An and η(ω) > n,

ξ(ω, u0(ω)

)if ω ∈ Ac

n.

Evidently ηn ∈ L1(Ω) and ηn ↗ η in μ-measure. Hence by passing toa suitable subsequence if necessary, we may assume that

ηn(ω) −→ η(ω) μ-almost everywhere on Ω

(see Proposition 3.131). Invoking the Lebesgue monotone convergencetheorem (see Theorem 3.92), we can find n0 � 1 such that

β <

∫

Ω

ηn0 dμ.

Let G : Ω −→ 2X be defined by

G(ω) = F (ω) ∩ {x ∈ X : ηn0(ω) � ξ(ω, x)

}.

By modifying G on a μ-null set, we may assume that

G(ω) = 0 ∀ ω ∈ Ω.

The joint measurability of ξ and the graph measurability of F implythat GrG ∈ Σμ⊗B(X). Invoking the Yankov–von Neumann–Aumannselection theorem (see Theorem 4.57), we obtain a Σμ-measurable se-lection g : Ω −→ X of G. Let

Cn = Ωn ∩ {ω ∈ Ω :

∥∥g(ω)

∥∥ � n

} ∈ Σμ ∀ n � 1

4.3. Solutions 793

and

fn = χCng + χ

Ccnu0 ∀ n � 1.

Evidently fn ∈ SpF for n � 1 and

Iξ(fn) =

∫

Cn

ξ(ω, g(ω)

)dμ+

∫

Ccn

ξ(ω, u0(ω)

)dμ

�∫

Ω

ηn0(ω) dμ +

∫

Ccn

(ξ(ω, u0(ω)

)− ηn0(ω))dμ.

Note that

μ(Ccn) ↘ 0

and recall that ∫

Ω

ηn0(ω) dμ > β.

So, for some n1 � 1, we will have β < Iξ(fn1) as desired.

Solution of Problem 4.115By Problem 4.113, the function ω �−→ ∣

∣F (ω)∣∣ is Σ-measurable. More-

over, Problem 4.114 implies that

∫

Ω

∣∣F (ω)

∣∣p dμ =

∫

Ω

supx∈F (ω)

‖x‖p dμ = supf∈Sp

F

‖f‖pp.

Therefore, the set SpF ⊆ Lp(Ω;X) is bounded if and only if

∣∣F (·)∣∣ ∈

Lp(Ω)+.

Solution of Problem 4.116By Theorem 4.51(b), it suffices to show that for every closed setC ⊆ X , we have F−(C) ∈ Σ. Since X is σ-compact (see Defini-tion 2.99), we can find a sequence {Kn}n�1 of compact subsets ofX such that

X =⋃

n�1

Kn.


We have

F−(C) = F−( ⋃

n�1

(C ∩Kn))

=⋃

n�1

F−(C ∩Kn) ∈ Σ.


“=⇒”: By Problem 4.116, it suffices to show that for every K ∈Pk(Y ), we have that G−(K) ∈ Σ⊗ B(X). By definition

G−(K) = {(ω, x) ∈ Ω×X : ∃y ∈ Y such that (x, y) ∈ F (ω)∩×(X ×K)}.

Let

AK ={ω ∈ Ω : F (ω) ∩ (X ×K) = ∅}.

Then Theorem 4.51(d) implies that AK ∈ Σ.Let S : AK −→ Pf (X × Y ) be defined by

S(ω) = F (ω) ∩ (X ×K).

Then

GrS = GrF ∩ (Ω×X ×K) ∈ Σ⊗ B(X)⊗ B(Y )

and so again by Theorem 4.51(d), we infer that S is a measur-able multifunction. Hence Theorem 4.55(b) implies that we can findtwo sequences of Σ-measurable functions

{fn : AK −→ X

}n�1

and{gn : AK −→ Y

}n�1

such that

S(ω) ={(

fn(ω), gn(ω))}

n�1∀ ω ∈ Ω.

Let

DK ={(ω, x) ∈ AK ×X : x ∈ {

fn(ω)}n�1

}.

We claim that G−(K) = DK . Clearly G−(K) ⊆ DK . We will showthat the opposite inclusion also holds. To this end, let (ω, x) ∈ DK .Then x = lim

k→+∞fnk

(ω) for some subsequence {nk}k�1 of {n}n�1.

Consider the corresponding subsequence of{gn(ω)

}n�1

. We have

4.3. Solutions 795

{gnk

(ω)}k�1

⊆ K and so by passing to a further subsequence if neces-sary, we may assume that

gnk(ω) −→ y ∈ K.

So (x, y) ∈ S(ω), which means that (ω, x) ∈ G−(K) and soG−(K) = DK . Now note that

DK ={(ω, x) ∈ AK ×X : inf

n�1d(x, fn(ω)

)= 0

} ∈ Σ⊗B(X).

Therefore, G−(K) ∈ Σ⊗ B(X) which proves the measurability of G.

“⇐=”: Note that GrG = GrF and by hypothesis

GrG ∈ Σ⊗ B(X)⊗ B(Y ) = Σ⊗ B(X × Y )

(see Theorem 4.6(b)). Invoking Theorem 4.51(c), we conclude that Fis measurable.


F (ω) = u−1(ω) ∀ ω ∈ Ω.

By hypothesis (a), the values of F are in Pf

(X). Moreover, hypoth-

esis (b) implies that for every open set V ⊆ X, we have F−(V ) =u(V ) ∈ Σ. Hence F is measurable (see Definition 4.49(a)). Therefore,we can apply the Kuratowski–Ryll Nardzewski selection theorem (seeTheorem 4.53) and find a Σ-measurable function f : Ω −→ X suchthat

f(ω) ∈ F (ω) ∀ ω ∈ Ω.

Then

u(f(ω)

)= ω ∀ ω ∈ Ω.


Solution of Problem 4.119Since U is a measurable multifunction, by Theorem 4.55, we can finda sequence

{un : Ω −→ X

}n�1

of Σ-measurable selections of U suchthat

U(ω) ={un(ω)

}n�1

∀ ω ∈ Ω.

Let V ⊆ Y be a nonempty open set. Because of the continuity ofg(ω, ·) for all ω ∈ Ω, we have

G−(V ) ={ω ∈ Ω : g

(ω,U(ω)

) ∩ V = ∅}

=⋃

n�1

{ω ∈ Ω : g

(ω, un(ω)

) ∈ V}.

But from Theorem 4.166 for every n � 1, the function ω �−→g(ω, un(ω)

)is (Σ,B(Y ))-measurable, hence

{ω ∈ Ω : g

(ω, un(ω)

)} ∈ Σ ∀ n � 1.

Therefore G−(V ) ∈ Σ.

Solution of Problem 4.120Let F : Ω −→ 2T \ {∅} be the multifunction, defined by

F (ω) ={t ∈ U(ω) : h(ω) = g(ω, t)

}.

We have

GrF = GrU ∩ {(ω, t) ∈ Ω× T : h(ω) = g(ω, t)

} ∈ Σ⊗ B(T ).

Invoking the Yankov–von Neumann–Aumann selection theorem(see Theorem 4.57), we can find a

(Σ,B(T ))-measurable function

u : Ω −→ T such that

u(ω) ∈ F (ω) ∀ ω ∈ Ω.

4.3. Solutions 797

Then

u(ω) ∈ U(ω) and h(ω) = g(ω, u(ω)

) ∀ ω ∈ Ω.

Solution of Problem 4.121Let U ⊆ X be a nonempty open set. We know that U is the countableunion of open balls. Moreover, every open ball is a countable unionof closed balls. So, U is a countable union of closed balls, i.e., U =⋃

n�1B(n) with each B(n) being a closed ball in X. Then

F−(U) =⋃

n�1

F−(B(n)) ∈ Σ,

hence F is measurable.

Solution of Problem 4.122The set U c is closed. Let

ξ(ω, x) = dist(f(ω, x), U c

) ∀ (ω, x) ∈ Ω×X.

Evidently ξ is a Caratheodory function, thus measurable (see Theo-rem 4.166). We have

GrF ={(ω, x) ∈ Ω×X : ξ(ω, x) > 0

}

and so GrF ∈ Σ⊗ B(X), i.e., F is graph measurable.

Solution of Problem 4.123Let {xn}n�1 be dense in X. We have

F−(C) ={ω ∈ Ω : F (ω) ∩ C = ∅}

={ω ∈ Ω : f(ω, x) ∈ U for some x ∈ C

}

={ω ∈ Ω : f(ω, xn) ∈ U for some n � 1

}

=⋃

n�1

{ω ∈ Ω : f(ω, xn) ∈ U

}


(since U is open and f(ω, ·) is continuous). Since for every n � 1, thefunction f(·, xn) is Σ-measurable, we have that

{ω ∈ Ω : f(ω, xn) ∈

U} ∈ Σ. Hence

F−(C) =⋃

n�1

{ω ∈ Ω : f(ω, xn) ∈ U

} ∈ Σ.

Solution of Problem 4.124Let D be a countable dense subset of X. For any open set V ⊆ X andeach open interval I = (a, b) ⊆ R, we have

E−(V × I) ={ω ∈ Ω : E(ω) ∩ (V × I) = ∅}

={ω ∈ Ω : epi f(ω, ·) ∩ (V × I) = ∅}

={ω ∈ Ω : f(ω, x) < b for some x ∈ D ∩ V

}

=⋃

u∈D∩V

{ω ∈ Ω : f(ω, u) < b

} ∈ Σ

(due to the continuity of f(ω, ·) and since f(·, x) is measurable).Now recall that each open set U ⊆ X ×R can be written as count-

able union of open sets of the form V × I, since R has countable basisconsisting of open intervals.

Solution of Problem 4.125Let f ∈ S1

F . By considering ω �−→ F (ω) − f(ω) if necessary, we mayassume that for all ω ∈ Ω, we have 0 ∈ F (ω). Let ϑ : Ω −→ R+ be thefunction, defined by

ϑ(ω) = dist(0, F (ω)c)

(recall that F (ω)c = X\F (ω)). First we show that ϑ is Σμ-measurable.To this end, let λ > 0 and let us set

Lλ ={ω ∈ Ω : ϑ(ω) < λ

}

G ={(ω, x) ∈ Ω×X : x ∈ Bλ ∩ F (ω)c

},

where Bλ ={x ∈ X : ‖x‖ < λ

}. Then G = (Ω × Bλ) ∩ GrF c ∈

Σ ⊗ B(X) (since F is graph measurable). Note that Lλ = projΩG.

4.3. Solutions 799

Hence by the Yankov–von Neumann–Aumann projection theorem (seeTheorem 4.65), we have Lλ ∈ Σμ and so ϑ is Σμ-measurable. Since Fhas open values, ϑ(ω) > 0 for all ω ∈ Ω. The Σμ-measurability of ϑimplies that we can find ε > 0 and A ∈ Σ with μ(A) > 0 such thatϑ(ω) � ε for all ω ∈ A. Then Bε ⊆ F (ω) for all ω ∈ A and so

μ(A)Bε ⊆∫

A

F dμ ⊆∫

Ω

F dμ.

Solution of Problem 4.126From the Riesz representation theorem (see Theorem 4.67), we haveLp(Ω)∗ = Lp′(Ω) (where 1

p + 1p′ = 1). Let h ∈ Lp′(Ω). For a given

ε > 0, we can find a simple function s0 ∈ Lp′(Ω) such that

‖h− s0‖p′ � ε4M .

Since ∫

A

fn dμ −→∫

A

f dμ ∀ A ∈ Σ,

from the linearity of the integral, it follows that

∫

Ω

fns0 dμ −→∫

Ω

fs0 dμ.

Then we can find n0 = n0(ε) � 1 such that

∣∣h(fn − f)∣∣ �

∣∣(h− s0)(fn − f)∣∣+

∣∣s0(fn − f)∣∣

� ‖h− s0‖p′‖fn − f‖p + ε2

� 2M ε4M + ε

2 = ε ∀ n � n0,

so∫

Ω

(fn−f)hdμ −→ 0 for all h ∈ Lp′(Ω) and thus fnw−→ f in Lp(Ω).


Solution of Problem 4.127Let {Ωn}n�1 ⊆ Σ be a sequence such that Ωn ↗ Ω and μ(Ωn) < +∞for every n � 1. Let

Cn ={ω ∈ Ωn :

∣∣f(z)∣∣ � n

} ∈ Σ ∀ n � 1.

We set fn = χCn

f and consider the linear functionals on Lp(Ω), definedby

Ln(g) =

∫

Ω

fng dμ and L(g) =

∫

Ω

fg dμ ∀ g ∈ Lp(Ω).

From the Riesz representation theorem (see Theorem 4.67), we havethat Ln is continuous. Note that

∣∣fn(ω)g(ω)

∣∣ �

∣∣f(ω)g(ω)

∣∣ μ-almost everywhere in Ω

with fg ∈ L1(Ω) and fng −→ fg μ-almost everywhere in Ω. So, bythe Lebesgue dominated convergence theorem (see Theorem 3.94), wehave ∫

Ω

fng dμ −→∫

Ω

fg dμ,

hence Ln(g) −→ L(g). Thus L is continuous either by invoking Corol-lary 5.42 or directly as follows. Let gk −→ g in Lp(Ω). Then wehave

Ln(gk) −→ Ln(g) as k → +∞(continuity of Ln) and

Ln(g) −→ L(g) as n → +∞.

Invoking Problem 1.175, we can find a sequence k �−→ n(k) such that

Ln(k)(gk) −→ L(g) as k → +∞.

Then∣∣L(gk)− L(g)

∣∣ �

∣∣L(gk)− Ln(k)(gk)

∣∣

+∣∣Ln(k)(gk)− Ln(k)(g)

∣∣ +

∣∣Ln(k)(g) − L(g)

∣∣.

We have

∣∣L(gk)−Ln(k)(gk)∣∣ =

∣∣∫

Ω

(f−fnk)gk dμ

∣∣ �∫

Ω

(1−χCn(k)

)|fgk| dμ −→ 0

4.3. Solutions 801

(by Theorem 3.94) and

∣∣Ln(k)(gk)−Ln(k)(g)

∣∣ =

∣∣∫

Ω

fnk(gk−g) dμ

∣∣ � ‖f‖p′‖gk−g‖p −→ 0

and ∣∣Ln(k) − L(g)

∣∣ −→ 0 as k → +∞.

Therefore, finally we have∣∣L(gk)− L(g)

∣∣ −→ 0 as k → +∞,

so L is continuous and so f ∈ Lp′(Ω) (by Theorem 4.67).

Solution of Problem 4.128For every n � 1, we set

μn(A) =

∫

A

fn dμ ∀ A ∈ Σ.

Then {μn}n�1 ⊆ M(Σ) and μn � μ for all n � 1 (see Definition 3.150).By hypothesis we know that for every A ∈ Σ, the limit lim

n→+∞μn(A)

exists and is finite. By Theorem 4.73, we have that the sequence{μn}n�1 is uniformly μ-absolutely continuous (see Definition 4.71).Hence, for a given ε > 0, we can find δ = δ(ε) > 0 such that for allA ∈ Σ, we have

if |μ|(A) < δ, then∣∣μn(A)

∣∣ =∣∣∫

A

fn dμ∣∣ < ε

2 for all n � 1.

For each such A, we set

A+n =

{a ∈ A : fn(x) � 0

}and A−

n ={a ∈ A : fn(x) < 0

}.

Evidently

A+n ∈ Σ, A−

n ∈ Σ, μ(A+n ) � μ(A) < δ, μ(A−

n ) � μ(A) < δ.

So, we have∫

A+n

|fn| dμ =∣∣∫

A+n

fn dμ∣∣ < ε

2 and

∫

A−n

|fn| dμ =∣∣∫

A−n

fn dμ∣∣ < ε

2 ,


hence∫

A

|fn| dμ =

∫

A+n

|fn| dμ +

∫

A−n

|fn| dμ < ε2 +

ε2 = ε.

Solution of Problem 4.129Let h ∈ L∞(X) = L1(X)∗ (see Theorem 4.67). We claim that

{〈fn, h〉 =

∫

X

fnhdμ

}

n�1

is a Cauchy sequence.

To this end let ε > 0 and let s be a simple function such that

‖h− s‖∞ < ε3M

(see Proposition 3.110). We have

∣∣∫

X

fnhdμ −∫

X

fns dμ∣∣ � ‖h− s‖∞‖fn‖1 < εM

3 ∀ n � 1.

Also, by hypothesis the limit

limn→+∞

∫

X

fns dμ

exists and is finite. So, we can find n0 � 1 such that

∣∣∫

X

fns dμ−∫

X

fms dμ∣∣ < ε

3 ∀ n,m � n0,

so∣∣∫

X

fnhdμ −∫

X

fmhdμ∣∣ < ε ∀ n,m � n0

and thus{〈fn, h〉 =

∫

X

fnhdμ

}

n�1

is a sequence Cauchy.

4.3. Solutions 803

So, for every h ∈ L∞(X), the limit

limn→+∞

∫

X

fnhdμ

exists and is finite. Let us set

ξ(h) = limn→+∞

∫

X

fnhdμ ∀ h ∈ L∞(X).

Hence ξ ∈ (L∞(X)

)∗(see hypothesis (ii)). Moreover, if

νn(A) =

∫

A

fn dμ,

then {νn}n�1 ⊆ M(Σ) and νn(A) −→ ν(A) = ξ(χA) and so ν ∈ M(Σ)

(see the Nikodym theorem; Theorem 4.72) and ν � μ (see Def-inition 3.150). Then by the Radon–Nikodym theorem (see Theo-rem 3.152), we can find f ∈ L1(X) such that

ν(A) =

∫

A

f dμ ∀ A ∈ Σ,

so

ξ(s) =

∫

X

fs dμ for every simple function s.

As simple functions are dense in L∞(X) (see Proposition 3.110), wehave

ξ(h) =

∫

X

fh dμ,

so

〈fn, h〉 −→ 〈f, h〉 ∀ h ∈ L∞(X)

and thus

fnw−→ f in L1(X).


Solution of Problem 4.130Arguing by contradiction, suppose that the sequence {μn}n�1 isnot uniformly μ-absolutely continuous. So, we can find a sequence{Ak}k�1 ⊆ Σ and a subsequence {nk}k�1 of {n}n�1 such that

μ(Ak) < 12k+1 and μnk

(Ak) � ε ∀ k � 1

(see Definition 4.71(c)). Let

Ck =⋃

i�k

Ai ∀ k � 1.

Then {Ck}k�1 is a decreasing Σ-sequence and

μ(Ck) �∑

i�k

μ(Ai) <∑

i�k

12i+1 = 1

2k.

LetC0 =

⋂

k�1

Ck

and let us setEk = Ck \ C0 ∀ k � 1.

Then {Ek}k�1 is a decreasing Σ-sequence and

⋂

k�1

Ek = ∅.

Since by hypothesis the sequence {μn}n�1 is Vitali equicontinuous, wecan find k0 � 1 such that

∣∣μnk

(Ek)∣∣ < ε ∀ k � k0.

Becauseμnk

� μ and μ(Ck) < 12k,

we have μ(C0) = 0. Therefore

∣∣μnk

(Ck)∣∣ < ε ∀ k � k0.

But we know that

μnk(Ck) � μnk

(Ak) � ε ∀ k � 1.

4.3. Solutions 805

Hence, we have reached a contradiction. This proves that the sequence{μn}n�1 is uniformly μ-absolutely continuous.

Solution of Problem 4.131Let ε > 0 and consider a decreasing sequence {An}n�1 ⊆ Σ such that

⋂

n�1

An = ∅.

By hypothesis, we can find δ = δ(ε) > 0 such that

if A ∈ Σ, μ(A) < δ, then∣∣μk(A)

∣∣ < ε ∀k � 1.

Since μ(X) < +∞, we have μ(An) −→ 0 (see Theorem 3.19(c)). So,we can find n0 � 1 such that

μ(An) < δ ∀ n � n0.

Hence ∣∣μk(An)

∣∣ < ε ∀ n � n0, k � 1

and this proves that the sequence {μn}n�1 is Vitali equicontinuous.

Solution of Problem 4.132Let [a, b] = [0, 2π], p = 2 and consider the sequence

fn(t) = cosnt ∀ n � 1.

Then2π∫

0

cos2 nt dt = π ∀ n � 1.

So, fn −→ 0 in Lp(a, b). On the other hand, let h = χ[c,d]

, with[c, d] ⊆ [0, 2π]. Then

2π∫

0

χ[c,d]

cosnt dt = 1n

(sin(nd)− sin(nc)

) −→ 0.


From the linearity of the integral, it follows that

2π∫

0

s(t)fn(t) dt −→ 0

for every step function

s(t) =n∑

k=1

βkχ[ck,dk].

Finally, recall that step functions are dense in L2(0, 2π), to concludethat

2π∫

0

hfn dt −→ 0 ∀ h ∈ L2(0, 2π),

sofn

w−→ 0 in L2(0, 2π).

Solution of Problem 4.133By the Dunford–Pettis theorem (see Theorem 4.75), we can find asubsequence {unk

}k�1 of {un}n�1 and a function h ∈ L1(Ω) such that

unk

w−→ h in L1(Ω),

so ∫

A

unkdμ −→

∫

A

hdμ ∀ A ∈ Σ.

On the other hand, by the Vitali theorem (see Theorem 3.128), wehave ∫

A

un dμ −→∫

A

u dμ ∀ A ∈ Σ.

4.3. Solutions 807

Therefore ∫

A

hdμ =

∫

A

u dμ ∀ A ∈ Σ,

hence h = u. Hence by the Urysohn criterion (see Problem 1.3), forthe original sequence, we have un

w−→ u = h in L1(Ω).

Solution of Problem 4.134By the Lusin theorem (see Theorem 3.77), for a given ε > 0, we canfind a continuous function g such that

λ({f = g}) < ε

(λ being the Lebesgue measure on R). Let C = {f = g}. Then C isLebesgue measurable and according to Proposition 4.91, almost everypoint of C is a density point. Let x ∈ C be a density point of C. Then

limx′ → xx′ ∈ C

f(x′) = limx′→x

g(x′) = g(x) = f(x),

so f is approximately continuous on C and λ(Cc) < ε. Since ε > 0 wasarbitrary, we conclude that f is continuous at almost every point.

Solution of Problem 4.135By Remark 4.97, we can find a Lebesgue-null set D ⊆ (0, 1) such thatevery x ∈ (0, 1) \ D is a Lebesgue point of f (see Definition 4.89).Hence

limx ∈ [a, b]b− a ↘ 0

1b−a

b∫

a

f(x) dλ = f(x) ∀ x ∈ (0, 1) \D.

By hypothesis, for every x ∈ (0, 1)\D and every integer n � 1, we canfind an open interval In ⊆ (0, 1) such that

x ∈ In, λ(In) < 1n , and

∫

In

f dλ = 0.


Hence, it follows that f(x) = 0 and so

f(y) = 0 for almost all y ∈ (0, 1).

Therefore∫

I

f dλ = 0 for all open interval I ⊆ (0, 1).

Solution of Problem 4.136No. By Proposition 4.91, for any Lebesgue measurable set A ⊆ R, wehave

λ(A∩(x−h,x+h))2h −→ 1 as h ↘ 0, for almost all x ∈ A

and

λ(A∩(x−h,x+h))2h −→ 0 as h ↘ 0, for almost all x ∈ A.

Solution of Problem 4.137According to Theorem 4.86, the set [0, 1] \ S is Lebesgue-null. Letε > 0 and let x ∈ S. We choose a rational r such that

∣∣r − f(x)

∣∣ < ε.

If h > 0, then

1h

x+h∫

x

∣∣f(t)− f(x)∣∣ dt � 1

h

x+h∫

x

∣∣f(t)− r∣∣ dt+ 1

h

x+h∫

x

∣∣r − f(x)∣∣ dt

� 1h

x+h∫

x

∣∣f(t)− r∣∣ dt+ ε.

Let h ↘ 0. Then

1h

x+h∫

x

∣∣f(t)− r∣∣ dt −→ ∣∣f(x)− r

∣∣ < ε,

4.3. Solutions 809

since x ∈ S. Since ε > 0 is arbitrary, we conclude that

limh→0+

1h

x+h∫

x

∣∣f(t)− f(x)∣∣ dt = 0.

Similarly, we also show that

limh→0+

1h

x∫

x−h

∣∣f(t)− f(x)

∣∣ dt = 0,

so, from Theorem 4.86, we get that x ∈ Lf . Thus S ⊆ Lf .

Solution of Problem 4.138Let ξn ↘ 0 and let An ⊆ A be the set of points of A at which theupper density of C is greater than ξn. Let ε > 0. Since by hypothesisA and C are metrically separated, we can find two open sets U, V ⊆ R

such thatA ⊆ U, C ⊆ V, λ(U ∩ V ) < ε.

For each x ∈ An there is a sequence of closed intervals {Ik}k�1 suchthat

x ∈ Ik, λ(Ik) ↘ 0 and λ(C∩Ik)λ(Ik)

> ξn ∀ k � 1.

The collection of such intervals for all x ∈ An is a Vitali cover ofAn (see Definition 4.77(b)). So, by the Vitali covering theorem (seeTheorem 4.79), we can find pairwise disjoint intervals {Ii}mi=1 such that

λ(An)− ε �m∑

i=1

λ(An ∩ Ii) �m∑

i=1

λ(Ii) � λ(An) + ε.

Since Ii ⊆ U and C ⊆ V , we have

ξn(λ(An)− ε

)� ξn

m∑

i=1

λ(Ii) <m∑

i=1

λ(C ∩ Ii) � λ(U ∩ V ) < ε.

Since ε > 0 is arbitrary, we let ε ↘ 0 to conclude that λ(An) = 0.


Note that⋃

n�1An is the subset of A at which the upper density of

C is nonzero. Evidently⋃

n�1An is a Lebesgue-null set. Since the lower

density is always less than or equal to the upper density, we concludethat at almost all points of A the density of C is zero. Similarly, if weinterchange the roles of A and C.

Solution of Problem 4.139Let D be the set of points at which the density of A exists. LetS ⊆ D be those points at which the density is equal to 0 or 1. FromProposition 4.91, we know that

λ(D \ S) = 0

(λ being the Lebesgue measure on R). For every positive integer n � 1,let

fn(x) = n2λ

(A ∩ (

x− 1n , x+ 1

n

)) ∀ x ∈ R.

Clearly fn is continuous and we have

f(x) = limn→+∞ fn(x) ∀ x ∈ D

and f(x) is the density of A at x.Hence by Problem 1.66, the discontinuity set of f |

Dis of first cate-

gory relative to D (see Definition 1.25). So, every interval I containinga point of D \ S also contains a point of S and at that point f is 0or 1. Since f |

D\S is never 0 or 1, it follows that D \ S ⊆ Disc (f) (thediscontinuity set of f). Therefore we conclude that D \ S is of firstcategory.

Solution of Problem 4.140By hypothesis, we have

1b−a

b∫

a

χAdx � ϑ.

4.3. Solutions 811

Invoking Proposition 4.96 (see also Remark 4.97), we infer that

χA(x) � ϑ for almost all x ∈ [0, 1],

so

χA(x) = 1 for almost all x ∈ [0, 1]

and thus λ(A) = 1.

Solution of Problem 4.141By the Radon–Nikodym theorem (see Theorem 3.152), we have f =dμdλ ∈ L1

([0, 1], λ

). Let x ∈ (0, 1) and h > 0 be such that 0 < x − h <

x+ h < 1 and let

Fh(x) =μ([0, 1] ∩ (x− h, x+ h))

λ([0, 1] ∩ (x− h, x+ h)).

We have

Fh(x) = 12h

x+h∫

x−h

f(t) dt.

Then Proposition 4.96 (see also Remark 4.97) implies that

limh→0

Fh(x) = f(x) for almost all x ∈ [0, 1].

Solution of Problem 4.142No. Because, if such set exists, then at every x ∈ A, the density of Awould be 1

2 (see Problem 4.138), which contradicts Proposition 4.91.

Alternative SolutionAnother explanation follows immediately from Problem 4.13.


Solution of Problem 4.143Suppose that

∣∣f(x)− f(u)∣∣ � k|x− u| ∀ x, u ∈ R,

for some k > 0. Since g is absolutely continuous, for a given ε > 0, wecan find δ > 0 such that

n∑

i=1

(bi − ai) < δ =⇒n∑

i=1

∣∣g(bi)− g(ai)

∣∣ < ε

k ,

where{(ai, bi)

}n

i=1are disjoint subintervals of [a, b]. We have

n∑

i=1

∣∣(f ◦ g)(bi)− (f ◦ g)(ai)∣∣ =

n∑

i=1

∣∣f(g(bi))− f(g(ai))∣∣

� k

n∑

i=1

∣∣g(bi)− g(ai)∣∣ < ε,

so f ◦ g is absolutely continuous.


(a) For every n � 1 and x ∈ [0, 1], we have

0 � fn(x) = n

x+ 1n∫

x

χA(t) dt � n 1

n = 1.

(b) For a, b ∈ [0, 1], a < b and every n � 1, we have

∣∣fn(b)− fn(a)∣∣ = n

∣∣

b+1n∫

a+1n

χA(t) dt−

b∫

a

χA(t) dt

∣∣

� n(

b+1n∫

a+1n

χA(t) dt+

b∫

a

χA(t) dt)

� 2n(b− a).

4.3. Solutions 813

So, if a1 < b1 � a2 < b2 � . . . � ak < bk � ak+1 < . . . � am < bm,ε > 0 and

m∑

k=1

(bk − ak) < ε2n ,

thenm∑

k=1

∣∣fn(bk)− fn(ak)

∣∣ < ε,

which proves that for every n � 1, the function fn is absolutely con-tinuous, i.e., fn ∈ AC

([0, 1]

).

(c) By Proposition 4.96 (see also Remark 4.97), we have

fn(x) −→ χA(x) for almost all x ∈ R.

(d) Using (c) and the Lebesgue dominated convergence theorem (seeTheorem 3.94), we have

∥∥fn − χA

∥∥1

=

∫

R

∣∣fn(x)− χA(x)

∣∣ dx −→ 0

and so we conclude that

fn −→ χA

in L1(R).


R0 = R \ {0} and C = A \ {0}.Then

qC = C and q(R0 \ C) = R0 \ C ∀ q ∈ Q \ {0}.

Suppose that

λ(R0 \ C) = λ(R \ A) > 0

(λ being the Lebesgue measure on R). Then since λ is a Radon measure(see Problem 4.28), we can find a compact set K ⊆ R0\C with positive


Lebesgue measure. For any compact subset E of C, we consider thefunction f : R0 −→ R, defined by

f(x) =

∫

R0

χK(y)χ

E−1

(xy

)1y dy ∀ x ∈ R0,

where E−1 ={1u : u ∈ E

}. Then f is continuous (as E−1 is compact)

and for any q ∈ Q \ {0}, we have

χE−1

( qy

)= χ

E

(yq

)= χ

qE(y),

so f(q) = 0 (as K ∩ qE ⊆ K ∩ C = ∅), thus f ≡ 0. Since

0 =

∫

R0

f(x) dx =

∫

R0

∫

R0

χK(y)χ

E−1

(xy

)1y dy dx

=

∫

R0

∫

R0

χK(y)χ

E−1 (t) dt dx =

∫

R0

χK(y) dy

∫

R0

χE−1 (t) dt,

we conclude that λ(E−1) = 0. As the function u �−→ 1u is lo-

cally Lipschitz, thus Lipschitz on E, from the N -property (see Def-inition 4.123), we infer that λ(E) = 0, from which we conclude thatλ(C) = λ(A) = 0.

Solution of Problem 4.146Let {Un}n�1 be a sequence of open sets in R such that

D ⊆ Un and λ(Un) < 12n ∀ n � 1

(λ being the Lebesgue measure on R). We set

h =∑

n�1

χUn

� 0.

Evidently h is integrable. We set

f(x) =

x∫

0

h(s) ds ∀ x ∈ [0, 1].

4.3. Solutions 815

Clearly f is increasing and absolutely continuous (see Theorem 4.127).Let x ∈ D. For every integer n � 1, there exists > 0 such that

if |y − x| < , then y ∈ Uk for all k ∈ {1, . . . , n}.Then, we have

f(y)− f(x)

y − x� n

and so f ′(x) = +∞.

Solution of Problem 4.147Clearly we need to check only the differentiability of f at x = 0. Wehave ∣

∣f(x)−f(0)x−0

∣∣ =

∣∣x cos

(1x2

)∣∣ � x ∀ x ∈ (0, 1]

so f ′(0) = 0.Consider the partition

Pn ={0,

(2

2nπ

) 12 ,

(2

(2n−1)π

) 12 , . . . ,

(23π

) 12 ,

(22π

) 12 , 1

}.

Then the variation of f with respect to this partition satisfies

cos 1 + 2π

n∑

k=2

1k � Var (f) ∀ n � 1,

hence Var (f) = +∞.


“⇐=”: By Theorem 4.127, we have

f(x)− f(a) =

x∫

a

f ′(t) dt = 0 ∀ x ∈ [a, b],

sof(x) = f(a) ∀ x ∈ [a, b],

i.e., f is a constant function.


“=⇒”: The function f being absolutely continuous, it is differentiablealmost everywhere on [a, b]. Since f is constant, we conclude that

f ′(x) = 0 almost everywhere on [a, b].

Solution of Problem 4.149Let g = χ

D. Then g ◦ f = χ

f−1(D). From the change of variables

formula, we have

b∫

a

χf−1(D)

f ′ dx = 0 for all intervals (a, b).

Letξ+ =

(χ

f−1(D)f ′)+ and ξ− =

(χ

f−1(D)f ′)−.

Thenb∫

a

ξ+ dx =

b∫

a

ξ− dx for all intervals (a, b),

soξ+ = ξ− for almost all x ∈ R,

thusf ′(x) = 0 for almost all x ∈ f−1(D).

Solution of Problem 4.150Without any loss of generality, we may assume that Var u > 0 (other-wise the conclusion is obvious). Let ϑ ∈ (0,Var u) and let P = {xk}mk=0

be a partition of [a, b] such that

ϑ <

m∑

k=1

∣∣u(xk)− u(xk−1)

∣∣

4.3. Solutions 817

(see Definition 4.9) Since un(xk) −→ u(xk) for all k ∈ {0, . . . ,m}, fora given ε > 0, we can find n0 � 1 such that

∣∣un(xk)− u(xk)

∣∣ � ε

2m ∀ k ∈ {0, . . . ,m}, n � n0.

Therefore for n � n0, we have

ϑ <

m∑

k=1

∣∣u(xk)− u(xk−1)

∣∣

�m∑

k=1

∣∣u(xk)− un(xk)

∣∣+

m∑

k=1

∣∣un(xk)− un(xk−1)

∣∣

+m∑

k=1

∣∣un(xk−1)− u(xk−1)

∣∣

�m∑

k=1

(∣∣un(xk)− un(xk−1)∣∣+ ε

m

)

=

m∑

k=1

∣∣un(xk)− un(xk−1)∣∣+ ε � Var un + ε,

soϑ � lim inf

n→+∞ Varun + ε.

Let ε ↘ 0 and ϑ ↗ Var u, to conclude that

Var u � lim infn→+∞ Var un.

Solution of Problem 4.151“=⇒”: Follows from Proposition 4.153.

“⇐=”: Clearly u is continuous and u′ ∈ L1(a, b) (see Proposi-tion 4.112). Let D ⊆ [a, b] be a Lebesgue-null set. Then by Prob-lem 3.48, we have

λ∗(u(D))

�∫

D

∣∣u′(t)

∣∣ dt = 0,


so the set u(D) is Lebesgue-null. Therefore u satisfies theLusin N -property. Invoking Corollary 4.126, we conclude that u ∈AC

([a, b]

).

Solution of Problem 4.152Since ϑ < Var u, we can find a partition

P ={a = t0 < t1 < . . . < tm = b

}

such that

ξ =

m∑

i=1

∣∣u(ti)− u(ti−1)∣∣ > ϑ.

Note that u is uniformly continuous. So, we can find δ > 0 such that

if |x− y| < δ, then∣∣u(x)− u(y)

∣∣ < ξ−ϑ

2m .

Consider a partition P ={a = x0 < x1 < . . . < xn = b

}such that

max1�k�n

|xk − xk−1| < δ. Then the contribution in the sum

n∑

k=1

∣∣u(xk)− u(xk−1)

∣∣

of the terms corresponding to k’s such that ti−1 � xk−1 < xk � ti isgreater than ∣

∣u(ti)− u(ti−1)∣∣− ξ−ϑ

m .

Hence, it follows that

n∑

k=1

∣∣u(xk)− u(xk−1)

∣∣ > ξ − (ξ − ϑ) = ϑ.

4.3. Solutions 819

Solution of Problem 4.153By Theorem 4.119, the Banach indicatrix function Nu ∈ L1(R). HereNu(y) ∈ R for almost all y ∈ R. This implies that λ(A) = 0 (λ beingthe Lebesgue measure on R).

Solution of Problem 4.154Since {u′n}n�1 ⊆ L1(0, 1) is uniformly integrable, by the Dunford–Pettis theorem (see Theorem 4.75) and by passing to a suitable sub-sequence if necessary, we may assume that u′n

w−→ v in L1(0, 1). FromTheorem 4.127, we know that

un(t)− un(s) =

t∫

s

u′n(τ) dτ ∀ n � 1, t, s ∈ [0, 1], s � t,

so

u(t)− u(s) =

t∫

s

v(τ) dτ

(since u′n −→ v in L1(0, 1)) and thus v = u′ and u ∈ AC([0, 1]

).

Solution of Problem 4.155The result follows from the following fact:

if x, y ∈ R and |x|, |y| � M, then∣∣|x|p − |y|p∣∣ � pMp−1|x− y|

(see Definition 4.120).


(a) Suppose that A ⊆ C. If s > H-dimC, then Theorem 4.155(f)implies that

Hs(A) � Hs(C) = 0

and so H- dimA � s (see Definition 4.156). Since s > H- dimC isarbitrary, we let s ↘ H-dimC, to conclude that

H- dimA � H- dimC.


(b) Let s > max{H- dimA, H-dimC}. Then s > H-dimA, s >H-dimC, hence

Hs(A) = Hs(C) = 0

(see Theorem 4.155(f)). We have

Hs(A ∪C) � Hs(A) +Hs(C) = 0

and so Hs(A ∪ C) = 0, which implies that

H- dim(A ∪ C) � s

(see Definition 4.156). Let s ↘ max{H- dimA, H-dimC}, to obtain

H- dim(A ∪ C) � max{H- dim, H- dimC}.On the other hand, from (a), we have

H- dimA � H- dim(A ∪ C) and H- dimC � H- dim(A ∪ C),

so finally

H- dim(A ∪C) = max{H- dimA, H- dimC}.


ξ(x) =∥∥x− γ(a)

∥∥ ∀ x ∈ R

N .

Then ξ : RN −→ R+ is nonexpansive (i.e., Lipschitz continuous withLipschitz constant 1) and so by Theorem 4.155(e), we have

H1(ξ(γ([a, b]

)))�H1

(γ([a, b]

)).

But H1 = λ (λ being the Lebesgue measure on R; see Theo-rem 4.155(a)) and the set γ

([a, b]

) ⊆ RN is connected (see Defini-

tion 2.104 and Theorem 1.90), therefore

ξ(γ([a, b]

))= [0, ϑ] ⊆ R.

So, we have

H1(ξ(γ([a, b]

)))= sup

t∈[a,b]ξ(γ(t)

)= sup

t∈[a,b]

∥∥γ(t)−γ(a)

∥∥ �

∥∥γ(b)−γ(a)

∥∥,

4.3. Solutions 821

thus ∥∥γ(b)− γ(a)

∥∥ � H1

(γ([a, b]

)).

Solution of Problem 4.158It is well known that such curve can be reparametrized by arc-lengthand so without any loss of generality we may assume that

γ′(t) = 0 almost everywhere on [a, b]

with a = 0 and b = Var γ. Moreover, γ is of bounded variation, sinceit is absolutely continuous, being Lipschitz continuous (hence γ is acontinuous, rectifiable simple curve).

For a given δ > 0, choose an integer n � 1 such that

1nVar γ < δ

and let

h = 1nVar γ and Tk =

[kh, (k + 1)h

]for k ∈ {0, . . . , n− 1}.

Since γ is 1-Lipschitz function (because we have reparametrized γ byarc-length), from Theorem 4.155(e), we have

H1δ

(γ([a, b]

))�

n−1∑

k=0

diamTk = Var γ.

Letting δ ↘ 0, we obtain

H1(γ([a, b]

))� Var γ.

On the other hand, let P ={a = t0 < t1 < . . . < tn = b

}be a

partition of [a, b]. From Problem 4.157, we have

n∑

k=1

∥∥γ(tk)− γ(tk−1)

∥∥ �

n∑

k=1

H1(γ([tk−1, tk]

))� H1

(γ([a, b]

))

(the last inequality being a consequence of the injectivity of γ andthe additivity of the Hausdorff measure). Since the partition P wasarbitrary, we obtain

Var γ � H1(γ([a, b]

)),


soH1

(γ([a, b]

))= Var γ.

Solution of Problem 4.159Let x ∈ C and consider the function ξ : RN −→ R+, defined by

ξ(y) = ‖y − x‖.Then ξ is nonexpansive and so by Theorem 4.155(e), we have

H1(ξ(C ∩Br(x)

))� H1

(C ∩Br(x)

).

moreover, we know that on R, we have H1 = λ (λ being the Lebesguemeasure on R; see Theorem 4.155(a)) and ξ

(C ∩Br(x)

)= [a, b] (being

connected and compact in R). Hence

H1(ξ(C∩Br(x)

))= sup

y,z∈C∩Br(x)

(‖y−x‖−‖z−x‖)= supy∈C∩Br(x)

‖y−x‖�r.

Solution of Problem 4.160Let {Uk}k�1 be a sequence of open convex subsets of RN such that

diamUk < ε ∀ k � 1, and A ⊆⋃

k�1

Uk.

Then(λN )∗(A) �

∑

k�1

λN (Uk) �∑

k�1

(diamUk)N ,

so(λN )∗(A) � HN (A)

(see Definition 4.151 and Remark 4.152).If (λN )∗(A) = +∞, then the conclusion of the problem is clearly

true.So, suppose that (λN )∗(A) < +∞. Then for a given δ > 0, we can

find an open set U ⊇ A such that

λN (U)− δ2 � (λN )∗(A).

4.3. Solutions 823

We can find a sequence {Ck}k�1 of half-open pairwise disjoint N -cubessuch that

U =⋃

k�1

Ck.

We have

λN (U) =∑

k�1

λN (Ck).

We can also find a sequence {Dk}k�1 of open cubes such that

Ck ⊆ Dk and λN (Dk) � λN (Ck) +δ

2n+1 ∀ k � 1.

Therefore ∑

k�1

λN (Dk)− δ2 � λN (U).

Note that

λN (Dk) = 1(√N)N

(diamDk)N .

Hence

(λN )∗(A) � λN (U)− δ2 �

∑

k�1

λN (Dk)− δ

= 1√N

N

∑

k�1

(diamDk)N − δ � cHN

ε (A)− δ,

with c = 2N√N

NωN

. Since δ, ε > 0 are arbitrary, we let δ, ε ↘ 0, to

conclude that

(λN )∗(A) � cHN (A).

Solution of Problem 4.161Let proj

Nbe the projection of RN × R

M onto RN . Note that

projN

(Gr f |A

)= A.

Also, by Theorem 4.155(e), the H-dimension does not increase underthe action of proj

N. So, H- dimGr f |A � H-dimA = N (recall that A

has a positive outer measure and see Problem 4.160).


If f is Lipschitz, then Gr f |A = ξ(A), where ξ : A −→ RN ×R

M isthe Lipschitz function, defined by

ξ(x) =(x, f(x)

) ∀ x ∈ A.

So, by the fact that Hs(RN ) = 0 for s > N and Theorem 4.155(e), weconclude that

H- dimGr f |A = N.

Solution of Problem 4.162Let {Ck}k�1 be an open cover of A. First we show that for everyx, u ∈ A, there exists a finite subfamily {Cki}ni=1 such that x ∈ Ck1 ,u ∈ Ckn and Cki ∩ Cki+1

= ∅ for all i ∈ {1, . . . , n − 1}. To see this,let us fix x ∈ A and consider the set D of all points u ∈ A for whichsuch a finite sequence exists. Then for every set Ck, we have Ck ⊆ Dor Ck ⊆ A \ D. So, D and A \ D are both open sets and since A isconnected (see Definition 2.104), we conclude that D = A.

Now, let x, u ∈ A and let {Cki}ni=1 be the finite sequence describedabove. For every i ∈ {1, . . . , n− 1}, pick a point xi ∈ Cki ∩ Cki+1

andalso let x0 = x, xn = u. We have

‖xi−1 − xi‖ � diamCki ∀ i ∈ {1, . . . , n}

(note that xi−1, xi ∈ Cki), so

∑

k�1

diamCk �n∑

i=1

diamCki �n∑

i=1

‖xi−1−xi‖ � ‖x0−xn‖ = ‖x−u‖

and thus

H1(A) � ‖x− u‖(see Definition 4.151 and recall that ω1

2 = 1). Since x, u ∈ A werearbitrary, we conclude that

H1(A) � diamA.

4.3. Solutions 825

Solution of Problem 4.163Let x, u ∈ A, x = u and let dx : R

N −→ R+ be defined by

dx(y) = ‖y − x‖.

Since dx is nonexpansive, from Theorem 4.155(e), we have

H- dim dx(A) � H- dimA < 1,

so

dx(A) ⊆ R is a Lebesgue-null set.

So, there exists > 0 such that

< dx(u) and ∈ dx(A).

Then

A ={y ∈ A : dx(y) <

} ∪ {y ∈ A : dx(y) >

}

and x is in one set and u in the other. Therefore x and u are in dif-ferent connected component (see Definition 2.111), hence A is totallydiscontinuous.

Solution of Problem 4.164The space C(X;Y ) topologized as above is a separable metric space.So, it has a countable basis consisting of balls centred at a countabledense subset of C(X;Y ) with radius in Q. For a fix g ∈ C(X;Y ) andε > 0 let us set

Bε(g) ={h ∈ C(X;Y ) : d(h, g) < ε

}.


It suffices to show that f−1(Bε(g)

) ∈ Σ. But X being a compactmetric space, it is separable. Hence

Bε(g) =⋂

n�1

{h ∈ C(X;Y ) : d

Y

(h(xn), g(xn)

)< ε

},

with {xn}n�1 being dense in X. Therefore, if we fix n � 1 and set

Bn ={h ∈ C(X;Y ) : d

Y

(h(xn), g(xn)

)< ε

},

it suffices to show that f−1(Bn) ∈ Σ. But note that

f−1(Bn) ={ω ∈ Ω : d

Y

(f(ω, xn), g(xn)

)< ε

} ∈ Σ,

so f is measurable as claimed.

Solution of Problem 4.165Clearly for all ω ∈ Ω, the function f(ω, ·) is continuous on X. Letx ∈ X. We need to show that the function f(·, x) is Σ-measurable.Let U ⊆ Y be an open set and let

U ={h ∈ C(X;Y ) : h(x) ∈ U

}.

Then the set U is open in C(X;Y ) (see Definition 2.174). Note that

{ω ∈ Ω : f(ω, x) ∈ U

}=

(ω ∈ Ω : f(ω) ∈ U

) ∈ Σ

(since f is Σ-measurable), so for all x ∈ X, the function ω �−→f(ω, x) is Σ-measurable and thus the function (ω, x) �−→ f(ω, x) isCaratheodory.


“(a) =⇒ (b)”: We know that E is norm bounded in Mb(X) (seeCorollary 5.45). So, for every compact set K ⊆ X, we can find ηK > 0such that

4.3. Solutions 827

∣∣μ(f)

∣∣ � ηK‖f‖∞ ∀ f ∈ Cc(X), supp f ⊆ K.

Now let K be a compact superset of K (i.e., K ⊆ K) andf0 ∈ Cc(K) with supp f0 ⊆ K, f0 � χ

K(see Problem 2.103). Since

‖μ‖ = ‖|μ|‖ � ηK, we have

|μ|(K) = |μ|(χK) � |μ|(f0) � η

K‖f0‖∞ = ηK ∀ μ ∈ E

“(b) =⇒ (a)”: For every compact set K ⊆ X and every f ∈ Cc(X)with supp f ⊆ K, we have

∣∣μ(f)∣∣ � |μ|(f) � |μ|(‖f‖∞χ

K

)= ηK‖f‖∞ ∀ μ ∈ E,

with ηK = |μ|(K). So E is norm bounded, hence w∗-bounded too.

Thus Ew∗

is w∗-compact (the Alaoglu theorem; see Theorem 5.66).

Solution of Problem 4.167By Problem 4.166, it suffices to show that E is w∗-closed. So, let

{μα}α∈J be a net in E and assume that μαw∗−→ μ. If f ∈ Cc(X)

with 0 � f � 1, then∣∣μα(f)

∣∣ � ϑ and so

∣∣μ(f)

∣∣ � ϑ, from which it

follows that ‖μ‖ � ϑ (see Theorem 4.23) and so we conclude that Eis w∗-closed, hence w∗-compact too.

In general, the set D is not w∗-closed (it is however relativelyw∗-compact). To see this, let X = [−1, 1] and let

μn = δ 1n− δ− 1

n∀ n.

then μnw∗−→ 0 but ‖μn‖ = 2 for all n � 1.

Finally suppose that X is compact and let

D+ ={μ ∈ Mb(X) : μ � 0, ‖μ‖ = ϑ

}.

Let

M+b (X) =

{μ ∈ Mb(X) : μ � 0

}


be furnished with the relative w∗-topology and consider the continuousfunction ξ : M+

b (X) −→ R+, defined by

ξ(u) = μ(1) ∀ μ ∈ M+b (X).

We see that D+ = ξ−1(ϑ) and so D+ is indeed w∗-compact.

Solution of Problem 4.168If xα −→ x in X, then for all f ∈ Cc(X), we have

δxα(f) = f(xα) −→ f(x) = δx(f)

and so we see that σ is continuous. Also, since X is locally compact(see Definition 2.92), it is completely regular (see Problem 2.42 andthe diagram at the end of Chap. 2). So, it follows that σ is an injection.Suppose that xα −→ x inX. Then we can find U ∈ N (x) and a subnet{xβ}β∈I of {xα}α∈I such that

xβ ∈ U ∀ β ∈ I.

Due to the local compactness of X, we may assume that U is compact.Then the complete regularity of X implies that we can find f ∈ C(X)such that

f(x) = 1 and f |Uc = 0,

sof ∈ Cc(X) and δx(f) = f(x) = 1,

while

δxβ(f) = f(xβ) = 0.

Thus δxβ−→ δx and so

δxα −→ δx in the w∗-topology.

This proves that σ is a homeomorphism of X into M+b (X).

If X is not compact, then we consider the Alexandrov one-pointcompactification X∗ = X ∪ {∞} of X (see Remark 2.97 and Theo-rem 2.98). Then clearly from the definition of the compact topology

4.3. Solutions 829

on X∗, we have that xα −→ ∞ implies δxα

w∗−→ 0 and so for everyε > 0 and f ∈ Cc(X), there is a compact set K ⊆ X such that

∣∣δx(f)

∣∣ < ε ∀ x ∈ K.

Solution of Problem 4.169“=⇒”: Since X is homeomorphic to a subset of M+

b (X) (see Prob-lem 4.168) and the latter is by hypothesis separable metrizable (hencesecond countable; see Definition 2.24), we infer that X is second count-able.“⇐=”: SinceX is by hypothesis second countable and locally compact,it is σ-compact and we have

X =⋃

n�1

Kn

with {Kn}n�1 being an increasing sequence of compact subsets suchthat

Kn ⊆ Kn+1 ∀ n � 1

(see Propositions 2.101 and 2.100). For such compact set Kn, let{fn,m}m�1 ⊆ C(Kn)+ be dense (recall that C(Kn) is separable sinceKn is compact and

C(Kn)+ ={f ∈ C(Kn) : f � 0

}).

The functionals ϕn,m : M+b (X) −→ R, defined by

ϕn,m(μ) = μ(fn,m)

are w∗-continuous and so, M+b (X) is homeomorphic to a subset of

RN×N which is separable metrizable, therefore so is M+

b (X).


Solution of Problem 4.170Let x ∈ suppμ (see Definition 4.13) and consider the open ball

B 1m(x) =

{y ∈ X : d(y, x) < 1

m

}.

Since μnw∗−→ μ, we have

0 � μ(B 1

m(x)

)� lim inf

n→+∞ μn

(B 1

m(x)

) ∀ m � 1.

We define i0 = 0 and

im = min{n � 1 : n > im−1, suppμj ∩B 1

n= ∅ for all j � n

}.

This is a well-defined, strictly increasing sequence. For n ∈ [ik, ik+1),let

xn ∈ suppμn ∩B 1k(x).

Then xn −→ x and so we conclude that

suppμ ⊆ lim infn→+∞ suppμn.

Solution of Problem 4.171From Problem 2.20(a), we know that we can find a sequence{fm : X −→ [0,+∞)

}n�1

of continuous function such that fm ↗ h.

Let ξ ∈ Cc(X) be such that 0 � ξ � 1. Then from the definition ofthe w∗-topology (see Problem 4.166), we have

limn→+∞

∫

X

ξfm dμn =

∫

X

ξfm dμ.

Taking supremum over the ξ’s as above and using the fact that fm � h,we have ∫

X

ξfm dμn �∫

X

fm dμn �∫

X

hdμn,

so ∫

X

ξfm dμ � lim infn→+∞

∫

X

hdμn.

4.3. Solutions 831

From the Lebesgue monotone convergence theorem (see Theo-rem 3.92), we have

∫

X

ξh dμ � lim infn→+∞

∫

X

hdμn

and taking supremum over the ξ’s as above, we have

∫

X

hdμ � lim infn→+∞

∫

X

hdμn.

Solution of Problem 4.172Since ϕ is continuous except possibly on a countable set, we haveϕn −→ ϕ almost everywhere on R (for the Lebesgue measure). Also

‖ϕn‖∞ � ‖μn‖ � M ∀ n � 1.

If f ∈ C1c (R), then integration by parts and the Lebesgue dominated

convergence theorem (see Theorem 3.94) imply that

∫

X

f dμn =

∫

X

f ′ϕn dx −→∫

X

f ′ϕdx =

∫

X

f dμ.

But the embedding C1c (R) ⊆ C0(X) is dense. Therefore, we conclude

thatμn

w∗−→ μ in Mb(R).

Bibliography

[1] Bourbaki, N.: Integration: Chapitres 1 a 4. Springer, Berlin(2007)

[2] Cohn, D.L.: Measure Theory. Birkhauser Boston, Boston (1993)

[3] Dellacherie, C., Meyer, P.A.: Probability and Potential. North-Holland Mathematics Studies, vol. 29. North-Holland PublishingCo., Amsterdam (1978)

[4] Folland, G.B.: Real Analysis. Modern Techniques and TheirApplications. Pure and Applied Mathematics. Wiley, New York(1999)

[5] Konig, H.: Measure and Integration. An Advanced Course in Ba-sic Procedures and Applications. Springer, Berlin (2009)

[6] Malliavin, P.G.: Integration and Probability. Graduate Texts inMathematics, vol. 157. Springer, New York (1995)

[7] Natanson, I.P.: Theory of Functions of a Real Variable. FrederickUngar Publishing Co., New York (1961)

[8] Oxtoby, J.C.: Measure and Category. A Survey of the Analo-gies between Topological and Measure Spaces. Graduate Texts inMathematics, vol. 2. Springer, New York (1971)

[9] Parthasarathy, K.R.: Probability Measures on Metric Spaces.Probability and Mathematical Statistics, vol. 3. Academic, NewYork (1967)

833

834 Bibliography

[10] Schwartz, L.: Radon Measures on Arbitrary TopologicalSpaces and Cylindrical Measures. Tata Institute of Fundamen-tal Research Studies in Mathematics, vol. 6. Oxford UniversityPress, London (1973)

[11] Stroock, D.W.: A Concise Introduction to the Theory of Integra-tion. Birkhauser Boston, Boston (1999)

[12] Topsøe, F.: Topology and Measure. Lecture Notes in Mathemat-ics, vol. 133. Springer, Berlin (1970)

[problem books in mathematics] exercises in analysis || measures and topology

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