Luigi Ambrosio, Giuseppe Da Prato, Andrea Mennucci Introduction to Measure Theory and Integration...

7/29/2019 Luigi Ambrosio, Giuseppe Da Prato, Andrea Mennucci Introduction to Measure Theory and Integration 2011

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10

APPUNTI

LECTURE NOTES


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Luigi Ambrosio, Giuseppe Da Prato and Andrea MennucciScuola Normale SuperiorePiazza dei Cavalieri, 756126 Pisa, Italy

Introduction to Measure Theory and Integration


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Luigi Ambrosio, Giuseppe Da Pratoand Andrea Mennucci

Introductionto Measure Theoryand Integration


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c 2011 Scuola Normale Superiore Pisa

ISBN: 978-88-7642-385-7e-ISBN: 978-88-7642-386-4


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Contents

Preface ix

Introduction xi

1 Measure spaces 1

1.1 Notation and preliminaries . . . . . . . . . . . . . . . . 1

1.2 Rings, algebras and algebras . . . . . . . . . . . . . . 21.3 Additive and additive functions . . . . . . . . . . . . 4

1.4 Measurable spaces and measure spaces . . . . . . . . . . 7

1.5 The basic extension theorem . . . . . . . . . . . . . . . 8

1.5.1 Dynkin systems . . . . . . . . . . . . . . . . . . 9

1.5.2 The outer measure . . . . . . . . . . . . . . . . 11

1.6 The Lebesgue measure in R . . . . . . . . . . . . . . . 14

1.7 Inner and outer regularity of measures on metric spaces . 18

2 Integration 23

2.1 Inverse image of a function . . . . . . . . . . . . . . . . 23

2.2 Measurable and Borel functions . . . . . . . . . . . . . 24

2.3 Partitions and simple functions . . . . . . . . . . . . . . 25

2.4 Integral of a nonnegative Emeasurable function . . . . 27

2.4.1 Integral of simple functions . . . . . . . . . . . 27

2.4.2 The repartition function . . . . . . . . . . . . . 28

2.4.3 The archimedean integral . . . . . . . . . . . . . 31

2.4.4 Integral of a nonnegative measurable function . . 32

2.5 Integral of functions with a variable sign . . . . . . . . . 35

2.6 Convergence of integrals . . . . . . . . . . . . . . . . . 36

2.6.1 Uniform integrability and Vitali convergence

t h e o r e m . . . . . . . . . . . . . . . . . . . . . . 38

2.7 A characterization of Riemann integrable functions . . . 39


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vi Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

3 Spaces of integrable functions 45

3.1 Spaces Lp(X,E, ) and Lp(X,E, ) . . . . . . . . . . 45

3.2 The Lp norm . . . . . . . . . . . . . . . . . . . . . . . 47

3.2.1 Holder and Minkowski inequalities . . . . . . . 48

3.3 Convergence in Lp(X,E, ) and completeness . . . . . 49

3.4 The space L(X,E, ) . . . . . . . . . . . . . . . . . . 52

3.5 Dense subsets ofLp(X,E, ) . . . . . . . . . . . . . . 56

4 Hilbert spaces 61

4.1 Scalar products, pre-Hilbert and Hilbert spaces . . . . . 61

4.2 The projection theorem . . . . . . . . . . . . . . . . . . 634.3 Linear continuous functionals . . . . . . . . . . . . . . . 66

4.4 Bessel inequality, Parseval identity and orthonormal

systems . . . . . . . . . . . . . . . . . . . . . . . . . . 67

4.5 Hilbert spaces on C . . . . . . . . . . . . . . . . . . . . 70

5 Fourier series 73

5.1 Pointwise convergence of the Fourier series . . . . . . . 75

5.2 Completeness of the trigonometric system . . . . . . . . 795.3 Uniform convergence of the Fourier series . . . . . . . . 80

6 Operations on measures 83

6.1 The product measure and FubiniTonelli theorem . . . . 83

6.2 The Lebesgue measure on Rn . . . . . . . . . . . . . . . 87

6.3 Countable products . . . . . . . . . . . . . . . . . . . . 90

6.4 Comparison of measures . . . . . . . . . . . . . . . . . 94

6.5 Signed measures . . . . . . . . . . . . . . . . . . . . . 1016.6 Measures inR . . . . . . . . . . . . . . . . . . . . . . . 105

6.7 Convergence of measures on R . . . . . . . . . . . . . . 107

6.8 Fourier transform . . . . . . . . . . . . . . . . . . . . . 112

6.8.1 Fourier transform of a measure . . . . . . . . . . 113

7 The fundamental theorem of the integral calculus 119

8 Measurable transformations 1298.1 Image measure . . . . . . . . . . . . . . . . . . . . . . 129

8.2 Change of variables in multiple integrals . . . . . . . . . 130

8.3 Image measure ofLn by a C1 diffeomorphism . . . . . 131

A 137

A.1 Continuity and differentiability of functions depending

on a parameter . . . . . . . . . . . . . . . . . . . . . . . 137


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vii Introduction to Measure Theory and Integration

A.2 The dual space of continuous functions . . . . . . . . . . 139

References 183


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Preface

This textbook collects the notes for an introductory course in measure

theory and integration taught by the authors to undergraduate students of

Scuola Normale Superiore in the last 10 years.

The goal of the course was to present, in a quick but rigorous way,

the modern point of view on measure theory and integration, putting Le-

besgues theory in Rn into a more general context and presenting the ba-

sic applications to Fourier series, calculus and real analysis. The text can

also pave the way to more advanced courses in probability, stochasticprocesses or geometric measure theory.

Prerequisites for the book are a basic knowledge of calculus in one and

several variables, metric spaces and linear algebra.

All results presented here, as well as their proofs, are classical. We

claim some originality only in the presentation and in the choice of the

exercises. Detailed solutions to the exercises are provided in the final part

of the book.

Pisa, July 2011

Luigi Ambrosio, Giuseppe Da Prato

and Andrea Mennucci


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Introduction

This course consists of an introduction to the modern theories of measure

and of integration. Historically, this has been motivated by the necessity

to go beyond the classical theory of Riemanns integration, usually taught

in elementary Calculus courses on the real line. It is therefore useful to

describe the reasons that motivate this extension.

(1) It is not possible to give a simple, handy, characterization of the class

of Riemanns integrable function, within Riemanns theory. This is indeed

possible within the stronger theory, due essentially to Lebesgue, that we

are going to introduce.

(2) The extensions of Riemanns theory to multiple integrals are very

cumbersome. This extension, useful to compute areas, volumes, etc., is

known as PeanoJordan theory, and it is sometimes taught in elementary

courses of integration in more than one variable. In addition to that, im-

portant heuristic principles like Cavalieris one can be proved only under

technical and basically unnecessary regularity assumptions on the do-

mains of integration.

(3) Many constructive processes typical of Analysis (limits, series, in-

tegrals depending on a parameter, etc.) cannot be handled well within

Riemanns theory of integration. For instance, the following statement is

true (it is a particular case of the so-called dominated convergence the-

orem):

Theorem 1. Let fh : [1, 1] R be continuous functions pointwise

converging to a continuous function f. Assume the existence of a con-stant M satisfying |fh(x)| M for all x [1, 1] and all h N. Then

limh

11

fh(x) dx =

11

f(x) dx .

Even though this statement makes perfectly sense within Riemanns the-

ory, any attempt to prove this result within the theory (try, if you dont


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xii Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

believe!) seems to fail, and leads (more or less explicitely, see [2]) to

a larger theory. In addition to that, the continuity assumption on the

limit function f is not natural, because a pointwise limit of continuous

functions need not be continuous, and we would like to give a sense to11

f(x) dx even without this assumption. This necessity emerges for

instance in the study of the convergence of Fourier series

f(x) =

i=0

ai cos(ix) + bi sin(ix) x [1, 1].

In this case the uniform convergence of the series, which implies thecontinuity of f as well, is ensured by the condition

i |ai | + |bi | 0.

(4) The spaces of integrable functions, as for instance

H :=

f : [1, 1] R :

11

f2(x) dx <

endowed with the scalar product

f, g :=1

1

f(x)g(x) dx

and with the (pseudo) induced distance d( f, g) = f g, f g1/2,

are not complete, if we restrict ourselves to Riemann integrable functions

only. In this sense, the path from Riemanns to Lebesgues theory is the

same one that led from the (incomplete) set of rational numbers Q to the

(complete) real line R.


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xiii Introduction to Measure Theory and Integration

Lebesgues theory extends Riemanns theory in two independent direc-

tions. The first one is concerned, as we already said, with more general

classes of functions, not necessarily continuous or piecewise continuous

(the so-called Borel or measurable functions). The second direction can

be better understood if we remind the very definition of Riemanns integ-

ral 11

f(x) dx

n1i=1

(ti+1 ti ) f(ti )

where t1 = 1, tn = 1 and the approximation is better and better as the

parameter supi


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Chapter 1

Measure spaces

In this chapter we shall introduce all basic concepts of measure theory,

adopting the point of view of measures as set functions. The domains

of measures may have different stability properties, and this leads to theconcepts of ring, algebra and algebra. The most basic tool developed

in the chapter is Caratheodorys theorem, which ensures in many cases

the existence and the uniqueness of a additive measure having some

prescribed values on a set of generators of the algebra. In the final

part of the chapter we will apply these abstract tools to the problem of

constructing a length measure on the real line, the so-called Lebesgue

measure, and we will study its main properties.

1.1. Notation and preliminaries

We denote by N = {0, 1, 2, . . .} the set of natural numbers, and by N the

set of positive natural numbers. Unless otherwise stated, sequences will

always be indexed by natural numbers.

We shall denote by X a non-empty set, by P(X) the set of all parts of

X and by the empty set. For any subset A of X we shall denote by Ac

its complement Ac := {x X : x / A}. If A, B P(X) we denote

by A \ B the relative complement A Bc, and by AB the symmetricdifference (A \ B) (B \ A).

Let (An) be a sequence in P(X). Then the following De Morgan

identity holds,

n=0

An

c=

n=0

Acn.

Moreover, we define (1)

lim supn

An :=

n=0

m=n

Am , lim infn

An :=

n=0

m=n

Am .

(1) Notice the analogy with liminf and limsup limits for a sequence (an ) of real numbers. We havelim sup

nan = inf

nNsup

mnam and lim inf

nan = sup

nN

infmn

am . This is something more than an analogy,

see Exercise 1.1.

L. Ambrosio et al., Introduction to Measure Theory and Integration

Scuola Normale Superiore Pisa 2011


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2 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

As it can be easily checked, lim supn An (respectively lim infn An) con-

sists of those elements of X that belong to infinitely many An (respect-

ively that belong to all but finitely many An).

It easy to check that if(An) is nondecreasing (i.e. An An+1, n N),

we have

lim infn

An = lim supn

An =

n=0

An,

whereas if(An) is nonincreasing (i.e. An An+1, n N), we have

lim infn

An = lim supn

An =

n=0

An.

In the first case we shall write An L , and in the second one An L .

1.2. Rings, algebras and algebras

Definition 1.1 (Rings and Algebras). A non empty subset AofP(X)

is said to be a ring if:

(i) belongs to A;

(ii) A, B A A B, A B A;

(iii) A, B A A \ B A.

We say that a ring is an algebra if X A.

Notice that rings are stable only with respect to relative complement,

whereas algebras are stable under complement in X.

Let K P(X). As the intersection of any family of algebras is still

an algebra, the minimal algebra includingK(that is the intersection of allalgebras including K) is well defined, and called the algebra generated

byK. A constructive characterization of the algebra generated byKcan

be easily achieved as follows: set F(0) = K {} and

F(i +1) :=

A B, Ac : A, B F(i )

i 0.

Then, the algebra Agenerated by K is given by i F(i ). Indeed, it is

immediate to check by induction on i thatA

F(i )

, and therefore theunion of the F(i )s is contained in A. On the other hand, this union is

easily seen to be an algebra, so the minimality ofAprovides the opposite

inclusion.

Definition 1.2 (algebras). A non-empty subset EofP(X) is said to

be a algebra if:

(i) E is an algebra;


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3 Introduction to Measure Theory and Integration

(ii) if(An) is a sequence of elements ofE then

n=0 An E.IfE is a algebra and (An) E we have

n An E by the De

Morgan identity. Moreover, both sets

lim infn

An, lim supn

An,

belong to E.

Obviously, {, X} andP

(X) are algebras, respectively the smal-lest and the largest ones. LetKbe a subset ofP(X). As the intersection

of any family of algebras is still a -algebra, the minimal algebra

including K (that is the intersection of all algebras including K) is

well defined, and called the algebra generated by K. It is denoted by

(K).

In contrast with the case of generated algebras, it is quite hard to give

a constructive characterization of the generated -algebras: this requires

the transfinite induction and it is illustrated in Exercise 1.18.

Definition 1.3 (Borel -algebra). If (E, d) is a metric space, the

algebra generated by all open subsets of E is called the Borel algebra

of E and it is denoted by B(E).

In the case when E = R the Borel -algebra has a particularly simple

class of generators.

Example 1.4 (B(R)). LetI be the set of all semiclosed intervals [a,b)

with a b. Then (I) coincides withB(R). In fact (I) contains allopen intervals (a, b) since

(a, b) =

n=n0

a +

1

n, b

with n0 >1

b a.

Moreover, any open set A in R is a countable union of open intervals. (2)

An analogous argument proves that B(R) is generated by semi-closed

intervals (a, b], by open intervals, by closed intervals and even by openor closed half-lines.

(2) Indeed, let (ak) be a sequence including all rational numbers of A and denote by Ik the largestopen interval contained in A and containing ak. We clearly have A

k=0 Ik, but also the opposite

inclusion holds: it suffices to consider, for any x A, r > 0 such that (x r,x + r ) A, and ksuch that ak (x r,x + r ) to obtain (x r,x + r ) Ik, by the maximality of Ik, and then x Ik.


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1.3. Additive and additive functions

Let A P(X) be a ring and let be a mapping from A into [0, +]such that () = 0. We say that is additive if

A, B A, A B = (A B) = (A) + (B).

If is additive, A, B Aand A B, we have (A) = (B) + (A \

B), so that (A) (B). Therefore any additive function is nondecreas-

ing with respect to set inclusion. Moreover, by applying repeatedly the

additivity property, additive measures satisfy

n

k=1

Ak

=

nk=1

(Ak)

for n N and mutually disjoint sets A1, . . . , An A.

A set function on A is called additive if () = 0 and for any

sequence (An) Aof mutually disjoint sets such that

n An Awe

have

n=0

An

=

n=0

(An).

Obviously additive functions are additive, because we can consider

countable unions in which only finitely many An are nonempty.

Another useful concept is the subadditivity: we say that is

subadditive if

(B)

n=0

(An),

for any B A and any sequence (An ) A such that B

n An.

Notice that, unlike the definition ofadditivity, the sets An need not be

disjoint here.

Remark 1.5 (additivity and subadditivity). Let be additive on

a ring Aand let (An) Abe mutually disjoint and such that

n An

A. Then by monotonicity we have

n=0

An

k

n=0

An

=

kn=0

(An), for all k N.

Therefore, letting k we get

n=0

An

n=0

(An ).


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Thus, to show that an additive function is additive, it is enough to

prove that it is subadditive.

Conversely, it is not difficult to show that additive set functions are

subadditive: indeed, if B nAn we can define A0 = B A0 and

An := B An \

m


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Proof. Setting Bn := A0 \An, B := A0 \A, we have Bn B, therefore the

previous proposition gives (Bn) (B). As (An) = (A0) (Bn)

and (A) = (A0) (B) the proof is achieved.

Corollary 1.8 (Upper and lower semicontinuity of the measure). Let

be additive on a algebra E and let(An) E. Then we have

lim inf

nAn

lim inf

n(An) (1.2)

and, if(X) < , we have also

lim supn

(An)

lim sup

n

An

. (1.3)

Proof. Set L := lim supn An. Then we can write

L =

n=0

Bn where Bn :=

m=n

Am . (1.4)

Now, assuming (X) < , by Proposition 1.7 it follows that

(L) = limn

(Bn) = infnN

(Bn) infnN

supmn

(Am ) = lim supn

(An).

Thus, we have proved (1.3). The inequality (1.2) can be proved similarly

using Proposition 1.6, thus without using the assumption (X) < .

The following result is very useful to estimate the measure of a lim sup

of sets.

Lemma 1.9. Let be additive on a algebra E and let (An) E.

Assume that

n=0

(An) < . Then

lim sup

n

An

= 0.

Proof. Set L = lim supn An and define Bn as in (1.4). Then the inclusion

L Bn gives

(L) (Bn)

m=n

(Am ) for all n N.

As n we find (L) = 0.


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1.4. Measurable spaces and measure spaces

Let E be a algebra of subsets of X. Then we say that the pair (X,E)is a measurable space. Let : E [0, +] be a additive function.

Then we call a measure on (X,E), and we call the triple (X,E, ) a

measure space.

The measure is said to be finite if(X) < , finite if there exists

a sequence (An) E such that

n An = X and (An) < for all

n N. Finally, is called a probability measure if(X) = 1.

The simplest (but fundamental) example of a probability measure is

the Dirac mass x , defi

ned by

x (B) :=

1 ifx B

0 ifx / B.

This example can be generalized as follows, see also Exercise 1.5 and

Exercise 1.23.

Example 1.10 (Discrete measures). Assume that Y X is afi

nite orcountable set. Given c : Y [0, +] we can define a measure on

(X,P(X)) as follows:

(B) :=

xBY

c(x) B X.

Clearly =

xY c(x)x is a finite measure if and only if

xY c(x ) 0 there exist Ai,j Asuch that

j=0

(Ai,j) < (Ei ) +

2i +1, Ei

j=0

Ai,j, i N.

Consequently

i, j=0

(Ai,j)

i =0

(Ei ) + .

Since E

i,j=0

Ai,j we have

(E)

i,j=0

(Ai,j)

i =0

(Ei ) +

and the first part of the statement follows from the arbitrariness of.

Now, let us assume that is -subadditive on Aand choose E A;

since E

i Ai then (E)

i (Ai ), so we deduce (E) (E);

but, by choosing A0 = E and An = for n 1, we obtain that (E) =

(E). This proves that extends .


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Let us now define the additive sets, according to Caratheodory. A set

A P(X) is called additive if

(E) = (E A) + (E Ac) for all E P(X). (1.7)

We denote by G the family of all additive sets.

Notice that, since is subadditive, (1.7) is equivalent to

(E) (E A) + (E Ac) for all E P(X). (1.8)

Obviously, the class G of additive sets is stable under complement;

moreover, by taking E = A B with A G and A B = , we

obtain the additivity property

(A B) = (A) + (B). (1.9)

Other important properties ofG are listed in the next proposition.

Theorem 1.17. Assume thatA is a ring and that is additive. Then G

is a algebra containing Aand is additive on G.

Proof. We proceed in three steps: we show that G containsA, that G is a

algebra and that

is additive on G. As pointed in Remark 1.5, if

is subadditive and additive on the algebra G, then is additive.

Step 1. A G. Let A A and E P(X), we have to show (1.8).

Assume (E) < (otherwise (1.8) trivially holds), fix > 0 and

choose (Bi ) Asuch that

E

i =0

Bi , (E) + >

i =0

(Bi ).

Then, by the definition of

, it follows that

(E) + >

i =0

(Bi ) =

i =0

[(Bi A) + (Bi Ac)]

(E A) + (E Ac).

Since is arbitrary we have (E) (EA) + (EAc), and (1.8)

follows.

Step 2. G is an algebra and is additive on G. We already know

that A G implies Ac G. Let us prove now that if A, B G thenA B G. For any E P(X) we have

(E) = (E A) + (E Ac)

= (E A) + (E Ac B) + (E Ac Bc)

= [(E A) + (E Ac B)] + (E (A B)c).

(1.10)


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Since

(E A) (E Ac B) = E (A B),

we have by the subadditivity of,

(E A) + (E Ac B) (E (A B)).

So, by (1.10) it follows that

(E) (E (A B)) + (E (A B)c),

and A B G as required. The additivity of on G follows directly

from (1.9).Step 3. G is a algebra. Let (An) G. We are going to show that

S :=

n An G. Since we know that G is an algebra, it is not restrictive

to assume that all sets An are mutually disjoint. Set Sn :=n

0 Ai , for

n N.

For any n N, by using the subadditivity of and by applying

(1.7) repeatedly, we get

(E Sc) + (E S) (E Sc) +

i =0

(E Ai )

= limn

(E Sc) +

ni =0

(E Ai )

= limn

(E Sc) + (E Sn )

.

Since Sc Scn it follows that

(E Sc) + (E S) lim supn

(E Sn) +

(E Scn )

= (E).

So, S G and G is a algebra.

Remark 1.18. We have proved that

(A) G P(X). (1.11)

One can show that the inclusions above are strict in general, for instance

when is the Lebesgue measure we shall consider in the next section.

In fact, in the case when X = R and (A) is the Borel -algebra, Exer-

cise 1.19 shows that (A) has the cardinality of continuum, while G has

the cardinality ofP(R), since it contains all subsets of Cantors middle

third set (see Exercise 1.8). An example of a non-additive set will be built

in Remark 1.23, so that also the second inclusion in (1.11) is strict.


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1.6. The Lebesgue measure in R

In this section we build the Lebesgue measure on the real line R. To thisaim, we consider first the set I of all bounded right open intervals ofR

I := {(a, b] : a, b R, a < b}

and the collection A containing and the finite unions of elements of

I. Our choice of half-open intervals ensures that A is a ring, because

I is stable under intersection and relative complement (the families of

open and closed intervals, instead, do not have this property).

We definelength((a, b]) := b a.

More generally, any non-empty A Acan be written, possibly in many

ways, as a disjoint finite union of intervals Ii , i = 1, . . . , N; we define

(A) :=

Ni =1

length(Ii ). (1.12)

Setting () = 0, it is not hard to show by elementary methods that is well defined (i.e. (A) does not depend on the chosen decomposition)

and additive onA.

In the next definition we introduce the notion of characteristic function,

which can be used to turn set-theoretic operations into algebraic ones:

for instance the intersection corresponds to the product, when seen at the

level of characteristic functions (see also Exercise 1.1).

Definition 1.19 (Characteristic function of a set). Let A X. The

characteristic function 1A : X {0, 1} is defined by

1A(x ) :=

1 ifx A;

0 ifx X \ A.

The reader already acquainted with Riemanns theory of integration can

also notice that (A) is the Riemann integral of the characteristic function

1A of A, and deduce the additivity property of directly by the additivity

properties of the Riemann integral. In the next theorem we shall rigor-ously prove these facts, and more. We first state an auxiliary lemma, a

simple consequence of the Bolzano-Weierstrass compactness theorem on

the real line.

Lemma 1.20. Any bounded and closed interval J contained in the union

of a sequence {An}nN of open sets is contained in the union offinitely

many of them.


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Proof. Assume with no loss of generality that I = N and An An+1,

and assume by contradiction that there exist xn J \ An for all n; by the

BolzanoWeierstrass theorem there exists a subsequence (xn(k)) conver-

ging to some x J. If n is such that x An, for k large enough xn(k)belongs to An , because An is open. But this is not possible, as soon as

n(k) n, because xn(k) / An(k) and An(k) An.

Theorem 1.21. The set function defined in (1.12) is additive on A.

Proof. ( is well defi

ned) Given disjoint partitions I1, . . . , In and J1, . . .. . . , Jm of A A, we say that J1, . . . , Jm is finer than I1, . . . , In if any

interval Ii is the disjoint union of some of the intervals Jj. Obviously,

given any two partitions, there exists a third partition finer than both: it

suffices to take all intersections of elements of the first partition with ele-

ments of the second partition, neglecting the empty intersections. Given

these remarks, to show that is well defined, it suffices to show that

i(Ii ) = j

(Jj) if J1, . . . , Jm is finer than I1, . . . , In. This state-

ment reduces to the fact that (I) = k

(Fk) if I I is the disjoint

union of some elements Fk I; this last statement can be easily proved,

starting from the identity (a, b] = (a, c] (c, b], by induction on the

number of the intervals Fk.

( is additive) If F, G Aand F G = , any disjoint decompositions

of F in intervals I1, . . . , In I and any disjoint decomposition of G in

intervals J1, . . . , Jm I provide a decomposition I1, . . . , In, J1, . . . , Jmof F G in intervals belonging to I. Using this decomposition to com-

pute (F G) the additivity easily follows.

( is additive) Let (Fn) Abe a sequence of disjoints sets in Aand

assume that

F :=

n=0

Fn (1.13)

also belongs to A.

We prove the additivity property first in the case when F = (x , y]

I. It is also not restrictive to assume that the series

n (Fn) is con-vergent. As any Fn is a finite union of intervals, say Nn, we can find,

given any > 0, a finite union Fn Fn of intervals in I such that

(Fn) (Fn) + /2n and the internal part of Fn contains Fn (just shift

the endpoints of each interval in Fn by a small amount, to obtain a lar-

ger interval in I, increasing the length at most by /(Nn2n)). Let also

Fn be the internal part of F

n, that still includes Fn, and let x (x, y].

Then, since [x , y]

nFn , Lemma 1.20 provides an integer k such


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that [x , y]

k0 F

n. Hence, the additivity of inAgives

y x

k

n=0

Fn

kn=0

(Fn)

kn=0

(Fn) +

2n 2 +

n=0

(Fn).

By letting first 0 and then letting x x we obtain that (F)

0 (Fn). The opposite inequality simply follows by the monotonicityand the additivity of , because the finite unions of the sets Fn are con-tained in F.

In the general case, let

F =

ki =1

Ii ,

where I1, . . . , Ik are disjoint sets inI. Then, since for any i {1, . . . , k}

we have that Ii is the disjoint union of Ii Fn, we know by the previous

step that

(Ii ) =

n=0

(Ii Fn).

Adding these identities for i = 1, . . . , k, commuting the sums on the

right hand side and eventually using the additivity of on Awe obtain

(F) =

k

i =1

(Ii F) =

n=0

k

i =1

(Ii Fn) =

n=0

(Fn).

We say that a measure in (R,B(R)) is translation invariantif(A +

h) = (A) for all A B(R) and h R (notice that, by Exercise 1.2,

the class of Borel sets is translation invariant as well). We say also that

is locally finite if(I) < for all bounded intervals I R.

Theorem 1.22 (Lebesgue measure in R). There exists a unique, up

to multiplication with constants, translation invariant and locally fi-

nite measure in (R,B(R)). The unique such measure satisfying([0, 1]) = 1 is called Lebesgue measure.

Proof. (Existence) Let A be the class of finite unions of intervals and

let : A [0, +) be the additive set function defined in (1.12).

According to Theorem 1.21 admits a unique extension, that we still

denote by , to (A) = B(R). Clearly is locally finite, and we can use

the uniqueness of the extension to prove translation invariance: indeed,


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for any h R also the additive measure A (A +h) is an extension

of|A. As a consequence (A) = (A + h) for all h R.

(Uniqueness) Let be a translation invariant and locally finite measure

in (R,B(R)) and set c := ([0, 1]). Notice first that the set of atoms of

is at most countable (Exercise 1.5), and since R is uncountable there

exists at least one x such that ({x}) = 0. By translation invariance this

holds for all x , i.e., has no atom.

Excluding the trivial case c = 0 (that gives 0 by translation in-

variance and additivity), we are going to show that = c on the

class Aoffinite unions of intervals; by the uniqueness of the extension

in Caratheodory theorem this would imply that = c onB(R).

By finite additivity and translation invariance it suffices to show that

([0, t)) = ct for any t 0 (by the absence of atoms the same holds for

the intervals (0, t), (0, t], [0, t]). Notice first that, for any integer q 1,

[0, 1) is the union of q disjoint intervals all congruent to [0, 1/q); as a

consequence, additivity and translation invariance give

[0, 1/q) = ([0, 1))q

=c

q.

Similarly, for any integer p 1 the interval [0, p/q) is the union of p

disjoint intervals all congruent to [0, 1/q); again additivity and transla-

tion invariance give

([0,p

q)) = p

[0,

1

q)

= cp

q.

By approximation we eventually obtain that ([0, t)) = ct for allt 0.

The completion of the Borel algebra with respect to is the so-

called -algebra of Lebesgue measurable sets. It coincides with the

class C of additive sets with respect to considered in the proof of

Caratheodory theorem (see Exercise 1.12).

Remark 1.23 (Outer Lebesgue measure and non-measurable sets).

The measure

used in the proof of Caratheodorys theorem is also calledouter Lebesgue measure, and it is defined on all parts ofR. The termin-

ology is slightly misleading here, since , though subadditive, fails

to be additive. In particular, there exist subsets ofR that are not Le-

besgue measurable. To see this, let us consider the equivalence relation

in R defined by x y if x y Q and let us pick a single element

x [0, 1] in any equivalence class induced by this relation, thus forming

a set A [0, 1]. Were this set Lebesgue measurable, all the sets A + h


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would still be measurable, by translation invariance, and the family of

sets {A + h}hQ would be a countable and measurable partition ofR,

with (A + h) = c independent of h Q. Now, ifc = 0 we reach a

contradiction with the fact that (R) = , while if c > 0 we consider

all sets A + h with h Q [0, 1] to obtain

2 = ([0, 2])

hQ[0,1]

(A + h) = ,

reaching again a contradiction.

Notice that this example is not constructive and strongly requires theaxiom of choice (also the arguments based on cardinality, see Exercise

1.19 and Exercise 1.20, have this limitation). On the other hand, one

can give constructive examples of Lebesgue measurable sets that are not

Borel (see for instance 2.2.11 in [3]).

The construction done in the previous remark rules out the existence of

locally finite and translation invariant additive measures defined on all

parts ofR. In Rn, with n 3, the famous BanachTarski paradox (see

for instance [6]) shows that it is also impossible to have a locallyfi

nite,invariant under rigid motions and finitely additive measure defined on all

parts ofRn.

1.7. Inner and outer regularity of measures on metric spaces

Let (E, d) be a metric space and let be a finite measure on (E,B(E)).

We shall prove a regularity property of.

Proposition 1.24. For any B B

(E) we have

(B) = sup{(C) : C B, closed} = inf{(A) : A B, open}.

(1.14)

Proof. Let us set

K= {B B(E) : (1.14) holds}.

It is enough to show that K is a algebra of parts of E including the

open sets of E. Obviously Kcontains E and . Moreover, if B K

then its complement Bc belongs to K. Let us prove now that (Bn) K

implies

n Bn K. Fix > 0. We are going to show that there exist a

closed set C and an open set A such that

C

n=0

Bn A, (A \ C) 2. (1.15)


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Let n N. Since Bn K there exist an open set An and a closed set Cnsuch that Cn Bn An and

(An \ Cn)

2n+1.

Setting A :=

n An, S :=

n Cn we have S

n Bn A and (A \

S) . However, A is open but S is not necessarily closed. So, we

approximate S by setting Sn :=n

0 Ck. The set Sn is obviously closed,

Sn S and consequently (Sn) (S). Therefore there exists n N

such that (S\ Sn ) < . Now, setting C = Sn we have C

n Bn Aand (A \ C) < (A \ S) + (S\ C) < 2. Therefore

n Bn K. We

have proved that K is a algebra. It remains to show that Kcontains

the open subsets of E. In fact, let A be open and set

Cn =

x E : d(x , Ac)

1

n

,

where d(x, Ac) := infyAc d(x , y) is the distance function from Ac. Then

Cn are closed subsets of A, and moreover Cn A, which implies (A \

Cn) 0. Thus the conclusion follows.

Notice that inner and outer approximation hold for measurable sets

B as well: one has just to notice that there exist Borel sets B1, B2 such

that B1 B B2 with (B2 \ B1) = 0, and apply inner approximation

to B1 and outer approximation to B2.

Remark 1.25 (Inner and outer approximation for -finite measures).

It is possible to extend the inner approximation property to -finite meas-ures: suffices to assume the existence of a sequence of closed sets Cn with

finite measure such that (X\ nCn) = 0. Indeed, assuming with no loss

of generality that Cn Cn+1, we know that for any Borel set B and any

n N it holds

(B Cn) = sup {(C) : C closed, C B Cn} ,

so that(B Cn) sup {(C) : C closed, C B} .

Letting n we recover the inner approximation property.

Analogously, if we assume the existence of a sequence of open sets Anwith finite measure satisfying X = nAn, we have the outer approxim-

ation property: indeed, for any n and any > 0 we can find (assuming

with no loss of generality (B) < +) open sets Bn An containing


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B An and such that

Bn \ (B An)

< 2n. It follows that nBn

contains B and

nN

Bn \ B

< 2.

Since Bn are also open in X, the set nBn is open and since is arbitrary

we get the outer approximation property.

We conclude this chapter with the following result, whose proof is a

straightforward consequence of Proposition 1.24 (alternatively, one can

use Dynkins argument, since the class of closed sets is a -system and

generates the Borel -algebra).

Corollary 1.26. Let , be finite measures in (E,B(E)), such that

(C) = (C) for any closed subsetC of E. Then = .

Exercises

1.1 Given A X, denote by 1A : X {0, 1} its characteristic function, equalto 1 on A and equal to 0 on Ac. Show that

1AB = max{1A, 1B }, 1AB = min{1A, 1B }, 1Ac = 1X 1A

and thatlim sup

nAn = A lim sup

n1An

= 1A,

lim infn

An = A lim infn

1An= 1A.

1.2 Let A Rn be a Borel set. Show that for h Rn and t R the sets

A + h := {a + h : a A} , t A := {ta : a A}

are Borel as well.

1.3 Find an example of a additive measure on a algebra A such thatthere exist An Awith An A and infn (An) > (A).

1.4 Let be additive and finite, on an algebra A. Show that is additive ifand only if it is continuous along nonincreasing sequences.

1.5 Let be a finite measure on (X,E). Show that the set of atoms of, definedby

A := {x X : {x} E and ({x}) > 0}

is at most countable. Show that the same is true for finite measures, andprovide an example of a measure space for which this property fails.

1.6 Let (X,E, ) be a measure space, with finite. We say that is diffuse iffor all A E with (A) > 0 there exists B A with 0 < (B) < (A). Showthat, if is diffuse, then (E) = [0, (X)].

1.7 Show that if X is a separable metric space and E is the Borel algebra,then a additive measure : E [0, +) is diffuse if and only if has noatom.


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1.8 Let be the Lebesgue measure in [0, 1]. Show the existence of a negli-gible set having the cardinality of the continuum. Hint: consider the classical

Cantors middle third set, obtained by removing the interval (1/3, 2/3) from[0, 1], then by removing the intervals (1/9, 2/9) and (7/9, 8/9), and so on.

1.9 Let be the Lebesgue measure in [0, 1]. Show the existence, for any > 0,of a closed set C [0, 1] containing no interval and such that (C) > 1 .

Hint: remove from [0, 1] a sequence of open intervals, centered on the rationalpoints of[0, 1].

1.10 Using the previous exercise, write [0, 1] = A B where A is negligiblein the measure-theoretic sense (i.e. (A) = 0) and B is negligible in the Bairecategory sense (i.e. it is the union of countably many closed sets with empty

interior). So, the two concepts of negligible should be never used at the sametime.

1.11 Let be the Lebesgue measure in [0, 1]. Construct a Borel set E (0, 1)such that

0 < (E I) < (I)

for any open interval I (0, 1).

1.12 Let (X,E, ) be a measure space and let : P(X) [0, +] bethe outer measure induced by . Show that the completed algebra E iscontained in the class C of additive sets with respect to .

1.13 Let (X,E, ) be a measure space and let : P(X) [0, +] be theouter measure induced by . Show that for all A X there exists B Econtaining A with (B) = (A).

1.14 Let (X,E, ) be a measure space. Check the following statements, madein Definition 1.12:

(i) E is a algebra;(ii) the extension (A) := (B), where B E is any set such that AB is

contained in a negligible set ofE, is well defined and additive onE

;(iii) negligible sets ofE are characterized by the property of being coin-tained in a negligible set ofE.

1.15 Let (X,E, ) be a measure space and let : P(X) [0, +] bethe outer measure induced by . Show that if (X) is finite, the class C ofadditive sets with respect to coincides with the class ofEmeasurable sets.

Hint: one inclusion is provided by Exercise 1.12. For the other one, given anadditive set A, by applying Exercise 1.13 twice, find first a set B E with(B \ A) = 0, and then a set C E with (C) = 0 and B \ A C.

1.16 Find a algebra E P(N) containing infinitely many sets and such thatany B E different from has an infinite cardinality.

1.17 Find : P(N) {0, +} that is additive, but not additive.

1.18 Let be the first uncountable ordinal and, for K P(X), define bytransfinite induction a family F(i ), i , as follows: F(0) := K {},

F(i ) :=

k=0

Ak, Bc : (Ak) F

(j), B F(j)

,


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ifi is the successor of j, and F(i ) :=

ji F

(j) otherwise.

Show that i

F(i ) = (K).

1.19 Show thatB(R) has the cardinality of the continuum. Hint: use the con-struction of the previous exercise, and the fact that has at most the cardinalityof continuum.

1.20 Show that the algebra L of Lebesgue measurable sets has the samecardinality ofP(R), thus strictly greater than the continuum. Hint: consider allsubsets of Cantors middle third set.

1.21 Show that the cardinality of any algebra is either finite or uncount-able.

1.22 Let X be a set and let A P(X) be an algebra with finite cardinality.Show that its cardinality is equal to 2n for some integer n 1.

1.23 Let (X,E, ) be a a measure space and suppose that X is finite or count-able. Show the existence of a measure on P(X) that extends , that is,(A) = (A) for all A E.

1.24 Find an example of an additive set function : P(N) {0, 1}, with(N) = 1 and ({n}) = 0 for all n N (in particular is not additive, theconstruction of this example requires Zorns lemma).

1.25 Let C B([0, 1]) with (C) > 0. Without using the continuum hypo-thesis, show that C has the cardinality of continuum.1.26 Let (K, d) be a compact metric space and let be as in Exercise 1.24.Lets say that a sequence (xn) K -converges to x K if

{n N : d(xn,x) > }

= 0 > 0.

Show that any sequence (xn) K is -convergent and that the -limit isunique.


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Chapter 2

Integration

This chapter is devoted to the construction of the integral ofEmeasur-

able functions in general measure spaces (,E, ), and its main con-

tinuity and lower semicontinuity properties. Having built in the previous

chapter the Lebesgue measure in the real line R, we obtain as a byproduct

the Lebesgue integral on R; in the last section we compare Lebesgue and

Riemann integral.

In the construction of the integral we prefer to empahsize two view-

points: the first, more traditional oneX

f d =

zIm(f)z({f = z})

is appropriate to deal with simple functions (i.e. functions whose range is

finite) and useful to show the additivity of the integral with respect to f.

The second one, for nonnegative functions is summarized by the formula

X

f d =

0

({f > t}) dt.

This second viewpoint is more appropriate to show the continuity prop-

erties of the integral with respect to f (the integral on the right side can

be elementarily defined, since t ({f > t}) is nonincreasing, seeSection 2.4.3). Of course we show that the two viewpoints are consistent

if we restrict ourselves to the class of simple functions.

2.1. Inverse image of a function

Let X be a non empty set. For any function : X Y and any I P(Y) we set

1(I) := {x X : (x ) I} = { I}.

The set 1(I) is called the inverse image of I.




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Let us recall some elementary properties of1 (the easy proofs are leftto the reader as an exercise):

(i) 1(Ic) = (1(I))c for all I P(Y);(ii) if{Ji}iI P(Y) we have

iI1(Ji ) = 1

iI

Ji

,iI

1(Ji ) = 1

iIJi

.

In particular, if I J = we have 1(I) 1(J) = . Also, ifE

P(Y) and we consider the family 1(E) of subset of X defined

by1(E) := 1(I) : I E , (2.1)

we have that 1(E) is a algebra whenever E is a algebra.

2.2. Measurable and Borel functions

We are given measurable spaces (X,E) and (Y,F). We say that a func-

tion : X Y is (E,F)measurable if 1(F) E. If (Y,F) =(R

,B

(R

)), we say that is a real valuedE

measurable function, and if(X, d) is a metric space and E is the Borel algebra, we say that is a

real valued Borel function.

The following simple but useful proposition shows that the measurab-

ility condition needs to be checked only on a class of generators.

Proposition 2.1 (Measurability criterion). Let G F be such that (G) = F. Then : X Y is (E,F)measurable if and only if1(I)

E for all I

G (equivalently, iff1(G)

E).

Proof. Consider the family D := {I F : 1(I) E}. By theabove-mentioned properties of 1 as an operator between P(Y) andP(X), it follows that D is a algebra including G. So, it coincides

with (G) = F.A simple consequence of the previous proposition is the fact that any

continuous function is a Borel function: more precisely, assume that :X

Y is continuous and that E

=B(X) and F

=B(Y). Then, the

algebra A Y : 1(A) B(X)

contains the open subsets of Y (as, by the continuity of , 1(A) isopen in X, and in particular Borel, whenever A is open in Y), and then it

contains the generated algebra, i.e. B(Y).

The following proposition, whose proof is straightforward, shows that

the class of measurable functions is stable under composition.


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Proposition 2.2. Let(X,E), (Y,F), (Z,G) be measurable spaces and

let

:X

Y and

:Y

Z be respectively (E,F)measurable and

(F,G)measurable. Then is (E,G)measurable.It is often convenient to consider functions with values in the extended

space R := R {+, }, the so-called extended functions. We saythat a mapping : X R is Emeasurable if1({}), 1({+}) E and 1(I) E , I B(R).

(2.2)

This condition can also be interpreted in terms of measurability between

E and a suitable Borel algebra in R, see Exercise 2.3. Analogously,when (X, d) is a metric space and E is the Borel algebra, we say that

: X R is Borel whenever the conditions above hold.The following proposition shows that extended Emeasurable func-

tions are stable under pointwise limits and countable supremum and in-

fimum.

Proposition 2.3. Let(n) be a sequence of extendedEmeasurable func-

tions. Then the following functions are Emeasurable:

supnN

n(x ), infnN

n(x ), lim supn

n(x), lim infn

n(x ).

Proof. Let us prove that (x ) := supn n(x ) is Emeasurable (all othercases can be deduced from this one, or directly proved by similar argu-

ments). For any a R we have

1([

, a

])

=

n=0

1n ([

, a

])

E.

In particular { = } E, so that 1((, a]) E for all a R;by letting a we get 1(R) E. As a consequence, the class

I B(R) : 1(I) Eis a algebra containing the intervals of the form (, a] with a R,and therefore coincides with B(R). Eventually, { = +} = X \

[1(R)

{

= }]belongs to E as well.

2.3. Partitions and simple functions

Let (X,E) be a measurable space. A function : X R is said to besimple if its range (X) is a finite set. The class of simple functions is

obviously a real vector space, as the range of + is contained in{a + b : a range(), b range()} .


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If (X) = {a1, . . . , an}, with ai = aj if i = j, setting Ai = 1({ai}),i

=1, . . . , n we can canonically represent as

(x) =n

k=1ak1Ak, x X. (2.3)

Moreover, A1, . . . , An is a finite partition of X (i.e. Ai are mutually

disjoint and their union is equal to X). However, a simple function has

many representations of the form

(x ) =N

k=1ak1Ak, x X,

where A1, . . . , AN need not be mutually disjoint and a

k need not be in

the range of. For instance

1[0,1) + 31[1,2] = 1[0,2] + 21[1,2].It is easy to check that a simple function is Emeasurable if, and only if,

all level sets Ak in (2.3) are Emeasurable; in this case we shall also saythat {Ak} is a finite Emeasurable partition of X.

Now we show that any nonnegative Emeasurable function can be ap-

proximated by simple functions; a variant of this result, with a different

construction, is proposed in Exercise 2.8.

Proposition 2.4. Let be a nonnegative extendedEmeasurable func-

tion. For any n N, define

n(x ) =

i 12n

if i 12n

(x ) < i2n

, i = 1, 2, . . . , n2n;

n if(x ) n.(2.4)

Then n are simple andEmeasurable, (n) is nondecreasing and con-

vergent to . If in addition is bounded the convergence is uniform.

Proof. It is not difficult to check that (n ) is nondecreasing. Moreover,

we have

0 (x ) n(x) 1

2nif(x ) < n, x X,

and

0 (x) n(x) = (x ) n if(x ) n, x X.So, the conclusion easily follows.


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2.4. Integral of a nonnegative Emeasurable function

We are given a measure space (X,E, ). We start to defi

ne the integralfor simple nonnegative functions.

2.4.1. Integral of simple functions

Let be a nonnegative simple Emeasurable function, and let us repres-

ent it as

(x ) =N

k

=1

ak1Ak, x X,

with N N, a1, . . . , aN 0 and A1, . . . , AN in E. Then we define(using the standard convention in the theory of integration that 0 = 0),

X

d :=N

k=1ak(Ak).

It is easy to see that the definition does not depend on the choice of the

representation formula for . Indeed, let {b1, . . . , bM} be the range ofand let =

M

1 bj1

Bj, with Bj := 1

(bj), be the canonical representa-tion of. We have to prove that

Nk=1

ak(Ak) =M

j=1bj(Bj). (2.5)

As the Bi s are pairwise disjoint, (2.5) follows by adding the M identities

N

k=1

ak(Ak

Bj)=

bj(Bj) j=

1, . . . , M. (2.6)

In order to show (2.6) we fix j and consider, for I {1, . . . , N}, the setsAI :=

x Bj : x Ai iffi I

,

so that {AI} are a Emeasurable partition of Bj and x AI iff the setof i s for which x Ai coincides with I. Then, using first the fact thatAI Ai if i I, and Ai AI = otherwise, and then the fact thatkI

ak =

bj

whenever AI =

(because N1

ak1

Akcoincides with b

j, the

constant value of on Bj), we have

Nk=1

ak(Ak Bj) =N

k=1

I

ak(Ak AI) =

I

Nk=1

ak(Ak AI)

=

I

kI

ak(AI) =

I

bj(AI) = bj(Bj).


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Proposition 2.5. Let , be simple nonnegative Emeasurable func-

tions on X and let ,

0. Then

+ is simple, Emeasurable

and we have X

( + ) d =

X

d +

X

d

Proof. Let

=n

k=1ak1Ak, =

mh=1

bh1Bh

with {Ak}, {Bh} finite Emeasurable partitions of X. Then {Ak Bh} is afinite Emeasurable partition of X and + is constant (and equalto ak + bh ) on any element Ak Bh of the partition. Therefore thelevel sets of + are finite unions of elements of this partition andthe Emeasurability of + follows (see also Exercise 2.2). Then,writing

(x )=

n

k=1

m

h=1

ak1Ak

Bh

(x), (x)=

n

k=1

m

h=1

bh1Ak

Bh

(x ), x

X,

we arrive at the conclusion.

2.4.2. The repartition function

Let : X R be Emeasurable. The repartition function F of, relat-ive to , is defined by

F(t) := ({ > t}), t R.The function F is nonincreasing and satisfies

limt

F(t) = limn

F(n) = limn

({ > n}) = ({ > }),

and, if is finite,

limt+ F(t) = limn F(n) = limn ({ > n}) = ({ = +}),since

{ > } =

n=1{ > n}, { = +} =

n=1

{ > n}.

Other important properties of F are provided by the following result.


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Proposition 2.6. Let : X R be Emeasurable and let F be its re-partition function.

(i) For any t0 R we have limtt+0

F(t) = F(t0), that is, F is right con-tinuous.

(ii) If is finite, for any t0 R we have limtt0

F(t) = ({ t0)}, that

is, F has left limits (1).

Proof. Let us prove (i). We have

limtt+0

F(t) = limn+

F

t0 + 1

n

= lim

n+

> t0 + 1

n

= ({ > t0}) = F(t0),since

{ > t0} =

n

=1

> t0 +

1

n

= lim

n

> t0 +

1

n

.

So, (i) follows. We prove now (ii). We have

limtt0

F(t) = limn+

F

t0

1

n

= limn+

> t0

1

n

= ({ t0}),

since

{ t0} =

n=1

> t0

1

n

= lim

n

> t0

1

n

and (ii) follows.

From Proposition 2.6 it follows that, in the case when is finite, F is

continuous at t0 iff({ = t0}) = 0.Now we want to extend the integral operator to nonnegative Emea-

surable functions. Let be a nonnegative, simple and Emeasurablefunction and let

(x ) =n

k=0ak1Ak, x X,

(1) In the literature F is called a cadlag function.


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with n N, 0 = a0 < a1 < a2 < < an < . Then the repartitionfunction F of is given by

F(t) =

(A1) + (A2) + + (An) = F(0) if 0 t < a1(A2) + (A3) + + (An) = F(a1) ifa1 t < a2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

(An) = F(an1) ifan1 t < an0 = F(an) if t an.

Consequently, we can write

X

(x) d(x ) = nk=1

ak(Ak) = nk=1

ak(F(ak1) F(ak))

=n

k=1akF(ak1)

nk=1

akF(ak)

=n1k=0

ak+1 F(ak) n1k=0

akF(ak)

= n1k=0

(ak+1 ak)F(ak) =

0

F(t) dt.

(2.7)

Example 2.7. We set X = R, = ,A1 = [1, 2] [10, 11], A2 = [2, 3], A3 = [3, 4],

A4 = [4, 6], A5 = [7, 10],a1

=5, a2

=, a3

=10, a4

=7, a5

=2

and := 5k=1 ak1Ak to be the simple function shown in Figure 2.1. It iseasy to verify that F has the graph shown in the right picture in Figure 2.1.

F

1

1

2

3

4

5

6

7

9

8

10

4 6 7 8 1091 2 3 5

Figure 2.1. a simple function , and its repartition F


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The color scheme used for the areas below the two graphs in 2.1 proves

graphically that the areas are identical.

Now, we want to define the integral of any nonnegative extended E

measurable function by generalizing formula (2.7). For this, we need

first to define the integral of any nonnegative nonincreasing function in

(0, +).

2.4.3. The archimedean integral

We generalize here the (inner) Riemann integral to any nonincreasing

function f : [0, +) [0, +]. The strategy is to consider the su-premum of the areas of piecewise constant minorants of f.Let be the set of all finite decompositions = {t1, . . . , tN} of

[0, +], where N N and 0 = t0 t1 < < tN < +.Let now f : [0, +) [0, +] be a nonincreasing function. For

any = {t0, t1, . . . , tN} we consider the partial sum

If( ) :=N1

k=0

f(tk+1)(tk+1 tk). (2.8)

We define 0

f(t) dt := sup{If( ) : }. (2.9)

The integral

0f(t) dt is called the archimedean integral of f. It enjoys

the usual properties of the Riemann integral (see Exercise 2.5) but, among

these, we will need only the monotonicity with respect to f in the sequel.

For our purposes the most relevant property of the Archimedean integral

is instead the continuity under monotonically nondecreasing sequences.

Proposition 2.8. Let fn f, with fn : [0, +) [0, +] nonin-creasing. Then

0

fn(t) dt

0

f(t) dt.

Proof. It is obvious that

0

fn(t) dt 0

f(t) dt.

To prove the converse inequality, fix L L .


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Since for n large enough

0

fn(t) dt N1k=0

fn(tk+1)(tk+1 tk) > L ,

letting n we find that

supnN

0

fn(t) dt L .

This implies

supnN

0

fn(t) dt

0

f(t) dt

and the conclusion follows.

2.4.4. Integral of a nonnegative measurable function

We are given a measure space (X,E, ) and an extended nonnegative

Emeasurable function . Having the identity (2.7) in mind, we define

X

d : =

0

({ > t}) dt. (2.10)

Notice that the function t ({ > t}) [0, +] is nonnegative andnonincreasing in [0, +), so that its archimedean integral is well definedand (2.10) extends, by the remarks made at the end of Section 2.4.2, the

integral elementarily defined on simple functions. If the integral is finite

we say that is integrable.

It follows directly from the analogous properties of the archimedeanintegral that the integral so defined is monotone, i.e.

X

d

X

d.

Indeed, implies { > t} { > t} and ({ > t}) ({ >t}) for all t > 0. Furthermore, the integral is invariant under modifica-tions of in negligible sets, that is

= a.e. in X

X

d =

X

d.

To show this fact it suffices to notice that = a.e. in X implies thatthe sets { > t} and { > t} differ in a negligible set for all t > 0,therefore ({ > t}) = ({ > t}) for all t > 0.

Let us prove the following basic Markov inequality.


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Proposition 2.9. For any a (0, +) we have

({ a}) 1a

X

(x) d(x ). (2.11)

Proof. For any a (0, +) we have, recalling the inclusion { a} { > t} for any t (0, a), that ({ > t}) ({ a}) for allt (0, a). The monotonicity of the archimedean integral gives

X

(x) d(x ) =

0

({ > t}) dt

0

1(0,a)(t)({ > t}) dt

a({ a}).

The Markov inequality has some important consequences.

Proposition 2.10. Let : X [0, +] be an extendedEmeasurablefunction.

(i)If is integrable then the set

{

= +}has measure

0, that

is, is finite a.e. in X.

(ii) The integral of vanishes iff is equal to 0 a.e. in X.

Proof. (i) Since

X d < we deduce from (2.11) that

lima+

({ > a}) = 0.

Since

{ = } = n=1

{ > n},

by applying the continuity along decreasing sequences in the space ({ >1} (with finite measure) we obtain

({ = }) = limn+

({ > n}) = 0.

(ii) If

X d

= 0 we deduce from (2.11) that(

{ > a

})

= 0 for alla > 0. Since

({ > 0}) = limn+

({ > 1n

}) = 0,

the conclusion follows. The other implication follows by the invariance

of the integral.


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Proposition 2.11 (Monotone convergence). Let (n) be a nondecreas-

ing sequence of extended nonnegative Emeasurable functions and set

(x ) := limn

n (x ) for any x X. Then

0

n(x) d(x )

0

(x ) d(x).

Proof. It suffices to notice that ({n > t}) ({ > t}) for all t > 0,and then to apply Proposition 2.8.

Now, by Proposition 2.4 we obtain the following important approxim-

ation property.

Proposition 2.12. Let : X [0, +] be an extendedEmeasurablefunction. Then there exist simple Emeasurable functions n : X [0, +) such thatn , so that

0

n(x) d(x )

0

(x ) d(x).

Remark 2.13 (Construction of Lebesgue and Riemann integrals).

Proposition 2.12 could be used as an alternative, and equivalent, defini-tion of the Lebesgue integral: we can just define it as the supremum of the

integral of minorant simple functions. This alternative definition is closer

to the definitions of Archimedean integrals and of inner Riemann integ-

ral: the only (fundamental) difference is due to the choice of the family of

simple functions. In all cases simple functions take finitely many val-

ues, but within the Lebesgue theory their level sets belong to a algebra,

and so the family of simple function is much richer, in comparison with

the other theories.We can now prove the additivity property of the integral.

Proposition 2.14. Let, : X [0, ] be Emeasurable functions.Then

X

( + ) d =

X

d +

X

d.

Proof. Let n, n be simple functions with n and n . Then,the additivity of the integral on simple functions gives

X

(n + n) d =

X

n d +

X

n d.

We conclude passing to the limit as n and using the monotoneconvergence theorem.

The following Fatous lemma, providing a semicontinuity property of

the integral, is of basic importance.


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Lemma 2.15 (Fatou). Let n : X [0, +] be extendedEmeasur-able functions. Then we have

X

lim infn

n(x) d(x) lim infn

X

n(x ) d(x ). (2.12)

Proof. Setting (x ) := lim infn n(x ), and n(x ) = infmn m (x ), wehave that n(x ) (x ). Consequently, by the monotone convergencetheorem,

X

(x ) d(x) = limn

X

n(x ) d(x ).

On the other handX

n(x ) d(x)

X

n(x ) d(x ),

so that X

(x ) d(x ) lim infn

X

n(x) d(x).

In particular, ifn are pointwise converging to , we haveX

(x ) d(x ) lim infn

X

n(x) d(x).

2.5. Integral of functions with a variable sign

Let : X R be an extended Emeasurable function. We say that is integrable if both the positive part +(x ) := max{(x ), 0} and thenegative part (x ) := max{(x), 0} of are integrable in X. As = + , in this case it is natural to define

X

(x ) d(x ) :=

X

+(x) d(x)

X

(x ) d(x).

As || = + + , the additivity properties of the integral give that

is integrable if and only if X |

|d 0 there exists > 0 such that (A) <

A

|| d < .(2.15)

The proof of this property is sketched in Exercise 2.9.


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2.6.1. Uniform integrability and Vitali convergence theorem

In this subsection we assume for simplicity that the measure isfi

nite.A family {i}iI ofRvalued integrable functions is said to be uniformly integrable if

lim(A)0

A

|i (x)| d(x ) = 0, uniformly in i I.

This means that for any > 0 there exists > 0 such that

(A) < A |i (x )| d(x) i I.This property obviously extends from single functions to families of func-

tions the absolute continuity property of the integral.

Notice that any family {i }iI dominated by a single integrablefunction (i.e. such that |i | || for any i I) is obviously uniformly integrable. Taking this remark into account, we are going to

to prove the following extension of the dominated convergence theorem,

known as Vitali Theorem.

Theorem 2.18 (Vitali). Assume that is a finite measure and let(n) be

a uniformly integrable sequence of functions pointwise converging to

a real valued function . Then is integrable and

limn

X

n d =

X

d.

To prove the Vitali theorem we need the following Egorov Lemma.

Lemma 2.19 (Egorov). Assume that is a finite measure and let (n)

be a sequence ofEmeasurable functions pointwise converging to a real

valued function . Then for any > 0 there exists a set A E such that(A) < andn uniformly in X \ A.Proof. For any integer m 1 we write X as the increasing union of thesets Bn,m , where

Bn,m :=

x X : |i (x ) (x)| < 1m

i n .Since is finite there exists n(m) such that (Bn(m),m ) > (X) 2m .We denote by A the union of X \ Bn(m),m , so that

(A)

m=1(X \ Bn(m),m ) 0, we can choose m > 1/ to obtain that

|n(x ) (x )| 1m

< for all x Bn(m),m , n n(m).

As X \ A Bn(m),m , this proves the uniform convergence of n to onX \ A.Proof of the Vitali Theorem. Fix > 0 and find > 0 such that

A|n| d < whenever (A) < . Again, Fatous Lemma yields that

A|| d whenever (A) < .

Assume now thatA

is given by Egorov Lemma, so that

n

uni-formly on X \ A. Then, writingX

( n) d =

X\A( n) d +

A

( n) d

and using the fact that limn supX\A

|n | = 0 we obtain

X

(

n) d

3

for n large enough. The statement follows letting 0.

2.7. A characterization of Riemann integrable functions

The integrals

Jf d, with J = [a, b] closed interval of the real line and

equal to the Lebesgue measure in R, are traditionally denoted with the

classical notation

b

af d x or with

Jf d x . This is due to the fact that

Riemanns and Lebesgues integral coincide on the class of Riemannsintegrable functions.

We denote by I(f) and I(f) the upper and lower Riemann integralof f respectively, the former defined by taking the supremum of the sumsn1

1 ai (ti+1 ti ) in correspondence of all step functions

h =n1i=1

ai1[ti ,ti+1) f a = t1 < < tn = b, (2.16)

and the latter considering the infimum in correspondence of all step func-tions h f. We denote by I(f) the Riemann integral, equal to the upperand lower integral whenever the two integrals coincide.

As the Lebesgue integral of the function h in (2.19) coincides withn1i ai (ti+1 ti ), we have

J

g d = I(g) for any step function g : J R.


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Now, if f : J R is continuous, we can choose a uniformly boundedsequence of step functions gh converging pointwise to f (for instance

splitting J into i equal intervals [xi ,xi+1[ and setting ai = min[xi ,xi+1] f)whose Riemann integrals converge to I(f). Therefore, passing to the

limit in the identity above with g = gh , and using the dominated conver-gence theorem we get

J

f d = I(f) for any continuous function f : J R.

We are going to generalize this fact, providing a full characterization,within the Lebesgue theory, of Riemmans integrable functions.

Theorem 2.20. Let f : J = [a, b] R be a bounded function. Thenf is Riemann integrable if and only if the set of its discontinuity points

is Lebesgue negligible. If this is the case, we have that f is B(J)

measurable and

J

f d

=I(f). (2.17)

Proof. Let

f(x) := inf

lim infh

f(xh) : xh x

f(x ) := sup

lim suph

f(xh) : xh x

.

(2.18)

It is not hard to show (see Exercise 2.6 and Exercise 2.7) that f is lowersemicontinuous and f is upper semicontinuous, therefore both f andf are Borel functions.

We are going to show that I(f) =

Jf d and I(f) =

J

f d.These two equalities yield the conclusion, as f is continuous at a.e.

point in J iff ff = 0 a.e. in J, and this holds iff (because ff 0)

J

(f f) d = 0.

Furthermore, if the set of discontinuity points of f is negligible, the

Borel function f differs from f only in a negligible set, thus f isB(J)measurable (because {f > t} differs from the Borel set {f > t}only in a negligible set, see also Exercise 2.4) and its integral coincides

with

Jf d =

Jf d; this leads to (2.17).


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Since I(f) = I(f) and f = (f), we need only to prove thefirst of the two equalities, i.e.

J

f d = I(f). (2.19)

In order to check the inequality in (2.19) we apply Exercise 2.11, find-ing a sequence of continuous functions gh f f and obtaining,thanks to the monotone convergence theorem,

J

f d = suphN

J

gh d = suphN I(gh) = suphN I(gh) I(f).

In order to prove in (2.19) we fix a step function h f in [a, b) as in(2.16) and we notice that f ai = h in (ti , ti+1) implies f ai in thesame interval. Hence f h in J \ {t1, . . . , tn} and, being the set of theti s Lebesgue negligible, we have

J

f

d

Jh d

=I(h).

Since h is arbitrary the inequality is achieved.

Exercises

2.1 Show that any of the conditions listed below is equivalent to the Emea-surability of : X R.

(i) 1((

, t

])

E for all t

R;

(ii) 1((, t)) E for all t R;(iii) 1([a, b]) E for all a, b R;(iv) 1([a, b)) E for all a, b R;(v) 1((a, b)) E for all a, b R.

2.2 Let , : X R be Emeasurable. Show that + and are Emeasurable. Hint: prove that

{ + < t} =r

Q

[{ < r} { < t r}]

and{2 > a} = { > a} { < a}, a 0.

2.3 Let us define a distance d in R by

d(x, y) := | arctanx arctan y|

where, by convention, arctan() = /2.


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(i) Show that (R, d) is a compact metric space (the so-called compactificationofR) and that A

R is open relative to the Euclidean distance if, and only

if, it is open relative to d;(ii) use (i) to show that, given a measurable space (X,E), f : X R is E

measurable according to (2.2) if and only if it is measurable between E and

the Borel algebra of(R, d).

2.4 Let (X,E, ) be a measure space and let E be the completion ofE inducedby . Show that f : X R is Emeasurable iff there exists a Emeasurablefunction g such that {f = g} is contained in a negligible set ofE.2.5 Let us define If as in (2.8) and let us endow with the usual partial ordering

= {t1, . . . , tN} = {s1, . . . , sM} if and only if . Show that If( ) is nondecreasing. Use this fact to show that f

0 f(t) dt is additive.

2.6 Let f : R R be a function. Show that the functions f, f defined in(2.18) are respectively lower semicontinuous and upper semicontinuous.

2.7 Let f : R R be a bounded function. Using Exercise 2.6 show that{f t} and {f t} are closed for all t R. In particular deduce that

= {x R : f is continuous at x}

belongs to B(R).

2.8 Let (an) (0, ) with

i=0ai = , lim

iai = 0.

Show that for any : X [0, +] Emeasurable there exist Ai E such that = i ai 1Ai . Hint: set 0 := , A0 := { a0} and 1 := 0 a01A0 0.Then, set A1 := {1 a1} and 2 := 0 a11A1 and so on.2.9 Let : X R be integrable. Show that the property (2.15) holds. Hint:assume by contradiction its failure for some > 0 and find Ai with (Ai ) < 2

iand

Ai

|| d . Then, notice that B := lim supi Ai is negligible, consider

Bn :=in

Ai \ B

and apply the dominated convergence theorem.

2.10 Prove that ifn in L1(, E, ), then (n) is uniformly integrable.In addition, find a space (X,E, ) and a sequence (n) that is uniformlyintegrable, for which there is no g L1(X,E, ) satisfying |n| g for alln N.2.11 Let (X, d) be a metric space and let g : X [0, ] be lower semicon-tinuous and not identically equal to . For any > 0 define

g(x) := infyX

{g(y) + d(x , y)} .


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Check that:

(a) |g(x) g(x )| d(x,x ) for all x, x X;(b) g g as .2.12 Let f : R2 R be satisfying the following two properties:(i) x f(x , y) is continuous in R for all y R;

(ii) y f(x , y) is continuous in R for all x R.Show that f is a Borel function. Hint: first reduce to the case when f isbounded. Then, for > 0 consider the functions

f(x , y) := 12

x+x

f(x , y) d x ,

proving that f are continuous and f f as 0.


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Chapter 3

Spaces of integrable functions

This chapter is devoted to the properties of the so-called Lp spaces, the

spaces of measurable functions whose p-th power is integrable. Through-

out the chapter a measure space (X,E, ) will be fixed.

3.1. Spaces Lp(X,E, ) and Lp(X,E, )

Let Y be a real vector space. We recall that a norm on Y is a non-

negative map defined on Y satisfying:

(i) y = 0 if and only if y = 0;

(ii) y = || y for all R and y Y;

(iii) y1 + y2 y1 + y2 for all y1, y2 Y.

The space Y, endowed with the norm , is called a normed space.

Y is also a metric space when endowed with the distance d(y1, y2) =

y1 y2 (the triangle inequality is a direct consequence of (iii)). If

(Y, d) is a complete metric space, we say that (Y, ) is a Banach space.

We denote by L1(X,E, ) the real vector space of all integrable

functions on (X,E). We define

1 :=

X

|(x)| d(x ), L1(X,E,).

We have clearly

1 = || 1 R, L1(X,E,),

and

+ 1 1 + 1 , L1(X,E, ),

so that conditions (ii) and (iii) in the definition of the norm are fulfilled.

However, 1 is not a norm in general, since 1 = 0 if and only if

= 0 a.e. in X, so (i) fails.




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Then, we can consider the following equivalence relation R on

L1(X,E, ),

= a.e. in X (3.1)

and denote by L1(X,E, ) the quotient space ofL1(X,E, ) with re-

spect to R. In other words, L1(X,E, ) is the quotient vector space

ofL1(X,E, ) with respect to the vector subspace made by functions

vanishing a.e. in X.

For any L1(X,E, ) we denote by the equivalence class de-

termined by and we set

+ := + , := . (3.2)It is easily seen that these definitions do no depend on the choice of rep-

resentatives in the equivalence class, and endow L1(X,E, ) with the

structure of a real vector space, whose origin is the equivalence class of

functions vanishing a.e. in X. Furthermore, setting

1 = 1, L1(X,E,),

it is also easy to see that this definition does not depend on the particular

element chosen in , and that (ii), (iii) still hold. Now, if 1 = 0

we have that the integral of || is zero, and therefore = 0. Therefore

L 1(X,E, ), endowed with the norm 1, is a normed space.

To simplify the notation typically is identified with whenever the

formula does not depend on the choice of the function in the equival-

ence class: for instance, quantities as ({ > t}) or X d have thisindependence, as well as most statements and results in Measure Theory

and Probability, so this slight abuse of notation is justified. It should be

noted, however, that formulas like ( x) = 0, for some fixed x X, do

not make sense in L1(X,E, ), since they depend on the representative

chosen (unless ({ x}) > 0).

More generally, if an exponent p (0, ) is given, we can apply a

similar construction to the space

Lp(X,E, ) :=

: is Emeasurable and

X

||p d <

.

Since |x + y|p |x |p + |y|p if p 1, and |x + y|p 2p1(|x |p +

|y|p) if p 1, it turns out that Lp(X,E, ) is a vector space, and we

shall denote by Lp(X,E, ) the quotient vector space, with respect to the

equivalence relation (3.1). Still we can define the sum and product by a


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real number as in (3.2), to obtain that Lp(X,E, ) has the structure of a

real vector space. The case p = 2 is particularly relevant for the theory,

as we will see.

Sometimes we will omit eitherEor , writing Lp(X,) or even Lp(X).

This typically happens when (X, d) is a metric space, and E is the Borel

-algebra, or when X R and is the Lebesgue measure.

3.2. The Lp norm

For any Lp(X,E, ) we define

p :=

X

||p d1/p

.

We are going to show that p is a norm for any p [1, +). Notice

that we already checked this fact when p = 1, and that the homogen-

eity condition (ii) trivially holds, whatever the value of p is. Further-

more, condition (i) holds precisely because Lp(X,E, ) consists, strictly

speaking, of equivalence classes induced by (3.1). So, the only condi-

tion that needs to be checked is the subadditivity condition (ii), and in thesequel we can assume p > 1.

The concept ofLegendre transform will be useful. Let f : R R be

a function; we define its Legendre transform f : R R {+} by

f(y) = supxR

{x y f(x )}, y R.

Then the following inequality clearly holds:

x y f(x) + f

(y) x, y R, (3.3)

and actually f could be equivalently defined as the smallest function

with this property.

Example 3.1. Let p > 1 and let

f(x ) =

x p

pifx 0,

0 ifx < 0.

Then, by an elementary computation, we find that

f(y) =

yq

qif y 0,

+ if y < 0,


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where q = p/(p 1) (equivalently, 1p

+ 1q

= 1). Consequently, the

following inequality, known as Young inequality, holds:

x y x p

p+

yq

q, x , y 0. (3.4)

Motivated by the previous example, we say that p and q are dual (or

conjugate) exponents if 1p

+ 1q

= 1, i.e. q = p/(p 1). The duality

relation is symmetric in (1, +), and obviously 2 is self-dual.

Example 3.2. Let f(x) = ex , x R. Then

f(y) := supxR

{x y ex } =+ if y < 0,0 if y = 0,

y log y y if y > 0.

Consequently, the following inequality holds:

x y ex + y log y y, x , y 0. (3.5)

3.2.1. Holder and Minkowski inequalities

Proposition 3.3 (Holder inequality). Assume that Lp(X,E, ) and Lq (X,E, ), with p and q dual exponents in (1, +). Then

L1(X,E, ) and

1 pq . (3.6)

Proof. If either p = 0 or q = 0 then one of the two functions

vanishes a.e. in X, hence vanishes a.e. and the inequality is

trivial. If both p and q are strictly positive, by the 1homogeneity

of the both sides in (3.6) with respect to and , we can assume with noloss of generality that the two norms are equal to 1.

Now we apply (3.4) to |(x )| and |(x )| to obtain

|(x )(x )| |(x )|p

p+

|(x)|q

q.

Integrating over X with respect to yields

X |(x )(x )| d(x ) 1p + 1q = 1.A particular case of the Holder inequality is

X

(x )(x ) d(x )

X

2(x ) d(x )

1/2X

2(x ) d(x)

1/2.

It also follows, as we shall see, from the Cauchy-Schwarz inequality of

scalar products.


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Proposition 3.4 (Minkowski inequality). Assume that p [1, +)

and, Lp(X,E, ). Then + Lp(X,E, ) and

+ p p + p. (3.7)

Proof. The cases p = 1 is obvious. Assume that p (1

Luigi Ambrosio, Giuseppe Da Prato, Andrea Mennucci Introduction to Measure Theory and Integration...

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