Introduction to Measure Theory and Integration

10

APPUNTI LECTURE NOTES

Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci Scuola Normale Superiore Piazza dei Cavalieri, 7 56126 Pisa, Italy Introduction to Measure Theory and Integration

Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Introduction to Measure Theory and Integration

c 2011 Scuola Normale Superiore Pisa  ISBN: 978-88-7642-385-7 e-ISBN: 978-88-7642-386-4

Contents

Preface

ix

Introduction

xi

1 Measure spaces 1.1 Notation and preliminaries . . . . . . . . . . . . . . . 1.2 Rings, algebras and σ –algebras . . . . . . . . . . . . . 1.3 Additive and σ –additive functions . . . . . . . . . . . 1.4 Measurable spaces and measure spaces . . . . . . . . . 1.5 The basic extension theorem . . . . . . . . . . . . . . 1.5.1 Dynkin systems . . . . . . . . . . . . . . . . . 1.5.2 The outer measure . . . . . . . . . . . . . . . 1.6 The Lebesgue measure in R . . . . . . . . . . . . . . 1.7 Inner and outer regularity of measures on metric spaces 2 Integration 2.1 Inverse image of a function . . . . . . . . . . . . . . 2.2 Measurable and Borel functions . . . . . . . . . . . 2.3 Partitions and simple functions . . . . . . . . . . . . 2.4 Integral of a nonnegative E –measurable function . . 2.4.1 Integral of simple functions . . . . . . . . . 2.4.2 The repartition function . . . . . . . . . . . 2.4.3 The archimedean integral . . . . . . . . . . . 2.4.4 Integral of a nonnegative measurable function 2.5 Integral of functions with a variable sign . . . . . . . 2.6 Convergence of integrals . . . . . . . . . . . . . . . 2.6.1 Uniform integrability and Vitali convergence theorem . . . . . . . . . . . . . . . . . . . . 2.7 A characterization of Riemann integrable functions .

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1 1 2 4 7 8 9 11 14 18

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23 23 24 25 27 27 28 31 32 35 36

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38 39

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vi Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

3 Spaces of integrable functions 3.1 Spaces L p (X, E , μ) and L p (X, E , μ) . . . . . 3.2 The L p norm . . . . . . . . . . . . . . . . . . 3.2.1 H¨older and Minkowski inequalities . . 3.3 Convergence in L p (X, E , μ) and completeness 3.4 The space L ∞ (X, E , μ) . . . . . . . . . . . . . 3.5 Dense subsets of L p (X, E , μ) . . . . . . . . .

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45 45 47 48 49 52 56

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61 61 63 66

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67 70

5 Fourier series 5.1 Pointwise convergence of the Fourier series . . . . . . . 5.2 Completeness of the trigonometric system . . . . . . . . 5.3 Uniform convergence of the Fourier series . . . . . . . .

73 75 79 80

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4 Hilbert spaces 4.1 Scalar products, pre-Hilbert and Hilbert spaces . . 4.2 The projection theorem . . . . . . . . . . . . . . . 4.3 Linear continuous functionals . . . . . . . . . . . . 4.4 Bessel inequality, Parseval identity and orthonormal systems . . . . . . . . . . . . . . . . . . . . . . . 4.5 Hilbert spaces on C . . . . . . . . . . . . . . . . .

6 Operations on measures 6.1 The product measure and Fubini–Tonelli theorem 6.2 The Lebesgue measure on Rn . . . . . . . . . . . 6.3 Countable products . . . . . . . . . . . . . . . . 6.4 Comparison of measures . . . . . . . . . . . . . 6.5 Signed measures . . . . . . . . . . . . . . . . . 6.6 Measures in R . . . . . . . . . . . . . . . . . . . 6.7 Convergence of measures on R . . . . . . . . . . 6.8 Fourier transform . . . . . . . . . . . . . . . . . 6.8.1 Fourier transform of a measure . . . . . . 7 The fundamental theorem of the integral calculus

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83 83 87 90 94 101 105 107 112 113 119

8 Measurable transformations 129 8.1 Image measure . . . . . . . . . . . . . . . . . . . . . . 129 8.2 Change of variables in multiple integrals . . . . . . . . . 130 8.3 Image measure of L n by a C 1 diffeomorphism . . . . . 131 A

137 A.1 Continuity and differentiability of functions depending on a parameter . . . . . . . . . . . . . . . . . . . . . . . 137

vii Introduction to Measure Theory and Integration

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

183

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 Lebesgues theory in Rn into a more general context and presenting the basic applications to Fourier series, calculus and real analysis. The text can also pave the way to more advanced courses in probability, stochastic processes 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

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 Riemann’s 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 Riemann’s integrable function, within Riemann’s theory. This is indeed possible within the stronger theory, due essentially to Lebesgue, that we are going to introduce. (2) The extensions of Riemann’s theory to multiple integrals are very cumbersome. This extension, useful to compute areas, volumes, etc., is known as Peano–Jordan theory, and it is sometimes taught in elementary courses of integration in more than one variable. In addition to that, important heuristic principles like Cavalieri’s one can be proved only under technical and basically unnecessary regularity assumptions on the domains of integration. (3) Many constructive processes typical of Analysis (limits, series, integrals depending on a parameter, etc.) cannot be handled well within Riemann’s theory of integration. For instance, the following statement is true (it is a particular case of the so-called dominated convergence theorem): Theorem 1. Let f h : [−1, 1] → R be continuous functions pointwise converging to a continuous function f . Assume the existence of a constant M satisfying | f h (x)| ≤ M for all x ∈ [−1, 1] and all h ∈ N. Then  lim

1

h→∞ −1

 f h (x) dx =

1

−1

f (x) dx.

Even though this statement makes perfectly sense within Riemann’s theory, any attempt to prove this result within the theory (try, if you don’t

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 to 1 −1 f (x) dx even without this assumption. This necessity emerges for instance in the study of the convergence of Fourier series f (x) =

∞ 

ai cos(iπ x) + bi sin(iπ x)

x ∈ [−1, 1].

i=0

In this case the uniform convergence of the series, which  implies the continuity of f as well, is ensured by the condition i |ai | + |bi | < ∞. On the other hand, we will see that the “natural” condition for the convergence (in a suitable sense) of the series is much weaker: ∞ 

ai2 + bi2 < ∞.

i=0

Under this condition the limit function f need not be continuous: for instance, if f (x) = 1 for x ∈ [−1/2, 1/2] and f (x) = 0 otherwise, then we will see that the coefficients of the Fourier series are given by bi = 0 for all i (because f is even) and by ⎧ 1 ⎪ ⎪ if i = 0; ⎪ ⎨2 ai = ⎪ ⎪ ⎪ ⎩ sin(πi/2) if i > 0. iπ (4) The spaces of integrable functions, as for instance 

 1 f 2 (x) dx < ∞ H := f : [−1, 1] → R : −1

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 Riemann’s to Lebesgue’s theory is the same one that led from the (incomplete) set of rational numbers Q to the (complete) real line R.

xiii Introduction to Measure Theory and Integration

Lebesgue’s theory extends Riemann’s theory in two independent directions. 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 Riemann’s integral  1 n−1  f (x) dx ∼ (ti+1 − ti ) f (ti ) −1

i=1

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

−1

f (x) dμ(x) ∼

n 

μ(Ai ) f (xi )

(1)

i=1

where A1 , . . . , An is a partition of [−1, 1], xi ∈ Ai and the approximation is expected to be better and better as the parameter supi diam(Ai ) tends to 0. We may think, for instance to [−1, 1] as a possibly non-homogeneous bar, and to μ(A) as the “mass” of the subset A of the bar: because of non-homogeneity, μ(A) need not be proportional to the length of A. Once we adopt this viewpoint, we will see that it is not hard to obtain a theory of integration in general metric spaces, and even in more general classes of spaces. On the other hand, the approximation (1), that in any case clarifies the intuitive meaning of the integral, will remain valid for continuous functions only.

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 the concepts of ring, algebra and σ –algebra. The most basic tool developed in the chapter is Carath´eodory’s theorem, which ensures in many cases the existence and the uniqueness of a σ –additive measure having some prescribed values on a set of generators of the σ –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 ∩ B c , and by AB the symmetric difference (A \ B) ∪ (B \ A). Let (An ) be a sequence in P (X). Then the following De Morgan identity holds, c ∞ ∞

 An = Acn . n=0

Moreover, we define lim sup An := n→∞

n=0

(1) ∞ ∞  n=0 m=n

Am ,

lim inf An := n→∞

∞  ∞

Am .

n=0 m=n

(1) Notice the analogy with liminf and limsup limits for a sequence (a ) of real numbers. We have n lim sup an = inf sup am and lim inf an = sup inf am . This is something more than an analogy, n→∞ n∈N m≥n n→∞ n∈N m≥n

see Exercise 1.1.

L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

2 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

As it can be easily checked, lim supn An (respectively lim infn An ) consists of those elements of X that belong to infinitely many An (respectively 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 inf An = lim sup An = An , n→∞

n→∞

n=0

whereas if (An ) is nonincreasing (i.e. An ⊃ An+1 , n ∈ N), we have lim inf An = lim sup An = n→∞

n→∞

∞ 

An .

n=0

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

1.2. Rings, algebras and σ –algebras Definition 1.1 (Rings and Algebras). A non empty subset A of P (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 including K (that is the intersection of all algebras including K ) is well defined, and called the algebra generated by K . A constructive characterization of the algebra generated by K can be easily achieved as follows: set F (0) = K ∪ {∅} and

  F (i+1) := ∀i ≥ 0. A ∪ B, Ac : A, B ∈ F (i)  Then, the algebra A generated by K is given by i F (i) . Indeed, it is immediate to check by induction on i that A ⊃ F (i) , and therefore the union 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 of A provides the opposite inclusion. Definition 1.2 (σ –algebras). A non-empty subset E of P (X) is said to be a σ –algebra if: (i) E is an algebra;

3 Introduction to Measure Theory and Integration

∞ 

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

An ∈ E .

n=0

If E is a σ –algebra and (An ) ⊂ E we have Morgan identity. Moreover, both sets lim inf An , n→∞

 n

An ∈ E by the De

lim sup An , n→∞

belong to E . Obviously, {∅, X} and P (X) are σ –algebras, respectively the smallest and the largest ones. Let K be a subset of P (X). As the intersection of any family of σ –algebras is still a σ -algebra, the minimal σ –algebra including K (that is the intersection of all σ –algebras including K ) is well defined, and called the σ –algebra generated by K . It is denoted by σ (K ). In contrast with the case of generated algebras, it is quite hard to give a constructive characterization of the generated σ -algebras: this requires the transfinite induction and it is illustrated in Exercise 1.18. Definition 1.3 (Borel σ -algebra). If (E, d) is a metric space, the σ – algebra generated by all open subsets of E is called the Borel σ –algebra of E and it is denoted by B (E). In the case when E = R the Borel σ -algebra has a particularly simple class of generators. Example 1.4 (B (R)). Let I be the set of all semi–closed intervals [a,b) with a ≤ b. Then σ (I ) coincides with B (R). In fact σ (I ) contains all open intervals (a, b) since (a, b) =

∞ 

1  a + ,b n n=n 0

with n 0 >

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 open or closed half-lines.

(2) Indeed, let (a ) be a sequence including all rational numbers of A and denote by I the largest k k  open 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 k such that ak ∈ (x − r, x + r) to obtain (x − r, x + r) ⊂ Ik , by the maximality of Ik , and then x ∈ Ik .

4 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

1.3. Additive and σ –additive functions Let A ⊂ P (X) be a ring and let μ be a mapping from A into [0, +∞] such that μ(∅) = 0. We say that μ is additive if A, B ∈ A , A ∩ B = ∅

⇒

μ(A ∪ B) = μ(A) + μ(B).

If μ is additive, A, B ∈ A and A ⊃ B, we have μ(A) = μ(B) + μ(A \ B), so that μ(A) ≥ μ(B). Therefore any additive function is nondecreasing with respect to set inclusion. Moreover, by applying repeatedly the additivity property, additive measures satisfy  n n

 Ak = μ(Ak ) μ k=1

k=1

for n ∈ N∗ and mutually disjoint sets A1 , . . . , An ∈ A . A set function μ on A is called σ –additive if μ(∅) =  0 and for any sequence (An ) ⊂ A of mutually disjoint sets such that n An ∈ A we have  ∞ ∞

 μ An = μ(An ). n=0

n=0

Obviously σ –additive functions are additive, because we can consider countable unions in which only finitely many An are nonempty. Another useful concept is the σ –subadditivity: we say that μ is σ – subadditive if ∞  μ(An ), μ(B) ≤ n=0

 for any B ∈ A and any sequence (An ) ⊂ A such that B ⊂ n An . Notice that, unlike the definition of σ –additivity, the sets An need not be disjoint here. Remark 1.5 (σ –additivity and σ –subadditivity). Let μ be additive on  a ring A and let (An ) ⊂ A be mutually disjoint and such that n An ∈ A . Then by monotonicity we have   ∞ k k

 An ≥ μ An = μ(An ), for all k ∈ N. μ n=0

n=0

n=0

Therefore, letting k ↑ ∞ we get  ∞ ∞

 An ≥ μ(An ). μ n=0

n=0

5 Introduction to Measure Theory and Integration

Thus, to show that an additive function is σ –additive, it is enough to prove that it is σ –subadditive. Conversely, it is not difficult to show that σ –additive set functions are σ –subadditive: indeed, if B ⊂ ∪n An we can define A0 = B ∩ A0 and  An := B ∩ An \ m
∞ 

μ(An ) ≤

n=0

∞ 

μ(An ).

n=0

Let μ be additive on A . Then σ –additivity of μ is equivalent to continuity of μ in the sense of the following proposition. Proposition 1.6 (Continuity on nondecreasing sequences). If μ is additive on a ring A , then (i) ⇐⇒ (ii), where: (i) μ is σ –additive; (ii) (An ) ⊂ A and A ∈ A , An ↑ A ⇒ μ(An ) ↑ μ(A). Proof. (i)⇒(ii). In the proof of this implication we can assume with no loss of generality that μ(An ) < ∞ for all n ∈ N. Let (An ) ⊂ A , A ∈ A , An ↑ A. Then ∞

A = A0 ∪ (An+1 \ An ), n=0

the unions being disjoint. Since μ is σ –additive, we deduce that ∞  (μ(An+1 ) − μ(An )) = lim μ(An ), μ(A) = μ(A0 ) + n→∞

n=0

and (ii) follows.  (ii)⇒(i). Let (An ) ⊂ A be mutually disjoint and such that A := n An ∈ A . Set m

Bm := Ak . k=0

Then Bm ↑ A and μ(Bm ) = implies (i).

m 

μ(Ak ) ↑ μ(A) by the assumption. This

0

Proposition 1.7 (Continuity on nonincreasing sequences). Let μ be σ –additive on a ring A . Then (An ) ⊂ A and A ∈ A , An ↓ A, μ(A0 ) < ∞

⇒

μ(An ) ↓ μ(A). (1.1)

6 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proof. Setting Bn := A0 \ An , B := A0 \ A, we have Bn ↑ B, therefore the previous proposition gives μ(Bn ) ↑ μ(B). As μ(An ) = μ(A0 ) − μ(Bn ) and μ(A) = μ(A0 ) − μ(B) the proof is achieved. Corollary 1.8 (Upper and lower semicontinuity of the measure). Let μ be σ –additive on a σ –algebra E and let (An ) ⊂ E . Then we have   μ lim inf An ≤ lim inf μ(An ) (1.2) n→∞

n→∞

and, if μ(X) < ∞, we have also   lim sup μ(An ) ≤ μ lim sup An . n→∞

(1.3)

n→∞

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

∞ 

Bn

Bn :=

where

∞

Am .

(1.4)

m=n

n=0

Now, assuming μ(X) < ∞, by Proposition 1.7 it follows that μ(L) = lim μ(Bn ) = inf μ(Bn ) ≥ inf sup μ(Am ) = lim sup μ(An ). n∈N

n→∞

n∈N m≥n

n→∞

Thus, we have proved (1.3). The inequality (1.2) can be proved similarly using Proposition 1.6, thus without using the assumption μ(X) < ∞. The following result is very useful to estimate the measure of a lim sup of sets. Lemma 1.9. Let μ be σ –additive on a σ –algebra E and let (An ) ⊂ E . ∞  μ(An ) < ∞. Then Assume that n=0

  μ lim sup An = 0. n→∞

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

n→∞

μ(L) ≤ μ(Bn ) ≤

∞  m=n

As n → ∞ we find μ(L) = 0.

μ(Am )

for all n ∈ N.

7 Introduction to Measure Theory and Integration

1.4. Measurable spaces and measure spaces Let E be a σ –algebra of subsets of X. Then we say that the pair (X, E ) is a measurable space. Let μ : E → [0, +∞] be a σ –additive function. Then we call μ a measure on (X, E ), and we call the triple (X, E , μ) a measure space. The measure μ is said to be finite if μ(X) < ∞, σ –finite if there exists a sequence (An ) ⊂ E such that n An = X and μ(An ) < ∞ for all n ∈ N. Finally, μ is called a probability measure if μ(X) = 1. The simplest (but fundamental) example of a probability measure is the Dirac mass δx , defined by ⎧ ⎨1 if x ∈ B δx (B) := ⎩0 if x ∈ / 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 a finite or countable set. Given c : Y → [0, +∞] we can define a measure on (X, P (X)) as follows:  c(x) ∀B ⊂ X. μ(B) := x∈B∩Y



 Clearly μ = x∈Y c(x)δx is a finite measure if and only if x∈Y c(x) < ∞, and it is σ –finite if and only if c(x) ∈ [0, +∞) for all x ∈ Y . More generally, the construction above works even when Y is uncountable, by replacing the sum with  c(x), sup c∈B∩Y 

where the supremum is made among the finite subsets Y  of Y . The measures arising in the previous example are called atomic, and clearly if X is either finite or countable then any measure μ in (X, P (X)) is atomic: it suffices to notice that  c(x)δx with c(x) := μ({x}). μ= x∈X

In the next section we will introduce a fundamental tool for the construction of non-atomic measures.

8 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Definition 1.11 (μ–negligible sets and μ–almost everywhere). Given a measure space (X, E , μ), we say that B ∈ E is μ–negligible if μ(B) = 0, and we say that a property P(x) holds μ–almost everywhere if the set {x ∈ X : P(x) is false} is contained in a μ–negligible set. Notice that the class of μ–negligible sets is stable under finite or countable unions. It is sometimes convenient to know that any subset of a μ–negligible set is still μ–negligible. Definition 1.12 (μ–completion of aσ–algebra and μ–measurable sets). Let (X, E , μ) be a measure space. We define E μ := {A ∈ P (X) : for some B, C ∈ E with μ(C) = 0, AB ⊂ C} . It is easy to check that E μ is still a σ –algebra, the so-called completion of E with respect to μ. The elements of E μ are called μ–measurable sets. It is also easy to check that μ can be extended to all A ∈ E μ simply by setting μ(A) = μ(B), where B ∈ E is any set such that AB is contained in a μ–negligible set of E . This extension is well defined (i.e. independent of the choice of B), still σ –additive and μ–negligible sets coincide with those sets that are contained in some B ∈ E with μ(B) = 0. As a consequence, any subset of a μ–negligible set is μ–negligible as well.

1.5. The basic extension theorem The following result, due to Carath´eodory, allows to extend a σ –additive function on a ring A to a σ –additive function on σ (A ). It is one of the basic tools in the construction of non-trivial measures in many cases of interest, as we will see. Theorem 1.13 (Carath´eodory). Let A ⊂ P (X) be a ring, and let E be the σ –algebra generated by A . Let μ : A → [0, +∞] be σ –additive. Then μ can be extended to a measure on E . If μ is σ –finite, i.e. there exist An ∈ A with An ↑ X and μ(An ) < ∞ for any n, then the extension is unique. To prove this theorem we need some preliminaries: for the uniqueness the Dynkin theorem and for the existence the concepts of outer measure and additive set.

9 Introduction to Measure Theory and Integration

1.5.1. Dynkin systems A non-empty subset K of P (X) is called a π–system if A, B ∈ K ⇒ A ∩ B ∈ K . A non-empty subset D of P (X) is called a Dynkin system if (i) X, ∅ ∈ D ; (ii) A ∈ D ⇒ Ac ∈ D ;  (iii) (Ai ) ⊂ D mutually disjoint ⇒ i Ai ∈ D . Obviously any σ –algebra is a Dynkin system. Moreover, if D is both a Dynkin system and a π–system then it is a σ –algebra. In fact, if (Ai ) is a sequence in D of not necessarily disjoint sets we have ∞

Ai = A0 ∪ (A1 \ A0 ) ∪ ((A2 \ A1 ) \ A0 ) ∪ · · ·

i=0

 and so i Ai ∈ D by (ii) and (iii). Let us prove now the following important result. Theorem 1.14 (Dynkin). Let K be a π–system and let D ⊃ K be a Dynkin system. Then σ (K ) ⊂ D . Proof. Let D 0 be the minimal Dynkin system including K . We are going to show that D 0 is a σ –algebra which will prove the theorem. For this it is enough to show, as remarked before, that the following implication holds: (1.5) A, B ∈ D 0 ⇒ A ∩ B ∈ D 0 . For any B ∈ D 0 we set H (B) = {F ∈ D 0 : B ∩ F ∈ D 0 }. We claim that H (B) is a Dynkin system. In fact properties (i) and (iii) are clear. It remains to show that if F ∩ B ∈ D 0 then F c ∩ B ∈ D 0 or, equivalently, F ∪ B c ∈ D 0 . In fact, since F ∪ B c = (F \ B c ) ∪ B c = (F ∩ B) ∪ B c and F ∩ B and B c are disjoint, we have that F ∪ B c ∈ D 0 as required. Notice first that if K ∈ K we have K ⊂ H (K ) since K is a π–system. Therefore H (K ) = D 0 , by the minimality of D 0 . Consequently, the following implication holds K ∈ K , B ∈ D 0 ⇒ K ∩ B ∈ D 0 ,

10 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

which implies K ⊂ H (B) for all B ∈ D 0 . Again, the fact that H (B) is a Dynkin system and the minimality of D 0 give that H (B) = D 0 for all B ∈ D 0 . By the definition of H (B), this proves (1.5). The uniqueness part in Caratheodory’s theorem is a direct consequence of the following coincidence criterion for measures; in turn, the proof of the criterion relies on Theorem 1.14. Proposition 1.15 (Coincidence criterion). Let μ1 , μ2 be measures in (X, E ) and assume that: (i) the coincidence set D := {A ∈ E : μ1 (A) = μ2 (A)} contains a π–system K with σ (K ) = E ; (ii) there exists a nondecreasing sequence (X i ) ⊂ K with μ1 (X i ) = μ2 (X i ) < ∞ and X i ↑ X. Then μ1 = μ2 . Proof. We first assume that μ1 (X) = μ2 (X) is finite. Under this assumption D is a Dynkin system including the π–system K (stability of D under complement is ensured precisely by the finiteness assumption). Thus, by the Dynkin theorem, D = E , which implies that μ1 = μ2 . Assume now that we are in the general case and let X i be given by assumption (ii). Fix i ∈ N and define the σ –algebra E i of subsets of X i by E i := {A ⊂ X i : A ∈ E } . We may obviously consider μ1 and μ2 as finite measures in the measurable space (X i , E i ). Since these measures coincide on the π–system K i := {A ⊂ X i : A ∈ K } we obtain, by the previous step, that μ1 and μ2 coincide on σ (K i ) ⊂ P (X i ). Now, let us prove the inclusion {B ∈ E : B ⊂ X i } ⊂ σ (K i ). Indeed



(1.6)

 B ⊂ X : B ∩ X i ∈ σ (K i )

is a σ –algebra containing K (here we use the fact that X i ∈ K ), and therefore contains E . Hence any element of E contained in X i belongs to σ (K i ). By (1.6) we obtain μ1 (B ∩ X i ) = μ2 (B ∩ X i ) for all B ∈ E and all i ∈ N. Passing to the limit as i → ∞, since B is arbitrary we obtain that μ1 = μ2 .

11 Introduction to Measure Theory and Integration

1.5.2. The outer measure Let μ be a set function defined on A ⊂ P (X). For any E ∈ P (X) we define:   ∞ ∞ 

μ(Ai ) : Ai ∈ A , E ⊂ Ai . μ∗ (E) := inf i=0

i=0

μ∗ is called the outer measure induced by μ. We can easily show that μ∗ is a nondecreasing set function, namely μ∗ (E) ≤ μ∗ (F) whenever E ⊂ F ⊂ X. We will obtain the proof of the existence part of Carath´eodory’s theorem by showing in the proposition below that μ∗ extends μ if μ is σ – subadditive, and that (Theorem 1.17) μ∗ is σ -additive on a σ -algebra containing σ (A ) if μ is A is a ring and μ is additive on A . In particular if μ is σ –additive on A we see that μ∗ provides the desired σ –additive extension to σ (A ). Proposition 1.16. The set function μ∗ is σ –subadditive on P (X) and extends μ if μ is σ –subadditive on A and μ(∅) = 0.   Proof. Let (E i ) ⊂ P (X) and set E := i E i . Assume that i μ∗ (E i ) are finite (otherwise the assertion is trivial). Then, since μ∗ (E i ) is finite for any i ∈ N, for any ε > 0 there exist Ai, j ∈ A such that ∞ 

∗

μ(Ai, j ) < μ (E i ) +

j=0

Consequently

∞ 

∞ 

2i+1

,

μ(Ai, j ) ≤

i, j=0

Since E ⊂

ε

Ei ⊂

∞

Ai, j ,

i ∈ N.

j=0

∞ 

μ∗ (E i ) + ε.

i=0

Ai, j we have

i, j=0

μ∗ (E) ≤

∞  i, j=0

μ(Ai, j ) ≤

∞ 

μ∗ (E i ) + ε

i=0

and the first part of the statement follows from the arbitrariness of ε. Now, let us assume that μ is  σ -subadditive on A and choose E ∈ A ; since E ⊂ i Ai then μ(E) ≤ i μ(Ai ), so we deduce μ∗ (E) ≥ μ(E); but, by choosing A0 = E and An = ∅ for n ≥ 1, we obtain that μ∗ (E) = μ(E). This proves that μ∗ extends μ.

12 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Let us now define the additive sets, according to Carath´eodory. A set A ∈ P (X) is called additive if μ∗ (E) = μ∗ (E ∩ A) + μ∗ (E ∩ Ac ) for all E ∈ P (X).

(1.7)

We denote by G the family of all additive sets. Notice that, since μ∗ is subadditive, (1.7) is equivalent to μ∗ (E) ≥ μ∗ (E ∩ A) + μ∗ (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 μ∗ (A ∪ B) = μ∗ (A) + μ∗ (B).

(1.9)

Other important properties of G are listed in the next proposition. Theorem 1.17. Assume that A is a ring and that μ is additive. Then G is a σ –algebra containing A and μ∗ is σ –additive on G . Proof. We proceed in three steps: we show that G contains A , that G is a σ –algebra and that μ∗ is additive on G . As pointed in Remark 1.5, if μ∗ is σ –subadditive and additive on the σ –algebra G , then μ∗ is σ –additive. Step 1. A ⊂ G . Let A ∈ A and E ∈ P (X), we have to show (1.8). Assume μ∗ (E) < ∞ (otherwise (1.8) trivially holds), fix ε > 0 and choose (Bi ) ⊂ A such that E⊂

∞

Bi ,

μ∗ (E) + ε >

i=0

∞ 

μ(Bi ).

i=0

Then, by the definition of μ∗ , it follows that μ∗ (E) + ε >

∞  i=0

μ(Bi ) =

∞  [μ(Bi ∩ A) + μ(Bi ∩ Ac )] i=0

≥ μ∗ (E ∩ A) + μ∗ (E ∩ Ac ). Since ε is arbitrary we have μ∗ (E) ≥ μ∗ (E ∩ A) + μ∗ (E ∩ Ac ), and (1.8) follows. Step 2. G is an algebra and μ∗ is additive on G . We already know that A ∈ G implies Ac ∈ G . Let us prove now that if A, B ∈ G then A ∪ B ∈ G . For any E ∈ P (X) we have μ∗ (E) = μ∗ (E ∩ A) + μ∗ (E ∩ Ac ) = μ∗ (E ∩ A) + μ∗ (E ∩ Ac ∩ B) + μ∗ (E ∩ Ac ∩ B c ) = [μ∗ (E ∩ A) + μ∗ (E ∩ Ac ∩ B)] + μ∗ (E ∩ (A ∪ B)c ).

(1.10)

13 Introduction to Measure Theory and Integration

Since

(E ∩ A) ∪ (E ∩ Ac ∩ B) = E ∩ (A ∪ B),

we have by the subadditivity of μ∗ , μ∗ (E ∩ A) + μ∗ (E ∩ Ac ∩ B) ≥ μ∗ (E ∩ (A ∪ B)). So, by (1.10) it follows that μ∗ (E) ≥ μ∗ (E ∩ (A ∪ B)) + μ∗ (E ∩ (A ∪ B)c ), and A ∪ B ∈ G as required. The additivity of μ∗ on G follows directly from (1.9). Step  3. G is a σ –algebra. Let (An ) ⊂ G . We are going to show that restrictive S := n An ∈ G . Since we know that G is an algebra, it is not to assume that all sets An are mutually disjoint. Set Sn := n0 Ai , for n ∈ N. For any n ∈ N, by using the σ –subadditivity of μ∗ and by applying (1.7) repeatedly, we get ∗

∗

∗

μ (E ∩ S ) + μ (E ∩ S) ≤ μ (E ∩ S ) + c

c

 = lim

n→∞

∞ 

μ∗ (E ∩ Ai )

i=0

μ∗ (E ∩ S c ) +

n 

 μ∗ (E ∩ Ai )

i=0

  = lim μ∗ (E ∩ S c ) + μ∗ (E ∩ Sn ) . n→∞

Since S c ⊂ Snc it follows that

  μ∗ (E ∩ S c ) + μ∗ (E ∩ S) ≤ lim sup μ∗ (E ∩ Sn ) + μ∗ (E ∩ Snc ) n→∞ ∗

= μ (E). So, S ∈ G and G is a σ –algebra. Remark 1.18. We have proved that σ (A ) ⊂ G ⊂ P (X).

(1.11)

One can show that the inclusions above are strict in general, for instance when μ is the Lebesgue measure we shall consider in the next section. In fact, in the case when X = R and σ (A ) is the Borel σ -algebra, Exercise 1.19 shows that σ (A ) has the cardinality of continuum, while G has the cardinality of P (R), since it contains all subsets of Cantor’s 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.

14 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

1.6. The Lebesgue measure in R In this section we build the Lebesgue measure on the real line R. To this aim, we consider first the set I of all bounded right open intervals of R 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 define length((a, b]) := b − a. More generally, any non-empty A ∈ A can be written, possibly in many ways, as a disjoint finite union of intervals Ii , i = 1, . . . , N ; we define λ(A) :=

N 

length(Ii ).

(1.12)

i=1

Setting λ(∅) = 0, it is not hard to show by elementary methods that λ is well defined (i.e. λ(A) does not depend on the chosen decomposition) and additive on A . 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 1 A : X → {0, 1} is defined by ⎧ ⎨1 if x ∈ A; 1 A (x) := ⎩0 if x ∈ X \ A. The reader already acquainted with Riemann’s theory of integration can also notice that λ(A) is the Riemann integral of the characteristic function 1 A of A, and deduce the additivity property of λ directly by the additivity properties of the Riemann integral. In the next theorem we shall rigorously 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 }n∈N of open sets is contained in the union of finitely many of them.

15 Introduction to Measure Theory and Integration

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 Bolzano–Weierstrass theorem there exists a subsequence (xn(k) ) converging 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 / An(k) and An(k) ⊃ An¯ . n(k) ≥ n, ¯ because xn(k) ∈ Theorem 1.21. The set function λ defined in (1.12) is σ –additive on A . Proof. (λ is well defined) 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 J j . 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 elements of the second partition, neglecting the empty intersections. Given these  to show that λ is well defined, it suffices to show that  remarks, λ(I ) = finer than I1 , . . . , In . This statei i j λ(J j ) if J1 , . . . , Jm is ment reduces to the fact that λ(I ) = k λ(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 . (λ is additive) If F, G ∈ A and 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 , . . . , Jm of F ∪ G in intervals belonging to I . Using this decomposition to compute λ(F ∪ G) the additivity easily follows. (λ is σ –additive) Let (Fn ) ⊂ A be a sequence of disjoints sets in A and assume that ∞

F := Fn (1.13) n=0

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 λ(Fn ) is convergent. As any Fn is a finite union of intervals, say Nn , we can find, given any ε > 0, a finite union Fn ⊃ Fn of intervals in I such that λ(Fn ) ≤ λ(Fn ) + ε/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 larger interval in I , increasing the length at most by ε/(Nn 2n )). Let also of Fn , that still includes Fn , and let x  ∈ (x, y]. Fn be the internal part   Then, since [x , y] ⊂ n Fn , Lemma 1.20 provides an integer k such

16 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

that [x  , y] ⊂

k

Fn . Hence, the additivity of λ in A gives  k k

 y − x ≤ λ Fn ≤ λ(Fn ) 0

n=0

≤

k 

n=0

λ(Fn ) +

n=0

∞  ε ≤ 2ε + λ(Fn ). 2n n=0

By letting first ε ↓ 0 and then letting x  ↓ x we obtain that λ(F) ≤  ∞ 0 λ(Fn ). The opposite inequality simply follows by the monotonicity and the additivity of λ, because the finite unions of the sets Fn are contained in F. In the general case, let k

F= Ii , i=1

where I1 , . . . , Ik are disjoint sets in I . 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 ∞  λ(Ii ∩ Fn ). λ(Ii ) = n=0

Adding these identities for i = 1, . . . , k, commuting the sums on the right hand side and eventually using the additivity of λ on A we obtain λ(F) =

k  i=1

λ(Ii ∩ F) =

∞  k  n=0 i=1

λ(Ii ∩ Fn ) =

∞ 

λ(Fn ).

n=0

We say that a measure μ in (R, B (R)) is translation invariant if μ(A + h) = μ(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 μ is locally finite if μ(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 finite measure λ in (R, B (R)). The unique such measure λ satisfying λ([0, 1]) = 1 is called Lebesgue measure. Proof. (Existence) Let A be the class of finite unions of intervals and let λ : A → [0, +∞) be the σ –additive set function defined in (1.12). According to Theorem 1.21 λ admits a unique extension, that we still denote by λ, to σ (A ) = B (R). Clearly λ is locally finite, and we can use the uniqueness of the extension to prove translation invariance: indeed,

17 Introduction to Measure Theory and Integration

for any h ∈ R also the σ –additive measure A → λ(A+h) is an extension of λ|A . As a consequence λ(A) = λ(A + h) for all h ∈ R. (Uniqueness) Let ν be a translation invariant and locally finite measure in (R, B (R)) and set c := ν([0, 1]). Notice first that the set of atoms of ν is at most countable (Exercise 1.5), and since R is uncountable there exists at least one x such that ν({x}) = 0. By translation invariance this holds for all x, i.e., ν has no atom. Excluding the trivial case c = 0 (that gives ν ≡ 0 by translation invariance and σ –additivity), we are going to show that ν = cλ on the class A of finite unions of intervals; by the uniqueness of the extension in Carath´eodory theorem this would imply that ν = cλ on B (R). By finite additivity and translation invariance it suffices to show that ν([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  c ν([0, 1)) = . ν [0, 1/q) = q 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 translation invariance give ν([0,

 1 p p )) = pν [0, ) = c . q q q

By approximation we eventually obtain that ν([0, t)) = ct for all t ≥ 0. The completion of the Borel σ –algebra with respect to λ is the socalled σ -algebra of Lebesgue measurable sets. It coincides with the class C of additive sets with respect to λ∗ considered in the proof of Carath´eodory theorem (see Exercise 1.12). Remark 1.23 (Outer Lebesgue measure and non-measurable sets). The measure λ∗ used in the proof of Carath´eodory’s theorem is also called outer Lebesgue measure, and it is defined on all parts of R. The terminology is slightly misleading here, since λ∗ , though σ –subadditive, fails to be σ –additive. In particular, there exist subsets of R that are not Lebesgue 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

18 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

would still be measurable, by translation invariance, and the family of sets {A + h}h∈Q would be a countable and measurable partition of R, with λ∗ (A + h) = c independent of h ∈ Q. Now, if c = 0 we reach a contradiction with the fact that λ∗ (R) = ∞, while if c > 0 we consider all sets A + h with h ∈ Q ∩ [0, 1] to obtain  λ∗ (A + h) = ∞, 2 = λ∗ ([0, 2]) ≥ h∈Q∩[0,1]

reaching again a contradiction. Notice that this example is not constructive and strongly requires the axiom 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 σ –additive measures defined on all parts of R. In Rn , with n ≥ 3, the famous Banach–Tarski paradox (see for instance [6]) shows that it is also impossible to have a locally finite, invariant under rigid motions and finitely additive measure defined on all parts of Rn .

1.7. Inner and outer regularity of measures on metric spaces Let (E, d) be a metric space and let μ be a finite measure on (E, B (E)). We shall prove a regularity property of μ. Proposition 1.24. For any B ∈ B (E) we have μ(B) = sup{μ(C) : C ⊂ B, closed} = inf{μ(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 σ –algebra of parts of E including the open sets of E. Obviously K contains E and ∅. Moreover, if B ∈ K then its complement B c belongs to K. Let us prove now that (Bn ) ⊂ K  implies n Bn ∈ K . Fix ε > 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,

μ(A \ C) ≤ 2ε.

(1.15)

19 Introduction to Measure Theory and Integration

Let n ∈ N. Since Bn ∈ K there exist an open set An and a closed set Cn such that Cn ⊂ Bn ⊂ An and μ(An \ Cn ) ≤

ε 2n+1

.

   Setting A := n An , S := n Cn we have S ⊂ n Bn ⊂ A and μ(A \ S) ≤ ε. However, A is open but  S is not necessarily closed. So, we approximate S by setting Sn := n0 Ck . The set Sn is obviously closed, Sn ↑ S and consequently μ(Sn ) ↑ μ(S). Therefore there exists  nε ∈ N ⊂ such that μ(S \ Snε ) < ε. Now, setting C = Snε we have C n Bn ⊂ A  and μ(A \ C) < μ(A \ S) + μ(S \ C) < 2ε. Therefore n Bn ∈ K . We have proved that K is a σ –algebra. It remains to show that K contains the open subsets of E. In fact, let A be open and set

 1 , Cn = x ∈ E : d(x, A ) ≥ n c

where d(x, Ac ) := inf y∈Ac d(x, y) is the distance function from Ac . Then Cn are closed subsets of A, and moreover Cn ↑ A, which implies μ(A \ Cn ) ↓ 0. Thus the conclusion follows. Notice that inner and outer approximation hold for μ–measurable sets B as well: one has just to notice that there exist Borel sets B1 , B2 such that B1 ⊂ B ⊂ B2 with μ(B2 \ B1 ) = 0, and apply inner approximation to B1 and outer approximation to B2 . Remark 1.25 (Inner and outer approximation for σ-finite measures). It is possible to extend the inner approximation property to σ -finite measures: suffices to assume the existence of a sequence of closed sets Cn with finite measure such that μ(X \∪n Cn ) = 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 μ(B ∩ Cn ) = sup {μ(C) : C closed, C ⊂ B ∩ Cn } , so that μ(B ∩ Cn ) ≤ sup {μ(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 An with finite measure satisfying X = ∪n An , we have the outer approximation property: indeed, for any n and any > 0 we can find (assuming with no loss of generality μ(B) < +∞) open sets Bn ⊂ An containing

20 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

 B ∩ An and such that μ Bn \ (B ∩ An ) < 2−n . It follows that ∪n Bn contains B and   Bn \ B < 2 . μ n∈N

Since Bn are also open in X, the set ∪n Bn 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 Dynkin’s argument, since the class of closed sets is a π-system and generates the Borel σ -algebra). Corollary 1.26. Let μ, ν be finite measures in (E, B (E)), such that μ(C) = ν(C) for any closed subset C of E. Then μ = ν.

Exercises 1.1 Given A ⊂ X, denote by 1 A : X → {0, 1} its characteristic function, equal to 1 on A and equal to 0 on Ac . Show that 1 A∪B = max{1 A , 1 B }, and that

1 Ac = 1 X − 1 A

lim sup An = A

⇐⇒

lim sup 1 An = 1 A ,

lim inf An = A

⇐⇒

lim inf 1 An = 1 A .

n→∞

n→∞

1.2 Let A ⊂

1 A∩B = min{1 A , 1 B },

Rn

n→∞

n→∞

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 σ –additive measure μ on a σ –algebra A such that there exist An ∈ A with An ↓ A and infn μ(An ) > μ(A). 1.4 Let μ be additive and finite, on an algebra A . Show that μ is σ –additive if and only if it is continuous along nonincreasing sequences. 1.5 Let μ be a finite measure on (X, E ). Show that the set of atoms of μ, defined by Aμ := {x ∈ X : {x} ∈ E and μ({x}) > 0} is at most countable. Show that the same is true for σ –finite measures, and provide an example of a measure space for which this property fails. 1.6 Let (X, E , μ) be a measure space, with μ finite. We say that μ is diffuse if for all A ∈ E with μ(A) > 0 there exists B ⊂ A with 0 < μ(B) < μ(A). Show that, if μ is diffuse, then μ(E ) = [0, μ(X)]. 1.7 Show that if X is a separable metric space and E is the Borel σ –algebra, then a σ –additive measure μ : E → [0, +∞) is diffuse if and only if μ has no atom.

21 Introduction to Measure Theory and Integration

1.8 Let λ be the Lebesgue measure in [0, 1]. Show the existence of a λ–negligible set having the cardinality of the continuum. Hint: consider the classical Cantor’s 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 λ be the Lebesgue measure in [0, 1]. Show the existence, for any ε > 0, of a closed set C ⊂ [0, 1] containing no interval and such that λ(C) > 1 − ε. Hint: remove from [0, 1] a sequence of open intervals, centered on the rational points of [0, 1]. 1.10 Using the previous exercise, write [0, 1] = A ∪ B where A is negligible in the measure-theoretic sense (i.e. λ(A) = 0) and B is negligible in the Baire category 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 same time. 1.11 Let λ be the Lebesgue measure in [0, 1]. Construct a Borel set E ⊂ (0, 1) such that 0 < λ(E ∩ I ) < λ(I ) for any open interval I ⊂ (0, 1). 1.12 Let (X, E , μ) be a measure space and let μ∗ : P (X) → [0, +∞] be the outer measure induced by μ. Show that the completed σ –algebra E μ is contained in the class C of additive sets with respect to μ∗ . 1.13 Let (X, E , μ) be a measure space and let μ∗ : P (X) → [0, +∞] be the outer measure induced by μ. Show that for all A ⊂ X there exists B ∈ E containing A with μ(B) = μ∗ (A). 1.14 Let (X, E , μ) be a measure space. Check the following statements, made in Definition 1.12: (i) E μ is a σ –algebra; (ii) the extension μ(A) := μ(B), where B ∈ E is any set such that AB is contained in a μ–negligible set of E , is well defined and σ –additive on E μ; (iii) μ–negligible sets of E μ are characterized by the property of being cointained in a μ–negligible set of E . 1.15 Let (X, E , μ) be a measure space and let μ∗ : P (X) → [0, +∞] be the outer measure induced by μ. Show that if μ(X) is finite, the class C of additive sets with respect to μ∗ coincides with the class of E μ –measurable sets. Hint: one inclusion is provided by Exercise 1.12. For the other one, given an additive set A, by applying Exercise 1.13 twice, find first a set B ∈ E with μ∗ (B \ A) = 0, and then a set C ∈ E with μ(C) = 0 and B \ A ⊂ C. 1.16 Find a σ –algebra E ⊂ P (N) containing infinitely many sets and such that any B ∈ E different from ∅ has an infinite cardinality. 1.17 Find μ : P(N) → {0, +∞} that is additive, but not σ –additive. 1.18 Let ω be the first uncountable ordinal and, for K ⊂ P (X), define by transfinite induction a family F (i) , i ∈ ω, as follows: F (0) := K ∪ {∅},   ∞

(i) c ( j) ( j) Ak , B : (Ak ) ⊂ F , B ∈ F F := , k=0

22 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

 if i is the successor of j, and F (i) := j∈i F ( j) otherwise.  Show that i∈ω F (i) = σ (K ). 1.19 Show that B (R) has the cardinality of the continuum. Hint: use the construction of the previous exercise, and the fact that ω has at most the cardinality of continuum. 1.20 Show that the σ –algebra L of Lebesgue measurable sets has the same cardinality of P(R), thus strictly greater than the continuum. Hint: consider all subsets of Cantor’s middle third set. 1.21 Show that the cardinality of any σ –algebra is either finite or uncountable. 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 , μ) be a a measure space and suppose that X is finite or countable. Show the existence of a measure μ˜ on P (X) that extends μ, that is, μ(A) = μ(A) ˜ for all A ∈ E . 1.24 Find an example of an additive set function μ : P (N) → {0, 1}, with μ(N) = 1 and μ({n}) = 0 for all n ∈ N (in particular μ is not σ –additive, the construction of this example requires Zorn’s lemma). 1.25 Let C ∈ B ([0, 1]) with λ(C) > 0. Without using the continuum hypothesis, show that C has the cardinality of continuum. 1.26 Let (K , d) be a compact metric space and let μ be as in Exercise 1.24. Let’s say that a sequence (xn ) ⊂ K μ-converges to x ∈ K if  μ {n ∈ N : d(xn , x) > ε} = 0 ∀ε > 0. Show that any sequence (xn ) ⊂ K is μ-convergent and that the μ-limit is unique.

Chapter 2 Integration

This chapter is devoted to the construction of the integral of E –measurable functions in general measure spaces ( , E , μ), and its main continuity 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 viewpoints: the first, more traditional one   f dμ = zμ({ f = z}) X

z∈Im( f )

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  ∞  f dμ = μ({ f > t}) dt. X

0

This second viewpoint is more appropriate to show the continuity properties of the integral with respect to f (the integral on the right side can be elementarily defined, since t → μ({ f > t}) is nonincreasing, see Section 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 ϕ : X → Y and any I ∈ P (Y ) we set ϕ −1 (I ) := {x ∈ X : ϕ(x) ∈ I } = {ϕ ∈ I }. The set ϕ −1 (I ) is called the inverse image of I . L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

24 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Let us recall some elementary properties of ϕ −1 (the easy proofs are left to the reader as an exercise): (i) ϕ −1 (I c ) = (ϕ −1 (I ))c for all I ∈ P (Y ); (ii) if {Ji }i∈I ⊂ P (Y ) we have

   ϕ −1 (Ji ) = ϕ −1 Ji , ϕ −1 (Ji ) = ϕ −1 Ji . i∈I

i∈I

i∈I

i∈I

In particular, if I ∩ J = ∅ we have ϕ −1 (I ) ∩ ϕ −1 (J ) = ∅. Also, if E ⊂ P (Y ) and we consider the family ϕ −1 (E ) of subset of X defined by   (2.1) ϕ −1 (E ) := ϕ −1 (I ) : I ∈ E , we have that ϕ −1 (E ) is a σ –algebra whenever E is a σ –algebra.

2.2. Measurable and Borel functions We are given measurable spaces (X, E ) and (Y, F ). We say that a function ϕ : X → Y is (E , F )–measurable if ϕ −1 (F ) ⊂ E . If (Y, F ) = (R, B (R)), we say that ϕ is a real valued E –measurable function, and if (X, d) is a metric space and E is the Borel σ –algebra, we say that ϕ is a real valued Borel function. The following simple but useful proposition shows that the measurability condition needs to be checked only on a class of generators. Proposition 2.1 (Measurability criterion). Let G ⊂ F be such that σ (G ) = F . Then ϕ : X → Y is (E , F )–measurable if and only if ϕ −1 (I ) ∈ E for all I ∈ G (equivalently, iff ϕ −1 (G ) ⊂ E ). Proof. Consider the family D := {I ∈ F : ϕ −1 (I ) ∈ E }. By the above-mentioned properties of ϕ −1 as an operator between P (Y ) and P (X), it follows that D is a σ –algebra including G . So, it coincides with σ (G ) = F . A simple consequence of the previous proposition is the fact that any continuous function is a Borel function: more precisely, assume that ϕ : X → Y is continuous and that E = B (X) and F = B (Y ). Then, the σ –algebra   A ⊂ Y : ϕ −1 (A) ∈ B (X) contains the open subsets of Y (as, by the continuity of ϕ, ϕ −1 (A) is open in X, and in particular Borel, whenever A is open in Y ), and then it contains the generated σ –algebra, i.e. B (Y ). The following proposition, whose proof is straightforward, shows that the class of measurable functions is stable under composition.

25 Introduction to Measure Theory and Integration

Proposition 2.2. Let (X, E ), (Y, F ), (Z , G ) be measurable spaces and let ϕ : X → Y and ψ : Y → Z be respectively (E , F )–measurable and (F , G )–measurable. Then ψ ◦ ϕ 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 say that a mapping ϕ : X → R is E –measurable if ϕ −1 ({−∞}), ϕ −1 ({+∞}) ∈ E

ϕ −1 (I ) ∈ E , ∀I ∈ B (R). (2.2) This condition can also be interpreted in terms of measurability between E and a suitable Borel σ –algebra in R, see Exercise 2.3. Analogously, when (X, d) is a metric space and E is the Borel σ –algebra, we say that ϕ : X → R is Borel whenever the conditions above hold. The following proposition shows that extended E –measurable functions are stable under pointwise limits and countable supremum and infimum. and

Proposition 2.3. Let (ϕn ) be a sequence of extended E–measurable functions. Then the following functions are E–measurable: sup ϕn (x), n∈N

inf ϕn (x),

n∈N

lim sup ϕn (x), n→∞

lim inf ϕn (x). n→∞

Proof. Let us prove that ϕ(x) := supn ϕn (x) is E –measurable (all other cases can be deduced from this one, or directly proved by similar arguments). For any a ∈ R we have −1

ϕ ([−∞, a]) =

∞ 

ϕn−1 ([−∞, a]) ∈ E .

n=0

In particular {ϕ = −∞} ∈ E , so that ϕ −1 ((−∞, a]) ∈ E for all a ∈ R; by letting a ↑ ∞ we get ϕ −1 (R) ∈ E . As a consequence, the class   I ∈ B (R) : ϕ −1 (I ) ∈ E is a σ –algebra containing the intervals of the form (−∞, a] with a ∈ R, and therefore coincides with B (R). Eventually, {ϕ = +∞} = X \ [ϕ −1 (R) ∪ {ϕ = −∞}] belongs to E as well.

2.3. Partitions and simple functions Let (X, E ) be a measurable space. A function ϕ : X → R is said to be simple if its range ϕ(X) is a finite set. The class of simple functions is obviously a real vector space, as the range of ϕ + ψ is contained in {a + b : a ∈ range(ϕ), b ∈ range(ψ)} .

26 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

If ϕ(X) = {a1 , . . . , an }, with ai = a j if i = j, setting Ai = ϕ −1 ({ai }), i = 1, . . . , n we can canonically represent ϕ as ϕ(x) =

n 

ak 1 A k ,

x ∈ X.

(2.3)

k=1

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 ϕ has many representations of the form ϕ(x) =

N  k=1

ak 1 Ak ,

x ∈ X,

where A1 , . . . , AN need not be mutually disjoint and ak need not be in the range of ϕ. For instance 1[0,1) + 31[1,2] = 1[0,2] + 21[1,2] .

It is easy to check that a simple function is E –measurable if, and only if, all level sets Ak in (2.3) are E –measurable; in this case we shall also say that {Ak } is a finite E –measurable partition of X. Now we show that any nonnegative E –measurable function can be approximated by simple functions; a variant of this result, with a different construction, is proposed in Exercise 2.8. Proposition 2.4. Let ϕ be a nonnegative extended E –measurable function. For any n ∈ N∗ , define ⎧ i −1 i i −1 ⎪ ⎨ n if ≤ ϕ(x) < n , i = 1, 2, . . . , n2n ; n 2 2 2 (2.4) ϕn (x) = ⎪ ⎩ n if ϕ(x) ≥ n. Then ϕn are simple and E –measurable, (ϕn ) is nondecreasing and convergent to ϕ. If in addition ϕ is bounded the convergence is uniform. Proof. It is not difficult to check that (ϕn ) is nondecreasing. Moreover, we have 0 ≤ ϕ(x) − ϕn (x) ≤ and

1 2n

if ϕ(x) < n, x ∈ X,

0 ≤ ϕ(x) − ϕn (x) = ϕ(x) − n

So, the conclusion easily follows.

if ϕ(x) ≥ n, x ∈ X.

27 Introduction to Measure Theory and Integration

2.4. Integral of a nonnegative E –measurable function We are given a measure space (X, E , μ). We start to define the integral for simple nonnegative functions. 2.4.1. Integral of simple functions Let ϕ be a nonnegative simple E –measurable function, and let us represent it as N  ϕ(x) = ak 1 Ak , x ∈ X, k=1

with N ∈ N, a1 , . . . , a N ≥ 0 and A1 , . . . , A N in E . Then we define (using the standard convention in the theory of integration that 0·∞ = 0),  N  ϕ dμ := ak μ(Ak ). X

k=1

It is easy to see that the definition does not depend on the choice of the representation formula for ϕ. Indeed, let {b1 , . . . , b M } be the range of ϕ and let ϕ = 1M b j 1 B j , with B j := ϕ −1 (b j ), be the canonical representation of ϕ. We have to prove that N 

ak μ(Ak ) =

k=1

M 

b j μ(B j ).

(2.5)

j=1

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

ak μ(Ak ∩ B j ) = b j μ(B j )

j = 1, . . . , M.

(2.6)

k=1

In order to show (2.6) we fix j and consider, for I ⊂ {1, . . . , N }, the sets   A I := x ∈ B j : x ∈ Ai iff i ∈ I , so that {A I } are a E –measurable partition of B j and x ∈ A I iff the set of i’s for which x ∈ Ai coincides with I . Then, using first the fact that and then the fact that A I ⊂ Ai if i ∈ I , and Ai ∩ A I = ∅ otherwise, N  ak = b j whenever A I = ∅ (because 1 ak 1 Ak coincides with b j , the k∈I

constant value of ϕ on B j ), we have N 

ak μ(Ak ∩ B j ) =

k=1

N   k=1

=

I

 I

ak μ(Ak ∩ A I ) =

k∈I

ak μ(A I ) =

 I

N  I

ak μ(Ak ∩ A I )

k=1

b j μ(A I ) = b j μ(B j ).

28 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proposition 2.5. Let ϕ, ψ be simple nonnegative E –measurable functions on X and let α, β ≥ 0. Then αϕ + βψ is simple, E –measurable and we have    (αϕ + βψ) dμ = α ϕ dμ + β ψ dμ X

X

Proof. Let ϕ=

n 

ak 1 A k ,

X

ψ=

k=1

m 

bh 1 Bh

h=1

with {Ak }, {Bh } finite E –measurable partitions of X. Then {Ak ∩ Bh } is a finite E –measurable partition of X and αϕ + βψ is constant (and equal to αak + βbh ) on any element Ak ∩ Bh of the partition. Therefore the level sets of αϕ + βψ are finite unions of elements of this partition and the E –measurability of αϕ + βψ follows (see also Exercise 2.2). Then, writing ϕ(x) =

n  m 

ak 1 Ak ∩Bh (x),

ψ(x) =

k=1 h=1

n  m 

bh 1 Ak ∩Bh (x),

x ∈ X,

k=1 h=1

we arrive at the conclusion. 2.4.2. The repartition function Let ϕ : X → R be E –measurable. The repartition function F of ϕ, relative to μ, is defined by F(t) := μ({ϕ > t}),

t ∈ R.

The function F is nonincreasing and satisfies lim F(t) = lim F(n) = lim μ({ϕ > −n}) = μ({ϕ > −∞}),

t→−∞

n→−∞

n→∞

and, if μ is finite, lim F(t) = lim F(n) = lim μ({ϕ > n}) = μ({ϕ = +∞}),

t→+∞

n→∞

n→∞

since {ϕ > −∞} =

∞

{ϕ > −n},

n=1

{ϕ = +∞} =

∞ 

{ϕ > n}.

n=1

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

29 Introduction to Measure Theory and Integration

Proposition 2.6. Let ϕ : X → R be E –measurable and let F be its repartition function. (i) For any t0 ∈ R we have lim+ F(t) = F(t0 ), that is, F is right cont→t0

tinuous. (ii) If μ is finite, for any t0 ∈ R we have lim− F(t) = μ({ϕ ≥ t0 )}, that t→t0

is, F has left limits

(1)

.

Proof. Let us prove (i). We have  

  1 1 = lim μ ϕ > t0 + lim F(t) = lim F t0 + n→+∞ n→+∞ n n t→t0+ = μ({ϕ > t0 }) = F(t0 ), since {ϕ > t0 } =

  ∞

1 1 ϕ > t0 + = lim ϕ > t0 + . n→∞ n n n=1

So, (i) follows. We prove now (ii). We have   1 lim F(t) = lim F t0 − n→+∞ n t→t0− 

 1 = lim μ ϕ > t0 − = μ({ϕ ≥ t0 }), n→+∞ n since

  ∞

 1 1 {ϕ ≥ t0 } = ϕ > t0 − = lim ϕ > t0 − n→∞ n n n=1

and (ii) follows. From Proposition 2.6 it follows that, in the case when μ is finite, F is continuous at t0 iff μ({ϕ = t0 }) = 0. Now we want to extend the integral operator to nonnegative E –measurable functions. Let ϕ be a nonnegative, simple and E –measurable function and let n  ak 1 Ak , x ∈ X, ϕ(x) = k=0

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

30 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

with n ∈ N∗ , 0 = a0 < a1 < a2 < · · · < an < ∞. Then the repartition function F of ϕ is given by ⎧ μ(A1 ) + μ(A2 ) + · · · + μ(An ) = F(0) if 0 ≤ t < a1 ⎪ ⎪ ⎪ ⎪ ⎨ μ(A2 ) + μ(A3 ) + · · · + μ(An ) = F(a1 ) if a1 ≤ t < a2 F(t) = . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ⎪ ⎪ if an−1 ≤ t < an μ(An ) = F(an−1 ) ⎪ ⎪ ⎩ 0 = F(a ) if t ≥ an . n Consequently, we can write  n n   ϕ(x) dμ(x) = ak μ(Ak ) = ak (F(ak−1 ) − F(ak )) X

k=1

= =

n 

k=1 n 

ak F(ak−1 ) −

k=1

k=1

n−1 

n−1 

ak+1 F(ak ) −

k=0

ak F(ak ) (2.7) ak F(ak )

k=0

 n−1  = (ak+1 − ak )F(ak ) =

∞

F(t) dt.

0

k=0

Example 2.7. We set X = R, μ = λ, 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 5 and ϕ := k=1 ak 1 Ak to be the simple function shown in Figure 2.1. It is easy to verify that F has the graph shown in the right picture in Figure 2.1. ϕ

F

10 9 8 7 6 5 4 3 2 1 1

1

2

3

Figure 2.1. a simple function ϕ, and its repartition F

4

5

6 7

8

9

10

31 Introduction to Measure Theory and Integration

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 supremum of the areas of piecewise constant minorants of f . Let  be the set of all finite decompositions σ = {t1 , . . . , t N } of [0, +∞], where N ∈ N∗ and 0 = t0 ≤ t1 < · · · < t N < +∞. Let now f : [0, +∞) → [0, +∞] be a nonincreasing function. For any σ = {t0 , t1 , . . . , t N } ∈  we consider the partial sum I f (σ ) :=

N −1 

f (tk+1 )(tk+1 − tk ).

(2.8)

k=0

We define



∞

f (t) dt := sup{I f (σ ) : σ ∈ }.

(2.9)

0

∞ The integral 0 f (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 f n ↑ f , with f n : [0, +∞) → [0, +∞] nonincreasing. Then  ∞  ∞ f n (t) dt ↑ f (t) dt. 0

Proof. It is obvious that  ∞ 0

0

 f n (t) dt ≤

∞

0

To prove the converse inequality, fix L < σ = {t1 , . . . , t N } ∈  such that N −1  k=0

f (t) dt. ∞ 0

f (t) dt. Then there exists

f (tk )(tk+1 − tk ) > L .

32 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Since for n large enough 

∞

f n (t) dt ≥

0

N −1 

f n (tk+1 )(tk+1 − tk ) > L ,

k=0

letting n → ∞ we find that



sup n∈N

This implies



∞

sup n∈N

∞

f n (t) dt ≥ L .

0

 f n (t) dt ≥

0

∞

f (t) dt

0

and the conclusion follows. 2.4.4. Integral of a nonnegative measurable function We are given a measure space (X, E , μ) and an extended nonnegative E –measurable function ϕ. Having the identity (2.7) in mind, we define   ∞ ϕ dμ : = μ({ϕ > t}) dt. (2.10) X

0

Notice that the function t → μ({ϕ > t}) ∈ [0, +∞] is nonnegative and nonincreasing in [0, +∞), so that its archimedean integral is well defined and (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 ϕ is μ–integrable. It follows directly from the analogous properties of the archimedean integral that the integral so defined is monotone, i.e.   ϕ dμ ≥ ψ dμ. ϕ ≥ ψ ⇒ X

X

Indeed, ϕ ≥ ψ implies {ϕ > t} ⊃ {ψ > t} and μ({ϕ > t}) ≥ μ({ψ > t}) for all t > 0. Furthermore, the integral is invariant under modifications of ϕ in μ–negligible sets, that is   ϕ = ψ μ–a.e. in X ⇒ ϕ dμ = ψ dμ. X

X

To show this fact it suffices to notice that ϕ = ψ μ–a.e. in X implies that the sets {ϕ > t} and {ψ > t} differ in a μ–negligible set for all t > 0, therefore μ({ϕ > t}) = μ({ψ > t}) for all t > 0. Let us prove the following basic Markov inequality.

33 Introduction to Measure Theory and Integration

Proposition 2.9. For any a ∈ (0, +∞) we have  1 ϕ(x) dμ(x). μ({ϕ ≥ a}) ≤ a X

(2.11)

Proof. For any a ∈ (0, +∞) we have, recalling the inclusion {ϕ ≥ a} ⊂ {ϕ > t} for any t ∈ (0, a), that μ({ϕ > t}) ≥ μ({ϕ ≥ a}) for all t ∈ (0, a). The monotonicity of the archimedean integral gives  ∞  ∞  ϕ(x) dμ(x) = μ({ϕ > t}) dt ≥ 1(0,a) (t)μ({ϕ > t}) dt X

0

0

≥ aμ({ϕ ≥ a}). The Markov inequality has some important consequences. Proposition 2.10. Let ϕ : X → [0, +∞] be an extended E –measurable function. (i) If ϕ is μ–integrable then the set {ϕ = +∞} has μ–measure 0, that is, ϕ is finite μ–a.e. in X. (ii) The integral of ϕ vanishes iff ϕ is equal to 0 μ–a.e. in X.  Proof. (i) Since X ϕ dμ < ∞ we deduce from (2.11) that lim μ({ϕ > a}) = 0.

a→+∞

Since {ϕ = ∞} =

∞ 

{ϕ > n},

n=1

by applying the continuity along decreasing sequences in the space ({ϕ > 1} (with finite μ measure) we obtain μ({ϕ = ∞}) = lim μ({ϕ > n}) = 0. n→+∞



(ii) If X ϕ dμ = 0 we deduce from (2.11) that μ({ϕ > a}) = 0 for all a > 0. Since μ({ϕ > 0}) = lim μ({ϕ > n→+∞

1 }) = 0, n

the conclusion follows. The other implication follows by the invariance of the integral.

34 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proposition 2.11 (Monotone convergence). Let (ϕn ) be a nondecreasing sequence of extended nonnegative E –measurable functions and set ϕ(x) := lim ϕn (x) for any x ∈ X. Then n→∞  ∞  ∞ ϕn (x) dμ(x) ↑ ϕ(x) dμ(x). 0

0

Proof. It suffices to notice that μ({ϕn > t}) ↑ μ({ϕ > t}) for all t > 0, and then to apply Proposition 2.8. Now, by Proposition 2.4 we obtain the following important approximation property. Proposition 2.12. Let ϕ : X → [0, +∞] be an extended E –measurable function. Then there exist simple E –measurable functions ϕn : X → [0, +∞) such that ϕn ↑ ϕ, so that  ∞  ∞ ϕn (x) dμ(x) ↑ ϕ(x) dμ(x). 0

0

Remark 2.13 (Construction of Lebesgue and Riemann integrals). Proposition 2.12 could be used as an alternative, and equivalent, definition 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 integral: the only (fundamental) difference is due to the choice of the family of “simple” functions. In all cases simple functions take finitely many values, but within the Lebesgue theory their level sets belong to a σ –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 ϕ, ψ : X → [0, ∞] be E –measurable functions. Then    (ϕ + ψ) dμ = ϕ dμ + ψ dμ. X

X

X

Proof. Let ϕn , ψn be simple functions with ϕn ↑ ϕ and ψn ↑ ψ. Then, the additivity of the integral on simple functions gives    (ϕn + ψn ) dμ = ϕn dμ + ψn dμ. X

X

X

We conclude passing to the limit as n → ∞ and using the monotone convergence theorem. The following Fatou’s lemma, providing a semicontinuity property of the integral, is of basic importance.

35 Introduction to Measure Theory and Integration

Lemma 2.15 (Fatou). Let ϕn : X → [0, +∞] be extended E –measurable functions. Then we have   lim inf ϕn (x) dμ(x) ≤ lim inf ϕn (x) dμ(x). (2.12) X n→∞

n→∞

X

Proof. Setting ϕ(x) := lim infn ϕn (x), and ψn (x) = infm≥n ϕm (x), we have that ψn (x) ↑ ϕ(x). Consequently, by the monotone convergence theorem,   ϕ(x) dμ(x) = lim ψn (x) dμ(x). n→∞

X

On the other hand



X

 ψn (x) dμ(x) ≤ X

ϕn (x) dμ(x), X



so that

 ϕ(x) dμ(x) ≤ lim inf n→∞

X

ϕn (x) dμ(x). X

In particular, if ϕn are pointwise converging to ϕ, we have   ϕ(x) dμ(x) ≤ lim inf ϕn (x) dμ(x). n→∞

X

X

2.5. Integral of functions with a variable sign Let ϕ : X → R be an extended E –measurable function. We say that ϕ is μ–integrable if both the positive part ϕ + (x) := max{ϕ(x), 0} and the negative part ϕ − (x) := max{−ϕ(x), 0} of ϕ are μ–integrable in X. As ϕ = ϕ + − ϕ − , in this case it is natural to define    ϕ(x) dμ(x) := ϕ + (x) dμ(x) − ϕ − (x) dμ(x). X

X

X

As |ϕ| = ϕ + + ϕ − , the additivity properties of the integral give that  |ϕ| dμ < ∞. ϕ is μ–integrable if and only if X

Let ϕ : X → R and let A ∈ E be such that 1 A ϕ is μ-integrable. We define also   ϕ(x) dμ(x) := 1 A (x)ϕ(x) dμ(x). A

X

In the following proposition we summarize the main properties of the integral.

36 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proposition 2.16. Let ϕ, ψ : X → R be μ–integrable functions. (i) For any α, β ∈ R we have that αϕ + βψ is μ–integrable and    (αϕ + βψ) dμ = α ϕ dμ + β ψ dμ. X

(ii) If ϕ ≤ ψ in X we have !  ! ! ! ! |ϕ| dμ. (iii) ! ϕ dμ!! ≤ X

X

X



 ϕ dμ ≤ X

ψ dμ. X

X

 + − − + Proof.  (i). Since (−ϕ) = ϕ and (−ϕ) = ϕ , we have X −ϕ dμ = − X ϕ dμ. So, possibly replacing ϕ by −ϕ and ψ by −ψ we can assume that α ≥ 0 and β ≥ 0. We have (αϕ + βψ)+ + αϕ − + βψ − = (αϕ + βψ)− + αϕ + + βψ + , so that we can integrate both sides and use the additivity on nonnegative functions to obtain    + − (αϕ + βψ) dμ + α ϕ dμ + β ψ − dμ X X X    − + = (αϕ + βψ) dμ + α ϕ dμ + β ψ + dμ. X

X

X

Rearranging terms we obtain (i). (ii). It follows by the monotonicity of the integral on nonnegative functions and from the inequalities ϕ + ≤ ψ + and ϕ − ≥ ψ − . (iii). Since −|ϕ| ≤ ϕ ≤ |ϕ| the conclusion follows from (ii). Another consequence of the additivity property of the integral is the additivity of the real-valued map  A ∈ E → ϕ dμ A

whenever ϕ is μ–integrable. We will see in the next section that, as a consequence of the dominated convergence theorem, this map is even σ –additive.

2.6. Convergence of integrals In this section we study the problem of commuting limit and integral; we have already seen that this can be done in some particular cases, as when the functions are nonnegative and monotonically converge to their supremum, and now we investigate some more general cases, relevant for the applications.

37 Introduction to Measure Theory and Integration

Proposition 2.17 (Lebesgue dominated convergence theorem). Let (ϕn ) be a sequence of E –measurable functions pointwise converging to ϕ. Assume that there exists a nonnegative μ–integrable function ψ such that |ϕn (x)| ≤ ψ(x)

∀x ∈ X, n ∈ N.

Then the functions ϕn and the function ϕ are μ–integrable and 

 ϕn dμ =

lim

n→∞

ϕ dμ.

X

X

Proof. Passing to the limit as n → ∞ we obtain that ϕ is E –measurable and |ϕ| ≤ ψ in X. In particular ϕ is μ–integrable. Since ϕ + ψ is nonnegative, by the Fatou lemma we have 

 (ϕ + ψ) dμ ≤ lim inf Consequently,

(ϕn + ψ) dμ.

n→∞

X



X

 ϕ dμ ≤ lim inf

ϕn dμ.

n→∞

X

(2.13)

X

In a similar way we have 

 (ψ − ϕ) dμ ≤ lim inf n→∞

X

Consequently,



(ψ − ϕn ) dμ. X

 ϕ dμ ≥ lim sup X

n→∞

ϕn dμ.

(2.14)

X

Now the conclusion follows by (2.13) and (2.14). An important consequence of the dominated convergence theorem is the absolute continuity property of the integral of μ–integrable functions ϕ:  |ϕ| dμ < ε.

for any ε > 0 there exists δ > 0 such that μ(A) < δ ⇒ A

The proof of this property is sketched in Exercise 2.9.

(2.15)

38 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

2.6.1. Uniform integrability and Vitali convergence theorem In this subsection we assume for simplicity that the measure μ is finite. A family {ϕi }i∈I of R–valued μ–integrable functions is said to be μ– uniformly integrable if  |ϕi (x)| dμ(x) = 0, uniformly in i ∈ I . lim μ(A)→0

A

This means that for any ε > 0 there exists δ > 0 such that  |ϕi (x)| dμ(x) ≤ ε ∀ i ∈ I. μ(A) < δ ⇒ A

This property obviously extends from single functions to families of functions the absolute continuity property of the integral. Notice that any family {ϕi }i∈I dominated by a single μ–integrable function ϕ (i.e. such that |ϕi | ≤ |ϕ| for any i ∈ I ) is obviously μ– 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 μ is a finite measure and let (ϕn ) be a μ–uniformly integrable sequence of functions pointwise converging to a real valued function ϕ. Then ϕ is μ–integrable and   ϕn dμ = ϕ dμ. lim n→∞

X

X

To prove the Vitali theorem we need the following Egorov Lemma. Lemma 2.19 (Egorov). Assume that μ is a finite measure and let (ϕn ) be a sequence of E –measurable functions pointwise converging to a real valued function ϕ. Then for any δ > 0 there exists a set Aδ ∈ E such that μ(Aδ ) < δ and ϕn → ϕ uniformly in X \ Aδ . Proof. For any integer m ≥ 1 we write X as the increasing union of the sets Bn,m , where

 1 Bn,m := x ∈ X : |ϕi (x) − ϕ(x)| < ∀i ≥ n . m Since μ is finite there exists n(m) such that μ(Bn(m),m ) > μ(X) − 2−m δ. We denote by Aδ the union of X \ Bn(m),m , so that μ(Aδ ) ≤

∞  m=1

μ(X \ Bn(m),m ) <

∞  δ = δ. 2m m=1

39 Introduction to Measure Theory and Integration

Now, given any ε > 0, we can choose m > 1/ε to obtain that 1 < ε for all x ∈ Bn(m),m , n ≥ n(m). m As X \ Aδ ⊂ Bn(m),m , this proves the uniform convergence of ϕn to ϕ on X \ Aδ . |ϕn (x) − ϕ(x)| ≤

Proof of the Vitali Theorem. Fix ε > 0 and find δ > 0 such that   A |ϕn | dμ < ε whenever μ(A) < δ. Again, Fatou’s Lemma yields that A |ϕ| dμ ≤ ε whenever μ(A) < δ. Assume now that A is given by Egorov Lemma, so that ϕn → ϕ uniformly on X \ A. Then, writing    (ϕ − ϕn ) dμ = (ϕ − ϕn ) dμ + (ϕ − ϕn ) dμ X

X\A

A

and using the fact that limn sup |ϕn − ϕ| = 0 we obtain X\A

! ! ! ! ! (ϕ − ϕn ) dμ! ≤ 3ε ! ! X

for n large enough. The statement follows letting ε ↓ 0.

2.7. A characterization of Riemann integrable functions

 The integrals J f dλ, with J = [a, b] closed interval of the real line and λ equal to the Lebesgue measure in R, are traditionally denoted with the  b classical notation a f dx or with J f dx. This is due to the fact that Riemann’s and Lebesgue’s integral coincide on the class of Riemann’s integrable functions. We denote by I∗ ( f ) and I ∗ ( f ) the upper and lower Riemann integral of f respectively, the former defined by taking the supremum of the sums n−1 ai (ti+1 − ti ) in correspondence of all step functions 1 h=

n−1 

ai 1[ti ,ti+1 ) ≤ f

a = t1 < · · · < tn = b,

(2.16)

i=1

and the latter considering the infimum in correspondence of all step functions h ≥ f . We denote by I ( f ) the Riemann integral, equal to the upper and lower integral whenever the two integrals coincide. the Lebesgue integral of the function h in (2.19) coincides with As n−1 ai (ti+1 − ti ), we have i  g dλ = I (g) for any step function g : J → R. J

40 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Now, if f : J → R is continuous, we can choose a uniformly bounded sequence 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 convergence theorem we get  f dλ = I ( f ) for any continuous function f : J → R. J

We are going to generalize this fact, providing a full characterization, within the Lebesgue theory, of Riemman’s integrable functions. Theorem 2.20. Let f : J = [a, b] → R be a bounded function. Then f 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 )λ – measurable and  f dλ = I ( f ). (2.17) J

Proof. Let

 ⎧ ⎪ ⎪ f ∗ (x) := inf lim inf f (x h ) : x h → x ⎪ ⎪ h→∞ ⎨

 ⎪ ⎪ ⎪ ∗ ⎪ ⎩ f (x) := sup lim sup f (x h ) : x h → x .

(2.18)

h→∞

It is not hard to show (see Exercise 2.6 and Exercise 2.7) that f ∗ is lower semicontinuous and f ∗ is upper semicontinuous, therefore both f ∗ and f ∗ are Borel functions.   We are going to show that I∗ ( f ) = J f ∗ dλ and I ∗ ( f ) = J f ∗ dλ. These two equalities yield the conclusion, as f is continuous at λ–a.e. point in J iff f ∗ − f ∗ = 0 λ–a.e. in J , and this holds iff (because f ∗ − f ∗ ≥ 0)  J

( f ∗ − f ∗ ) dλ = 0.

Furthermore, if the set of discontinuity points of f is λ–negligible, the Borel function f ∗ differs from f only in a λ–negligible set, thus f is B (J )λ –measurable (because { f > t} differs from the Borel set { f ∗ > t} only in set, see also Exercise 2.4) and its integral coincides   a λ–negligible with J f ∗ dλ = J f ∗ dλ; this leads to (2.17).

41 Introduction to Measure Theory and Integration

Since I ∗ ( f ) = −I∗ (− f ) and f ∗ = −(− f )∗ , we need only to prove the first of the two equalities, i.e.  f ∗ dλ = I∗ ( f ). (2.19) J

In order to check the inequality ≤ in (2.19) we apply Exercise 2.11, finding a sequence of continuous functions gh ↑ f ∗ ≤ f and obtaining, thanks to the monotone convergence theorem,   f ∗ dλ = sup gh dλ = sup I (gh ) = sup I∗ (gh ) ≤ I∗ ( f ). J

h∈N

h∈N

J

h∈N

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 the same interval. Hence f ∗ ≥ h in J \ {t1 , . . . , tn } and, being the set of the ti ’s Lebesgue negligible, we have   f ∗ dλ ≥ h dλ = I (h). J

J

Since h is arbitrary the inequality is achieved.

Exercises 2.1 Show that any of the conditions listed below is equivalent to the E –measurability of ϕ : X → R. (i) (ii) (iii) (iv) (v)

ϕ −1 ((−∞, t]) ⊂ E for all t ∈ R; ϕ −1 ((−∞, t)) ⊂ E for all t ∈ R; ϕ −1 ([a, b]) ⊂ E for all a, b ∈ R; ϕ −1 ([a, b)) ⊂ E for all a, b ∈ R; ϕ −1 ((a, b)) ⊂ E for all a, b ∈ R.

2.2 Let ϕ, ψ : X → R be E –measurable. Show that ϕ + ψ and ϕψ are E – measurable. Hint: prove that

{ϕ + ψ < t} = [{ϕ < r} ∩ {ψ < t − r}] r∈Q

and

{ϕ 2 > a} = {ϕ >

√ √ a} ∪ {ϕ < − a},

2.3 Let us define a distance d in R by d(x, y) := | arctan x − arctan y| where, by convention, arctan(±∞) = ±π/2.

a ≥ 0.

42 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

(i) Show that (R, d) is a compact metric space (the so-called compactification of R) 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 σ –algebra of (R, d). 2.4 Let (X, E , μ) be a measure space and let E μ be the completion of E induced by μ. Show that f : X → R is E μ –measurable iff there exists a E –measurable function g such that { f  = g} is contained in a μ–negligible set of E . 2.5 Let us define I f as in (2.8) and let us endow  with the usual partial ordering σ = {t1 , . . . , t N } ≤ ζ = {s1 , . . . , s M } if and only if σ ⊂ ζ . Show that σ  → ∞ I f (σ ) 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 ∞ 

ai = ∞,

i=0

lim ai = 0.

i→∞

Show that for any ϕ : X → [0, +∞] E –measurable there exist Ai ∈ E such that ϕ = i ai 1 Ai . Hint: set ϕ0 := ϕ, A0 := {ϕ ≥ a0 } and ϕ1 := ϕ0 − a0 1 A0 ≥ 0. Then, set A1 := {ϕ1 ≥ a1 } and ϕ2 := ϕ0 − a1 1 A1 and so on. 2.9 Let ϕ : X → R be μ–integrable. Show that the property (2.15) holds. Hint: −i assume  by contradiction its failure for some ε > 0 and find Ai with μ(Ai ) < 2 and Ai |ϕ| dμ ≥ ε. Then, notice that B := lim supi Ai is μ–negligible, consider Bn :=

Ai \ B ↓ ∅

i≥n

and apply the dominated convergence theorem. 2.10 Prove that if ϕn → ϕ in L 1 ( , E , μ), then (ϕn ) is μ–uniformly integrable. In addition, find a space (X, E , μ) and a sequence (ϕn ) that is μ–uniformly integrable, for which there is no g ∈ L 1 (X, E , μ) satisfying |ϕn | ≤ g for all n ∈ N. 2.11 Let (X, d) be a metric space and let g : X → [0, ∞] be lower semicontinuous and not identically equal to ∞. For any λ > 0 define gλ (x) := inf {g(y) + λd(x, y)} . y∈X

43 Introduction to Measure Theory and Integration

Check that: (a) |gλ (x) − gλ (x  )| ≤ λd(x, x  ) for all x, x  ∈ X; (b) gλ ↑ g as λ ↑ ∞. 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 is bounded. Then, for ε > 0 consider the functions  x+ε 1 f (x  , y) dx  , f ε (x, y) := 2ε x−ε proving that f ε are continuous and f ε → f as ε ↓ 0.

Chapter 3 Spaces of integrable functions

This chapter is devoted to the properties of the so-called L p spaces, the spaces of measurable functions whose p-th power is integrable. Throughout the chapter a measure space (X, E , μ) will be fixed.

3.1. Spaces L p (X, E , μ) and L p (X, E , μ) Let Y be a real vector space. We recall that a norm  ·  on Y is a nonnegative map defined on Y satisfying: (i) y = 0 if and only if y = 0; (ii) αy = |α| y for all α ∈ 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 L 1 (X, E , μ) the real vector space of all μ–integrable functions on (X, E ). We define  |ϕ(x)| dμ(x), ϕ ∈ L 1 (X, E , μ). ϕ1 := X

We have clearly αϕ1 = |α| ϕ1

∀α ∈ R, ∀ϕ ∈ L 1 (X, E , μ),

and ϕ + ψ1 ≤ ϕ1 + ψ1

∀ϕ, ψ ∈ L 1 (X, E , μ),

so that conditions (ii) and (iii) in the definition of the norm are fulfilled. However,  · 1 is not a norm in general, since ϕ1 = 0 if and only if ϕ = 0 μ–a.e. in X, so (i) fails. L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

46 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Then, we can consider the following equivalence relation R on L 1 (X, E , μ), ϕ∼ψ

⇐⇒

ϕ = ψ μ–a.e. in X

(3.1)

and denote by L 1 (X, E , μ) the quotient space of L 1 (X, E , μ) with respect to R . In other words, L 1 (X, E , μ) is the quotient vector space of L 1 (X, E , μ) with respect to the vector subspace made by functions vanishing μ–a.e. in X. For any ϕ ∈ L 1 (X, E , μ) we denote by ϕ˜ the equivalence class determined by ϕ and we set ϕ˜ + ψ˜ := ϕ + ψ,

α ϕ˜ := α" ϕ.

(3.2)

It is easily seen that these definitions do no depend on the choice of representatives in the equivalence class, and endow L 1 (X, E , μ) with the structure of a real vector space, whose origin is the equivalence class of functions vanishing μ–a.e. in X. Furthermore, setting ϕ ˜ 1 = ϕ1 ,

ϕ˜ ∈ L 1 (X, E , μ),

it is also easy to see that this definition does not depend on the particular element ϕ chosen in ϕ, ˜ and that (ii), (iii) still hold. Now, if ϕ ˜ 1 = 0 we have that the integral of |ϕ| is zero, and therefore ϕ˜ = 0. Therefore L 1 (X, E , μ), endowed with the norm  · 1 , is a normed space. To simplify the notation typically ϕ˜ is identified with ϕ whenever the formula does not depend on the choice of the function  in the equivalence class: for instance, quantities as μ({ϕ > t}) or X ϕ dμ have this independence, 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 ϕ(x) ¯ = 0, for some fixed x¯ ∈ X, do 1 not make sense in L (X, E , μ), since they depend on the representative chosen (unless μ({x}) ¯ > 0). More generally, if an exponent p ∈ (0, ∞) is given, we can apply a similar construction to the space

  p p |ϕ| dμ < ∞ . L (X, E , μ) := ϕ : ϕ is E –measurable and X

Since |x + y| p ≤ |x| p + |y| p if p ≤ 1, and |x + y| p ≤ 2 p−1 (|x| p + |y| p ) if p ≥ 1, it turns out that L p (X, E , μ) is a vector space, and we shall denote by L p (X, E , μ) the quotient vector space, with respect to the equivalence relation (3.1). Still we can define the sum and product by a

47 Introduction to Measure Theory and Integration

real number as in (3.2), to obtain that L p (X, E , μ) 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 either E or μ, writing L p (X,μ) or even L p (X). This typically happens when (X, d) is a metric space, and E is the Borel σ -algebra, or when X ⊂ R and μ is the Lebesgue measure.

3.2. The L p norm For any ϕ ∈ L p (X, E , μ) we define  1/ p p ϕ p := |ϕ| dμ . X

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 homogeneity condition (ii) trivially holds, whatever the value of p is. Furthermore, condition (i) holds precisely because L p (X, E , μ) consists, strictly speaking, of equivalence classes induced by (3.1). So, the only condition that needs to be checked is the subadditivity condition (ii), and in the sequel we can assume p > 1. The concept of Legendre transform will be useful. Let f : R → R be a function; we define its Legendre transform f ∗ : R → R ∪ {+∞} by f ∗ (y) = sup{x y − f (x)}, x∈R

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 ⎧ p x ⎪ ⎪ ⎨ p if x ≥ 0, f (x) = ⎪ ⎪ ⎩ 0 if x < 0. Then, by an elementary computation, we find that ⎧ q y ⎪ ⎪ if y ≥ 0, ⎨q ∗ f (y) = ⎪ ⎪ ⎩ +∞ if y < 0,

48 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

where q = p/( p − 1) (equivalently, 1p + q1 = 1). Consequently, the following inequality, known as Young inequality, holds: xy ≤

yq xp + , p q

x, y ≥ 0.

(3.4)

Motivated by the previous example, we say that p and q are dual (or conjugate) exponents if 1p + q1 = 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) = e x , x ∈ R. Then ⎧ ⎪ if y < 0, ⎨+∞ ∗ x f (y) := sup{x y − e } = 0 if y = 0, ⎪ x∈R ⎩ y log y − y if y > 0. Consequently, the following inequality holds: x y ≤ e x + y log y − y,

x, y ≥ 0.

(3.5)

3.2.1. H¨older and Minkowski inequalities Proposition 3.3 (H¨older inequality). Assume that ϕ ∈ L p (X, E , μ) and ψ ∈ L q (X, E , μ), with p and q dual exponents in (1, +∞). Then ϕψ ∈ L 1 (X, E , μ) and ϕψ1 ≤ ϕ p ψq .

(3.6)

Proof. If either ϕ p = 0 or ψq = 0 then one of the two functions vanishes μ–a.e. in X, hence ϕψ vanishes μ–a.e. and the inequality is trivial. If both ϕ p and ψq are strictly positive, by the 1–homogeneity of the both sides in (3.6) with respect to ϕ and ψ, we can assume with no loss of generality that the two norms are equal to 1. Now we apply (3.4) to |ϕ(x)| and |ψ(x)| to obtain |ϕ(x)ψ(x)| ≤

|ψ(x)|q |ϕ(x)| p + . p q

Integrating over X with respect to μ yields  1 1 |ϕ(x)ψ(x)| dμ(x) ≤ + = 1. p q X A particular case of the H¨older inequality is !  ! 1/2  1/2 ! ! 2 2 ! ϕ(x)ψ(x) dμ(x)! ≤ ϕ (x) dμ(x) ψ (x) dμ(x) . ! ! X

X

X

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

49 Introduction to Measure Theory and Integration

Proposition 3.4 (Minkowski inequality). Assume that p ∈ [1, +∞) and ϕ, ψ ∈ L p (X, E , μ). Then ϕ + ψ ∈ L p (X, E , μ) and ϕ + ψ p ≤ ϕ p + ψ p .

(3.7)

Proof. The cases p = 1 is obvious. Assume that p ∈ (1, +∞). Then we have    |ϕ + ψ| p dμ ≤ |ϕ + ψ| p−1 |ϕ| dμ + |ϕ + ψ| p−1 |ψ| dμ. X

X

X

Since |ϕ + ψ| p−1 ∈ L q (X, E , μ) where q = p/( p − 1), using the H¨older inequality we find that 



1/q |ϕ + ψ| dμ (ϕ p + ψ p ),

|ϕ + ψ| dμ ≤ p

X

p

X

and the conclusion follows. By the previous proposition it follows that · p is a norm on L p (X, E , μ).

3.3. Convergence in L p (X, E , μ) and completeness We have seen in the previous section that L p (X, E , μ) is a normed space for all p ∈ [1, +∞). In this section we prove some properties of the convergence in these spaces, obtaining as a byproduct the following result. Theorem 3.5. L p (X, E , μ) is a Banach space for any p ∈ [1, +∞). This theorem will be a direct consequence of the following proposition, that provides also a relation between convergence in L p and convergence μ–a.e. in X. Proposition 3.6. Let p ∈ [1, +∞) and let (ϕn ) be a Cauchy sequence in L p (X, E , μ). Then: (i) there exists a subsequence (ϕn(k) ) converging μ–a.e. to a function ϕ in L p (X, E , μ); (ii) (ϕn ) is converging to ϕ in L p (X, E , μ), so that L p (X, E , μ) is a Banach space. Proof. Let (ϕn ) be a Cauchy sequence in L p (X, E , μ). Choose a subsequence (ϕn(k) ) such that ϕn(k+1) − ϕn(k)  p < 2−k

∀k ∈ N.

50 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Next, set g(x) :=

∞ 

|ϕn(k+1) (x) − ϕn(k) (x)|,

x ∈ X.

k=0

By the monotone convergence theorem and the subadditivity of the L p norm it follows that !p 1/ p  ! 1/ p  N −1 ! ! ! ! p g (x) dμ(x) = lim |ϕn(k+1) − ϕn(k) |! dμ ! N →∞ ! X X ! k=0 ≤ lim

N →∞

N −1 

2−k = 2 < ∞.

k=0

Therefore, g is finite μ–a.e., that is, there exists B ∈ E such that μ(B) = 0 and g(x) < ∞ for all x ∈ B c . Set now ϕ(x) := ϕn(0) (x) +

∞  (ϕn(k+1) (x) − ϕn(k) (x)),

x ∈ Bc.

k=0 c The series above is absolutely convergent for any x ∈ B re;N moreover, placing the series in the definition of ϕ by the finite sum 0 −1 (ϕn(k+1) (x) − ϕn(k) (x)) we obtain ϕ(x) = limk ϕn(k) (x). Therefore, if we define (for instance) ϕ = 0 on the μ–negligible set B, we obtain that ϕn(k) → ϕ μ–a.e. on X. The inequality |ϕ| ≤ |ϕn(0) | + g gives that |ϕ| p is μ–integrable, so that ϕ ∈ L p (X, E , μ). So, (i) is proved. In order to prove (ii), we first claim that ϕn(k) → ϕ in L p (X, E , μ) as k → ∞. In fact, since

|ϕ(x) − ϕn(h) (x)| ≤

∞ 

|ϕn(k+1) (x) − ϕn(k) (x)|,

x ∈ X,

k=h

we have, again by monotone convergence and subadditivity of the norm, 

1/ p |ϕ(x) − ϕn(h) (x)| dμ(x) X 1/ p  ∞  ∞  p ≤ |ϕn(k+1) (x) − ϕn(k) (x)| dμ(x) ≤ 2−k , p

k=h

X

and the claim follows.

k=h

51 Introduction to Measure Theory and Integration

Since (ϕn ) is Cauchy, for any ε > 0 there exists n ε ∈ N such that n, m > n ε

⇒

ϕn − ϕm  p < ε.

Now choose k ∈ N such that n(k) > n ε and ϕ − ϕn(k)  p < ε. For any n > n ε we have ϕ − ϕn  p ≤ ϕ − ϕn(k)  p + ϕn(k) − ϕn  p ≤ 2ε. Remark 3.7 (L p convergence versus μ–a.e. convergence). The argument used in the previous proof applies also to converging sequences (as these sequences are obviously Cauchy), and proves that any sequence (ϕn ) strongly converging to ϕ in L p (X,E , μ) admits a subsequence (ϕn(k) ) converging μ–a.e. to ϕ: precisely, this happens whenever ∞ 

ϕn(k+1) − ϕn(k)  p < ∞ .

0

In general, however, convergence in L p does not imply convergence μ– a.e.: the functions ⎧ ϕ0 = 1[0,1] ⎪ ⎪ ⎪ ⎨ ϕ1 = 1[0,1/2] , ϕ2 = 1[1/2,1] ⎪ ϕ3 = 1[0,1/3] , ϕ4 = 1[1/3,2/3] , ϕ5 = 1[2/3,1] ⎪ ⎪ ⎩ ... converge to 0 in L p (0, 1), but are nowhere pointwise converging. The previous remark shows that we can expect to infer pointwise convergence from convergence in L p only modulo the extraction of a subsequence. Now, we ask ourselves about the converse implication: given a sequence (ϕn ) in L p (X, E , μ) pointwise converging to a function ϕ ∈ L p (X, E , μ), we want to find conditions ensuring the convergence of (ϕn ) to ϕ in L p (X, E , μ). This is not true in general, as the following example shows. Example 3.8. Let X = [0, 1], E = B ([0, 1]) and let μ = λ be the Lebesgue measure. Set

n if x ∈ [0, 1/n], ϕn (x) = 0 if x ∈ [1/n, 1]. Then ϕn (x) → 0 for all x ∈ (0, 1] but ϕn 1 = 1.

52 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

In the next proposition we assume that μ is a finite measure, since we defined μ–uniform integrability only for finite measures μ. Proposition 3.9. Let (ϕn ) be a sequence in L p (X, E , μ) pointwise convergent to a function ϕ ∈ L p (X, E , μ), with (|ϕn | p ) μ–uniformly integrable. Then ϕn → ϕ in L p (X, E , μ). Proof. The functions h n := |ϕn − ϕ| p are pointwise converging to 0 and, because of the inequality h n ≤ 2 p−1 (|ϕn | p + |ϕ| p ), they are also easily seen to be uniformly μ–integrable. Therefore, by applying Vitali Theorem 2.18 to h n we obtain the conclusion.

3.4. The space L ∞ (X, E , μ) Let ϕ : X → R be a E –measurable function. We say that ϕ is μ– essentially bounded if there exists a real number M > 0 such that μ({|ϕ| > M}) = 0. If ϕ is μ–essentially bounded there exists a nonnegative number, denoted by ϕ∞ , such that ϕ∞ = min {t ≥ 0 : μ({|ϕ| > t}) = 0} .

(3.8)

This easily follows from the fact that the function t → μ({|ϕ| > t}) is right continuous (Proposition 2.6), so the infimum is attained. Notice also that ϕ∞ is characterized by the property ϕ∞ ≤ M

⇐⇒

|ϕ| ≤ M μ–a.e. in X.

(3.9)

We shall denote by L ∞ (X, E , μ) the space of all equivalence classes of μ–essentially bounded functions with respect to the equivalence relation ∼ in (3.1), thus identifying functions that coincide μ–a.e. in X. Several properties of the L p spaces extend up to the case p = ∞: first of all L ∞ (X, E , μ) is a real vector space and we have the Minkowski inequality (3.10) ϕ + ψ∞ ≤ ϕ∞ + ψ∞ . Indeed, by (3.9) and the triangle inequality, |ϕ(x) + ψ(x)| ≤ ϕ∞ + ψ∞ μ–a.e. in X, therefore (3.8) provides (3.10). As a consequence, L ∞ (X, E , μ) endowed with the norm  · ∞ , is a normed space.

53 Introduction to Measure Theory and Integration

The H¨older inequality takes the form   |ϕψ| dμ ≤ ϕ∞ |ψ| dμ. X

(3.11)

X

Indeed, we have just to notice that |ϕ(x)ψ(x)| ≤ ϕ∞ |ψ(x)| for μ–a.e. x ∈ X, and then integrate with respect to μ. This inequality can be still written as (3.6), provided we agree that q = 1 is the dual exponent of p = ∞ (and conversely). For finite measures we can apply H¨older’s inequality to obtain that the p L spaces are nested; in particular L ∞ is the smaller one and L 1 is the larger one. Remark 3.10 (Inclusions between L p spaces). Assume that μ is finite. Then, if 1 ≤ r ≤ s ≤ ∞ we have L r (X, E , μ) ⊃ L s (X, E , μ). In fact, if r < s and ϕ ∈ L s (X, E , μ) we have, in view of the H¨older inequality (with p = s/r and q = s/(s − r)),  r/s  1−r/s  |ϕ(x)|r dμ(x) ≤ |ϕ(x)|s dμ(x) 1 X dμ(x) , X

and so

X

X

ϕr ≤ (μ(X))(s−r)/rs ϕs .

(3.12)

−1/ p

ϕ p is nondecreasing for ϕ in By (3.12) we obtain that p → μ(X) p the intersection of the spaces L (X, E , μ), so that it has a limit as p → ∞. Since μ(X)−1/ p → 1 as p → ∞ we obtain that lim p→∞ ϕ p exists, finite or infinite. The following proposition characterizes L ∞ (X, E , μ) and the L ∞ norm in terms of this limit. Proposition 3.11. Assume that μ is finite and let ϕ be in the intersection  L p (X, E , μ). p<∞

Then ϕ ∈ L ∞ (X, E , μ) if and only if the limit lim p→∞ ϕ p is finite. If this is the case, we have that ϕ∞ coincides with the value of the limit. Proof. If p ≥ 1 we have by the Markov inequality μ({|ϕ| ≥ a}) = μ({|ϕ| p ≥ a p }) ≤ a − p ϕ pp . Consequently, ϕ p ≥ aμ({|ϕ| ≥ a})1/ p , which yields lim p ϕ p ≥ a whenever μ({ϕ ≥ a}) > 0. So, if the limit is finite, we have ϕ ∈

54 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

L ∞ (X, E , μ) and ϕ∞ ≤ lim p ϕ p . The converse inequality follows directly from (3.11); the same inequality also proves that if the limit is not finite, then ϕ ∈ / L ∞ (X, E , μ). In the next remark we characterize the convergence in L ∞ , proving also that L ∞ (X, E , μ) is a Banach space: as a matter of fact, convergence in L ∞ (X, E , μ) differs from the convergence in supremum norm only because a μ–negligible set is neglected. Remark 3.12 (L ∞ (X, E , μ) is a Banach space). Assume that (ϕn ) ⊂ L ∞ (X, E , μ) is a Cauchy sequence, and let us consider the μ–negligible set ∞

{x ∈ X : |ϕn (x) − ϕm (x)| > ϕn − ϕm ∞ } . n, m=0

Then sup B c |ϕn − ϕm | ≤ ϕn − ϕm ∞ ; as a consequence, the completeness of the space of bounded functions defined in B c provides a bounded function ϕ : B c → R such that ϕn → ϕ uniformly in B c . Extending ϕ in an arbitrary E –measurable way (for instance with the 0 value) to the whole of X, we get ϕn → ϕ in L ∞ (X, E , μ). A similar argument proves that ϕn → ϕ in L ∞ (X, E , μ) if and only if there exists a μ–negligible set B ∈ E satisfying ϕn → ϕ uniformly in Bc. ! !  We know that ! X ϕ dμ! does not exceed X |ϕ| dμ. A nice and useful generalization of this fact is the so-called Jensen inequality. Recall that, if J ⊂ R is an interval, a continuous function g : J → R is said to be convex if   x+y g(x) + g(y) g ≤ ∀x, y ∈ J. (3.13) 2 2 By several approximations (see Exercise 3.7) one can prove that a convex function f satisfies g(t x +(1−t)y) ≤ tg(x)+(1−t)g(y) for all x, y ∈ J and t ∈ [0, 1], and even that  n n n    ti xi ≤ ti g(xi ) whenever ti ≥ 0, xi ∈ J and ti = 1. g i=1

i=1

i=1

(3.14) In the proof we use an elementary property of convex functions g : R → R satisfying g(t) → +∞ as |t| → +∞, namely the existence of a minimum point t0 ; moreover, the function g is nondecreasing in [t0 , +∞) and nonincreasing in (−∞, t0 ] (see Exercise 3.8).

55 Introduction to Measure Theory and Integration

Proposition 3.13 (Jensen). Assume that μ is a probability measure. Let g : R → R be convex and bounded from below and let ϕ ∈ L 1 (X, E , μ). Then we have    g ϕ dμ ≤ g(ϕ) dμ. (3.15) X

X

Proof. Let us first show (3.15) when ϕ is simple. Let ϕ=

n 

αi 1 Ai ,

i=1

where n ≥ 1 is an integer, α1 , . . . , αn ∈ R and A1 , . . . , An are mutually disjoint sets in E whose union is X, so that n 

μ(Ai ) = 1.

i=1

Then, from (3.14) we infer     n n   g ϕ dμ = g αi μ(Ai ) ≤ g(αi )μ(Ai ) = g(ϕ) dμ. X

i=1

X

i=1

In the general case, let us first assume that g(t) → +∞ as |t| → +∞. Then, by Exercise 3.8 we know that g has a minimum point t0 , and that g is nondecreasing in [t0 , +∞), and nonincreasing in (−∞, t0 ]. We can assume with no loss of generality (possibly replacing g(t) by g(t − t0 ) and  ϕ by ϕ + t0 ) that g attains its minimum value at t0 = 0, and that X g(ϕ) dμ is finite. Furthermore, replacing g by g−g(0), we can assume that the minimum value of g is 0. Let ϕn± be nonnegative simple functions satisfying ϕn± ↑ ϕ ± ; the simple functions ϕn+ − ϕn− converge to ϕ + − ϕ − = ϕ in L 1 (X, E , μ). In addition, since g is monotone in (−∞, 0] and [0, +∞), the monotone convergence theorem gives     g(ϕn+ ) dμ ↑ g(ϕ + ) dμ, g(−ϕn− ) dμ ↑ g(−ϕ − ) dμ, X

X

X



+ − + − + − X g(ϕn − ϕn ) dμ so that +(since g(0)  = 0, ϕ−n ϕn = 0 and ϕ ϕ = 0) + −  X g(ϕn ) dμ + X g(−ϕn ) converges to X g(ϕ ) dμ + X g(−ϕ )

= = X g(ϕ) dμ. Passing to the limit as n → ∞ in Jensen’s inequality for the simple functions ϕn+ − ϕn−    + − g (ϕn − ϕn ) dμ ≤ g(ϕn+ − ϕn− ) dμ X

we get (3.15).

X

56 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Finally, the assumption that g(t) → +∞ as t → +∞ can be removed by considering the functions gε (t) := g(t) + ε|t|, which converge to +∞ as |t| → ∞, thanks to the fact that g is bounded from below: we obtain ! !     ! ! ! ! ϕ dμ + ε ! ϕ dμ! ≤ g(ϕ) dμ + ε |ϕ| dμ. g X

X

X

X

and Jensen’s inequality follows by letting ε ↓ 0. An alternative proof of Jensen’s inequality is based on another viewpoint, namely the representation of g as the supremum of a family {L i }i∈I of affine functions. Since μ is a probability measure, for all i ∈ I it is easy to check that L i ( ϕ dμ) = L i (ϕ) dμ, so that    ϕ dμ) ≤ L(ϕ) dμ ∀i ∈ I. Li X

X

Taking the supremum in the right hand side we obtain Jensen’s inequality. Both viewpoints are important in the theory of convex functions. To be more precise, Jensen’s inequality holds provided g is convex on an interval containing the image of ϕ. The next example is very important in Probability and Information theory. Example 3.14 (Entropy functional). By applying Jensen’s inequality with the convex function g(z) = z ln z in [0, +∞) we obtain     ϕ ln ϕ dμ ≥ ϕ dμ ln ϕ dμ (3.16) X

X



X

for  all ϕ ∈ L (X, E , μ) nonnegative. If X ϕ dμ = 1 we obtain that X ϕ ln ϕ dμ ≥ 0 even though the function g has a variable sign (it attains the minimum value −1/e at z = 1/e). 1

3.5. Dense subsets of L p (X, E , μ) Proposition 3.15. For any p ∈ [1, +∞], the space of all simple μ– integrable functions is dense in L p (X, E , μ). Proof. Let f ∈ L p (X, E , μ) with f ≥ 0. Then the conclusion follows from Proposition 2.12 (by Proposition 2.4 in the case p = ∞) and the dominated convergence theorem. In the general case we write f as f + − f − and approximate in L p both parts by simple functions. We consider now the special situation when X is a metric space, E is the σ –algebra of all Borel subsets of X and μ is any finite measure on (X, E ). We denote by Cb (X) the space of all continuous bounded functions on X. Clearly, Cb (X) ⊂ L p (X, E , μ) for all p ∈ [1, +∞].

57 Introduction to Measure Theory and Integration

Proposition 3.16. For any p ∈ [1, +∞) and any finite measure μ, Cb (X) is dense in L p (X, E , μ). Proof. Let C be the closure of Cb (X) in L p (X, E , μ); obviously C is a vector space, as Cb (X) is a vector space. In view of Proposition 3.15 it is enough to show that for any Borel set I ∈ B (X) there exists a sequence (ϕn ) ⊂ Cb (X) such that ϕn → 1 I in L p (X, E , μ). Assume first that I is closed. Set ⎧ ⎨ 1 − n d(x, I ) if d(x, I ) ≤ n1 ϕn (x) = ⎩ 0 if d(x, I ) ≥ n1 , where d(x, I ) := inf{|x − y| : y ∈ I }. It is easy to see that ϕn are continuous, that 0 ≤ ϕn ≤ 1 and that ϕn (x) →

1 I (x), hence the dominated convergence theorem implies that ϕn → 1 I in

L p (X, E , μ). Now, let

G := {I ∈ B (X) : 1 I ∈ C }. It is easy to see that G is a Dynkin system (which includes the π–system of closed sets), so that by the Dynkin theorem we have G = B (X). Remark 3.17. Cb (X) (or more precisely, the equivalence classes of continuous bounded functions) is a closed subspace of L ∞ (X, E , μ), and therefore it is not dense in general. Indeed, if (ϕn ) ⊂ Cb (X) is Cauchy in L ∞ (X, E , μ), then it uniformly converges, up to a μ-negligible set B (just take in Remark 3.12 as B the union of the μ–negligible sets {|ϕn −ϕm | > ϕn −ϕm }). Therefore (ϕn ) uniformly converges on B c and on its closure K . Denoting by ϕ ∈ Cb (K ) its uniform limit, by Tietze’s exension theorem we may extend ϕ to a function, that we still denote by ϕ, in Cb (X). As X \ K ⊂ B is μ–negligible, it follows that ϕn → ϕ in L ∞ (X, E , μ).

Exercises 3.1 Assume that μ is σ –finite, but not finite. Provide examples showing that no inclusion holds between the spaces L p (X, E , μ) in general. Nevertheless, show that for any E –measurable function ϕ : X → R the set   p ∈ [1, ∞] : ϕ ∈ L p (X, E , μ) is an interval. Hint: consider for instance the Lebesgue measure on R.

58 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

3.2 Let 1 ≤ p ≤ q < ∞ and f ∈ L q (X, E , μ). Show that for any δ ∈ (0, 1) we can write f = g + f˜, with g ∈ L q (X, E , μ), f˜ ∈ L p (X, E , μ) and gq ≤ δ f q (notice that if μ is finite we can take g = 0). 3.3 Let p ∈ (1, ∞), ϕ ∈ L p and ψ ∈ L q , with q = p , be such that ϕψ1 = ϕ p ψq . Show that either ψ = 0 or there exists a constant λ ∈ [0, +∞) such that |ϕ| = λ|ψ|q−1 μ–a.e. in X. Hint: first investigate the case of equality in Young’s inequality. 3.4 Prove the following variant of H¨older’s inequality, known as Young’s inequality: if ϕ ∈ L p , ψ ∈ L q and 1p + q1 = r1 , with r ≥ 1, we have that ϕψ ∈ L r and ϕψr ≤ ϕ p ψq . 3.5 Let (ϕn ) ⊂ L 1 (X, E , μ) be nonnegative and satisfying lim infn ϕn ≥ ϕ μ– a.e. in X. Show that    ϕn dμ = ϕ dμ = 1 ⇒ |ϕ − ϕn | dμ → 0. X

X

X

Hint: notice that the positive part and the negative part of ϕ − ϕn have the same integral to obtain   |ϕ − ϕn | dμ = 2 (ϕ − ϕn )+ dμ. X

X

Then, apply the dominated convergence theorem. 3.6 Show that the following extension of Fatou’s lemma: if ϕn ≥ −ψn , with ψn ∈ L 1 (X) nonnegative, ψn → ψ in L 1 (X), then   ϕn dμ ≥ lim inf ϕn dμ. lim inf n→∞

X

X n→∞

Hint: prove first the statement under the additional assumption that ψn → ψ μ–a.e. in X. 3.7 Show that (3.13) implies g(t x + (1 − t)y) ≤ g(x) + (1 − t)g(y) for all x, y ∈ J and t ∈ [0, 1]. Then, deduce from this property (3.14). Hint: it is useful to consider dyadic numbers t = k/2m , with k ≤ 2m integer. 3.8 Let g : R → R be a convex function such that g(z) → +∞ as |z| → +∞. Show the existence of z 0 ∈ R where g attains its minimum value. Then, show that g is nondecreasing in [z 0 , +∞) and nonincreasing in (−∞, z 0 ]. 3.9 Let (ϕn ) ⊂ L 1 (X, E , μ) be nonnegative functions. Show that the conditions   lim sup ϕn dμ ≤ ϕ dμ < ∞ lim inf ϕn ≥ ϕ μ–a.e. in X, n→∞

n→∞

X

X

imply the convergence of ϕn to ϕ in L 1 (X, E , μ). Hint: use Exercise 3.5. 3.10 Let {ϕi }i∈I be a family of functions satisfying  (|ϕi |) dμ = M < +∞ sup i∈I

X

59 Introduction to Measure Theory and Integration

and assume that (c)/c is nondecreasing and tends to +∞ as c → +∞. Show that {ϕi }i∈I is μ–uniformly integrable. Hint: use the inequalities    (ϕi ) M |ϕi | dμ ≤ |ϕi | dμ ≤ dμ + + cμ(A), (c) A A∩{|ϕi |≥c} (c) A∩{|ϕi | 0 and any f ∈ L 1 (X, E , μ), there exists a closed set C ⊂ X such that μ(X \ C) < ε and f |C is continuous and bounded. Hint: use the density of Cb (X) in L 1 and Egorov’s theorem.

Chapter 4 Hilbert spaces

In this chapter we recall the basic facts regarding real vector spaces endowed with a scalar product. We introduce the concept of Hilbert space and show that, even for the infinite-dimensional ones, continuous linear functionals are induced by the scalar product. Moreover, we see that even in some classes of infinite dimensional spaces (the so-called separable ones) there exists a well-defined notion of basis (the so-called complete orthonormal systems), obtained replacing finite sums with converging series. Even though the presentation will be self-contained, we assume that the reader has already some familiarity with these concepts (basis, scalar product, representation of linear functionals) in finite-dimensional spaces.

4.1. Scalar products, pre-Hilbert and Hilbert spaces A real pre–Hilbert space is a real vector space H endowed with a mapping H × H → R, (x, y) → x, y, called scalar product, such that: (i) x, x ≥ 0 for all x ∈ H and x, x = 0 if and only if x = 0; (ii) x, y = y, x for all x, y ∈ H ; (iii) αx + βy, z = αx, z + βy, z for all x, y, z ∈ H and α, β ∈ R. In the following H represents a real pre–Hilbert space. The scalar product allows us to introduce the concept of orthogonality. We say that two elements x and y of H are orthogonal if x, y = 0. We are going to prove that the function # x := x, x, x∈H is a norm in H . For this we need the following Cauchy–Schwartz inequality. L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

62 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proposition 4.1. For any x, y ∈ H we have |x, y| ≤ x y.

(4.1)

In (4.1) equality holds if and only if x and y are linearly dependent. Proof. Set F(λ) = x + λy2 = λ2 y2 + 2λx, y + x2 ,

λ ∈ R.

Since F(λ) ≥ 0 for all λ ∈ R we have |x, y|2 − x2 y2 ≤ 0, which yields (4.1). If x and y are linearly dependent, it is clear that |x, y| = x y. Assume conversely that x, y = ±x y and that y  = 0. Then we have F(λ) = (x ± λy)2 so that, choosing λ = ∓x/y, we find F(λ) = 0. This implies x + λy = 0, so that x and y are linearly dependent. Now we can prove easily that  ·  is a norm in H . In fact, it is clear that αx = |α|x for all α ∈ R and all x ∈ H . Moreover, taking into account (4.1), we have for all x, y ∈ H , x + y2 = x + y, x + y = x2 + y2 + 2x, y ≤ x2 + y2 + 2x y = (x + y)2 , so that x + y ≤ x + y. Therefore a pre–Hilbert space H is a normed space and, in particular, a metric space. If H , endowed with the distance induced by the norm, is complete we say that H is a Hilbert space. Example 4.2. (i). Rn is a Hilbert space with the canonical scalar product x, y :=

n 

xk yk ,

k=1

inducing the Euclidean distance, where x = (x1 , . . . , xn ), y = (y1 , . . . . . . , yn ) ∈ Rn . (ii). Let (X, E , μ) be a measure space. Then L 2 (X, E , μ), endowed with the scalar product  ϕ(x)ψ(x) dμ(x) ϕ, ψ ∈ L 2 (X, E , μ), ϕ, ψ := X

is a Hilbert space (completeness follows from Proposition 3.5).

63 Introduction to Measure Theory and Integration

(iii). Let 2 be the space of all sequences of real numbers x = (xk ) such ∞  xk2 < ∞. 2 is a vector space with the usual operations, that k=0

a(xk ) = (axk ) a ∈ R,

(xk ) + (yk ) = (xk + yk ),

(xk ), (yk ) ∈ 2 .

The space 2 , endowed with the scalar product x, y :=

∞ 

xk yk ,

x = (xk ), y = (yk ) ∈ 2

k=0

is a Hilbert space. This follows from (ii) taking X = N, E = P (X) and μ({x}) = 1 for all x ∈ X. (iv). Let X = C([0, 1]) be the linear space of all real continuous functions on [0, 1]. X is a pre–Hilbert space with the scalar product  f (t)g(t) dt.  f, g := X

However, X is not a Hilbert space: indeed, X is dense, but strictly contained, in L 2 (0, 1). Finite-dimensional pre-Hilbert spaces H are always Hilbert spaces: indeed, if {v1 , . . . , vn }, with n = dim H , is a basis of H , the GramSchmidt orthonormalization process (recalled in Exercise 4.3) provides an orthonormal basis {e1 , . . . , en } of H (i.e. ei  = 1 and ei is orthogonal to e j for i = j), and the map x=

n  x, ei ei → (x, e1 , x, e2 , . . . , x, en ) i=1

(mapping x to the Euclidean vector of its coordinates with respect to this basis) is easily seen to provide an isometry with Rn : indeed, 

n n n    x, ei ei 2 = x, ei x, e j ei , e j  = (x, ei )2 . i=1

i, j=1

i=1

Thus, being R complete, H is complete. n

4.2. The projection theorem It is useful to notice that for any x, y ∈ H the following parallelogram identity holds: x + y2 + x − y2 = 2x2 + 2y2 ,

x, y ∈ H.

(4.2)

One can show that identity (4.2) characterizes pre-Hilbert spaces among normed spaces, and Hilbert among Banach spaces, see Exercise 4.1.

64 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Theorem 4.3 (Projection on closed subspaces). Let H be a Hilbert space and let Y be a closed subspace of H . Then for any x ∈ H there exists a unique y ∈ Y , called projection of x on Y and denoted by πY (x), such that x − y = min x − z. z∈Y

Moreover, y is characterized by the property x − y, z = 0 for all z ∈ Y.

(4.3)

Proof. Set d := infz∈Y x − z and choose (yn ) ⊂ Y such that x − yn  ↓ d. We are going to show that (yn ) is a Cauchy sequence. For any m, n ∈ N we have, by the parallelogram identity (4.2), (x −yn )+(x −ym )2 +(x −yn )−(x −ym )2 = 2x −yn 2 +2x −ym 2 . Consequently $ $2 $ yn + ym $ $ $ . yn − ym  = 2x − yn  + 2x − ym  − 4 $x − 2 $ 2

2

2

Taking into account that (yn + ym )/2 ∈ Y we find yn − ym 2 ≤ 2x − yn 2 + 2x − ym 2 − 4d 2 , so that yn − ym  → 0 as n, m → ∞. Thus, (yn ) is a Cauchy sequence and, since the space is complete and Y is closed, it is convergent to an element y ∈ Y. Since x − yn  → x − y we find that x − y = d. Existence is thus proved. Uniqueness follows again by the parallelogram identity, that gives $ $2 $ y + y $  2 2  2 $ $ y − y  ≤ 2x − y + 2x − y  − 4 $x − 2 $ ≤ 2d 2 + 2d 2 − 4d 2 = 0 whenever y and y  are minimizers. Let us prove (4.3). Define F(λ) = x − y − λz2 = λ2 z2 − 2λx − y, z + x − y2 ,

λ ∈ R.

Since F attains a minimum at λ = 0, we have F  (0) = x − y, z = 0, as claimed. Conversely, if (4.3) holds for all z ∈ Y , we have x − y − z2 = z2 + x − y2 ≥ x − y2 .

65 Introduction to Measure Theory and Integration

Remark 4.4 (Projection on convex closed sets). The previous proof works, with absolutely no modification, to show that for any convex closed set K ⊂ H and any x ∈ H there exists a unique solution y = π K (x) to the problem min x − z. z∈K

In this case, however, π K (x) is not characterized by (4.3), but by a onesided condition, namely x − π K (x), z − π K (x) ≤ 0 for all z ∈ K , see Exercise 4.2. Corollary 4.5. Let Y be a closed proper subspace of H . Then there exists x0 ∈ H \ {0} such that x0 , y = 0 for all y ∈ Y . Proof. It is enough to choose an element z 0 in H which does not belong to Y and set x0 = z 0 − πY (z 0 ). Fix an integer n ≥ 1, a n-dimensional subspace Hn ⊂ H and an orthonormal basis {e1 , . . . , en } of it. The following result characterizes the projection on Hn , giving the best approximation of an element x by a linear combination of {e1 , . . . , en }. Proposition 4.6. The projection of an element x ∈ H on Hn is given by n  π Hn (x) = x, ek ek . k=1

Proof. We have to show that for any y1 , . . . , yn ∈ R we have $2 $ $2 $ n n $ $ $ $   $ $ $ $ xk ek $ ≤ $x − yk ek $ , $x − $ $ $ $ k=1 k=1

(4.4)

where xk = x, ek . We have in fact $ $2 n n n $ $    $ $ yk ek $ = x2 + yk2 − 2 xk yk $x − $ $ k=1 k=1 k=1 = x2 −

n  k=1

xk2 +

n  (xk − yk )2 . k=1

This quantity is clearly minimal when xk = yk , and $2 $ n n $ $   $ $ xk ek $ = x2 − xk2 . $x − $ $ k=1 k=1

(4.5)

An alternative proof of the Proposition, based on the characterization (4.3) of π Hn (x), is proposed in Exercise 4.4.

66 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

4.3. Linear continuous functionals A linear functional F on H is a mapping F : H → R such that F(αx + βy) = α F(x) + β F(y)

∀x, y ∈ H, ∀α, β ∈ R.

F is said to be bounded if there exists K ≥ 0 such that |F(x)| ≤ K x

for all x ∈ H .

Proposition 4.7. A linear functional F is continuous if, and only if, it is bounded. Proof. It is obvious that if F is bounded then it is continuous (even Lipschitz continuous). Assume conversely that F is continuous and, by contradiction, that it is not bounded. Then for any n ∈ N there exists x n ∈ H such that |F(xn )| ≥ n 2 xn . Setting yn = n1 xn /xn  we have yn  = n1 → 0, whereas F(yn ) ≥ n, which is a contradiction. The following basic Riesz theorem, gives an intrinsic representation formula of all linear continuous functionals. Proposition 4.8. Let F be a linear continuous functional on H . Then there exists a unique x0 ∈ H such that F(x) = x, x0 

∀x ∈ H.

(4.6)

Proof. Assume that F = 0 and let Y = F −1 (0) = Ker F. Then Y  = H is closed (because F is continuous) and a vector space (because F is linear), so that by Corollary 4.5 there exists z 0 ∈ H such that F(z 0 ) = 1 and z 0 , z = 0 for all z ∈ Ker F. On the other hand, for any x ∈ H the element z = x − F(x)z 0 belongs to KerF since F(z) = F(x) − F(x)F(z 0 ) = 0. Therefore z 0 , x − F(x)z 0  = 0 for all x ∈ H, so that

x, z 0  − F(x)z 0 2 = 0

and (4.6) follows setting x0 = z 0 /z 0 2 . It remains to prove the uniqueness. Let y0 ∈ H be such that F(x) = x, x0  = x, y0 ,

x ∈ H.

Then, choosing x = x0 − y0 we find that x0 − y0 2 = 0, so that x0 = y0 .

67 Introduction to Measure Theory and Integration

4.4. Bessel inequality, Parseval identity and orthonormal systems Let us discuss the concept of basis in a Hilbert space H , assuming with no loss of generality that the dimension of H is not finite. We use Kronecker’s notation δhk , equal to 1 for h = k and equal to 0 if h = k. Definition 4.9 (Orthonormal system). A sequence (ek )k∈N ⊂ H is called an orthonormal system if eh , ek  = δh,k ,

h, k ∈ N.

Proposition 4.10. Let (ek )k∈N be an orthonormal system in H . (i) For any x ∈ H we have ∞ 

|x, ek |2 ≤ x2 .

(4.7)

k=0

(ii) For any x ∈ H the series

∞ 

x, ek ek is convergent in H (1) .

k=0

(iii) Equality holds in (4.7) holds if and only if x=

∞  x, ek ek .

(4.8)

k=0

Inequality (4.7) is called Bessel inequality and when the equality holds, Parseval identity. Proof. (i) Let n ∈ N. Then by (4.5) we have $2 $ n n $ $   $ $ 2 x, ek ek $ = x − |x, ek |2 , $x − $ $ k=0 k=0

(4.9)

so that (4.7) follows by the arbitrariness of n. (ii) Let n, p ∈ N and set sn =

n  x, ek ek . k=0

∞ (1) A series  x of vectors in a Banach space E is said to be convergent if the sequence of the i k=0 n  finite sums xi is convergent in E k=0

68 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Then $2 $ n+ p n+ p $ $  $ $ sn+ p − sn 2 = $ x, ek ek $ = |x, ek |2 . $ $k=n+1 k=n+1 Since the series

∞ 

|x, ek |2 is convergent by (i), the sequence (sn ) is

k=0

Cauchy and the conclusion follows. Passing to the limit as n → ∞ in (4.9) we find $ $2 ∞ ∞ $ $   $ $ 2 x, ek ek $ = x − |x, ek |2 . $x − $ $ k=0 k=0 This proves statement (iii). Definition 4.11 (Complete orthonormal system). An orthonormal system (ek )k∈N is called complete if x=

∞  x, ek ek

∀x ∈ H.

k=0

Example 4.12. Let H = 2 as in Example 4.2(iii). Then, it is easy to see that the system (ek ), where ek := (0, 0, . . . , 0, 1, 0, 0, . . .)

(with the digit 1 in the k-th position)

is complete. Indeed, if x = (xk ) ∈ 2 we have that x, ei  = xi (the i-th component of the sequence x), so that x −

n ∞   x, ei ei 2 = xk2 → 0. k=0

k=n+1

We already noticed that Rn is the canonical model of n-dimensional Hilbert spaces H , because any choice of an orthonormal basis {v1 , . . . ,vn} of H induces the linear isometry a →

n 

ai ei

i=1

from Rn to H (which, as a consequence, preserves also the scalar product, by the parallelogram identity). For similar reasons, 2 is the canonical

69 Introduction to Measure Theory and Integration

model of all spaces H having a complete orthonormal system (ek )k∈N : in this case, the linear map from 2 to H given by a →

∞ 

ai ei

i=0

is an isometry, thanks to Parseval’s identity. Proposition 4.13 (Completeness criterion). Let (en) be an orthonormal system. Then (en ) is complete if and only if the vector space E spanned by (en ) is dense in H . Proof. If (en ) is complete we have that any x ∈ H is the limit of the finite sums 1N x, ei ei , which all belong to E, therefore E is dense. Conversely,  if E is dense, for any x ∈ H and any ε > 0 we can find a n ai ei with z − x < ε. By applying Proposition 4.6 vector z = i=1 twice (first to the vector space spanned by {e1 , . . . , em }, and then to the vector space spanned by {e1 , . . . , en }) we get x −

m n n    x, ei ei  ≤ x − x, ei ei  ≤ x − ai ei  < ε i=1

i=1

i=1

for m ≥ n. Since ε is arbitrary this proves that the sum of the series is equal to x. The following proposition provides a necessary and sufficient condition for the existence of a complete orthonormal system. We recall that a metric space (X, d) is said to be separable if there exists a countable dense subset D ⊂ X. Theorem 4.14. A Hilbert space H admits a complete orthonormal system (ek )k∈N if and only if H , as a metric space, is separable. Proof. If H admits a complete orthonormal system (ek )k∈N then H is separable, because the collection D of finite sums with rational coefficients of the vectors ek provides a countable dense subset (indeed, the closure of D contains the finite linear combinations of the vectors ek and then the whole space). Conversely, assume that H is separable and let (vn ) be a dense sequence. We define e0 = v0 , e1 = vk1 where k1 is the first k > k0 = 0 such that vk is linearly independent from v0 , e2 = vk2 where k2 is the first k > k1 such that vk is linearly independent from {e0 , e1 }, and so on. In this way we have built a sequence (ei ) of linearly independent vectors

70 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

generating the same vector space generated by (vn ). Let S be this vector space, and let us represent it as ∪n Sn , where Sn is the vector space generated by {e0 , . . . , en }. Notice that S is dense, as all vn belong to S. By applying the Gram-Schmidt process to ei , an operation that does not change the vector spaces Sn generated by the vectors e0 , . . . , en , we can also assume that (ei ) is an orthonormal system. Then, Proposition 4.13 gives that (ei ) is complete.

4.5. Hilbert spaces on C In this section we illustrate briefly how the concepts introduced so far extend to complex vector spaces H . A pre–Hilbert space is a complex vector space H endowed with a mapping H × H → C,

(x, y) → x, y,

called scalar product, such that: (i) x, x ≥ 0 for all x ∈ H and x, x = 0 if and only if x = 0; (ii) x, y = y, x for all x, y ∈ H ; (iii) αx + βy, z = αx, z + βy, z for all x, y, z ∈ H and α, β ∈ C. √ It turns out that x := x, x is still a norm, because the CauchySchwarz inequality still holds. Hence, we can define Hilbert spaces as those spaces for which the norm induces a complete distance. The canonical model of n-dimensional Hilbert space is Cn . Given a measure space (X, F , μ), a basic example of Hilbert space is the space of F -measurable and square integrable functions f : X → C. In this context F -measurable means that both the real and the imaginary part of f are F -measurable. In this space one can define the scalar product  f (x)g(x) dμ(x)  f, g := X

and prove that it induces an Hilbert space structure. The space 2 (C) of complex-values sequences (z n ) with (|z n |) ∈ 2 (R) is a particular case. The norm still satisfies the parallelogram identity, so that we can still prove the existence of orthogonal projections on closed subspaces and its characterization in terms of  ∀z ∈ Y. Re x − πY (x), z = 0 Analogously, in Remark 4.4, one has to replace the scalar product by its real part.

71 Introduction to Measure Theory and Integration

Riesz representation theorem still holds (now for continuous and C-linear functionals) and the concepts of orthonormal system and complete orthonormal system make sense. We have Bessel’s inequality for orthonormal systems and Parseval’s identity for complete orthonormal systems. Finally, 2 (C) is the canonical model of all separable Hilbert spaces; as in the real case the correspondence is induced by the choice of a complete orthonormal system, which provides coordinates of a vector. We conclude this chapter providing a natural example, considered in the literature, of non-separable Hilbert space. Example 4.15 (Quasi-periodic functions). We define the space A P(R) of almost periodic functions as the closure, with respect to uniform convergence in R, of the vector space generated by complex-valued periodic functions (of arbitrary period). This space has been extensively studied by Bochner and Bohr. It is easy to show that the space of almost periodic functions is not only a vector space (it is a subspace of C(R, C)), but also an algebra, i.e. f g ∈ A P(R) whenever f, g ∈ A P(R). If f is almost periodic one can also show (by approximation, taking into account that this property is linear with respect to f and holds for periodic functions) that there exists the limit 1 T →+∞ 2T



M( f ) := lim

T

−T

f (x + t) dt.

In addition, it is easily seen that the limit is independent of x. The space A P(R) of all almost periodic functions is a pre-Hilbert space when endowed with the following inner product ¯  f, g A P := M( f g)

f, g ∈ A P(R).

For any λ ∈ R define eλ (t) = eiλt ,

t ∈ R.

Then eλ ∈ A P(R), eλ , eλ  A P = 1 and ei T (λ−ν) − e−i T (λ−ν) =0 T →+∞ T i(λ − ν)

eλ , eν  A P = lim

whenever λ  = ν,

so that (eλ )λ∈R is an orthonormal system in A P(R) having the cardinality of continuum. One can also characterize the (abstract) Hilbert completion of A P(R) (the so-called Bohr almost periodic functions) and prove that the system {eλ }λ∈R is complete. For more details see e.g. [4].

72 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercises 4.1 Let (X,  · ) be a normed space, and assume that the norm satisfies the parallelogram identity (4.2). Set 1 1 x + y2 − x − y2 , x, y ∈ X. 4 4 Show that ·, · is a scalar product whose induced norm is  · . Use this identity to show that any linear isometry between pre-Hilbert spaces preserves also the scalar product. 4.2 Show that, in the situation considered in Remark 4.4, π K (x) is characterized by the property x, y :=

x − π K (x), z − π K (x) ≤ 0

∀z ∈ K .

4.3 Let H be a finite dimensional pre-Hilbert space and let {v1 , . . . , vn }, with n = dim H , be a basis of it. Define v2 , f 1  v3 , f 1  v3 , f 2  f 1 = v1 , f 2 = v2 − f 1 , f 3 = v3 − f1 − f 2 , ......  f1, f1  f1, f1  f2, f2 Show that ei = f i / f i  is an orthonormal system in H (notice that vk − f k is the projection of vk on the vector space generated by {v1 , . . . , vk−1 }). 4.4 Let H be a Hilbert space, and let X be an infinite-dimensional separable subspace. Show that π X (x) =

∞  x, ek ek

∀x ∈ H,

k=0

where (ek ) is any complete orthonormal system of X. Hint: show that the vector x − k x, ek ek is orthogonal to all vectors of X. 4.5 Let X be the space of functions f : [0, 1] → R such that f (x)  = 0 for at  most countably many x, and x f 2 (x) < +∞. Show that X, endowed with the scalar product  f (x)g(x),  f, g := x∈[0,1]

is a non-separable Hilbert space. 4.6 Let (ek )k∈N be a complete orthonormal system of H . Show that, for any x, y ∈ H we have ∞  x, ek y, ek  = x, y. (4.10) k=0

4.7 Show that for any Hilbert space H there exists a family (not necessarily finite or countable) of vectors {ei }i∈I such that: (i) ei , e j  is equal to 1 if i = j, and to 0 otherwise; (ii) for any vector x ∈ H there exists a countable set J ⊂ I with  x, ei ei . x= i∈J

Hint: use Zorn’s lemma.

Chapter 5 Fourier series

In this chapter we study the problem of representing a given T -periodic function as a superposition, for a suitable choice of the coefficients, of more “elementary” ones. This problem was first studied by J. Fourier in the case when the elementary functions are the trigonometric ones (nowadays we know that many different choices are indeed possible). Thanks to the theory of L 2 spaces and of Hilbert spaces developed in the previous chapters, the problem can be formalized by looking for complete orthonormal systems in L 2 made by trigonometric functions. We shall mostly be concerned with the case of 2π-periodic functions, but a simple change of scale (see Remark 5.1) easily provides the translation of the results to arbitrary periods.  We are concerned with the measure space (−π, π), B ((−π, π)), λ , where λ is the Lebesgue measure. As usual, we shall write for brevity L 2 (−π, π). We shall denote by ·, · the canonical scalar product given by π

 f (x)g(x) dλ =

 f, g := (−π,π)

f (x)g(x) dx,

f, g ∈ L 2 (−π, π).

−π

Let us consider, as a family of elementary functions, the trigonometric system, given by: 1 √ ; 2π

1 √ cos kx, k ∈ N, k ≥ 1; π

1 √ sin kx, k ∈ N, k ≥ 1. π (5.1) It is easy to check with integration by parts that this is an orthonormal system in L 2 (−π, π), see Exercise 5.1. Thus, in view of Proposition 4.10, the series of functions S(x) =

∞  1 (ak cos kx + bk sin kx), a0 + 2 k=1

L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

(5.2)

74 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

is convergent in L 2 (−π, π) for any f ∈ L 2 (−π, π), where  π 1 f (y) cos kydy, k ∈ N, ak := π −π and

1 bk := π



π

−π

f (y) sin kydy,

k ∈ N, k ≥ 1.

Notice that a0 /2 is the mean value of f on (−π, π), in agreement with the fact that all terms in the series (5.2) have mean value 0 on (−π, π). To recognize (5.2) in terms of scalar products, we see that the term a0 /2 corresponds to % & 1 1 f, √ √ 2π 2π and the terms ak cos kx, bk sin kx for k ≥ 1, correspond respectively to % & % & 1 1 1 1 f, √ cos kx √ cos kx, f, √ sin kx √ sin kx. π π π π Formula (5.2) is called the trigonometric Fourier series of f . The Bessel inequality (4.7) reads, in this context, as follows: 1 π



∞ 1 2  | f (x)| dx ≥ a0 + (ak2 + bk2 ). 2 −π k=1 π

2

(5.3)

√ Indeed, it is easily seen that a02 π/2 = ( f, 1/ 2π)2 and, for k ≥ 1, % ak2 π =

1 f, √ cos kx π

%

&2 ,

bk2 π =

1 f, √ sin kx π

&2 .

First, we shall find sufficient conditions on f ensuring the pointwise convergence of the series S(x) to f (x) in (−π, π). Then, we shall show that the trigonometric system is complete, so that the inequality above is actually an equality. As shown in Exercise 5.4 and Exercise 5.5, the trigonometric system, the trigonometric series and the form of the coefficients become much more nice and symmetric in the complex-valued Hilbert space L 2 (−π, π); C : f (x) =

 n∈Z

an e

inx

1 where an := 2π



π

−π

f (x)e−inx dx.

75 Introduction to Measure Theory and Integration

Remark 5.1 (2T -periodic functions). If f ∈ L 2 (−T, T ) we can write instead ∞ π π a0  ak cos kx + bk sin kx f (x) = + 2 T T k=1 with

ak :=

⎧  T 1 ⎪ ⎪ f (x) dx ⎪ ⎪ ⎨ T −T

if k = 0;

⎪  ⎪ ⎪ 1 T π ⎪ ⎩ f (x) cos kx d x T −T T

if k > 0,

1 bk := T



T

f (x) sin

−T

π kx dx. T

5.1. Pointwise convergence of the Fourier series For any integer N ≥ 1 we consider the partial sum SN (x) :=

N  1 (ak cos kx + bk sin kx), a0 + 2 k=1

x ∈ [−π, π).

Since the functions cos kx and sin kx are 2π–periodic, it is natural to extend f to the whole of R as a 2π–periodic function ' f , setting ' f (x + 2πn) = f (x),

x ∈ [−π, π), n = ±1, ±2, . . . .

(5.4)

We shall denote in the sequel by Hl,r (z) the “Heaviside” function  l Hl,r (z) := r

if z ≤ 0; if z > 0.

Lemma 5.2. For any integer N ≥ 1 and x, l, r ∈ R we have l +r 1 S N (x) − = 2 2π



π

−π

  '  f (x + τ ) − Hl,r (τ ) 1  sin N + τ dτ. sin(τ/2) 2 (5.5)

76 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proof. Write N  1 (ak cos kx + bk sin kx) a0 + 2 k=1

S N (x) =

1 = π 1 = π



π

−π



π

−π



 N 1  f (y) (cos kx cos ky + sin kx sin ky) dy + 2 k=1 

 N 1  f (y) cos k(x − y) dy. + 2 k=1

To evaluate the sum, we notice that for any z ∈ R   N   1 + cos kz sin 12 z 2 k=1 N    1  1 sin k + sin 2 z + 2 k=1   1 = sin N + 12 z . 2

=

Therefore

1 2

  z − sin k −

1 2

z

  N 1  1 sin N + 12 z  cos kz = + 2 k=1 2 sin 12 z

and so, 1 S N (x) = 2π



π

−π

f (y)

sin

 N + 12 (x − y)  dy. sin 12 (x − y)



(5.6)



(5.7)

Now, setting τ = y − x we get    π−x sin N + 12 τ 1 '  f (x + τ ) dτ SN (x) = 2π −π−x sin 12 τ 1 = 2π



π

−π

' f (x + τ )

sin

 N + 12 τ  dτ sin 12 τ



since the function under the integral is 2π–periodic. Now, integrating (5.6) over [−π, π] yields    π sin N + 12 τ 1  dτ, 1= 2π −π sin 12 τ

77 Introduction to Measure Theory and Integration

so that 1 π

 0

π

sin

    N + 12 τ 1 0 sin N + 12 τ   dτ = 1 = dτ. π −π sin 12 τ sin 12 τ



If we multiply both sides by l and r, and subtract the resulting identities from (5.7), (5.5) follows. Proposition 5.3 (Dini’s test). Let x, l, r ∈ R be such that  π ' | f (x + τ ) − Hl,r (τ )| dτ < ∞. | sin(τ/2)| −π

(5.8)

Then the Fourier series of f converges to (l + r)/2 at x. Dini’s test shows a remarkable property of the Fourier series: while the specific value of the coefficients ak and bk depends on the behaviour of f on the whole interval (−π, π), and the same holds for the Fourier series, the character of the series (convergent or not) at a given point x depends only on the behaviour of f in the neighbourhood of x: indeed, it is this behaviour that influences the integrability of ( ' f (x +τ )− Hl,r (τ ))/ sin(τ/2) (the only singularity being at τ = 0). In the next example we provide sufficient conditions for the convergence of the Fourier series. Example 5.4. Assume that f : [−π, π] → R is L-Lipschitz continuous, i.e. | f (x) − f (y)| ≤ L|x − y| ∀ x, y ∈ [−π, π] for some L ≥ 0. Then Dini’s test is fulfilled at any x ∈ R \ Zπ choosing f (x+ ) (1) . l =r = ' f (x), and at any x ∈ Zπ choosing l = ' f (x− ) and r = ' Indeed, with these choices of l and r, the quotient ' f (x + τ ) − Hl,r (τ ) sin(τ/2) is bounded in a neighbourhood of 0. The same conclusions hold when f is α–H¨older continuous for some α ∈ (0, 1], i.e. | f (x) − f (y)| ≤ L|x − y|α ,

∀ x, y ∈ [−π, π]

for some L ≥ 0: in this case the quotient is bounded from above, near 0, by the function L|τ |α /| sin(τ/2)| ∼ 2L|τ |α−1 which is integrable.

(1) here we denote by g(x ), g(x ) the left and right limits of g at x − +

78 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

More generally, the argument of the previous example can be used to show that the Fourier series is pointwise convergent for piecewise C 1 functions f : at continuity points x the series converges to f (x), and at (jump) discontinuity points x it converges to ( f (x− ) + f (x+ ))/2. However, the mere continuity of f is not sufficient to ensure pointwise convergence of the Fourier series. In order to prove Proposition 5.3, we need the following Riemann– Lebesgue lemma, a tool interesting in itself. Lemma 5.5. Let (ek ) be an orthonormal system in L 2 (−π, π). Assume that there exists M > 0 such that ek ∞ ≤ M for all k ∈ N. Then for any f ∈ L 1 (−π, π) we have  π lim f (x)ek (x) dx = 0. (5.9) k→∞ −π

Proof. Notice first that if f ∈ L 2 (−π, π) the conclusion of the lemma is trivial. We have in fact in this case  π f (x)ek (x) dx =  f, ek  −π

 2 and, since by Bessel’s inequality the series ∞ 1 | f, ek | is convergent, we have limk  f, ek  = 0. Let us now consider the general case. We know that bounded continuous functions are dense in L 1 (−π, π), hence for any ε > 0 we can find g ∈ Cb (−π, π) such that  f − g1 < ε. As a consequence | f, ek | = | f − g, ek | + |g, ek | ≤ Mε + |g, ek | and letting k → ∞ we obtain lim supk | f, ek | ≤ Mε. Since ε is arbitrary the proof is achieved. Proof of Proposition 5.3. Set g(τ ) :=

' f (x + τ ) − Hl,r (τ ) ∈ L 1 (−π, π). sin(τ/2)

(5.10)

Then, writing 1 1 1 sin[(N + )t] = sin N t cos t + cos N t sin t 2 2 2 and applying the Riemann–Lebesgue lemma to g cos t/2 (with e N = sin N t) and to g sin(t/2) (with e N = cos N t) we obtain from (5.5) that SN (x) converge to (l + r)/2.

79 Introduction to Measure Theory and Integration

5.2. Completeness of the trigonometric system Proposition 5.6. The trigonometric system (5.1) is complete. In particular equality holds in (5.3) and  π | f (x) − S N f (x)|2 dx = 0 ∀ f ∈ L 2 (−π, π). (5.11) lim N →∞ −π

Proof. We show that the vector space E generated by the trigonometric system is dense in L 2 (−π, π). Let H  be the closure, in the L 2 (−π, π) norm, of E, that is easily seen to be still a vector space as well. We will prove in a series of steps that H  contains larger and larger classes of functions. Let f : [−π, π] → [0, +∞) be a Lipschitz function, and let us prove that it belongs to H  . Indeed, we know from Example 5.4 that S N → f pointwise in (−π, π). On the other hand, we already know from Proposition 4.10(ii) that the Fourier series is convergent in L 2 (−π, π) to some function g (which is indeed, by Exercise 4.4, the orthogonal projection of f on H  ), therefore a subsequence (S N (k) ) is converging λ-almost everywhere to g. It follows that g = f and S N → f in L 2 (−π, π). If now g : [−π, π] → [0, +∞) is continuous, we know that g can be monotonically approximated by the Lipschitz functions  x ∈ [−π, π] gλ (x) := min g(y) + λ|x − y| , y∈[−π,π]

(see Exercise 2.11), that converge to g also in L 2 (−π, π) by the dominated convergence theorem. As a consequence also g belongs to H  . Since H  is invariant by addition of constants, we proved that all continuous functions in [−π, π] belong to H  . We conclude using the density of this class of functions in L 2 (−π, π). Remark 5.7. Let f ∈ L 2 (−π, π). Then, the Parseval identity reads as follows  π ∞  1 1 | f (x)|2 dx = a02 + (ak2 + bk2 ). (5.12) π −π 2 k=1 For instance, taking f (x) = x one finds the following nice relation between π and the harmonic series with exponent 2: ∞  1 π2 = . k2 6 k=1

Notice that (5.11) provides, for any f ∈ L 2 (−π, π), the existence of a subsequence N (k) such that S N (k) f (x) → f (x) for L 1 –a.e. x ∈

80 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

(−π, π). Is it true that the whole sequence S N f converges a.e. to f ? This problem, surprisingly difficult, has been solved by L.Carleson only in 1966, see [1]. Finally, we notice that there exist other important examples of complete orthonormal systems, besides the trigonometric one. Some of them are illustrated in the exercises.

5.3. Uniform convergence of the Fourier series We conclude by studying the uniform convergence of the Fourier series. ∞ space E is said to be totally We recall that a series 0 xn in a Banach  x  is convergent. Using the convergent if the numerical series ∞ n 0 completeness of E it is not difficult to check (see Exercise 5.2) that any totally convergent series is convergent  (as we have seen in the previous chapter, this means that the finite sums 0N xn converge in E to a vector, denoted by ∞ 0 x n ). Now we show that the Fourier series of C 1 functions f with f (−π) = f (π) are uniformly convergent: the proof highlights two important principles, whose validity extend to higher order derivatives (see Exercise 5.9) and to Fourier transforms: first, the Fourier coefficients of the derivative of a function are linked to the Fourier coefficients of the function; second, higher regularity of f implies a faster decay of the Fourier coefficients, and therefore a convergence in stronger norms of the Fourier series. Proposition 5.8. Assume that f ∈ C 1 ([−π, π]) and that f (−π) = f (π). Then the Fourier series of f converges uniformly to f in [−π, π]. Proof. We first notice that ' f in (5.4) is Lipschitz continuous, so that by Proposition 5.3 we have ∞  1 (ak cos kx + bk sin kx) f (x) = a0 + 2 k=1

∀x ∈ [−π, π].

Let us consider the Fourier series of the derivative f  of f , ∞  (ak cos kx + bk sin kx)

x ∈ [−π, π],

k=1

where, for k ≥ 1 integer,  π 1  ak = f  (y) cos ky dy, π −π

bk

1 = π



π

−π

f  (y) sin ky dy. (5.13)

81 Introduction to Measure Theory and Integration

Notice that a0 = 0 because f (−π) = f (π) implies that the mean value of f  on (−π, π) is 0. As easily checked through an integration by parts (using again the fact that f (−π) = f (π)), we have ak = kbk and bk = −kak . Then, by the Bessel inequality it follows that ∞ 

k

2

(ak2 + bk2 )

k=1

 π ∞  1  2  2 = (ak ) + (bk ) ≤ | f  (x)|2 dx < ∞. (5.14) π −π k=1

Therefore the Fourier series of f is totally convergent in C([−π, π]) and therefore uniformly convergent. We have indeed ∞  k=1

max |ak cos kx + bk sin kx|

x∈[−π,π]

≤

∞  (|ak | + |bk |) k=1

≤

∞ 

k 2 (|ak | + |bk |)2

1/2 ∞ 

k=1

1/2 k −2

< ∞.

k=1

Exercises 5.1 Check that the trigonometric system (5.1) is orthogonal.  5.2 Let E be a Banach space. Show that any totally convergent series n xn , with (xn ) ⊂ E, is convergent. Moreover, ∞ ∞ $  $ $ $ xn $ ≤ xn . $ n=0

Hint: estimate 

N 0

xn −

M 0

(5.15)

n=0

xn  with the triangle inequality.

5.3 Prove that the following systems on L 2 (0, π) are orthonormal and complete ( 2 sin kx, k ≥ 1, π and 1 √ ; π 5.4 Show that

(

2 cos kx, π

1 ek (x) := √ eikx , 2π

k ≥ 1.

k∈Z

is a complete orthonormal system in L 2 ((−π, π); C). Hint: in order to show completeness, consider first the cases where f is real-valued or i f is real-valued.

82 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

5.5 Let (ek ) be as in Exercise 5.4. Using the Parseval identity show that 

π

−π

| f (x)|2 dx =

2  π 1  f (x)e−ikx dx 2π k∈Z −π

∀ f ∈ L 2 ((−π, π); C) .

N 5.6 Let f ∈ L 2 ((−π, π); C) and let S N f = −N  f, ek ek , with N ≥ 1, be the Fourier sums corresponding to the complete orthonormal system in Exercise 5.4. Show that  π G N (x − y)( f (x) − f (y)) dy f (x) − S N f (x) = −π

with

sin((N + 1/2)z) . sin(z/2)   Hint: use the identities 0N eiky = 0N (eiy )k = (ei(N +1)z − 1)/(eiy − 1).  −4 = π 4 /90. Hint: consider the 5.7 Arguing as in Remark 5.7, show that ∞ 1 k 2 function f (x) = x . 5.8 Chebyschev polynomials Cn in L 2 (a, b), with (a, b) bounded interval, are the ones obtained by applying the Gram-Schmidt procedure to the vectors 1, x, x 2 , x 3 , . . .. They are also called Legendre polynomials when (a, b) = (−1, 1). G N (z) :=

(a) Compute explicitly the first three Legendre polynomials. (b) Show that {Cn }n∈N is a complete orthonormal system. Hint: use the density of polynomials in C([a, b]). (c) Show that the n-th Legendre polynomial Pn is given by ( 2n + 1 1 d n 2 Pn (x) = (x − 1)n . 2 2n n! d n x  5.9 Let f ∈ C m [−π, π]; C with f ( j) (−π) = f ( j) (π) for all j = 0, . . . , m − 1. Show that ck(m) , the k-th Fourier coefficient of f (m) is linked to ck , the k-th Fourier coefficient of f , by ck(m) = (ik)m ck .

Chapter 6 Operations on measures

In this chapter we collect many useful tools in Analysis and Probability that will be widely used in the following chapters. We will study the product of measures (both finite and countable), the product of measures by L 1 functions, the Radon–Nikod´ym theorem, the convergence of measures on the real line R and the Fourier transform.

6.1. The product measure and Fubini–Tonelli theorem Let (X, F ) and (Y, G ) be measurable spaces. Let us consider the product space X ×Y . A set of the form A× B, where A ∈ F and B ∈ G , is called a measurable rectangle. We denote by R the family of all measurable rectangles. R is obviously a π–system. The σ –algebra generated by R is called the product σ –algebra of F and G . It is denoted by F × G . Given σ –finite measures μ in (X, F ) and ν in (Y, G ), we are going to define the product measure μ × ν in (X × Y, F × G ). First, for any E ∈ F × G we define the sections of E, setting for x ∈ X and y ∈ Y , E x := {y ∈ Y : (x, y) ∈ E},

E y := {x ∈ X : (x, y) ∈ E}.

Proposition 6.1. Assume that μ and ν are σ –finite and let E ∈ F × G . Then the following statements hold. (i) E x ∈ G for all x ∈ X and E y ∈ F for all y ∈ Y . (ii) The functions x → ν(E x ),

y → μ(E y ),

are F –measurable and G –measurable respectively. Moreover,   ν(E x ) dμ(x) = μ(E y ) dν(y). (6.1) X

Y

L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

84 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proof. Assume first that E = A × B is a measurable rectangle. Then, if (x, y) ∈ X × Y we have



B if x ∈ A A if y ∈ B y E = Ex = ∅ if x ∈ / A, ∅ if y ∈ / B. Consequently, ν(E x ) = 1 A (x)ν(B),

μ(E y ) = 1 B (y)μ(A),

so that (6.1) clearly holds. Now, let D be the family of all E ∈ F × G such that (i) is fulfilled. Clearly, D is a Dynkin system including the π–system R . Therefore, (i) follows from the Dynkin theorem. Now, if both μ are ν are finite, let D be the family of all E ∈ F × G such that (ii) is fulfilled. Clearly, D is a Dynkin system including the π–system R (stability under complement follows by the identities ν((E c )x ) = ν(Y ) − ν(E x ) and μ((E c ) y ) = μ(X) − μ(E y )). Therefore, (ii) follows from the Dynkin theorem as well. In the general σ –finite case we argue by approximation: if E ∈ F ×G , F  X h ↑ X and G  Yh ↑ Y satisfy μ(X h ) < ∞ and ν(Yh ) < ∞, we define the finite measures μh (A) = μ(A ∩ X h ),

νh (B) = ν(B ∩ Yh )

to obtain that x → νh (E y ) is F –measurable and y  → μh (E x ) is G – measurable for all E ∈ E × G . Passing to the limit as h → ∞ in the identity    νh (E x ) dμ(x) = νh (E x ) dμh (x) = μh (E y ) dνh (y) Xh X Y  = μh (E y ) dν(y) Yh

the continuity properties of measures and integrals give (6.1) as well. Theorem 6.2 (Product measure). If μ and ν are σ –finite, there exists a unique measure λ in (X × Y, F × G ) satisfying λ(A × B) = μ(A)ν(B)

for all A ∈ F , B ∈ G .

The measure λ is σ -finite and denoted by μ × ν. Furthermore μ × ν is finite (resp. a probability measure) if both μ and ν are finite (resp. probability measures).

85 Introduction to Measure Theory and Integration

Proof. Existence is easy: we set   ν(E x ) dμ(x) = μ(E y ) dν(y), λ(E) = X

E ∈ F × G . (6.2)

Y

Using the continuity and additivity properties of the integral, it is immediate to check that λ is a measure on (X ×Y, F ×G ). In the case of σ –finite measures, uniqueness follows by the the coincidence criterion for positive measures stated in Proposition 1.15: indeed, the value of the product measure is uniquely determined on the π–system K made by rectangles A × B with μ(A) and ν(B) finite, and thanks to the σ –finiteness assumption there exist E n = An × Bn ∈ K with E n ↑ X × Y . Corollary 6.3. Let E ∈ F × G be such that μ × ν(E) = 0. Then μ(E y ) = 0 for ν–almost all y ∈ Y and ν(E x ) = 0 for μ–almost all x ∈ X. Proof. It follows directly from (6.2). We consider here the measure space (X × Y, F × G , λ), where λ = μ × ν and μ and ν are σ –finite. Theorem 6.4 (Fubini–Tonelli). Let F : X ×Y → [0, +∞] be a F ×G – measurable map. Then the following statements hold. (i) For any x ∈ X (respectively y ∈ Y ), the function y  → F(x, y) (respectively x → F(x, y)) is G –measurable (resp. F –measurable). (ii) The functions   F(x, y) dν(y), y → F(x, y) dμ(x) x → Y

X

are respectively F –measurable and G –measurable. (iii) We have *  )  F(x, y) dλ(x, y) = F(x, y) dν(y) dμ(x) X×Y X Y *  ) = F(x, y) dμ(x) dν(y). Y

(6.3)

X

Proof. Assume first that F = 1 E , with E ∈ F × G . Then we have F(x, y) = 1 E x (y),

x ∈ X,

F(x, y)(x) = 1 E y (x),

y ∈ Y,

so (i), (ii) and (iii) follow from Proposition 6.1. Consequently, by linearity, (i)–(iii) hold when F is a simple function. If F is general, it

86 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

is enough to approximate it by a monotonically increasing sequence of simple functions and then pass to the limit using the monotone convergence theorem. Remark 6.5 (The definition of integral revisited). We noticed in Remark 2.13 that the integral of nonnegative functions can also be defined without using the archimedean integral, by considering minorant simple functions. If we follow this approach, the identity that we used to define the integral can be derived by applying the Fubini–Tonelli theorem to the subgraph E := {(x, t) ∈ X × R : 0 < t < f (x)} , with the product measure μ × λ, λ being the Lebesgue measure. Indeed, it is not difficult to show that E is F × B (R)–measurable whenever f is F -measurable, so that  ∞   ∞ μ({ f > t}) dt = μ(E t ) dt = μ × λ(E) = λ(E x ) dμ(x) X 0 0  = f (x) dμ(x). X

Of course, splitting F in positive and negative parts, also the case of extended real valued maps can be considered: Corollary 6.6. Let F : X × Y → [−∞, +∞] be a F × G –measurable map. Then F is μ × ν–integrable if and only if: (i) for μ–a.e. x ∈ X the  function y → F(x, y) is ν–integrable; (ii) the function x → Y |F(x, y)| dν(y) is μ–integrable. If these conditions hold, we have *   ) F(x, y) d(μ × ν)(x, y) = F(x, y) dν(y) dμ(x). (6.4) X×Y

X

Y

Notice that, strictly speaking, the function in (ii) is defined only out of a μ–negligible set; by μ–integrability of it we mean μ–integrability of any F –measurable extension of it (for instance we may set it equal to 0  wherever Y |F(x, y)| dν(y) is not finite). Remark 6.7 (Finite products). The previous constructions extend without any difficulty to finite products of measurable spaces (X i ,F i ). Namen ly, the product σ -algebra F := i F i in the cartesian product X := n 1 X i is generated by the rectangles

×

×

{A1 × · · · × An : Ai ∈ F i , 1 ≤ i ≤ n} .

87 Introduction to Measure Theory and Integration

Furthermore, if μi are σ –finite measures in (X i , F i ), integrals with ren spect to the product measure μ = 1 μi are defined by     F(x) dμ(x) = · · · F(x1 , . . . , xn ) dμn (xn ) · · · dμ2 (x2 ) dμ1 (x1 ),

×

X

X1 X2

Xn

and any permutation in the order of the integrals would produce the same result. Finally, the product measure is uniquely determined, in the σ – finite case, by the product rule n

μ (A1 × · · · × An ) =  μi (Ai ) i=1

Ai ∈ F i , 1 ≤ i ≤ n.

It is also not hard to show that the product is associative, both at the level of σ –algebras and measures, see Exercise 6.1.

6.2. The Lebesgue measure on Rn This section is devoted to the construction, the characterization and the main properties of the Lebesgue measure in Rn , i.e. the length measure in R1 , the area measure in R2 , the volume measure in R3 and so on. Definition 6.8 (Lebesgue measure in Rn ). Let us consider the measure space (R, B (R), L 1 ), where L 1 is the Lebesgue measure on (R,B (R)). n

Then, we can define the measure space (Rn ,

×

n 1

× B(R), L

n

i=1

) with L n :=

L 1 . We say that L n is the Lebesgue measure on Rn .

Since (see Exercise 6.2) n

B (Rn ) =

× B (R), i=1

we can equivalently consider L n as a measure in (Rn , B (Rn )), forgetting its construction as a product measure (indeed, there exist alternative and direct constructions of L n independent of the concept of product measure). As in the one-dimensional case, we will keep using the classical notation   f (x) dx := f 1 E dL n E ∈ B (Rn ), f : Rn → R Borel E

Rn

for integrals with respect to Lebesgue measure L n (or Riemann integrals in more than one independent variable).

88 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

In the computation of Lebesgue integrals, a particular role is sometimes played by the dimensional constant ωn = L n (B(0, 1)) (so that ω1 = 2, ω2 = π, ω3 = 4π/3,. . . ). A general formula for the computation of ωn can be given using Euler’s  function:  ∞ t z−1 e−t dt z > 0. (z) := 0

Indeed, we have ωn =

π n/2 . ( n2 + 1)

(6.5)

A proof of this formula, based on the identity (z + 1) = z(z) (which gives also (n) = (n − 1)! for n ≥ 1 integer) is proposed in Exercise 6.7. We are going to show that L n is invariant under translations and rotations. For this we need some notation. For any a ∈ Rn and any δ > 0 we set   Q(a, δ) : = x ∈ Rn : ai ≤ xi < ai + δ, ∀ i = 1, . . . , n n

=

×[a , a + δ). i=1

i

i

Q(a, δ) is called the δ–box with corner at a. For all N ∈ N we consider the family Q N = {Q(2−N k, 2−N ) : k = (k1 , . . . , kn ) ∈ Zn }. It is also clear that each box in Q N is Borel and that its Lebesgue measure is 2−n N . Now we set ∞

Q = Q N. N =0

It is clear that all boxes in Q N are mutually disjoint and that their union is Rn . Furthermore, if N < M, Q ∈ Q N and Q  ∈ Q M , then either Q  ⊂ Q or Q ∩ Q  = ∅. If follows that if Q, Q  ∈ Q intersect, then one of the two sets is contained in the other one. Lemma 6.9. Let U be a non empty open set in Rn . Then U is the disjoint union of boxes in Q . Proof. For any x ∈ U , let Q x ∈ Q be the biggest box such that x ∈ Q x ⊂ U . This box is uniquely defined: indeed, fix an x; for any m there is only one box Q x,m ∈ Q m such that x ∈ Q x,m ; moreover, since U is open, for m large enough Q x,m ⊂ U ; we can then define Q x = Q x,m˜ where m˜ is the smallest integer m such that Q x,m ⊂ U .

89 Introduction to Measure Theory and Integration

This family {Q x }x∈U is a partition of U , that is, for any x, y ∈ U , either Q x = Q y or Q x ∩ Q y = ∅; indeed, if we suppose that Q x ∩ Q y  = ∅, then one of the two boxes is contained in the other, say Q x ⊂ Q y . This leads to x ∈ Q x ⊂ Q y ⊂ U , contradicting the definition of Q x unless Qx = Q y. From Lemma 6.9 it follows easily that the σ –algebra generated by Q coincides with B (Rn ). Proposition 6.10 (Properties of the Lebesgue measure). The following statements hold. (i) (translation invariance) For any E ∈ B (Rn ), x ∈ Rn we have L n (E + x) = L n (E), where E + x = {y + x : y ∈ Rn }. (ii) If μ is a translation invariant measure on (Rn , B (Rn )) such that μ(K ) < ∞ for any compact set K , there exists a number Cμ ≥ 0 such that μ(E) = Cμ L n (E)

∀ E ∈ B (Rn ).

(iii) (rotation invariance) For any orthogonal matrix R ∈ L(Rn ; Rn ) we have ∀ E ∈ B (Rn ). L n (R(E)) = L n (E) (iv) For any T ∈ L(Rn ; Rn ) we have L n (T (E)) = |det T |L n (E)

∀ E ∈ B (Rn ).

Proof. Fix x ∈ Rn . The measures L n (E) and L n (E + x) coincide on the π–system of boxes; thanks to Lemma 6.9, this π–system generates the Borel σ –algebra, so that the coincidence criterion for measures stated in Proposition 1.15 gives that L n (E) = L n (E + x) for all Borel sets E. Let us prove (ii). Let Q 0 ∈ Q 0 and set Cμ = μ(Q 0 ). Since Q 0 is included in a compact set, we have Cμ < ∞. Since μ is translation invariant, all boxes in Q 0 have the same μ measure. Now, let Q N ∈ Q N . Since Q 0 is the disjoint union of 2−n N boxes in Q N which have all the same μ measure (again by the translation invariance) we have that μ(Q N ) = Cμ L n (Q N ). So, Lemma 6.9 gives that μ(A) = Cμ L n (A) for any open set, and therefore for any Borel set.

90 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Let us now prove (iii). By the translation invariance of L n , the measure μ(E) = L n (R(E)) is easily seen to be translation invariant (because R(E +z) = R(E)+ R(z)), hence L n (R(E)) = CL n (E) for some contant C. We can identify the constant C choosing E equal to the unit ball, finding C = 1. Finally, let us prove √ (iv). By polar decomposition we can write T = R ◦ S with S = T ∗ ◦ T symmetric and nonnegative definite, and R orthogonal. Notice that on one hand |det T | = det S (because det R ∈ {−1, 1}) and on the other hand, by (iii) we have L n (T (E)) = L n (R(S(E))) = L n (S(E)). Hence, it suffices to show that L n (S(E)) = det SL n (E) for any symmetric and nonnegative definite matrix S. By the translation invariance of L n (S(E)) there exists a constant C such that L n (S(E)) = CL n (E) for any Borel set E. In this case we can identify the constant C choosing as E a suitable n-dimensional cube: denoting by (ei ) an orthonormal basis of eigenvectors of S, with eigenvalues αi ≥ 0 (whose product is det S), choosing     n n   E= ci ei : |ci | ≤ 1 , so that S(E) = αi ci ei : |ci | ≤ 1 , i=1

i=1

the rotation invariance of L gives L (E) = 1 and L n (S(E)) = α1 · · · αn . Therefore C = det S and the proof is complete. n

n

6.3. Countable products We are here concerned with a sequence (X i , F i , μi ), i = 1, 2, . . ., of probability spaces. We denote by X the product space ∞

X :=

×X k=1

k

and by x = (xk ) the generic element of X. We are going to define a σ –algebra of subsets of X. Let us first introduce the cylindrical sets in X. A cylindrical set In,A is a set of the following form In,A = {x : (x1 , . . . , xn ) ∈ A}, where n ≥ 1 is an integer and A ∈ unique; however, since

×

n 1

F k . This representation is not

∞

In,A = A ×

×

k=n+1

Xk

we have that In,A = Im,B with n < m implies B = A × X n+1 × · · · × X m .

91 Introduction to Measure Theory and Integration

We denote by C the family of all cylindrical sets of X. Notice also that c In,A = In,Ac ,

so that C is stable under complement. If In,A and Im,B belong to C we can assume by the previous remarks that m = n, so that In,A ∪ In,B = In,A∪B belongs to C . Therefore C is an algebra. The σ –algebra generated by C is called the product σ –algebra of the σ –algebras F i . It is denoted by ∞

×F

k.

k=1

Now we define a function μ on C , setting n  μ(In,A ) =

×μ

k

k=1

(A),

In,A ∈ C .

(6.6)

This definition is well posed, again thanks to the fact that In,A = Im,B with n < m when B = A × X n+1 × · · · × X m . It is easy to check that μ is additive: indeed, if In,A and Im,B are disjoint, using the previous remark we can assume with no loss of generality that n = m, and therefore the equality μ(In,A ∪ In,B ) = μ(In,A ) + μ(In,B ) follows by n  n  n 

×μ k=1

k

(A ∪ B) =

×μ

k

k=1

(A) +

×μ k=1

k

(B).

Theorem 6.11. The set function μ defined in (6.6) is σ –additive on C and therefore, by the Carath´eodory theorem, it has a unique extension to ∞ a probability measure on (X, 1 F k ) that is denoted by

×

∞

×μ k=1

k

Proof. To prove the σ –additivity of μ it is enough to show the continuity of μ at ∅, or equivalently the implication (E j ) ⊂ C , (E j ) nonincreasing, μ(E j ) ≥ ε0 > 0 ⇒

∞  n=1

E j  = ∅.

(6.7) In the following we are given a nonincreasing sequence (E j ) on C such that μ(E j ) ≥ ε0 > 0. To prove (6.7), we need some more notation. We set X (n) =

∞

×

k=n+1

Xk

92 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

and we define μ(n) on cylindrical sets of X (n) as in (6.6). Then, we consider the sections of E j defined as   E j (x1 ) = x (1) ∈ X (1) : (x1 , x (1) ) ∈ E j , x1 ∈ X 1 . E j (x1 ) is a cylindrical subset of X (1) and by the Fubini theorem we have  μ(1) (E j (x1 )) dμ1 (x1 ) ≥ ε0 > 0, j ≥ 1. (6.8) μ(E j ) = X1

Set now

+ ε0 , F j,1 = x1 ∈ X 1 : μ(1) (E j (x1 )) ≥ , 2

Then F j,1 is not empty and by (6.8) we have   (1) μ (E j (x1 )) dμ1 (x1 ) + μ(E j ) = F j,1

≤ μ1 (F j,1 ) +

c F j,1

j ≥ 1.

μ(1) (E j (x1 )) dμ1 (x1 )

ε0 . 2

Therefore μ1 (F j,1 ) ≥ ε0 /2 for all j ≥ 1. Obviously (F j,1 ) is a nonincreasing sequence of subsets of X 1 . Since μ 1∞is σ –additive, it is continuous at 0. Therefore, there exists α1 ∈ 1 F j,1 and so ε0 μ(1) (E j (α1 )) ≥ , j ≥ 1. (6.9) 2 Consequently we have E j (α1 ) = ∅,

j ≥ 1.

(6.10)

Now we iterate the procedure: for any x2 ∈ X 2 we consider the section   E j (α1 , x2 ) = x (2) ∈ X (2) : (α1 , x2 , x (2) ) ∈ E j , j ≥ 1. By the Fubini theorem we have  (1) μ(2) (E j (α1 , x2 )) dμ2 (x2 ). μ (E j (α1 )) = X2

We set

+ ε0 , , F j,2 = x2 ∈ X 2 : μ(2) (E j (α1 , x2 )) ≥ 4

j ≥ 1.

(6.11)

93 Introduction to Measure Theory and Integration

Then by (6.9) and (6.10) we have  ε0 (1) μ(2) (E j (α1 , x2 )) dμ2 (x2 ) ≤ μ (E j (α1 )) = 2 X2   (2) = μ (E j (α1 , x2 )) dμ2 (x2 ) + μ(2) (E j (α1 , x2 )) dμ2 (x2 ) [F j,2 ]c

F j,2

ε0 ≤ μ2 (F j,2 ) + . 4

Therefore μ2 (F j,2 ) ≥ ε0 /4. Since (F j,2 ) is nonincreasing and μ2 is σ – additive, there exists α2 ∈ X 2 such that μ2 (E j (α1 , α2 )) ≥

ε0 , j ≥ 1, 4

and consequently we have E j ((α1 , α2 )) = ∅.

(6.12)

Arguing in a similar way we see that there exists a sequence (αk ) ⊂ X such that (6.13) E j (α1 , . . . , αn ) = ∅, for all j, n ≥ 1, where   E j (α1 , . . . , αn ) = x ∈ X (n) : (α1 , . . . , αn , x (n) ) ∈ E j , j, n ≥ 1.  SinceE j are cylindrical, this easily implies that (αn ) ∈ ∞ 1 E j . ThereE is not empty, as required. fore ∞ j 1

Exercises 6.1 Let (X 1 , F 1 ), (X 2 , F 2 ), (X 3 , F 3 ) be measurable spaces. Show that (F 1 × F 2 ) × F 3 = F 1 × (F 2 × F 3 ). If we are given measures μi in F i , i = 1, 2, 3, show also that (μ1 ×μ2 ) ×μ3 = μ1 × (μ2 × μ3 ). 6.2 Let us consider the measurable spaces (R, B (R)), (Rn , B (Rn )). Show that n

B (Rn ) =

× B (R). i=1

Hint: to show the inclusion ⊂, use Lemma 6.9. 6.3 Let L n be the σ –algebra of Lebesgue measurable sets in Rn . Show that L 1 × L 1  L 2.

94 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Hint: to show the strict inclusion, consider the set E = F × {0}, where F ⊂ R is not Lebesgue measurable. 6.4 Show that the product σ –algebra is also generated by the family of products

×

∞ 1

Ai where Ai ∈ F i and Ai  = X i only for finitely many i.

6.5 Writing properly L 3 as a product measure, compute L 3 (T ), where + , T = (x, y, z) : x 2 + y 2 < r 2 and y 2 + z 2 < r 2 . 6.6 [Computation of ωn ] Find a recursive formula linking ωn to ωn−2 , and use it to show that ω2k = π k /k! and ω2k+1 = 2k+1 π k /(2k + 1)!!, where (2k + 1)!! is the product of all odd integers between 1 and 2k + 1. Hint: use the Fubini– Tonelli theorem. √ 6.7 Use Exercise 6.6 and the identities (1) = 1, (1/2) = π and (z + 1) = z(z) to show (6.5). 6.8 Let μ and ν be σ –finite measures on (X, F ) and (Y, G ) respectively and let λ = μ × ν. Let E = (F × G )λ , as defined in Definition 1.12, and let ζ be the extension of λ to E . Show this version of the Fubini–Tonelli Theorem 6.4: for any E –measurable function F : X × Y → [0, +∞] the following statements hold: (i) for μ–a.e. x ∈ X thefunction y  → F(x, y) is ν–measurable; F(x, y) dν(y), set to zero at all points x such that (ii) the function x  → Y

 F(x, y) is not ν–integrable,  is μ–measurable; y → (iii) X Y F(x, y) dμ(x) dμ(y) = X×Y F(x, y) dζ(x, y). 6.9 Using the notation of the Fubini-Tonelli theorem, let X = Y = [0, 1], F = G = B ([0, 1]), let μ be the Lebesgue measure and let ν be the counting measure. Let D = {(x,   x) y: x ∈ [0, 1]} be the diagonal in X × Y ; check that ν(D ) dμ(x)  = x X Y μ(D ) dν(y).

6.10 Let ( f h ) be converging to f in L 1 (X × Y, μ × ν). Show the existence of a subsequence h(k) such that f h(k) (x, ·) converge to f (x, ·) in L 1 (Y, ν) for μ–a.e. x ∈ X. Show by an example that, in general, this property is not true for the whole sequence.

6.4. Comparison of measures In this section we study some relations between measures in a measurable space (X, F ). The first (immediate) one is the order relation: viewing measures as set functions, we say that μ ≤ ν if μ(B) ≤ ν(B) for all B ∈ F . It is not hard to see that the space of measures endowed with this order relation is a complete lattice (see Exercise 6.13): in particular μ∨ν(B) = sup {μ(A1 ) + ν(A2 ) : A1 , A2 ∈ F , (A1 , A2 ) partition of B}

95 Introduction to Measure Theory and Integration

and μ∧ν(B) = inf {μ(A1 ) + ν(A2 ) : A1 , A2 ∈ F , (A1 , A2 ) partition of B} . Another relation between measures is linked to the concept of product of a function by a measure. Definition 6.12. Let μ be a measure in (X, F ) and let f ∈ L 1 (X, F , μ) be nonnegative. We set  f dμ ∀B ∈ F . (6.14) f μ(B) := B

It is immediate to check, using the additivity and the continuity properties of the integral, that f μ is a finite measure. Furthermore, the following simple rule provides a way for the computation of integrals with respect to f μ:   h d( f μ) = h f dμ, (6.15) X

X

whenever h is F –measurable and nonnegative (or h f is μ–integrable, see Exercise 6.11). It suffices to check the identity (6.15) on characteristic functions h = 1 B (and in this case it reduces to (6.14)), and then for simple functions. The monotone convergence theorem then gives the general result. Notice also that, by definition, f μ(B) = 0 whenever μ(B) = 0. We formalize this relation between measures in the next definition. Definition 6.13 (Absolute continuity). Let μ, ν be measures in F . We say that ν is absolutely continuous with respect to μ, and write ν # μ, if all μ–negligible sets are ν–negligible, i.e. μ(A) = 0

⇒

ν(A) = 0.

For finite measures, the absolute continuity property can also be given in a (seemingly) stronger way, see Exercise 6.14. The following theorem shows that absolute continuity of ν with respect to μ is not only necessary, but also sufficient to ensure the representation ν = f μ. Theorem 6.14 (Radon–Nikod´ym). Let μ and ν be finite measures on (X, F ) such that ν # μ. Then there exists a unique nonnegative ρ ∈ L 1 (X, F , μ) such that  ν(E) = ρ(x) dμ(x) ∀E ∈ F . (6.16) E

96 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

We are going to show a more general result, whose statement needs two more definitions. We say that a measure μ is concentrated on a F – measurable set A if μ(X \ A) = 0. For instance, the Dirac measure δa is concentrated on {a}, and the Lebesgue measure in R is concentrated on the irrational numbers, and f μ is concentrated (whatever μ is) on { f  = 0}. Definition 6.15 (Singular measures). Let μ, ν be measures in (X, F ). We say that μ is singular with respect to ν, and write μ ⊥ ν, if there exist disjoint F –measurable sets A, B such that μ is concentrated on A and ν is concentrated on B. The relation of singularity, as stated, is clearly symmetric. However, it can also be stated in a (seemingly) asymmetric way, by saying that μ ⊥ ν if μ is concentrated on a ν–negligible set A (just take B = Ac to see the equivalence with the previous definition). Example 6.16. Let X = R, F = B (R), μ the Lebesgue measure on (X, F ) and ν = δx0 the Dirac measure at x0 ∈ R. Then μ is concentrated on A := R \ {x0 }, whereas ν is concentrated on B := {x0 }. So, μ and ν are singular. Theorem 6.17 (Lebesgue). Let μ and ν be measures on (X, F ), with μ σ –finite and ν finite. Then the following assertions hold. (i) There exist two unique finite measures νa and νs on (X, F ) such that ν = νa + νs ,

νa # μ,

νs ⊥ μ.

(6.17)

(ii) There exists a unique ρ ∈ L 1 (X, F , μ) such that νa = ρμ. (6.17) is called the Lebesgue decomposition of ν with respect to μ. The function ρ in (ii) is called the density of ν with respect to μ and it is sometimes denoted by dν ρ: = . dμ Radon–Nikod´ym theorem simply follows by the Lebesgue theorem noticing that, in the case when ν # μ the uniqueness of the decomposition gives νa = ν and νs = 0, so that ν = νa = ρμ. Proof of Theorem 6.17. . We assume first that also μ is finite. Set λ = μ + ν and notice that, obviously, μ # λ and ν # λ. Define a linear functional F on L 2 (X, F , λ) setting  F(ϕ) := ϕ(x) dν(x), ϕ ∈ L 2 (X, F , λ). X

97 Introduction to Measure Theory and Integration

The functional F is well defined and bounded (and consequently continuous) since, in view of the H¨older inequality, we have   |ϕ(x)| dν(x) ≤ |ϕ(x)| dλ(x) ≤ [λ(X)]1/2 ϕ L 2 (X,F ,λ) . |F(ϕ)| ≤ X

X

Now, thanks to the Riesz theorem, there exists a unique function f ∈ L 2 (X, F , λ) such that   ϕ(x) dν(x) = f (x)ϕ(x) dλ(x) ∀ϕ ∈ L 2 (X, F , λ). (6.18) X

X

Setting ϕ = 1 E , with E ∈ F , yields  f (x) dλ(x) ≥ 0, ν(E) = E

which implies, by the arbitrariness of E, f (x) ≥ 0, λ–a.e. and, in particular, both μ–a.e. and ν–a.e. In the sequel we shall assume, possibly modifying f in a λ–negligible set, and preserving the validity of (6.18), that f ≥ 0 everywhere. By (6.18) it follows   ϕ(x)(1− f (x)) dν(x) = f (x)ϕ(x) dμ(x) ∀ϕ ∈ L 2 (X, F , λ). X

X

Setting ϕ = 1 E , with E ∈ F , yields   (1 − f (x)) dν(x) = f (x) dμ(x) ≥ 0 E

(6.19)

E

because f ≥ 0. Thus, being E arbitrary, we obtain that f (x) ≤ 1 for ν–a.e. x ∈ X. Set now A := {x ∈ X : 0 ≤ f (x) < 1},

B := {x ∈ X : f (x) ≥ 1},

so that (A, B) is a F –measurable partition of X, and νa (E) := ν(E ∩ A),

νs (E) := ν(E ∩ B)

∀E ∈ F ,

so that νa = 1 A ν is concentrated on A, νs = 1 B ν is concentrated on B and ν = νa + νs . Then, setting in (6.19) ϕ = 1 B , we see that   f dμ = (1 − f ) dν = 0 μ(B) ≤ B

B

98 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

because f = 1 ν–a.e. on B. It follows that νs is singular with respect to μ. We show now that the existence of ρ such that νa = ρμ. Heuristically, this can be obtained choosing in (6.19) the function ϕ = (1 − f )−1 1 E∩A , but since this function need not to be in L 2 (X, F , λ) we argue by approximation: set in (6.19) ϕ(x) = (1 + f (x) + · · · + f n (x))1 E∩A (x) where n ≥ 1 and E ∈ F . Then we obtain   n+1 (1− f (x)) dν(x) = [ f (x)+ f 2 (x)+· · ·+ f n+1 (x)] dμ(x). E∩A

E∩A

Set ρ(x) = 0 for x ∈ B and ρ(x) := lim [ f (x) + f 2 (x) + · · · + f n+1 (x)] = n→∞

f (x) , 1 − f (x)

x ∈ A.

Then, by the monotone convergence theorem it follows that   ρ(x) dμ(x) = ρ(x) dμ(x). νa (E) = ν(E ∩ A) = E∩A

E

Setting E = X we see that ρ ∈ L 1 (X, F , μ), and the arbitrariness of E gives that νa = ρμ. Now we consider the case when μ is σ –finite. In this case there exists a sequence of pairwise disjoint sets (X n ) ⊂ F such that X=

∞

Xn

with μ(X n ) < ∞.

n=0

Let us apply Theorem 6.17 to the finite measures μn = 1 X n μ, νn = 1 X n ν. For any n ∈ N let νn = (νn )a + (νn )s = ρn μn + (νn )s be the Lebesgue decomposition of νn with respect to μn . Now, set νa :=

∞  (νn )a , n=0

Since

νs :=

∞  (νn )s ,

ρ :=

n=0

n=0

k k   (νn )a + (νn )s = νn = 1∪k X n ν, n=0

∞ 

n=0

0

ρn 1 X n .

99 Introduction to Measure Theory and Integration

we can pass to the limit as k → ∞ to obtain that νa and νs are finite measures, and ν = νa + νs . Moreover, for any E ∈ F we have, using the monotone convergence theorem, ∞ ∞    (νn )a (E) = ρn (x) dμn (x) νa (E) = n=0

=

  ∞ E n=0

n=0

E



ρn (x)1 X n dμ(x) =

ρ(x) dμ(x). E

So, νa # μ, and setting E = X we see that ρ is integrable with respect to μ. Finally, it is easy to see that νs ⊥ μ, because if we denote by Bn ∈ F μ–negligible sets where (νn )s are concentrated, we have that νs is concentrated on the μ–negligible set ∪n Bn . Finally, let us prove the uniqueness of νa and νs : assume that ν = νa + νs = νa + νs and let B, B  be μ–negligible sets where νs and νs are respectively concentrated. Then, as B ∪ B  is μ–negligible and both νs and νs are concentrated on B ∪ B  , for any set E ∈ F we have νs (E) = νs (E ∩(B ∪ B  )) = ν(E ∩(B ∪ B  )) = νs (E ∩(B ∪ B  )) = νs (E). It follows that νs = νs and therefore νa = νa . The interested reader can have a look at a different proof of Theorem 6.17 independent of Hilbert space theory, and based on three auxiliary variational principles; it turns out that the density f of ν a is the maximizer in the problem 

 f dμ : f μ ≤ ν . (6.20) sup X

See Exercise 6.17 and Exercise 6.18 for more details. Remark 6.18. If μ is not σ –finite then the Lebesgue decomposition does not hold in general. Consider for instance the case when X = [0, 1], F = B ([0, 1]), μ is the counting measure and ν = L 1 . Then ν # μ (as the only μ–negligible set is the empty set) but there is no ρ : [0, 1] → [0, ∞] satisfying  ρ dμ. ν(E) = E

Indeed, this function should be μ-integrable and therefore it can be nonzero only in a set at most countable.

100 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercises 6.11 Show that a F –measurable function h is f μ–integrable if and only if f h is μ–integrable. 6.12 Show that ( f μ)∨(gμ) = ( f ∨g)μ and ( f μ)∧(gμ) = ( f ∧g)μ whenever f, g ∈ L 1 (X, F , μ) are nonnegative. 6.13 Let {μi }i∈I be a family of measures in (X, F ). Show that

 ∞ μ(B) := inf μi(k) (Bk ) : i : N → I, k=0



(Bk ) countable F –measurable partition of B is the greatest lower bound of the family {μi }i∈I , i.e. μ ≤ μi for all i ∈ I and it is the largest measure with this property. Show also that

 ∞ μ(B) := sup μi(k) (Bk ) : i : N → I, k=0



(Bk ) countable F –measurable partition of B is the smallest upper bound of the family {μi }i∈I , i.e. μ ≥ μi for all i ∈ I and it is the smallest measure with this property. 6.14 Let μ, ν be measures in (X, F ) with ν finite. Then ν # μ if and only if for all ε > 0 there exists δ > 0 such that A ∈ F , μ(A) < δ

⇒

ν(A) < ε.

6.15 Assume that ν # μ and that ν ⊥ μ. Show that ν = 0. 6.16 Assume that σ ≤ μ + ν and that σ ⊥ ν. Show that σ ≤ μ. 6.17 Prove Theorem 6.14 in the following two steps: (1) Show that a maximizer f in (6.20) exists. (2) Setting σ = ν − f μ ≥ 0, σ satisfies t > 0, B ∈ F, t1 B μ ≤ σ

⇒

μ(B) = 0.

(6.21)

Then, apply Exercise 6.18 to conclude that σ ⊥ μ. 6.18 Let μ, σ be nonnegative finite measures satisfying (6.21). Show that σ ⊥ μ. Hint: first show that inf {μ(A) : A ∈ F, σ is concentrated on A} has a solution A. Assuming by contradiction that μ(A) > 0 (otherwise we are done), show that F  B ⊆ A, μ(B) > 0

⇒

σ (B) > 0.

(6.22)

101 Introduction to Measure Theory and Integration

Then, show that the numbers + , ξh := sup μ(B) : F  B ⊆ A, 1 B μ ≥ 2h 1 B σ are infinitesimal as h → ∞, that the supremum is attained at Bh , and that μ(C) ≤ 2h σ (C)

for all sets C ⊂ A \ Bh .

(6.23)

Finally choose t = 2−h , with h sufficiently large so that ξh < μ(A) and B = A \ Bh , to get a contradiction with (6.21).

6.5. Signed measures Let (X, F ) be a measurable space. In this section we see how the concept of measure, still viewed as a set function, can be extended dropping the nonnegativity assumption on A → μ(A). We recall that sequence (E i ) ⊂ F of pairwise disjoint sets such that ∞ 0 E i = E is called a countable F –measurable partition of E. Definition 6.19 (Signed measures and total variation). A signed measure μ in (X, F ) is a map μ : F → R such that μ(E) =

∞ 

μ(E i )

i=0

for all countable F –measurable partitions (E i ) of E. Notice that the series above is absolutely convergent by the arbitrariness of (E i ): indeed, if σ : N → N is a permutation, then (E σ (i) ) is still a partition of E, hence ∞  i=0

μ(E i ) =

∞ 

μ(E σ (i) ).

i=0

This implies that the series is absolutely convergent. Let μ be a signed measure. Then we define the total variation |μ| of μ as follows:   ∞  |μ(E i )| : (E i ) F –measurable partition of E , |μ|(E) = sup i=0

E ∈ F. Proposition 6.20. Let μ be a signed measure and let |μ| be its total variation. Then |μ| is a finite measure on (X, F ).

102 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Proof. It is immediate to check that |μ| is a nondecreasing set function. Step 1. If A, B ∈ F are disjoint, we have |μ|(A ∪ B) = |μ|(A) + |μ|(B). Indeed, let E = A ∪ B and let (E i ) be a countable F –measurable partition of E. Set Aj = A ∩ E j,

Bj = B ∩ E j,

j ∈ N.

Then (A j ) is a countable F –measurable partition of A and (B j ) a countable F –measurable partition of B and we have E j = A j ∪ B j . Moreover, ∞ 

|μ(E j )| ≤

j=0

∞ 

|μ(A j )| +

j=1

∞ 

|μ(B j )| ≤ |μ|(A) + |μ|(B),

j=0

which yields |μ|(A ∪ B) ≤ |μ|(A) + |μ|(B). Let us prove the converse inequality, assuming with no loss of generality that |μ|(A ∪ B) < ∞. Since both |μ|(A) and |μ|(B) are finite, for any ε > 0 there exist countable F –measurable partitions (Aεk ) of A and (Bkε ) of B such that ∞ 

ε |μ(Aεk )| ≥ |μ|(A) − , 2 k=0

∞ 

ε |μ(Bkε )| ≥ |μ|(B) − . 2 k=0

Since (Aεk , Bkε ) is a countable F –measurable partition of A ∪ B, we have that |μ|(A ∪ B) ≥

∞  (|μ(Aεk )| + |μ(Bkε )|) ≥ |μ|(A) + |μ|(B) − ε. k=1

By the arbitrariness of ε we have |μ|(A ∪ B) ≥ |μ|(A) + |μ|(B). Step 2. |μ| is σ –additive. Since |μ| is additive by  Step 1, it is enough to show that |μ| is σ –subadditive, i.e. |μ(A)| ≤ ∞ 0 |μ|(Ai ) whenever (Ai ) ⊂ F is a partition of A. This can be proved arguing as in the first part of Step 1, i.e. building from a partition (E j ) of A partitions (E j ∩ Ai ) of all sets Ai . Step 3. |μ|(X) < ∞. Assume by contradiction that |μ|(X) = ∞. Then we claim that there exists a partition X = A ∪ B such that |μ(A)| ≥ 1 and |μ|(B) = ∞.

(6.24)

103 Introduction to Measure Theory and Integration

By the claim the conclusion follows since we can use it to construct by recurrence (replacing X with B and so on), a disjoint sequence (An ) ⊂ F such that |μ(An )| ≥ 1. Assume, to fix the ideas, that μ(An ) ≥ 1 for infinitely many n, and denote by E the union of these sets: then, the σ –additivity of μ forces μ(E) = +∞, a contradiction. Analogously, if μ(An ) ≤ −1 for infinitely many n, we find a set E such that μ(E) = −∞. Let us prove (6.24). By the assumption |μ|(X) = ∞ it follows the existence of a partition (X n ) of X such that ∞ 

|μ(X n )| > 2(1 + |μ(X)|).

n=0

Then either the sum of those μ(X n ) which are nonnegative or the absolute value of the sum of those μ(X n ) which are nonpositive is greater than 1 + |μ(X)|. To fix the ideas, assume that for a subsequence (X n(k) ) we have μ(X n(k) ) ≥ 0 and ∞ 

μ(X n(k) ) > 1 + |μ(X)|.

k=0

Set A =

∞ 0

X n(k) and B = Ac . Then we have |μ(A)| > 1 + |μ(X)| and

|μ(B)| = |μ(X) − μ(A)| ≥ |μ(A)| − |μ(X)| > 1. Since |μ|(X) = |μ|(A) + |μ|(B) = ∞, either |μ|(B) = +∞ or |μ|(A) = +∞. In the first case we are done, in the second one we exchange A and B. So, the claim is proved and the proof is complete. Let μ be a signed measure on (X, F ). We define μ+ := so that

1 (|μ| + μ), 2

μ = μ+ − μ−

and

μ− :=

1 (|μ| − μ), 2

|μ| = μ+ + μ− .

(6.25)

The measure μ+ (respectively μ− ) is called the positive part (respectively negative part) of μ and the first equation in (6.25) is called the Jordan representation of μ.

104 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Remark 6.21. It is easy to check that Theorems 6.17 and 6.14 hold when ν is a signed measure: it suffices to split it into its positive and negative part, see also Exercise 6.19. The following theorem proves also that μ+ and μ− are singular, and provides a canonical representation of μ± as suitable restrictions of ±μ. Theorem 6.22 (Hahn decomposition). Let μ be a signed measure on (X, F ) and let μ+ and μ− be its positive and negative parts. Then there exists a F –measurable partition (A, B) of X such that μ+ (E) = μ(A∩E) and μ− (E) = −μ(B∩E)

∀E ∈ F . (6.26)

Proof. Let us first notice that μ # |μ|. Thus, by the Radon–Nikod´ym theorem, there exists h ∈ L 1 (X, F , |μ|) such that  h d|μ| ∀E ∈ F . (6.27) μ(E) = E

Let us prove that |h(x)| = 1 for |μ|–a.e. x ∈ X. Indeed, set E 1 := {x ∈ X : h(x) > 1},

F1 := {x ∈ X : h(x) < −1}

We first show that |μ|(E 1 ) = |μ|(F1 ) = 0. Since we have  h d|μ| ≥ |μ|(E 1 ), |μ|(E 1 ) ≥ μ(E 1 ) = E1

and the second inequality is strict if |μ|(E 1 ) > 0, we have that |μ|(E 1 ) = 0. In a similar way one can prove that |μ|(F1 ) = 0, so that |h| ≤ 1 |μ|– a.e. in X. Now, let r ∈ (0, 1) and set G r := {x ∈ X : |h(x)| < r}. Let (G r,k ) be a countable F –measurable partition of G r . Then we have ! !  ! ! |μ(G r,k )| = !! h d|μ|!! ≤ |h| d|μ| ≤ r|μ|(G r,k ). G r,k

Therefore

∞ 

G r,k

|μ(G r,k )| ≤ r|μ|(G r ),

k=0

which yields, by the arbitrariness of the partition of G r , |μ|(G r ) ≤ r|μ|(G r ). Thus |μ|(G r ) = 0 and letting r ↑ 1 we obtain that |μ|({|h| <

105 Introduction to Measure Theory and Integration

1}) = 0. Hence, possibly modifying h in |μ|–negligible set, we can assume with no loss of generality that h takes its values in {−1, 1}. Now, to conclude the proof, we set A := {x ∈ X : h(x) = 1},

B := {x ∈ X : h(x) = −1}.

Then for any E ∈ F we have 1 1 μ (E) = (|μ|(E) + μ(E)) = 2 2  = hd|μ| = μ(E ∩ A), +

 (1 + h)d|μ| E

E∩A

and

 1 1 (1 − h)d|μ| μ (E) = (|μ|(E) − μ(E)) = 2 2 E =− hd|μ| = −μ(E ∩ B). −

E∩B

Exercises 6.19 Using the decomposition of ν in positive and negative part, show that Lebesgue decomposition is still possible when μ is σ –finite and ν is a signed measure. Using the Hahn decomposition extend this result to the case when even μ is a signed measure. Are these decompositions unique? 6.20 Show that | f μ| = | f |μ for any f ∈ L 1 (X, E , μ).

6.6. Measures in R In this section we estabilish a 1-1 correspondence between finite Borel measures in R and a suitable class of nondecreasing functions. In one direction this correspondence is elementary, and based on the concept of repartition function. Given a finite measure μ in (R, B (R)), we call repartition function of μ the function F : R → [0, +∞) defined by F(x) := μ ((−∞, x])

x ∈ R.

Notice that obviously (1) F is nondecreasing, right continuous, and satisfies lim F(x) ∈ [0, +∞). (6.28) lim F(x) = 0, x→−∞

x→+∞

Moreover, F is continuous at x if and only if x is not an atom of μ.

(1) The arguments are similar to those used in Section 2.4.2, in connection with the properties of the

function t → μ({ϕ > t})

106 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

The following result shows that this list of properties characterizes the functions that are repartition functions of some finite measure μ; in addition the measure is uniquely determined by its repartition function. Theorem 6.23. Let F : R → [0, +∞) be a nondecreasing and right continuous function satisfying (6.28). Then there exists a unique finite measure μ in (R, B (R)) such that F is the repartition function of μ. Proof. The proof follows the same lines of the construction of the Lebesgue measure in Section 1.6, with a simplification due to the fact that we can also consider unbounded intervals (because we are dealing with finite measures). We set I := {(a, b] : a ∈ [−∞, +∞), b ∈ R, a < b} and denote by A the ring generated by I : it consists, as it can be easily checked, of all finite disjoint unions of intervals in I . We define, with the convention F(−∞) = 0, μ((a, b]) := F(b) − F(a)

∀(a, b] ∈ I .

(6.29)

This definition is justified by the fact that, if μ were a measure and F were its repartition function, (6.29) would be valid, because (a, b] = (−∞, b] \ (−∞, a]. Then we extend μ to A with the same mechanism used in the proof of Theorem 1.21, and check that μ is additive on A . Also, the same argument used in that proof shows that μ is even σ – additive: in order to prove that μ(F) = i μ(Fi ) whenever F and all Fi belong to A one first reduces to the case when F = (a, b] belongs to I ; then, one enlarges Fi to Fi ∈ A with μ(Fi ) < μ(Fi ) + δ2−i and, using the fact that all intervals [a  , b] with a  > a are contained in a finite union of the sets Fi , obtains μ((a  , b]) ≤

∞ 

μ(Fi ) ≤ 2δ +

i=0

∞ 

μ(Fi ).

i=0

 Letting first  δ ↓ 0 and then a ↓ a we obtain the σ –subadditivity property μ(F) ≤ i μ(Fi ), and the opposite inequality follows by monotonicity. By the Carath´eodory theorem μ has a unique extension, that we still denote by μ, to B (R) = σ (A ). Setting a = −∞ and letting b tend to +∞ in the identity (6.29) we obtain that μ(R) = F(+∞) ∈ R. From (6.29) with a = −∞ we obtain that the repartition function of μ is F. Given a nondecreasing and right continuous function F satisfying (6.28), the Stieltjes integral  f dF

R

107 Introduction to Measure Theory and Integration

 is defined as f dμ F , where μ F is the finite measure built in the previous  theorem. The notation d F is justified by the fact that, when f = i z i 1(ai ,bi ] , we have (by the very definition of μ F ) 

 R

f dF =

R

f dμ F =



z i (F(bi ) − F(ai )).

i

This approximation of the Stieltjes integral will play a role in the proof of Theorem 6.28.

6.7. Convergence of measures on R In this section we study a notion of convergence for measures on the real line that is quite useful, both from the analytic and the probabilistic viewpoints. Definition 6.24 (Weak convergence). Let (μh ) be a sequence of finite measures on R. We say that (μh ) weakly converges to a finite measure μ on R if the repartition functions Fh of μh are pointwise converging to the repartition function F of μ on a co-countable set, i.e. if lim μh (−∞, x]) = μ ((−∞, x]) with at most countably many exceptions.

h→∞

(6.30)

Since the repartition function is right continuous, it is uniquely determined by (6.30). Then, since the measure is uniquely determined by its repartition function, we obtain that the weak limit, if exists, is unique. The following fundamental example shows why we admit at most countably many exceptions in the convergence of the repartition functions. Example 6.25. [Convergence tothe Dirac mass] Let ρ ∈ C ∞ (R) be a nonnegative function such that R ρ dx = 1 (an important example is 2 the Gauss function (2π)−1/2 e−x /2 ). We consider the rescaled functions ρh (x) = hρ(hx) and the induced measures μh = ρh L 1 , all probability measures. Then, it is immediate to check that μh weakly converge to δ0 : for x > 0 we have indeed  x  hx ρh (y) dy = ρ(y) dy → 1 μh ((−∞, x]) = −∞

−∞

because hx → +∞ as h → +∞. An analogous argument shows that μh ((−∞, x]) → 0 for any x < 0. If ρ is even, at x = 0 we don’t have pointwise convergence of the repartition functions: all the repartition functions Fh satisfy Fh (0) = 1/2, while F(0) = 1.

108 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Weak convergence is a quite flexible tool, because it allows also an opposite behaviour, the approximation of continuous measures (i.e. with no atom) by purely atomic ones, see for instance Exercise 6.21. From now on we will consider only, for the sake of simplicity, the case of weak convergence of probability measures. Before stating a compactness theorem for the weak convergence of probability measures, we introduce the following terminology. Definition 6.26 (Tightness). We say that a family of probability measures {μi }i∈I in R is tight if for any ε > 0 there exists a closed interval J ⊂ R such that μi (R \ J ) ≤ ε

∀i ∈ I.

Clearly any finite family of probability measures is tight. One can also check (see Exercise 6.24) that {μi }i∈I is tight if and only if lim Fi (x) = 0, lim Fi (x) = 1

x→−∞

x→+∞

uniformly with respect to i ∈ I ,

(6.31) where Fi are the repartition functions of μi . Furthermore, (see Exercise 6.25) any weakly converging sequence is tight. Conversely, we have the following compactness result for tight sequences: Theorem 6.27 (Compactness). Let (μh ) be a tight sequence of probability measures on R. Then there exists a subsequence (μh(k) ) weakly converging to a probability measure μ. Proof. We denote by Fh the repartition functions of μh . By a diagonal argument we can find a subsequence (Fh(k) ) pointwise converging on Q. We denote by G the pointwise limit, obviously a nondecreasing function. We extend G by monotonicity setting G(x) := sup {G(q) : q ∈ Q, q ≤ x}

x ∈R

and let E be the co-countable set of the discontinuity points of G. Let us check that Fh(k) is pointwise converging to G on R \ E: for x∈ / E we have indeed lim sup Fh(k) (x) ≤ k→∞

inf

q∈Q, q>x

lim sup Fh(k) (q) = k→∞

inf

G(q) = G(x),

sup

G(q) = G(x).

q∈Q, q>x

and analogously lim inf Fh(k) (x) ≥ k→∞

sup

lim inf Fh(k) (q) =

q∈Q, q<x k→∞

q∈Q, q<x

109 Introduction to Measure Theory and Integration

Since (μh ) is tight, we have also lim Fh (x) = 0,

lim Fh (x) = 1

x→−∞

x→+∞

uniformly with respect to h, hence G(−∞) = 0 and G(+∞) = 1. Notice now that the nondecreasing function F(x) := lim G(y) y↓x

is right continuous, and still satisfies F(−∞) = 0 and F(+∞) = 1, therefore (according to Theorem 6.23) F is the repartition function of a probability measure μ. Since F = G on R \ E, we have Fh(k) → F pointwise on R \ E, and this proves the weak convergence of μh(k) to μ. The following theorem provides a characterization of the weak convergence in terms of convergence of the integrals of continuous and bounded functions. Theorem 6.28. Let μh , μ be probability measures in R. Then μh weakly converge to μ if and only if   g dμh = g dμ ∀g ∈ Cb (R). (6.32) lim h→∞ R

R

Proof. Assuming that μh → μ weakly, we denote by Fh and F the corresponding repartition functions and fix g ∈ Cb (R). Let M = sup |g| and ε > 0. By Exercise 6.25 the sequence (μh ) is tight, so that we can find t > 0 satisfying μh (R \ (−t, t]) < ε for any h ∈ N; we may assume (possibly choosing a larger t) that also μ (R \ (−t, t]) < ε and that both −t and t are points where the repartition functions are converging. Thanks to the uniform continuity of g in [−t, t] we can find δ > 0 such that x, y ∈ [−t, t], |x − y| < δ

⇒

|g(x) − g(y)| < ε. (6.33)

Hence, we can find points t1 , . . . , tn in [−t, t] such that t1 = −t, tn = t, there is convergence of the repartition functions in all points ti , and ti+1 − ti < δ for i = 1, . . . , n −1. By (6.33) it follows that sup(−t,t] |g − f | < ε, where n−1  g(ti )1(ti ,ti+1 ] . f := i=1

110 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Splitting the integrals on R as the sum of an integral on (−t, t] and an integral on (−t, t]c we have ! !  ! ! ! g dμh − f dμh !! ≤ Mε + ε = (M + 1)ε ∀h ∈ N, (6.34) ! R

(−t,t]

and analogously !  ! ! g dμ − ! R

! ! f dμ!! ≤ Mε + ε = (M + 1)ε.

(−t,t]

(6.35)

Since  (−t,t]

f dμh =

n−1 

g(ti ) [Fh (ti+1 ) − Fh (ti )]

i=1

→

n−1 

 g(ti ) [F(ti+1 ) − F(ti )] =

i=1

(−t,t]

f dμ,

 adding and subtracting (−t,t] f dμh , and using (6.34) and (6.35), we conclude that ! !  ! ! g dμ!! ≤ (M + 1)ε. lim sup !! g dμh − h→∞

R

R

Since ε is arbitrary, (6.32) is proved. Conversely, assume that (6.32) holds. Given x ∈ R, define the open set A = (−∞, x); we can easily find (gk ) ⊆ Cb (R) monotonically converging to 1 A and deduce from (6.32) the inequality   lim inf μh (A) ≥ sup lim inf gk dμh = sup gk dμ = μ(A). h→∞

k∈N h→∞

R

k∈N

R

Analogously, using a sequence (gk ) ⊆ Cb (R) such that gk ↓ 1C , with C = (−∞, x], we deduce from (6.32) the inequality   gk dμ = μ(C). lim sup μh (C) ≤ inf lim sup gk dμh = inf h→∞

k∈N h→∞

R

k∈N R

Therefore we have convergence of the repartition functions for any x ∈ R such that μ(A) = μ(C), i.e. for any x that is not an atom of μ. We conclude thanks to Exercise 1.5. Notice that in (6.32) there is no mention to the order structure of R, and only the metric structure (i.e. the space Cb (R)) comes into play. In

111 Introduction to Measure Theory and Integration

a general context, of probability measures on a metric space (X, d) endowed with the Borel σ –algebra B (X), we say that μh weakly converge to μ if   lim g dμh = g dμ for any function g ∈ Cb (X). h→∞

X

X

Exercises 6.21 Show that the probability measures μh :=

h 1 δi h i=1 h

weakly converge to the probability measure 1[0,1] L 1 . 6.22 Let Fh : R → R be nondecreasing functions pointwise converging to a nondecreasing function F : R → R on a dense set D ⊂ R. Show that Fh (x) → F(x) at all points x where F is continuous. 6.23 Consider all atomic measures of the form h2  i=−h 2

ai δ i , h

where h ∈ N and a−h , . . . , ah ≥ 0. Show that for any finite Borel measure μ in R there exists a sequence of measures (μh ) of the previous form that weakly converges to μ. 6.24 Show that a family {μi }i∈I of probability measures in R is tight if and only if (6.31) holds. 6.25 Show that any sequence (μh ) of probability measures weakly convergent to a probability measure is tight. Hint: if μ is the weak limit and ε > 0 is given, choose an integer n ≥ 1 such that μ([1 − n, n − 1]) > 1 − ε and points x ∈ (−n, 1 − n) and y ∈ (n − 1, n) where the repartition functions of μh are converging to the repartition function of μ. 6.26 We want to extend what was shown in this section from the realm of probability measures to that of finite measures. Let (μh ), μ be finite positive Borel measures on R, and let Fh , F be their repartition functions. Consider the following implications:   (a) limh R g dμh = R g dμ ∀g ∈ Cb (R) (that is, (6.32)); (b) limh R g dμh = R g dμ ∀g ∈ Cc (R); (c) Fh converge to F at all points where F is continuous; (d) Fh converge to F on a dense subset of R; (e) limh μh (R) = μ(R); (f) (μh ) is tight. Find an example where (b) holds but (a), (c), (e) do not hold and prove the following implications: a ⇒ b, e, a ⇒ c, d ⇔ c, b ∧ e ⇒ c, d ∧ e ⇒ f , d ∧ f ⇒ e, d ∧ f ⇒ a. As a corollary, if (e) holds (as it happens in the case when all μh and μ are probability measures) we obtain that a ⇔ b ⇔ c ⇔ d ⇒ f .

112 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

6.8. Fourier transform The Fourier transform is a basic tool in Pure and Applied Mathematics, Physics and Engineering. Here we just mention a few basic facts, focussing on the use of this transform in Measure Theory and Probability. Definition 6.29 (Fourier transform of a function). Let f ∈ L 1 (R, C). We set  ˆ f (x)e−i xξ dx ∀ξ ∈ R. f (ξ ) := R

The function fˆ is called Fourier transform of f . Since the map ξ → f (x)e−iξ x is continuous, and bounded by | f (x)|, the dominated convergence theorem gives that fˆ(ξ ) is continuous. The same upper bound also shows that fˆ is bounded, and sup | fˆ| ≤  f 1 . More generally, the following result holds:  Theorem 6.30. Let k ∈ N be such that R |x|k | f |(x) dx < ∞. Then fˆ ∈ C k (R, C) and p f (ξ ) D p fˆ(ξ ) = (−i) p x-

∀ p = 0, . . . , k.

The proof of Theorem 6.30 is a straightforward consequence of the differentiation theorem for integrals depending on a parameter (in this case, the ξ variable, see the Appendix):   p p −i xξ f (x)e dx = Dξ f (x)e−i xξ dx Dξ R R  = (−i) p x p f (x)e−i xξ dx. R

According to the previous result, the Fourier transform allows to transform differentiations (in the ξ variable) into multiplications (in the x variable), thus allowing an algebraic solution of many linear differential equations. In the sequel we need an explicit expression of the Fourier transform of a Gaussian function. For σ > 0, let 2

2

e−|x| /(2σ ) ρσ (x) := (2πσ 2 )1/2

(6.36)

be the rescaled Gaussian functions, already considered in Example 6.25. Then  2 2 ρσ (x)e−iξ x dx = e−ξ σ /2 ∀ξ ∈ R. (6.37) R

The proof of this identity is sketched in Exercise 6.27.

113 Introduction to Measure Theory and Integration

Remark 6.31. (Discrete Fourier transform) If f : R → R is a 2T periodic function, then we can write the Fourier series (corresponding, up to a linear change of variables, to those considered in Chapter 5 for 2π-periodic functions) f =



π

an ein T x ,

in L 2 ((−T, T ); C),

(6.38)

n∈Z

with 1 an = 2T



T

−T

π

f (x)e−in T x dx,

π

ein T x = cos n

π π x + i sin n x. T T (6.39)

Remark 6.32. (Inverse Fourier transform) For g ∈ L 1 (R, C) we define inverse Fourier transform of f the function  1 g(ξ )ei xξ dξ x ∈ R. g(x) ˜ := 2π R It can be shown (see for instance Chapter VI.1 in [7]) that the maps f  → fˆ and g → g˜ are each the inverse of the other in the so-called Schwarz space S(R, C) of smooth and rapidly decreasing functions at infinity: 

∞ k i S(R, C) := f ∈ C (R, C) : lim |x| |D f |(x) = 0 ∀k, i ∈ N . |x|→∞

In particular we have     f aξ ei xξ dξ (x) = f (x) = 2π 2π R  1 aξ : = f (x)e−iξ x dx. 2π R

with

These formulas can be viewed as the continuous counterpart of the discrete Fourier transform (6.38), (6.39). In this sense, aξ are generalized Fourier coefficients, corresponding to the “frequency” ξ . The difference with Fourier series is that any frequency is allowed, not only the integer multiples nπ/T of a given one. 6.8.1. Fourier transform of a measure In this section we are concerned in particular with the concept of Fourier transform of a measure.

114 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Definition 6.33 (Fourier transform of a measure). Let μ be a finite measure on R. We set  e−i xξ dμ(x) ∀ξ ∈ R. μ(ξ ˆ ) := R

The function μˆ : R → C is called Fourier transform of μ. Notice that Definition 6.29 is consistent with Definition 6.33, because μˆ = fˆ whenever μ = f L 1 . Notice also that, by the dominated convergence theorem, the function μˆ is continuous. Moreover μ(0) ˆ = μ(R) and, by estimating from above the modulus of the integral with the integral of the modulus (see also Exercise 6.29), we obtain that |μ(ξ ˆ )| ≤ μ(R) for all ξ ∈ R. Still using the differentiation theorems under the integral sign, one can check that for k ∈ N the following implications hold:  |x|k dμ(x) < ∞ ⇒ μˆ ∈ C k (R, C) and (6.40) R p μ(ξ ) ∀ p = 0, . . . , k. D p μ(ξ ˆ ) = (−i) p xLet us see other basic examples of Fourier transforms of probability measures: Example 6.34. (1) If μ = δx0 then μ(ξ ˆ ) = e−i x0 ξ . (2) If μ = pδ1 + qδ0 (with p + q = 1) is the Bernoulli measure with parameter p, then μ(ξ ˆ ) = q + pe−iξ . (3) If n    n μ= pi q n−i δi i i=0 is the binomial measure with parameters n, p then μ(ξ ˆ ) = (q + pe−iξ )n

∀ξ ∈ R.

(4) If μ = e−x 1(0,∞) (x)L 1 is the exponential measure, then μ(ξ ˆ )=

1 1 + iξ

∀ξ ∈ R.

(5) If μ = (2a)−1 1(−a,a) L 1 is the uniform measure in [−a, a], then μ(ξ ˆ )=

sin(aξ ) aξ

∀ξ ∈ R \ {0}.

(6) If μ = [π(1 + x 2 )]−1 L 1 is the Cauchy measure, then (2) μ(ξ ˆ ) = e−|ξ |

∀ξ ∈ R.

(2) This computation can be done using the residue theorem in complex analysis

115 Introduction to Measure Theory and Integration

Theorem 6.35. Any finite measure μ in R is uniquely determined by its Fourier transform μ. ˆ Proof. For σ > 0 we denote by ρσ the rescaled Gaussian functions in (6.36). According to Exercise 6.27 we have  2 2 ρσ (w)e−i zw dw. e−z σ /2 = R

Setting z = (x − y)/σ , dividing both sides by (2πσ 2 )1/2 we deduce that  1 2 ρσ (x − y) = ρσ (w)e−iw(x−y)/σ dw. 2 1/2 (2πσ ) R 2

Using Fubini-Tonelli theorem we obtain     1 −iw(x−y)/σ 2 ρσ (x − y)dμ(x) = ρσ (w)e dw dμ(x) 2 1/2 R R (2πσ ) R  w ρσ (w) 2 = μ ˆ eiyw/σ dy. 2 )1/2 2 (2πσ σ R (6.41)  As a consequence, the integrals h σ (y) = R ρσ (y−x) dμ(x) are uniquely determined by μ. ˆ But, still using the Fubini-Tonelli theorem, one can check the identity     g(y)ρσ (x − y) dy dμ(x) = h σ (y)g(y) dy ∀g ∈ Cb (R). R

R

R

(6.42) Passing to the limit as σ ↓ 0 and noticing that (by Example 6.25, that provides the weak convergence of ρσ λ to δ0 as σ ↓ 0, or a direct verification)   g(y)ρσ (x − y) dy = g(x − z)ρσ (z) dz → g(x) ∀x ∈ R R

R

from the dominated convergence theorem we obtain that all integrals  g dμ, for g ∈ Cb (R), are uniquely determined. Hence μ is uniquely R determined by its Fourier transform. Remark 6.36. It is also possible to show an explicit inversion formula for the Fourier transform. Indeed, (6.42) holds not only for continuous functions, but also for bounded Borel functions; choosing a < b that are  not atoms of μ and g = 1(a,b) , we have that R g(x)ρσ (x − y) dy → g(x) for μ–a.e. x (precisely for x ∈ / {a, b}), so that (6.42) and (6.41) give  b  b −w2 /2σ 2 e w 2 h σ (y) dy = lim μ( ˆ 2 )eiyw/σ dwdy. μ((a, b)) = lim 2 σ ↓0 a σ ↓0 a R 2πσ σ

116 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

The change of variables w = tσ 2 and Fubini theorem give  1 eitb − eita 2 2 e−t σ /2 μ(t) ˆ μ((a, b)) = lim dt, σ ↓0 2π R it

(6.43)

for all points a < b that are not atoms of μ. According to Theorem 6.28 we have the implication: μh → μ weakly

⇒

μˆ h → μˆ pointwise in R.

(6.44)

The following theorem, due to L´evy, gives essentially the converse implication, allowing to deduce the weak convergence from the convergence of the Fourier transforms. Theorem 6.37 (L´evy). Let (μh ) be probability measures in R. If f h = μˆ h pointwise converge in R to some function f , and if f is continuous at 0, then f = μˆ for some probability measure μ in R and μh → μ weakly. Proof. Let us show first that (μh ) is tight. Fixed a > 0, taking into account that sin ξ is an odd function and using the Fubini theorem we get  a   a  a −i xξ σˆ (ξ ) dξ = e dσ (x)dξ = cos(xξ ) dξ dσ (x) R −a −a −a R  2 sin(ax) dσ (x) = R x for any probability measure σ . Hence, using the inequalities | sin t| ≤ |t| for all t and | sin t| ≤ |t|/2 for |t| ≥ 2, we get   sin(ax) 1 a 1 − σˆ (ξ ) dξ = 2 − 2 dσ (x) a −a ax R    sin(ax) =2 1− dσ (x) (6.45) ax R *  ) 2 2 . ≥ σ R\ − , a a For ε > 0 we can find, by the continuity of f at 0, a > 0 such that  a (1 − f (ξ )) dξ < εa. −a

By the dominated convergence theorem we get h 0 ∈ N such that  a  1 − μˆ h (ξ ) dξ < εa ∀h ≥ h 0 . (6.46) −a

117 Introduction to Measure Theory and Integration

a As a −1 −a (1 − μˆ h (ξ )) dξ → 0 as a ↓ 0 for any fixed h, we infer that we can find b ∈ (0, a] such that (6.46) holds with b replacing a for all h ∈ N. From (6.45) we get μh (R \ [−n, n]) < ε for all h ∈ N, as soon as n > 2/b. Being the sequence tight, we can extract a subsequence (μh(k) ) weakly converging to a probability measure μ and deduce from (6.44) that f = μ. ˆ It remains to show that the whole sequence (μh ) weakly converges to μ: if this is not the case there exist ε > 0, g ∈ Cb (R) and a subsequence h  (k) such that ! !  ! ! ! g dμh  (k) − g dμ!! ≥ ε ∀k ∈ N. ! R

R

But, possibly extracting one more subsequence, we can assume that μh  (k) weakly converge to a probability measure σ ; in particular ! !  ! ! ! ≥ ε > 0. ! g dσ − g dμ (6.47) ! ! R

R

As we are assuming that f h = μˆ h converge pointwise to f = μˆ we ˆ hence μˆ = σˆ . From Theorem 6.35 we obtain that σˆ = limk μˆ h  (k) = μ, obtain that μ = σ , contradicting (6.47). Notice that just pointwise convergences of the Fourier transforms is not enough to conclude the weak convergence, unless we know that the limit function is continuous: let us consider, for instance, the rescaled Gaussian kernels used in the proof of Theorem 6.35 and let us consider the behaviour of the Gaussian measures μσ = ρσ L 1 as σ ↑ ∞; in this case, from Exercise 6.27 we infer that the Fourier transforms are pointwise converging in R to the discontinuous function equal to 1 at ξ = 0 and equal to 0 elsewhere. In this case we don’t have weak convergence of the measures: we have, instead, the so-called phenomenon of dispersion of the whole mass at infinity  n n  lim μσ (R \ [−n, n]) = lim μ1 R \ [− , ] = μ1 (R \ {0}) = 1 σ ↑∞ σ ↑∞ σ σ ∀n ∈ N and the family of measures μσ is far from being tight as σ ↑ ∞.

Exercises 6.27 Check the identity (6.37). 6.28 Show that μˆ is uniformly continuous in R for any finite measure μ.

118 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

6.29 Let μ be a probability measure in R. Show that if |μ| ˆ attains its maximum at ξ0 = 0, then there exist x0 ∈ R and cn ∈ [0, ∞) such that μ=

 n∈Z

cn δxn

with

xn = x0 +

2nπ . ξ0

Use this fact to show that |μ| ˆ ≡ 1 in R if and only if μ is a Dirac mass.

Chapter 7 The fundamental theorem of the integral calculus

In this section we give a closer look at a classical theme, namely the fundamental theorem of the integral calculus, looking for optimal conditions on f ensuring the validity of the formula  x f  (s) ds. f (x) − f (y) = y

Notice indeed that in the classical theory of the Riemann integration there is agap between the conditions imposed to give a meaning to the integx ral a g(s) ds (i.e. Riemann integrability of g) and those that ensure its differentiability as a function of x (for instance, typically one requires the continuity of g). We will see that this gap basically disappears in Lebesgue’s theory, and that there is aprecise characterization of the class x of functions representable as c + a g(s) ds for a suitable (Lebesgue) integrable function g and for some constant c. The following definition is due to Vitali. Definition 7.1 (Absolutely continuous functions). Let I ⊂ R be an interval. We say that f : I → R is absolutely continuous if for any ε > 0 there exists δ > 0 for which the implication n  (bi − ai ) < δ

⇒

i=1

n 

| f (bi ) − f (ai )| < ε

(7.1)

i=1

holds for any finite family {(ai , bi )}1≤i≤n of pairwise disjoint intervals contained in I . An absolutely continuous function is obviously uniformly continuous, but the converse is not true, see Example 7.7. Let f : [a, b] → R be absolutely continuous. For any x ∈ [a, b] define F(x) = sup

n 

σ ∈a,x i=1

| f (xi ) − f (xi−1 )|,

L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

120 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

where a,x is the set of all decompositions σ = {a = x0 < x1 < · · · < xn = x} of [a, x]. F is called the total variation of f . Let us check that F is finite: let δ > 0 be satisfying the implication (7.1) with ε = 1 and let us estimate from above a sum in the definition of F. Without loss of generality we can assume that |xi − xi−1 | < δ/2 for all i = 1, . . . , n − 1, possibly adding more points (which increases the sum). Then, we can split the sum in families of intervals with total length larger than δ/2 and less than δ (just keep adding a new interval to a family if the total length does not exceed δ and notice that if it exceeds δ, the total length is at least δ/2); the number of these families is less than 2δ (x − a) and, as a consequence, (7.1) gives 2 F(x) ≤ (x − a) + 1. δ We set f + (x) =

1 (F(x) + f (x)), 2

f − (x) =

1 (F(x) − f (x)), 2

so that f (x) = f + (x) − f − (x),

F(x) = f + (x) + f − (x),

x ∈ [a, b].

Lemma 7.2. Let f : [a, b] → R be absolutely continuous and let F be its total variation. Then F, f + , f − are nondecreasing and absolutely continuous. Proof. Let x ∈ [a, b), y ∈ (x, b] and σ = {a = x0 < x1 < · · · < xn = x}. Then we have F(y) ≥ | f (y) − f (x)| +

n 

| f (xi ) − f (xi−1 )|.

i=1

Taking the supremum over all σ ∈ a,x , yields F(y) ≥ | f (y) − f (x)| + F(x), which implies that F, f + , f − are nondecreasing. It remains to show that F is absolutely continuous. Let ε > 0 and let δ = δ(ε) > 0 be such that the implication (7.1)  holds for all finite families (ai , bi ) of pairwise disjoint intervals with  i (bi − ai ) < δ. Let now (ai , bi ) bea family of disjoint intervals with i (bi − ai ) < δ and let us prove that i |F(bi ) −

121 Introduction to Measure Theory and Integration

F(ai )| < 2ε. For any i = 1, . . . , n we can find σi = {ai = x0,i < x1,i < · · · < xni ,i = bi } such that F(bi ) − F(ai ) <

ni ε  | f (xk,i ) − f (xk−1,i )|, + n k=1

1 ≤ i ≤ n. (7.2)

Indeed, if a = y0 < y1 < · · · < ym i = bi is a partition such that F(bi ) <

mi ε  | f (yk ) − f (yk−1 )| + n k=1

we can assume with no loss of generality (adding one more element to the partition if necessary) that yk = ai for some k; then, it suffices to estimate the first k terms of the above sum with F(ai ), and to call x0,i = yk , . . . , xm i −k+1,i = ym i to obtain (7.2) with n i = m i − k + 1. Adding the inequalities (7.2) and taking into account that the union of the disjoint intervals (xk,i−1 , xk,i ) (for 1 ≤ i ≤ n, 0 ≤ k ≤ n i ) has length less than δ, from the absolute continuity property of f we get n  (F(bi ) − F(ai )) < ε + ε = 2ε. i=1

This proves that F is absolutely continuous. The absolute continuity property characterizes integral functions, as the following theorem shows. Theorem 7.3. Le I = [a, b] ⊂ R. A function f : I → R is representable as  x f (x) = f (a) + g(t) dt ∀x ∈ I (7.3) a

for some g ∈ L 1 (I ) if and only if f is absolutely continuous. Proof. (Sufficiency) If f is representable as in (7.3), we have  y | f (x) − f (y)| ≤ |g(s)| ds ∀x, y ∈ I, x ≤ y. x

Hence, setting A = ∪i (ai , bi ), the absolute continuity property follows by the implication  ⇒ |g| ds < ε. L 1 (A) < δ A

122 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

The existence, given δ > 0, of ε > 0 with this property is ensured by Exercise 6.14 (with μ = L 1 and ν = gL 1 ). (Necessity) According to Lemma 7.2, we can write f as the difference of two nonincreasing absolutely continuous functions. Hence, we can assume with no loss of generality that f is nonincreasing, and possibly adding to f a constant we shall assume that f (a) = 0. We extend f to the whole of R setting f ≡ 0 in (−∞, a) and f ≡ f (b) in (b, ∞). It is clear that this extension, that we still denote by f , retains the monotonicity and absolute continuity properties. By Theorem 6.23 we obtain a unique finite measure ν on (R, B (R)) without atoms (because f is continuous) such that f is the repartition function of ν. Since f is constant on (−∞, a) and on (b, +∞), we obtain that ν is concentrated on I , so that f (x) = ν ((−∞, x]) = ν ((a, x])

∀x ∈ R.

(7.4)

Now, if we were able to show that ν # 1 I L 1 , by the Radon–Nikodym theorem we would find g ∈ L 1 (I ) such that ν = gL 1 , so that (7.4) would give  x

f (x) =

g(s) ds

∀x ∈ I.

a

Hence, it remains to show that ν # 1 I L 1 . Taking into account the identity ν((a, b)) = f (b) − f (a), the absolute continuity property can be rewritten as follows: for any ε > 0 there exists δ > 0 such that L 1 (A) < δ

⇒

ν(A) ≤ ε

for any finite union of open intervals A ⊂ I . But, by approximation, the same implication holds for all open sets, because any such set is the countable union of open intervals. By Proposition 1.24, ensuring an approximation from above with open sets, the same implication holds for Borel sets B ⊂ I as well. This proves that ν # 1 I L 1 and concludes the proof. We will need the following nice and elementary covering theorem. Theorem 7.4 (Vitali covering theorem). Let {Bri (xi )}i∈I be a finite family of balls in a metric space (X, d). Then there exists J ⊂ I such that the balls {Bri (xi )}i∈J are pairwise disjoint, and

Bri (xi ) ⊂ B3ri (xi ). (7.5) i∈I

i∈J

123 Introduction to Measure Theory and Integration

Proof. We proceed as follows: first we pick a ball with largest radius, then we remove all balls that intersect the first chosen ball and choose a second ball of largest radius among the remaining ones. We continue removing all balls that intersect the second chosen ball and picking a third ball of largest radius among the remaining ones, and so on. The process stops when either there is no ball left, i.e. when the remaining balls intersect at least one of the already chosen balls. The family of chosen balls is disjoint by construction. If x ∈ Bri (xi ) and the ball Bri (xi ) has not been chosen, then there is a chosen ball Br j (x j ) intersecting it, so that d(xi , x j ) < ri + r j . Moreover, if Br j (x j ) is the first chosen ball with this property, then r j ≥ ri (otherwise, if ri > r j , either the ball Bri (xi ) or a ball with larger radius would have been chosen, instead of Br j (x j )), so that d(xi , x j ) < 2r j . It follows that d(x, x j ) ≤ d(x, xi ) + d(xi , x j ) < ri + 2r j ≤ 3r j . As x is arbitrary, this proves (7.5). It is natural to think that the function g in (7.3) is, as in the classical fundamental theorem of integral calculus, the derivative of f . This is true, but far from being trivial, and it follows by the following weak continuity result (due to Lebesgue) of integrable functions. We state the result even in more then one variable, as the proof in this case does not require any extra difficulty. Theorem 7.5 (Continuity in mean). Let f ∈ L 1 (Rn ). Then, for L n – a.e. x ∈ Rn we have  1 | f (y) − f (x)| dy = 0. lim r↓0 ωn r n B (x) r The terminology “continuity in mean” can be explained as follows: it is easy to show that the integral means  1 f (y) dy ωn r n Br (x) of a continuous function f converge to f (x) as r ↓ 0 for any x ∈ Rn , because they belong to the interval [min f, max f ]. B r (x)

B r (x)

The previous theorem tells us that the same convergence occurs, for L n – a.e. x ∈ Rn , for any integrable function f . This simply follows by the

124 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

inequality ! !  ! ! 1 != 1 ! f (y) dy − f (x) ! !ω rn ωn r n n Br (x) 1 ≤ ωn r n

! ! ! ! ! ! f (y) − f (x) dy ! ! Br (x)  | f (y) − f (x)| dy. Br (x)

By the local nature of this statement, the same property holds for locally integrable functions. Proof of Theorem 7.5. Given ε, δ > 0 and an open ball B = B R (0), it suffices to check that the set    1 | f (y) − f (x)| dy > 2ε A := x ∈ B : lim sup ωn r n Br (x) r↓0 has Lebesgue measure less than (3n + 1)δ. To this aim, we write f as the sum of a “good” part g and a “bad”, but small, part h, i.e. f = g + h with g : B → R bounded and continuous, and h L 1 (B  ) < εδ, with B  = B R+1 (0); this decomposition is possible, because Proposition 3.16 ensures the density of bounded continuous functions in L 1 (B). The continuity of g gives  1 |g(y) − g(x)| dy = 0 ∀x ∈ B. lim r↓0 ωn r n B (x) r Hence, as f = g + h, we have A ⊂ A1 , where    1 A1 := x ∈ B : lim sup |h(y) − h(x)| dy > 2ε . ωn r n Br (x) r↓0 Then, it suffices to show that L n (A1 ) ≤ (3n + 1)δ. By the triangle inequality, we have also A1 ⊂ A2 ∪ A3 with A2 := {x ∈ B : |h(x)| > ε} and



1 A3 := x ∈ B : sup n r∈(0,1) ωn r



 Br (x)

|h(y)| dy > ε .

Markov inequality ensures that L n (A2 ) ≤ h L 1 (B) /ε < δ, so that we n n need only to show  that L (A3 ) ≤ 3 δ. Since x  → Br (x) |h(y)| dy is continuous, we have that  1 x → sup |h(y)| dy n r∈(0,1) ωn r Br (x)

125 Introduction to Measure Theory and Integration

is lower semi continuous, hence A3 is open. Notice also that for any x ∈ A3 there exists r ∈ (0, 1), depending on x, such that  |h(y)| dy > εωn r n . Br (x)

Let K ⊂ A3 be a compact set and let {B(xi , ri )}i∈I be a finite family of these balls whose union covers K . By applying Vitali’s covering theorem to this family of balls, we can find a disjoint subfamily {Bri (xi )}i∈J such that the union of the enlarged balls B3ri (xi ) still covers K . Adding the previous inequalities with x = xi and r = ri and summing in i ∈ J , since all balls Bri (xi ) are contained in B  we get   3n  n n L (K ) ≤ ωn (3ri ) ≤ |h(y)| dy ε i∈J Bri (xi ) i∈J  3n |h(y)| dy ≤ 3n δ. ≤ ε B As K is arbitrary we obtain that L n (A3 ) ≤ 3n δ. Since the continuity in mean is a local property, it is not difficult to extend the previous result to locally integrable functions. By applying this extended theorem to a characteristic function f = 1 E we get ⎧ L n (E ∩ Br (x)) ⎪ ⎪ = 1 for L n –a.e. x ∈ E ⎨lim r↓0 ω rn n

⎪ L n (E ∩ Br (x)) ⎪ ⎩lim = 0 for L n –a.e. x ∈ Rn \ E r↓0 ωn r n for any E ∈ B (Rn ); points of the first type are called density points, whereas points of the second type are called rarefaction points. Using the continuity in mean of integrable functions we obtain the fundamental theorem of calculus within the (natural) class of absolutely continuous functions. Theorem 7.6. Let I ⊂ R be an interval and let f : I → R be absolutely continuous. Then f is differentiable at L 1 –a.e. point of I . In addition f  is Lebesgue integrable in I and  x f (x) = f (a) + f  (s) ds ∀x ∈ I. (7.6) a

Proof. Let g be as in (7.3), let x0 ∈ I be a point where  1 x0 +r lim |g(s) − g(x0 )| ds = 0 r↓0 r x −r 0

(7.7)

126 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

and notice that f (x0 + r) − f (x0 ) 1 = r r



x0 +r

g(s) ds x0

= g(x0 ) +

1 r



x0 +r

g(s) − g(x0 ) ds

x0

for r > 0. Hence, passing to the limit as r ↓ 0, from (7.7) we get f + (x0 ) = g(x0 ); a similar argument shows that f − (x0 ) = g(x0 ). As, according to the previous theorem, L 1 –a.e. point x0 satisfies (7.7), we obtain that f is differentiable, with derivative equal to g, L 1 –a.e. in I . It suffices to replace g with f  in (7.3) to obtain (7.6). One might think that differentiability L 1 –a.e. and integrability of the derivative are sufficient for the validity of (7.6) (these are the minimal requirements to give a meaning to the formula). However, this is not true, as the Heaviside function 1(0,∞) fulfils these conditions but fails to be (absolutely) continuous. Then, one might think that one should require also the continuity of f to have (7.6). It turns out that not even this is enough: we build in the next example the Cantor-Vitali function, also called devil’s staircase: a continuous function having derivative equal to 0 L 1 –a.e., but not constant. This example shows why a stronger condition, namely the absolute continuity, is needed. Example 7.7 (Cantor–Vitali function). Let X := { f ∈ C([0, 1]) : f (0) = 0, f (1) = 1} . This is a closed subspace of the complete metric space C([0, 1]), hence X is complete as well. For any f : [0, 1] → R we set ⎧ ⎪ if 0 ≤ 3x ≤ 1, ⎨ f (3x)/2 T f (x) := 1/2 (7.8) if 1 < 3x < 2, ⎪ ⎩ 1/2 + f (3x − 2)/2 if 2 ≤ 3x ≤ 3. It is easy to see that T maps X into X, and that T is a contraction (with Lipschitz constant equal to 1/2). Hence, by the contraction principle, there is a unique f ∈ X such that T f = f . Let us check that f has zero derivative L 1 –a.e. in [0, 1]. As f = T f , f is constant, and equal to 1/2, in (1/3, 2/3). Inserting this information again in the identity f = T f we obtain that f is locally constant (equal to 1/4 and to 3/4) on (1/9, 2/9) ∪ (7/9, 8/9). Continuing in this way, one finds that f is locally constant on the union of 2n−1 intervals, each of length 3−n , n ≥ 1. The complement C = [0, 1] \ A of the union A

127 Introduction to Measure Theory and Integration

of these intervals is Cantor’s middle third set (see also Exercise 1.8), and since ∞ ∞  n  2 2n−1 1 L 1 (A) = = =1 n 3 2 3 n=1 n=1 we know that L 1 (C) = 0. At any point of A the derivative of f is obviously 0. In connection with the previous example, notice also that f maps A, a set of full Lebesgue measure in [0, 1], into the countable set {2−n }n≥1 . On the other hand, it maps C, a Lebesgue negligible set, into [0, 1], a set with strictly positive Lebesgue measure.

Exercises 7.1 Let H : R → R be satisfying the Lipschitz condition |H (x) − H (y)| ≤ C|x − y|

∀x, y ∈ R

and let f : [a, b] → R be an absolutely continuous function. Show that H ◦ f is absolutely continuous in [a, b]. 7.2 Let E ⊆ R be a Borel set and assume that any t ∈ R is either a point of density or a point of rarefaction of E. Show that either L 1 (E) = 0 or L 1 (R \ E) = 0. (Remark: the same result is true in Rn , but with a much harder proof, see [3], 4.5.11). 7.3[Lipschitz change of variables] Let f : I = [a, b] → R be absolutely continuous (resp. Lipschitz). Show that 

f (b) f (a)

 ϕ(y) dy =

a

b

ϕ( f (x)) f  (x) dx

for any bounded (resp. integrable) Borel function ϕ : f (I ) → R. 7.4 Use the previous exercise to show that, for any Lipschitz function f : R → R and any L 1 –negligible set N ∈ B (R), the derivative f  vanishes L 1 –a.e. on f −1 (N ).

Chapter 8 Measurable transformations

In this chapter we study the classical problem of the change of variables in the integral from a new viewpoint. We will compute how the Lebesgue measure in Rn changes under a sufficiently regular transformation, generalizing what we have already seen for linear, or affine, maps. As a byproduct we obtain a quite general change of variables formula for integrals with respect to the Lebesgue measure.

8.1. Image measure We are given two measurable spaces (X, E ) and (Y, F ), a measure μ on (X, E ) and a (E , F )–measurable mapping F : X → Y . We define a measure F# μ in (Y, F ) by setting F# μ(I ) := μ(F −1 (I )),

I ∈ F.

(8.1)

It is easy to see that F# μ is well defined, by the measurability assumption on F, and σ -additive on F . F# μ is called the image measure of μ by F. The following change of variable formula is simple, but of a basic importance. Proposition 8.1. Let ϕ : Y → [0, ∞] be a F –measurable function. Then we have   ϕ(F(x)) dμ(x) = ϕ(y) d F# μ(y). (8.2) X

Y

Proof. By monotone approximation it is enough to prove (8.2) when ϕ is a simple function. By linearity of both sides we need only to consider functions ϕ of the form ϕ = 1 I , where I ∈ F . In this case we have ϕ ◦ F = 1 F −1 (I ) , hence (8.2) reduces to (8.1). In the following example we discuss the relation between the change of variables formula (8.2), that even on the real line involves no derivative, and the classical one. The difference is due to the fact that in (8.2) we are not using the density of F# μ with respect to L 1 . It is precisely in this density that the derivative of F shows up. L. Ambrosio et al., Introduction to Measure Theory and Integration © Scuola Normale Superiore Pisa 2011

130 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Example 8.2. Let F : R → R be of class C 1 and such that F  (t) > 0 for all t ∈ R. Let A be the image of F (an open interval, by the assumptions made on F) and let ψ : A → R be continuous. Then for any interval [a, b] ⊂ A the following elementary formula of change of variables holds (just put y = F(x) in the right integral):  b  F −1 (b) 1 ψ(F(x)) dx = ψ(y)  −1 dy. F (F (y)) F −1 (a) a On the other hand, choosing ϕ = ψ 1 I with I = [a, b] in (8.2), we have  b  F −1 (b) ψ(F(x)) dx = ψ(y) d F# L 1 . F −1 (a)

a

Hence, comparing the two expressions, we find  b  b 1 ψ(y)  −1 ψ(y) d F# L 1 . dy = F (F (y)) a a

(8.3)

Since a, b and ψ are arbitrary, (8.3) can be interpreted by saying that F# L 1 # L 1 and 1 F# L 1 =  L 1. F ◦ F −1 In the next section, we shall generalize this formula to Rn , and even in one space dimension we will see that the assumption that F  > 0 everywhere can be weakened (see also Exercise 8.3).

8.2. Change of variables in multiple integrals We consider here the measure space (Rn , B (Rn ), L n ), where L n is the Lebesgue measure. We recall a few basic facts from calculus with several variables: given an open set U ⊂ Rn and a mapping F : U → Rn , F is said to be differentiable at x ∈ U if there exists a linear operator D F(x) ∈ L(Rn ; Rn ) (1) such that |F(x + h) − F(x) − D F(x)h| lim = 0. |h|→0 |h| The operator D F(x) if exists is unique, and is called the differential of F at x. If F is affine, i.e. F(x) = T x + a for some T ∈ L(Rn ; Rn ) and a ∈ Rn , we have D F(x) = T for all x ∈ U .

(1) L(Rn ; Rm ) is the Banach space of all linear mappings T : Rn → Rm endowed with the sup norm T  = sup{|T x| : x ∈ Rn , |x| = 1}

131 Introduction to Measure Theory and Integration

If F is differentiable at x ∈ U we define the Jacobian determinant JF (x) of F at x by setting JF (x) = det D F(x). If F is differentiable at any x ∈ U and if the mapping D F : U → L(Rn ; Rn ) is continuous, we say that F is of class C 1 . If, in addition, F is bijective between U and an open domain A and F −1 is of class C 1 in A, we say that F is a C 1 diffeomorphism of U onto A. In this case we have that D F(x) is invertible and D(F −1 )(F(x)) = (D F(x))−1

∀x ∈ U.

Finally, by Proposition 6.10 we know that if T ∈ L(Rn ; Rn ) we have L n (T (E)) = | det T | L n (E)

∀E ∈ B (Rn ).

(8.4)

8.3. Image measure of L n by a C 1 diffeomorphism In this section we study how the Lebesgue measure changes under the action of a C 1 map F. The relevant quantity will be the function |JF |, which really corresponds to the distorsion factor of the measure. Let U ⊂ Rn be open. The critical set C F of F ∈ C 1 (U ; Rn ) is defined by C F := {x ∈ U : JF (x) = 0} . Lemma 8.3. The image F(C F ) of the critical set is Lebesgue negligible. Proof. Let K ⊂ C F be a compact set and ε > 0; for any x ∈ K the set D F(x)(B 1 (0)) is Lebesgue negligible (because D F is singular at x, so that D F(x)(Rn ) is contained in a (n − 1)-dimensional subspace of Rn ), hence we can find δ = δ(ε, x) > 0 such that  L n {z ∈ Rn : dist (z − F(x), D F(x)(B 1 (0))) < δ} < ε. By a scaling argument we get  L n {z ∈ Rn : dist (z − F(x), D F(x)(B r (0))) < δr} < εr n

∀r > 0.

On the other hand, since |F(y) − F(x) − D F(x)(y − x)| < δr in Br (x), provided r is small enough, we get   F(Br (x)) ⊂ z ∈ Rn : dist (z − F(x), D F(x)(B r (0)) < δr . It follows that Br (x) ⊂ U and L n (F(Br (x))) < εr n for r > 0 small enough, depending on x.

132 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Since the family of balls {Br/3 (x)}x∈K covers the compact set K , we can find a finite family {Bri /3 (xi )}i∈I whose union still covers K and extract from it, thanks to Vitali’s covering theorem, a subfamily {Bri /3 (xi )}i∈J made by pairwise disjoint balls such that the union of the enlarged balls {Bri (xi )}i∈J covers K . In particular, covering F(K ) by the union of F(Bri (xi )) for i ∈ J , we get  3n ε   ri n 3n ε n L n (F(K )) ≤ εrin = ωn ≤ L (U ). ωn i∈J 3 ωn i∈J Letting ε ↓ 0 we obtain that L n (F(K )) = 0. Since K is arbitrary, by approximation (recall that C F , being a closed subset of U , can be written as the countable union of compact subsets of U ) we obtain that L n (F(C F )) = 0. The following theorem provides a necessary and sufficient condition for the absolute continuity of F# L n with respect to L n , assuming a C 1 regularity of F. Theorem 8.4. Let U ⊂ Rn be an open set and let F : U → Rn be of class C 1 , whose restriction to U \ C F is injective. Then: (i) F# (1U L n ) is absolutely continuous with respect to L n if and only if C F is Lebesgue negligible. (ii) If F# (1U L n ) # L n we have F# (1U L n ) =

1 1 L n. |JF |(F −1 ) F(U \C F )

(8.5)

Proof. (i) If L n (C F ) > 0, we have F# (1U L n )(F(C F )) ≥ L n (C F ) > 0 and F# (1U L n ) fails to be absolutely continuous with respect to L n , because we proved in Lemma 8.3 that F(C F ) is Lebesgue negligible. Let G be the inverse of the restriction of F to the open set U \ C F . The local invertibility theorem ensures that the domain A = F(U \C F ) of G is an open set, that G is of class C 1 in A and that DG(y) = (D F)−1 (G(y)) for all y ∈ A. Let us assume now that C F is Lebesgue negligible and let us show that F −1 (E) is Lebesgue negligible whenever E ⊂ F(U ) is Lebesgue negligible. Since we already know that C F is L n –negligible set, we can assume with no loss of generality that E ⊂ A and show that G(E) is Lebesgue negligible. Let A M be the open sets A M := {y ∈ A : DG(y) < M} . We will prove that L n (G(K )) ≤ (3M)n L n (K )

(8.6)

133 Introduction to Measure Theory and Integration

for any compact set K ⊂ A M . So, F# L n ≤ (3M)n L n on the compact sets of A M and therefore on the Borel sets; in particular L n (G(E ∩ A M )) ≤ (3M)n L n (E ∩ A M ) = 0, and letting M ↑ ∞ we obtain that L n (G(E)) = 0, because E ⊂ A. In order to show (8.6) we consider a bounded open set B contained in A M and containing K , and the family of balls Br (y) ⊂ B with y ∈ K and r > 0. For any of these balls the mean value theorem gives (with t = t (y, z) ∈ (0, 1)) |G(z)−G(y)| = |DG((1−t)y +t z)(z − y)| ≤ M|z − y|

∀z ∈ Br (y),

therefore G(Br (y)) ⊂ B Mr (G(y)) for any of these balls. Since the family of balls {Br/3 (y)} y∈F covers K , we can find a finite family {Bri /3 (yi )}i∈I whose union still covers K and extract from it, thanks to Vitali’s covering theorem, a subfamily {Bri /3 (yi )}i∈J made by pairwise disjoint balls such that the union of the enlarged balls {Bri (yi )}i∈J covers K . In particular, by our choice of the radii of the balls, the family {B Mri (G(yi ))}i∈J covers G(K ). We have then   ri n  L n (G(K )) ≤ ωn (Mri )n = (3M)n ωn ≤ (3M)n L n (B). 3 i∈J i∈J Letting B ↓ K we obtain (8.6). Let us prove (ii). We denote by h the Radon–Nikodym derivative of F# (1U L n ) with respect to L n ; by Theorem 7.5 we have that  1 L n (G(Br (y))) h(z) dz = lim , h(y) = lim r↓0 ωn r n r↓0 ωn r n Br (y) for L n –a.e. y ∈ A. Taking into account that F# (1U L n ) is concentrated on A, and that 1/|JF | ◦ F −1 = |JG |, it remains to prove that for all y0 ∈ A we have lim r↓0

L n (G(Br (y0 ))) = |JG |(y0 ). ωn r n

(8.7)

For the sake of simplicity we only consider the case when y0 = 0 and G(0) = 0 (this is not restrictive, up to a translation in the domain and in the codomain). We divide the rest of the proof in two steps. Step 1. We assume in addition that DG(0) = I and show that lim r↓0

L n (G(Br (0))) = 1, ωn r n

which is equivalent to (8.7) in this case.

(8.8)

134 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Since D F(0) = DG(0) = I we have by the definition of derivative, |F(x) − x| = 0, |y|→0 |x| lim

|G(y) − y| =0 |y|→0 |y| lim

So, for any ε ∈ (0, 1) there exists δε > 0 such that if |x| < δε we have x ∈ U \C F and |F(x)−x| < ε|x| and if |y| < δε we have y ∈ F(U \C F ) and |G(y) − y| < ε|y|. It follows that r < δε implies |F(x)| < r

∀x ∈ B(1−ε)r (0),

|G(y)| < (1 + ε)r

∀y ∈ Br (0). (8.9)

In particular B(1−ε)r (0) ⊂ G(Br (0)) ⊂ B(1+ε)r (0)

∀r < δε .

(8.10)

Now, by (8.10) it follows that (1 − ε)n ≤

L n (G(Br (0))) ≤ (1 + ε)n , ωn r n

provided r < δε , and this proves that (8.8) holds. Step 2. Set T = DG(0) and H (x) = T −1 G(x), so that D H (0) = I . Then we have G(Br (0)) = T (H (Br (0))) and so, thanks to (8.4), L n (G(Br (0))) = L n (T (H (Br (0)))) = | det T | L n (H (Br (0))), which implies lim r↓0

L n (G(Br (0))) L n (H (Br (0))) = | det T | lim = | det T |. r↓0 ωn r n ωn r n

The proof is complete. Example 8.5 (Polar and spherical coordinates). Let us consider the polar coordinates (ρ, θ) → (ρ cos θ, ρ sin θ). Here U = (0, ∞) × (0, 2π) and the critical set is empty, because the modulus of the Jacobian determinant is ρ. In the case of the spherical coordinates (ρ, θ, φ) → (ρ cos θ sin φ, ρ sin θ sin φ, ρ cos φ) we have U = (0, ∞) × (0, 2π) × (0, π) and the critical set is empty, because the modulus of the Jacobian determinant is −ρ 2 sin φ.

135 Introduction to Measure Theory and Integration

Theorem 8.6 (Change of variables formula). Let U ⊂ Rn be an open set and let F : U → Rn of class C 1 , injective on U \ C F . Then   ϕ(y) dy = ϕ(F(x))|JF |(x) dx (8.11) F(U )

U

for any Borel function ϕ : F(U ) → [0, +∞]. Proof. We first see that it is not restrictive to assume that C F = ∅; indeed, F(C F ) is Lebegue negligible and so images of points in C F do not affect the left hand side, while obviously points in C F do not affect the right hand side. So, possibly replacing U with U \ C F , we can assume that C F = ∅. By (8.2) and (8.5) we have   ψ(y) ψ(F(x)) dx. dy = −1 F(U ) |J F |(F (y)) U for any nonnegative Borel function ψ. We conclude choosing ψ(y) = ϕ(y)|JF |(F −1 (y)).

Exercises 8.1 Let (X, F ), (Y, G ) and (Z , H ) be measurable spaces and let f : X → Y , g : Y → Z be measurable maps. Show that g# ( f # μ) = (g ◦ f )# μ for any measure μ in (X, F ). N → [0, 1] be the map associating to a sequence (a ) ⊂ {0, 1} 8.2 Let f : {0, 1} i the real number i ai 2−i−1 ∈ [0, 1]. Show that

f#

 1 1 ( δ0 + δ1 ) = 1[0,1] L 1 . 2 i=0 2 ∞

×

8.3 Show the existence of a strictly increasing and C 1 function F : R → R such that F# L 1 is not absolutely continuous with respect to L 1 . 8.4 Remove the injectivity assumption in Theorem 8.4, showing that F# (1U L n ) =

 x∈F −1 (y)\C F

1 L n. 1 |J F|(x) F(U \C F )

for any C 1 function F : U → Rn with Lebesgue negligible critical set.

Appendix A

A.1. Continuity and differentiability of functions depending on a parameter In this section we consider the following problem: we are given a metric space (X, d) and a measure space (Y, F , μ). Given f : X × Y → R, we assume that for all x ∈ X the function f (x, ·) is μ–integrable, so that the function F : X → R given by  f (x, y) dμ(y) x∈X F(x) := Y

is well defined. We would like to understand under which conditions F, an integral depending on the parameter x, is continuous. When X is an open subset of Rn endowed with the Euclidean distance, it is also natural to investigate the differentiability properties of F. Theorem A.1 (Continuity of F). Assume that f (·, y) is continuous in X for μ-almost all y ∈ Y and that there exists m ∈ L 1 (Y, μ) satisfying sup | f (x, y)| ≤ m(y) x∈X

for μ–a.e. y ∈ Y .

(A.1)

Then F is bounded and continuous in X. Proof. It is clear that |F(x)| ≤ m1 for all x ∈ X. Continuity is a simple consequence of the dominated convergence theorem: indeed, if xn ∈ X converge to x, then f (xn , y) converge to f (x, y) for μ-almost every y and the convergence is dominated because of (A.1). It follows that F(xn ) → F(x). A more expressive way to state the continuity of F is to say that limit and integral commute, namely   lim f (x h , y) dμ(y) = lim f (x h , y) dμ(y). h→∞ Y

Y h→∞

138 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

The following example shows that if no uniform upper bound is imposed on f , then continuity might fail: Example A.2. Let X = Y = R, μ = L 1 and ⎧ ⎪ ⎨|x|(1 − |y||x|) if |y||x| < 1; f (x, y) := ⎪ ⎩ 0 if |y||x| ≥ 1. Then F(x) = 1 for all x = 0, while F(0) = 0. In this case the smallest possible function satisfying (A.1) is |y|−1 which is not integrable. Next, we assume that X is an open set of Rn endowed with the Euclidean distance and we investigate the differentiability of F. Under suitable assumption, we can commute derivative and integral, namely   ∂f ∂ f (x, y) dμ(y) = (x, y) dμ(y) ∀x ∈ X, i = 1, . . . , n. ∂ xi Y Y ∂ xi (A.2) Theorem A.3 (Differentiability of F). Assume that for μ-almost all y ∈ Y the function f (·, y) is differentiable in X with a continuous gradient ∇x f (x, y) and that, for any ball Br (x0 ) ⊂ X, there exists m ∈ L 1 (Y, μ) satisfying | f (x0 , y)| + sup |∇x f |(x, y) ≤ m(y) x∈Br (x0 )

for μ–a.e. y ∈ Y . (A.3)

Then F ∈ C 1 (X) and (A.2) holds. Proof. We fix x0 ∈ X, i ∈ {1, . . . , n} and xi = x + ti ei with ti  = 0 and ti → 0. The mean value theorem, applied for any y such that f (·, y) ∈ C 1 (X), gives θi (y) ∈ (0, 1) satisfying  ∂f F(x0 + ti ei ) − F(x0 ) (x0 + θi (y)ti ei , y) dμ(y) = ti Y ∂ xi For i large enough (as soon as |ti | < r) the functions of y inside the integral are dominated by the function m in (A.3), hence we can pass to the limit with the dominated convergence theorem to get (notice that the measurability of ∂ f /∂ xi (x0 , ·) follows by the same limiting process)  ∂F ∂f (x0 ) = (x0 , y) dμ(y). ∂ xi Y ∂ xi Finally, continuity of partial derivatives of F is a consequence of the previous theorem.

139 Introduction to Measure Theory and Integration

Of course similar statements can be given for k-th order derivatives of F, provided f (·, y) is k times differentiable and, for any ball Br (x0 ) ⊂ Rn there exists m ∈ L 1 (Y, μ) satisfying | f (x0 , y)| + sup

sup |D p f |(x, y) ≤ m(y)

x∈Br (x0 ) | p|≤k

for μ–a.e. y ∈ Y

(here p = ( p1 , . . . , pn ) and | p| = p1 + · · · + pn ). Under this assumption one obtains that   p f (x, y) dμ(y) = Dxp f (x, y) dμ(y) whenever | p| ≤ k. D Y

Y

A.2. The dual space of continuous functions In this section we want to characterize the space (C(X))∗ , dual space of C(X), with (X, d) compact metric space. Recall that C(X) is a Banach space, when endowed with the sup norm, regardless of any assumption on (X, d). Some knowledge of the basic terminology of Banach spaces (dual space, dual norm) is needed for this section. We start with some notation: we shall denote by M (X) the space of signed measures μ, i.e. the real-valued and σ -additive set functions μ, defined on B (X), of the form μ = μ+ − μ− with μ± positive and finite Borel measures satisfying μ+ ⊥ μ− . This orthogonality condition ensures uniqueness of the decomposition of μ, as we will see in a moment; existence, instead, is just a consequence of the σ -additivity (see Section 6.5), but we shall not use this fact in the sequel. For μ ∈ M (X) we denote |μ| = μ+ + μ− its total variation measure, as in Section 6.5, and set μ := |μ|(X) = μ+ (X) + μ− (X).

(A.4)

In the next proposition we show that the decomposition μ = μ+ − μ− is unique, so that (A.4) is well posed, and that M (X) is a normed space. The completeness of M (X) will be a consequence of Theorem A.6, since any dual space is complete. Proposition A.4. For any μ ∈ M (X) the decomposition μ = μ+ − μ− is unique. In addition M (X), endowed with the norm (A.4), is a normed space. Proof. Assume that μ = μ+ − μ− = μ˜ + − μ˜ − , with orthogonal decompositions. Let A be a Borel set where μ+ is concentrated, so that μ− is concentrated on X \ A, and let A˜ be an analogous Borel set for μ˜ ± . Since

140 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

μ ≥ 0 (respectively μ ≤ 0) on the subsets of A (respectively of X \ A) ˜ we obtain that μ (and therefore μ± and the same property holds for A, ± and μ˜ ) vanishes on subsets of A \ A˜ and of A˜ \ A. On the other hand, if B ⊂ A ∩ A˜ we have μ− (B) = μ˜ − (B) = 0 and μ+ (B) = μ(B) = μ˜ + (B). ˜ we have μ+ (B) = μ˜ + (B) = 0 Analogously, if B ⊂ (X \ A) ∩ (X \ A) − − and μ (B) = μ˜ (B). This proves that μ± = μ˜ ± . Now, stability of M (X) under multiplication with real constants and 1-homogeneity of the norm are obvious. Let us prove stability under addition and subadditivity of the norm: if μ = μ+ −μ− and ν = ν + −ν − we can write as before μ = f |μ| and ν = g|ν| with f, g : X → [−1, 1]. Then, setting σ = |μ|+|ν|, the Radon–Nikod´ym theorem gives |μ| = aσ and |ν| = bσ for suitable a, b : X → [0, 1], so that μ + ν = f |μ| + g||ν| = ( f a + gb)σ and we may take ( f a + gb)± σ as positive and negative parts of μ + ν. We obtain also   | f a + gb| dσ ≤ |a| + |b| dσ = μ + ν. μ + ν = X

X

This completes the proof of the proposition. We shall also denote by A (X) the collection of open subsets of X and use the following characterization of set functions defined on A (X) which are restrictions of σ -additive measures defined on the Borel σ algebra. Proposition A.5. Let (X, d) be a compact metric space and let α : A (X) → [0, +∞] be a nondecreasing set function satisfying α(∅) = 0 and: (i) (continuity) if An ∈ A (X), n ∈ N, monotonically converge from below to A, then α(An ) ↑ α(A); (ii) (subadditivity) α(A1 ∪ A2 ) ≤ α(A1 ) + α(A2 ) for all A1 , A2 ∈ A (X); (iii) (additivity on disjoint sets) α(A1 ∪ A2 ) = α(A1 ) + α(A2 ) whenever A1 ∈ A (X) and A2 ∈ A (X) are disjoint. Then

α(B) ˜ := inf {α(A) : A ∈ A (X), A ⊃ B}

is a σ -additive extension of α to B (X).

(A.5)

141 Introduction to Measure Theory and Integration

Proof. Notice first that α is σ –subadditive on A (X): indeed, if A ⊂ ∪i Ai and B is an open set with compact closure in A, then B is contained in the union of finitely many Ai ’s, so that (ii) gives α(B) ≤

∞ 

α(Ai ).

i=1

 Since B is arbitrary, (i) gives α(A) ≤ i α(Ai ). Now, if we take (A.5) as the definition of α˜ for all subsets of X, Proposition 1.16 gives that α˜ extends α and is σ –subadditive. Then, Theorem 1.17 gives that α is σ –additive on the Borel σ –algebra, provided we are able to show that any Borel set is α–additive. ˜ Since the class of additive sets is a σ –algebra, suffices to show that any closed set is α– ˜ additive. To this aim, we first show that α˜ is additive on distant sets, namely (recall that dist(U, V ) is the infimum of the distances d(x, y) for x ∈ U and y ∈ V ) ˜ 1 ) + α(B ˜ 2) α(B ˜ 1 ∪ B2 ) = α(B

whenever dist(B1 , B2 ) > 0. (A.6)

Indeed, if A ⊃ B1 ∪ B2 is open we can consider the disjoint open sets A1 := {x ∈ A : dist(x, B1 ) < dist(x, B2 )} , A2 := {x ∈ A : dist(x, B2 ) < dist(x, B2 )} containing B1 and B2 respectively to get α(A) ≥ α(A1 ∪ A2 ) = α(A1 ) + α(A2 ) ≥ α(B ˜ 1 ) + α(B ˜ 2 ). Since A is arbitrary the inequality ≥ in (A.6) follows, while the converse one is a consequence of subadditivity. Let F ⊂ X be closed, B ⊂ X and let us prove that α(B ˜ ∩ F) + α(B ˜ \ F) ≤ α(B) ˜ (the opposite inequality follows by subadditivity). Assuming with no loss of generality α(B) ˜ < ∞ and setting   Bh := x ∈ B : 2h > dist(x, F) ≥ 2h−1 the additivity on distant sets gives  α(B ˜ 2h ) ≤ α(B) ˜ < ∞, h∈Z

 h∈Z

h∈Z

α(B ˜ 2h+1 ) ≤ α(B) ˜ <∞

142 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

because all finite sums are made on distant sets, all contained in B. We ˜ h ) is convergent and, since the sets Bh are a have then that h∈Z α(B partition of B \ F, using once more the additivity on distant sets we get α(B ˜ ∩ F) + α(B ˜ \ F) ≤ α(B ˜ ∩ F) + α˜

 N

 Bh +

∞ 

α(B ˜ h)

h=N +1

h=−∞

  N ∞

 Bh + α(B ˜ h) = α˜ (B ∩ F) ∪ h=N +1

h=−∞

≤ α(B) ˜ +

∞ 

α(B ˜ h)

h=N +1

for any N ≥ 1. Letting N → ∞ the inequality follows. For g ∈ C(X) we can define    g dμ := g dμ+ − g dμ− . X

In this way



X

X

g dμ is linear w.r.t. g; in addition, since



 hμ=

∞

0 ∞

X

=



+

μ ({h > t}) dt −

∞

μ− ({h > t}) dt

0

μ({h > t}) dt

0

whenever h is nonnegative, splitting g in positive and negative part we obtain that X g dμ is also linear w.r.t. to μ. Since !  !  ! ! + ! g dμ! ≤ |g| dμ + |g| dμ− ≤ max |g|μ = gμ ! ! X

X

X

∀g ∈ C(X) the functional

 L μ (g) :=

g dμ

g ∈ C(X)

(A.7)

X

belongs to (C(X))∗ and satisfies L μ  ≤ μ. The remarkable fact is that any element in the dual is representable in this form, and that equality holds. This will also prove that M (X) is a Banach space (with the definition of M (X) given above, independent of Section 6.5, it is not even totally obvious that it is a linear space!).

143 Introduction to Measure Theory and Integration

Theorem A.6 (Riesz). Let (X, d) be a compact metric space. The space (C(X))∗ is, via (A.7), isomorphic and isometric to M(K ). That is: all functionals L μ belong to (C(X))∗ and, for any L ∈ (C(X))∗ , there exists a unique μ ∈ M(K ) satisfying L = L μ . Finally, L μ  = μ. Proof. The proof will be achieved in three steps. In the first one we build an auxiliary positive finite measure μ∗ and prove in the second one that μ∗ provides the desired representation of L when L is nondecreasing. In the last one we achieve the general case and provide equality of the norms. Step 1. Let α ∗ : A (X) → [0, +∞) be defined by α ∗ (A) := sup {|L(g)| : |g| ≤ 1, supp g ⊂ A} . Notice that α ∗ (X) ≤ L and that α ∗ (∅) = 0. Notice also that we can equivalently replace |L(g)| with L(g) inside the supremum and that a simple approximation argument gives α ∗ (A) ≥ |L(g)|

whenever |g| ≤ 1 A .

(A.8)

Indeed, if |g| ≤ 1 A we can find continuous functions gn : X → [−1, 1] convergent to g and with support contained in A. In addition, if L is monotone we have also α ∗ (A) ≤ L(χ)

whenever 1 A ≤ χ.

(A.9)

We claim that α ∗ satisfies all the assumption of Proposition A.5. Indeed, if g ∈ C(X) has support contained in A, since the support is compact we have K ⊂ Ai for i large enough; it follows that L(g) ≤ α ∗ (Ai ) ≤ sup j α ∗ (A j ) and since g is arbitrary the continuity follows. In order to prove the subadditivity, given a continuous g : X → [−1, 1] with support K contained in A1 ∩ A2 , we can consider the disjoint compact sets K \ A1 and K \ A2 and a continuous function χ : X → [0, 1] identically equal to 1 in a neighbourhood of K \ A1 and identically equal to 0 in a neighbourhood of K \ A2 . It follows that (1 − χ)g has support contained in A1 and χg has support contained in A2 , hence L(g) = L((1 − χ)g) + L(χg) ≤ α ∗ (A1 ) + α ∗ (A2 ). Since g is arbitrary, the subadditivity of α ∗ follows. Finally, to prove the additivity on disjoint sets it suffices to notice that, given gi with support in Ai and |gi | ≤ 1, the function g = g1 + g2 has support in A1 ∪ A2 and satisfies L(g) = L(g1 ) + L(g2 ) and |g| ≤ 1.

144 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

By Proposition A.5 we obtain that α ∗ is the restriction to A (X) of a positive measure μ∗ . Notice also that μ∗ is finite, since μ∗ (X) = α ∗ (X) = L.

(A.10)

Step 2. Now we claim that L μ∗ ≥ |L|, namely L μ∗ (g) ≥ |L(g)| for any nonnegative g ∈ C(X). Also, we shall prove that if L is nondecreasing, namely L(g) ≥ 0 whenever g ∈ C(X) is nonnegative, then L μ∗ coincides with L. This proves already Riesz theorem for positive functionals. By homogeneity, in the proof of the inequality L μ∗ (g) ≥ |L(g)|, it is not restrictive to assume 0 ≤ g ≤ 1. Given an integer N ≥ 1, let us consider the open sets Ai := {g > i/N }, i = 0, . . . , N − 1, and notice that N −1 N −1   1 1 1 Ai ≥ g ≥ 1 Ai . (A.11) + N N i=1 i=1 Now, given continuous functions χi : X → [0, 1] satisfying 1 Ai ≤ χi ≤

1 Ai−1 , i = 1, . . . , N , we can use (A.8) to estimate

N −1 N −1 !   1 ∗ 1 ! L μ∗ (g) ≥ μ (Ai ) ≥ |L(χi+1 )| ≥ ! L N N i=1 i=1

But, since



 N ! 1  ! χi !. N i=2

N −1 N −1   1 1 1 + χi ≥ g ≥ χi+1 N N N i=1 i=1

(A.12)

we can let N → ∞ and use the continuity of L to get L μ∗ (g) ≥ |L(g)|. If L is also monotone we can use the inequality (A.9) to get  N −1 N −1 N −1   1  1 ∗ 1 1 χi . L μ∗ (g) − ≤ μ (Ai ) ≤ L(χi ) = L N N N N i=1 i=1 i=1 Again we can let N → ∞ and use (A.12) to get L μ∗ (g) = L(g). Step 3. Now we define linear continuous functionals L ± : C(X) → R by L + (g) :=

L μ∗ (g) + L(g) , 2

L − (g) :=

L μ∗ (g) − L(g) . 2

We have L + + L − = L μ∗ and L + − L − = L. In addition, by Step 2, L ± are monotone. Now we can apply the construction of Step 1 and use monotonicity in Step 2 to find positive finite measures μ± such that L ± = L μ± . It follows that L = L + − L − = L μ+ − L μ− = L μ

145 Introduction to Measure Theory and Integration

and the representation of L follows. Analogously, we obtain that L μ∗ = L + + L − = L μ+ + L μ− = L μ+ +μ− so that μ∗ = μ+ + μ− . To conclude, we identify μ with L and show that μ+ and μ− are orthogonal. The bound on μ follows by (A.10): μ = μ+ (X) + μ− (X) = L μ+ (1) + L μ− (1) = L μ∗ (1) = L. In order to show that μ+ ⊥ μ− , write μ± = a ± μ∗ and use the identity μ∗ = μ+ + μ− to get a + + a − = 1 μ∗ –a.e. in X. On the other hand the density of C(X) in L 2 (X, μ∗ ) and a truncation argument provide a sequence of continuous functions gn : X → [−1, 1] convergent in L 2 (X, μ∗ ) to the sign of a + − a − , so that ! !  ! ! + − ∗ L = sup |L μ (g)| = sup !! (a − a )g dμ !! = |a + − a − | dμ∗ . |g|≤1

Hence

|g|≤1



X

(1 − |a + − a − |) dμ∗ = μ∗ (X) − L ≤ 0. X

Since |a + − a − | ≤ 1 it must be |a + − a − | = 1 μ∗ –a.e. in X. Since a ± ∈ [0, 1] μ–a.e., this can only happen if a + a − = 0 μ∗ –a.e. in X, which means that μ+ is orthogonal to μ− . Remark A.7. A similar result holds, with minor changes in the proof, if (X, d) is locally compact and separable, namely there exists an nondecreasing sequence of open sets with compact closure whose union is the whole of X. In this case C(X) has to be replaced by C0 (X), namely the closure in C(X) of the space Cc (X) of compactly supported functions, while M (X) remains unchanged.

Solutions of some exercises

In this chapter we provide solutions to the main exercises proposed in the text, and in particular of those marked with one or two .

Chapter 1 Exercise 1.1. All verifications are very simple and we omit them. Exercise 1.2. We prove the statement for the translations, the proof for the dilations being similar. Fix h ∈ R and consider the class F := {A ∈ B (R) : A + h ∈ B (R)} . Then F is a σ –algebra containing the intervals, because the class I of intervals is invariant under translations. Therefore F ⊃ σ (I ) = B (R). This proves that A + h is Borel whenever A is Borel.  Exercise 1.3. Set X = N and μ := n δn . Then the sets An := {n, n + 1, . . .} satisfy μ(An ) = +∞, but their intersection is empty. Exercise 1.4. Let An ↑ A with An , A ∈ A . Then the sets Bn := A \ An satisfy Bn ↓ ∅, so that by assumption μ(Bn ) ↓ μ(∅) = 0. Since μ is finite, μ(Bn ) = μ(A) − μ(An ), so that μ(An ) ↑ μ(A). Exercise 1.5. For any n ∈ N∗ the set An of all atoms x such that μ({x}) ≥ 1/n has at most cardinality nμ(X): indeed, if we choose k elements x1 , . . . , xk in this sets, adding the inequalities μ({xi }) ≥ 1/n we find k/n ≤ μ(X), whence the upper bound on the cardinality of An follows. If μ is σ –finite, we choose X i ↑ X with X i ∈ E and μ(X i ) < ∞ and repeat the previous argument with the sets Ai,n := {x ∈ Aμ ∩ X i : μ({x}) ≥ 1/n}, whose union gives Aμ . If not finiteness assumption is made, the statement fails: take X = R, E = P (R) and μ(A) = 0 if A = ∅ and μ(A) = +∞ otherwise. Exercise 1.6. Let μ be diffuse. First we prove that for all τ ∈ (0, 1) and all A ∈ E there exists a subset B ∈ E with 0 < μ(B) < τ μ(A). Indeed,

148 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

if this property fails for some τ and A, for all subsets B either μ(B) = 0 or μ(B) ≥ τ μ(A). Now, choose B1 ⊂ A with μ(B1 ) ∈ (0, μ(A)) (this is possible by assumption), then B2 ⊂ A \ B1 with μ(B2 ) ∈ (0, μ(B1 )) and so on. Since all these sets are contained in A, we have μ(Bi ) ≥ τ μ(A), and this contradicts the fact that they are disjoint. Now, given t ∈ (0, μ(X)) we define a sequence of pairwise disjoint sets Bi and numbers si as follows: first set s1 := sup {μ(B) : μ(B) ≤ t} and then choose B1 with t ≥ μ(B1 ) > s1 /2; then recursively set   sn+1 := sup μ(B) : B ⊂ Bnc , μ(B) ≤ t − μ(Bn ) and choose Bn+1 ⊂ Bnc with t − μ(Bn ) ≥ μ(Bn+1 ) > sn+1 /2. We now claim that μ(∪i Bi ) = t. If this property fails, then i μ(Bi ) < t and the convergence of the series implies that si → 0. On the other hand  

 Bi , μ(B) ≤ t − μ(Bi ) si ≥ sup μ(B) : B ⊂ X \ i

i



The previous property with A = X \ ∪i Bi and τ = (t − i μ(Bi ))/μ(A) shows that the supremum in the right hand side (independent of i) is positive, contradicting the fact that si → 0. Exercise 1.7. Let X be a separable metric space and let E = B (X). If μ({x}) > 0 for some x ∈ X, obviously μ is not diffuse. Conversely, if A ∈ B (X) is given, with μ(A) > 0 and μ(B) ∈ {0, μ(A)} for all B ⊂ A, we can fix a countable dense set (xi ) ⊂ X and define   r0 := sup r ≥ 0 : μ(A ∩ B r (x0 )) = 0 . Since r  → μ(A ∩ B r (x0 )) is right continuous, the maximality of r0 easily implies that μ(A ∩ B r0 (x0 )) > 0, and therefore μ(A ∩ B r0 (x0 )) = μ(A). Now we iterate this construction, setting A1 := A ∩ B r0 (x0 ), defining   r1 := sup r ≥ 0 : μ(A1 ∩ B r (x1 )) = 0 , so that μ(A1 ∩ B r1 (x1 )) = μ(A1 ) = μ(A). Continuing in this way, we have a nonincreasing family of sets (Ai ) with μ(Ai ) = μ(A); it follows   that μ( i Ai ) = μ(A) > 0. On the other hand, any point x ∈ i Ai satisfies d(x, xi ) = ri ∀i ∈ N.

149 Introduction to Measure Theory and Integration

By the density of the family (xi ), this intersection contains at most one point (and at least one, because the measure is positive). It follows that this point is an atom of μ. Exercise 1.8. Cantor’s middle third set can be obtained as follows: let C0 = [0, 1], let C1 the set obtained from C0 by removing the interval (1/3, 2/3), let C2 be the set obtained from C1 by removing the intervals (1/9, 2/9) and (7/9, 8/9), and so on. Each set Cn consists of 2n disjoint closed intervals with length 3−n , so that λ(Cn ) = (2/3)n → 0. If follows that the intersection C of all sets Cn is a closed and λ–negligible set. In order to show that C has the cardinality of continuum (at this stage it is not even obvious that C = ∅!) we recall that numbers x ∈ [0, 1] can be represented with a ternary, instead of a decimal, expansion: this means that we can write  ai 3−i = 0, a1 a2 a3 . . . x= i≥1

with the ternary digits ai ∈ {0, 1, 2}. As for decimal expansions, this representation is not unique; for instance 1/3 can be written either as 0.1 or as 0.0222 . . ., and 2/3 can be written either as 0.2 or as 0.1222 . . .. It is easy to check that C1 corresponds to the set of numbers that can be expressed by a ternary representation not having 1 as first digit, C2 corresponds to the set of numbers that admit a representation not having 1 as a first or second digit, and so on. It follows that C is the set of numbers that admit a ternary representation not using the digit 1: since the map ∞  ∗ ai 3−i (a1 , a2 , . . .) ∈ {0, 2}N → x = i=1 ∗

∗

provides a bijection of {0, 2}N with C, and the cardinality of {0, 2}N is the continuum, this proves that C has the cardinality of continuum. Exercise 1.9. Let {qn }n∈N be an enumeration of the rational numbers in [0, 1], and set ∞

ε ε (qn − 2−n , qn + 2−n ). A := 4 4 n=0  Then A ⊂ R is open and λ(A) < n ε2−n−1 = ε (why is the inequality strict ?). Therefore [0, 1] \ A has Lebesgue measure strictly less than ε and an empty interior, because [0, 1] \ A does not intersect Q. Exercise 1.11 Let {In }n∈N be an enumeration of the open intervals with rational endpoints of (0, 1). By the construction in Exercise (1.9), for any

150 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

interval I and any δ ∈ (0, λ(I )) we can find a compact set C ⊂ I with an empty interior such that 0 < λ(C) < δ. We will define E :=

∞

Ci

i=0

where Cn ⊂ In are compact sets with an empty interior, λ(Cn ) > 0 and λ(Cn ) < δn . The choice of Cn and δn will be done recursively. Notice first that ∀n ∈ N, λ(E ∩ In ) ≥ λ(Cn ) > 0 so we have only to take care of the condition λ(E ∩ In ) < λ(In ). Set βn = λ(In \ ∪n0 Ci ) and notice that βn > 0 because all Ci have an empty interior. Since λ(In ∩ E) ≤ λ(In ∩

n

0

Ci ) +

∞  i=n+1

δi = λ(In ) − βn +

∞ 

δi

i=n+1

 it suffices to choose δn (and Cn ) in such a way that ∞ n+1 δi < βn . This is possible, choosing for instance δn+1 > 0 satisfying

 1 1 1 δn+1 < max βn , βn−1 , . . . , n+1 β0 , 2 4 2 to get δi < 2n−i βn for i > n. Exercise 1.12. Let A be μ–measurable and let B, C ∈ E be satisfying AB ⊂ C and μ(C) = 0. For any set D ⊂ X we have, by monotonicity of μ∗ , μ∗ (D ∩ A) + μ∗ (D \ A) ≤ μ∗ (D ∩ (B ∪ C)) + μ∗ ((D \ B) ∪ C). Since μ∗ (D ∩C) ≤ μ∗ (C) = μ(C) = 0, by using twice the subadditivity of μ∗ and then the additivity of B we get μ∗ (D ∩ A) + μ∗ (D \ A) ≤ μ∗ (D ∩ B) + μ∗ (D \ B) = μ∗ (D). Since D is arbitrary, this proves that A is additive. Exercise 1.13. The statement is trivial if μ∗ (A) = ∞. If not, for any n ∈ N∗ we can find, by the definition of μ∗ , a countable union An of sets of N such that An ⊃ A and μ(An ) ≤ μ∗ (A) + 1/n. Then, setting B := n An we have B ⊃ A and μ(B) ≤ infn μ∗ (A) + 1/n = μ∗ (A). The inequality μ(B) ≥ μ∗ (B) follows by the monotonicity of μ∗ , taking into account that μ∗ (B) = μ(B).

151 Introduction to Measure Theory and Integration

Exercise 1.14. E μ is a σ –algebra: stability under complement isimmedi c c B = AB; if A B ⊂ C , then ( A )( ate, because A i i i i i i Bi ) ⊂  C , and since μ–negligible sets are stable under countable unions, this i i proves that E μ is stable under countable unions. The extension μ(A) := μ(B), where B ∈ E is any set such that AB is contained in a μ–negligible set of E , is well defined and σ –additive on E μ : if AB ⊂ C and AB  ⊂ C  , then BB  ⊂ C ∪ C  ; consequently, if μ(C) = μ(C  ) = 0 it must be μ(B) = μ(B  ). The σ –additivity can be proven with an argument analogous to the one used to show that E μ is a σ –algebra. μ–negligible sets of E μ are characterized by the property of being contained in a μ–negligible set of E : if A ∈ E μ is μ–negligible, there exist μ–negligible sets B, C ∈ E with AB ⊂ C; as a consequence A is contained in the μ–negligible set B ∪ C ∈ E . Conversely, if A ⊂ X is contained in a μ–negligible set C ∈ E we may take B = ∅ to conclude that A ∈ E μ and μ(A) = 0. Exercise 1.15. Let A be additive; by Exercise 1.13 we can find a set B ∈ E containing A with μ(B) = μ∗ (A). The additivity of A and the equality μ∗ (B) = μ(B) give μ(B) = μ∗ (A) + μ∗ (B \ A). As a consequence μ∗ (B \ A) = 0. Now we apply Exercise 1.13 again, to find a μ–negligible set C ∈ E containing A \ B. It follows that AB is contained in C, and therefore A is μ–measurable. Exercise 1.16. Let us first build a family of pairwise disjoints sets {Ai }i∈I ⊂ P (N), with I and all sets Ai having an infinite cardinality and i Ai = N (the construction of the σ –algebra will be more clear if we keep I and N distinct). The family {Ai } can be obtained, for instance, through a bijective correspondence S between N × N and N, setting Ai := S({i} × N). Then, we define π : N → I by π(n) = i, where i ∈ I is the unique index such that n ∈ Ai and (with the convention π −1 (∅) = ∅)   F := π −1 (J ) : J ⊂ I . It is immediate to check that F is a σ –algebra, that Ai = π −1 ({i}) ∈ F and that any nonempty set in F contains one of the sets Ai . Therefore F contains infinitely many sets, and all of them except ∅ have an infinite cardinality.

152 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercise 1.17. It suffices to define μ(A) = 0 if A has a finite cardinality, and +∞ otherwise. A finite union of sets has an infinite cardinality if and only if at least one of the sets has an infinite cardinality, and this shows that μ is additive. The solutions of the next exercises require a more advanced knowledge of set theory, and in particular the theory of ordinals, the transfinite induction, the behavior of cardinality under unions and products, and Zorn lemma. We shall denote by ω the smallest uncountable ordinal and by χ the cardinality of continuum. Exercise 1.18. Notice that F ( j) ⊂ σ (K ) implies   ∞

c ( j) ( j) Ak , B : (Ak ) ⊂ F , B ∈ F ⊂ σ (K ). k=0

Therefore, if i is the successor of j, we obtain F (i) ⊂ σ (K ); analogously, if i has no predecessor, and F ( j) ⊂ σ (K ) for all j ∈ i, then  ( j) , namely F (i) , is contained in σ (K ). Using these two facts, j∈i F one obtains by transfinite induction that F (i) ⊂ σ (K ) for all i ∈ ω. An analogous induction argument shows that F (i) ⊂ F ( j) whenever i ∈ j.  So, the union U := i∈ω F (i) is contained in σ (K ) and, to prove that equality holds, it suffices to show that this union is a σ –algebra. Let (Bk ) ⊂ U and let i k ∈ ω be such thatBk ∈ F (ik ) . Since i k are countable and ω is uncountable we have i := k i k ∈ ω and all sets Bk belong to F (i) . It follows that their union belongs to F ( j) , where j is the successor of i, and therefore to U . An analogous (and simpler) argument proves that U is stable under complement. Exercise 1.19. Obviously B (R) has at least the cardinality of continuum, so we need only to show an upper bound on the cardinality of B (R). The  proof is based on the fact that a union i∈J X i and a product i∈J X i have cardinality not greater than χ if the index set J and all sets X i have cardinality not greater than χ. Let F (i) be defined as in Exercise 1.19, with K having at most the cardinality of continuum. Using the previous property of products, with J even countable, one can prove by transfinite induction that, for all i ∈ ω, F (i) has at most cardinality χ. If we choose as K the class of intervals, whose cardinality is (at most) χ, we find

B (R) = σ (K ) = F (i) .

×

i∈ω

Now we use the above mentioned property of unions, with J = ω and X i = F (i) , to conclude that B (R) has at most the cardinality of continuum.

153 Introduction to Measure Theory and Integration

Exercise 1.20. Obviously L has a cardinality not greater than the cardinality of P (R); by Bernstein theorem (1) it suffices to show that the cardinality of P (R) is not greater than the cardinality of L : if C is the Cantor set of Exercise 1.8, we know that P (R) is in one-to-one correspondence of P (C), because C has the cardinality of continuum; on the other hand, any subset of C obviously belongs to L , because C has null Lebesgue measure. Exercise 1.21. Let E ⊂ P (X) be a σ –algebra. Assume by contradiction that E is infinite and countable. We define the equivalence relation y ∼ y

if and only if ((y ∈ B ⇔ y  ∈ B) ∀B ∈ E )

and let F be the partition of X in equivalence classes. We now prove that F ⊂ E . Indeed, let F ∈ F , fix f ∈ F, for any x ∈ / F we have f  ∼ x so there must be B ∈ E such that f ∈ B and x  ∈ B (or the opposite, but then we may consider B c ); given this set B, for any g ∈ F we have that g ∼ f implies g ∈ B, so that F ⊂ B. Since x is arbitrary we conclude that  B. F= B∈E ,F⊂B

Now, since E is countable, it follows that F ∈ E . We eventually note that any set in E is union of sets in F : but then, if F were finite then E would be finite, whereas if F were infinite then E would be uncountable. Exercise 1.22 We define F as in the solution of the previous exercise, in this case it has finite cardinality, say n; consequently, there are 2n sets in E. Exercise 1.23 We define F as in the solution of Exercise 1.21; we also adapt the above argument to show again that F ⊂ E . Indeed, let F ∈ F , fix f ∈ F, for any x ∈ / F we have f ∼ x so there must be B = B F,x ∈ E such that f ∈ B and x ∈ B; and again F ⊂ B F,x . Hence  F= B F,x x∈X, x ∈ F

and this proves that F ⊂ A , since X is countable. We then use the Axiom of Choice to define  a function φ : F → X such that φ(F) ∈ F, and eventually define μ˜ = F∈F μ(F)δφ(x) .

(1) If A has cardinality not greater than B, and B has cardinality not greater than A, then there exists

a bijection between A and B

154 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercise 1.24. We begin our construction with an algebra τ0 in P (N) and μ0 : τ0 → {0, 1} which is additive but not σ –additive. For instance we may take as τ0 the algebra generated by singletons {x} with x ∈ N (i.e. the sets A ⊂ N such that either A or Ac are finite) and set  0 if A is finite; μ0 (A) := 1 if Ac is finite. We will extend μ0 to an additive function, that we still denote by μ0 , defined on the whole of P (N). If such an extension exists, it can’t be σ –additive, because μ0 ({n}) = 0 for all n ∈ N, while μ0 (N) = 1. In the class C of pairs (τ, μ) with τ algebra and μ : τ → {0, 1} additive, we define the partial order relation (τ, μ) ≤ (τ  , μ ) by τ ⊂ τ  and μ|τ = μ; then we consider the class C 0 of all (τ, μ) satisfying ¯ in this (τ, μ) ≥ (τ0 , μ0 ). By Zorn lemma, we can find a maximal (τ¯ , μ) class: indeed, it is easy to check that any totally ordered chain I ⊂ C 0 has an upper bound (τ  , μ ), defined by

τ  := τ and μ (A) := μ(A) where A ∈ τ, (τ, μ) ∈ I. (τ,μ)∈I

We will show that the maximality of (τ¯ , μ) ¯ forces τ¯ to coincide with P (N), so that μ¯ will be the desired extension of μ0 . Let us assume by contradiction that τ¯  P (N) and choose Z ⊂ N with Z ∈ / τ¯ . We notice that   (A1 ∩ Z ) ∪ (A2 ∩ Z c ) : A1 , A2 ∈ τ¯ is the algebra generated by τ¯ ∪ {Z }. Moreover, either Z or Z c satisfy the following property for all A ∈ τ¯ with μ(A) ¯ = 1, Z ∩ A  = ∅.

(A.13)

If not, we would be able to find A1 , A2 ∈ τ¯ with A1 ∩ Z = A2 ∩ Z c = ∅ ¯ 2 ) = 1, so that A1 and A2 would be disjoint and and μ(A ¯ 1 ) = μ(A μ(A ¯ 1 ∪ A2 ) = 2, contradicting the fact that μ¯ maps τ¯ into {0, 1}. Possibly replacing Z by its complement we shall assume that Z fulfils (A.13). Now we extend μ¯ to the algebra generated by τ¯ ∪ {Z }, as follows: μ(B) ˜ := μ(A ¯ 1 ) whenever A1 , A2 ∈ τ¯ and B = (A1 ∩ Z ) ∪ (A2 ∩ Z c ). (A.14) Let us check that μ˜ is well defined and additive.

155 Introduction to Measure Theory and Integration

1. μ˜ is well defined: if B = (A1 ∩ Z ) ∪ (A2 ∩ Z c ) = (A3 ∩ Z ) ∪ (A4 ∩ Z c ) then (A1 ∩ Z ) = (A3 ∩ Z ), and if μ(A ¯ 1 ) = μ(A ¯ 3 ) then one of the ¯ 3 ) = 0. Defining A := two numbers, say μ(A ¯ 1 ), equals 1, while μ(A ¯ = 1 and A ∩ Z = ∅, contradicting (A.13). A1 \ A3 we have μ(A) 2. Suppose B, B  ∈ τ¯ are disjoint. Let B = (A1 ∩ Z ) ∪ (A2 ∩ Z c ) and B  = (A1 ∩ Z ) ∪ (A2 ∩ Z c ). Then A1 ∩ A1 ∩ Z = ∅. Setting A1 := A1 \ A1 we still have B  = (A1 ∩ Z ) ∪ (A2 ∩ Z c ), and then we can use the additivity of μ¯ to conclude that μ(B ˜ ∪ B  ) = μ(A ¯ 1 ∪ A1 ) = μ(A ¯ 1 ) + μ(A ¯ 1 ) = μ(B) ˜ + μ(B ˜  ). If B ∈ τ we can choose A1 = A2 = B in (A.14) to obtain that μ(B) ˜ = μ(B), ¯ so that μ˜ extends μ¯ to the algebra generated by τ¯ ∪ {Z }. This violates the maximality of (τ¯ , μ). ¯ Exercise 1.25 We obviously need only to show that the cardinality of C is at least equal to the continuum. By the inner regularity of λ we can assume with no loss of generality that C is closed. Now, we define A = (0, 1) \ C and  g(t) := λ [0, t] ∩ C

t ∈ [0, 1].

This continuous function maps continuously [0, 1] onto [0, λ(C)], and it is constant in any connected component of A, so that g(A) is at most countable. Since g(C) contains [0, λ(C)] \ g(A) we obtain that C has cardinality at least equal to the continuum (one can actually see that g(C) = g([0, 1])). Exercise 1.26 Since K is totally bounded, for all > 0 there exist finitely many balls B1 , . . . , B N with radius whose union covers K . The properties of μ imply the existence of an index i such that μ({n : xn ∈ Bi }) = 1. Now we start with = 1 and find a closed ball B (1) with radius 1 such that μ({n : xn ∈ B (1) }) = 1. Repeating this construction in B (1) we find a closed ball B (2) with radius 1/2 contained in B (1) with μ({n : xn ∈ B (2) }) = 1. Continuing in this way, if z is the common point of the balls B (i) , we find xn μ-converges to z.

Chapter 2 Exercise 2.1 The verification is straightforward and is omitted.

156 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercise 2.2 Let ϕ, ψ : X → R be E –measurable. If ϕ(x) + ψ(x) < t we can find a rational number r such that ϕ(x) < r and ψ(x) < t − r, hence

{ϕ + ψ < t} = [{ϕ < r} ∩ {ψ < t − r}] . r∈Q

This proves that ϕ + ψ is E –measurable. Analogously, since √ √ {ϕ 2 > a} = {ϕ > a} ∪ {ϕ < − a}, a ≥ 0 we obtain that ϕ 2 is measurable. Considering the difference (ϕ + ψ)2 − (ϕ − ψ)2 we obtain that ϕψ is E –measurable. Exercise 2.3. (i) The verification of the axioms of distance is immediate. In order to prove the compactness of R, let us consider a sequence (xn ) ⊂ R. If supn xn = +∞ we can find for any k an index n(k) such that xn(k) ≥ k; it follows that d(xnk , +∞) = | arctan xn(k) − π/2| tends to 0, so that xn(k) → +∞ in the metric space. Analogously, if infn xn = −∞ we can find a subsequence converging to −∞ in (R, d). Finally, if both supn xn and infn xn are finite, the sequence (xn ) is bounded and we can extract, thanks to the Bolzano–Weierstrass theorem, a subsequence xn(k) converging to x ∈ R. The continuity of z  → arctan z implies that xn(k) → x in (R, d). To prove the equivalence of the two topologies, let us work with closed sets: if C ⊂ R is closed with respect to the (R, d) topology, then it is closed with respect to the Euclidean topology, because |xn − x| → 0 implies | arctan xn − arctan x| → 0. On the other hand, if | arctan xn − arctan x| → 0 then for n large enough arctan xn belongs to an interval I := (arctan x − ε, arctan x + ε) ⊂ (−π/2, π/2); the continuity of y → tan y in I implies that xn → x. This proves the converse implication, and the equivalence of the two topologies. (ii) We notice first that, according to (i), B (R) and {−∞}, {+∞} belong to B (R). Therefore, if f is measurable between E and the Borel σ – algebra of (R, d), then it is E –measurable according to (2.2). According to the measurability criterion, in order to prove the converse implication it suffices to show that B (R) is generated by B (R) ∪ {−∞} ∪ {+∞}: this follows by the fact that if C ⊂ R is closed, then C = (C ∩ R) ∪ (C ∩ {−∞}) ∪ (C ∩ {+∞}) (again by (i)) belongs to the algebra generated by B (R)∪{−∞}∪{+∞}, therefore the σ –algebra generated by this family of sets contains B (R). Exercise 2.4. If { f = g} is contained in a μ–negligible set C of E , for some E –measurable function g, then { f > t}{g > t} ⊂ C for all t ∈ R, and since {g > t} ∈ E it follows that { f > t} ∈ E μ ; this means

157 Introduction to Measure Theory and Integration

that f is E μ –measurable. Conversely, assume that f is E μ –measurable and find for all q ∈ Q a set Bq ∈ E and a μ–negligible set Cq ∈ E with { f > q}Bq ⊂ Cq . We define

  g(x) := sup q ∈ Q : x ∈ Bq , C := Cq . q∈Q



Since {g ≤ t} = q≤t Bq we have that g is E –measurable. Let us prove that f (x) = g(x) for all x ∈ / C: for any such x we have x ∈ Bq for all q < f (x), therefore g(x) ≥ f (x); if the inequality were strict, there would exist q ∈ Q with x ∈ Bq and q > f (x), therefore x would be in Bq \ { f > q} ⊂ Cq ⊂ C. Exercise 2.5. If σ ≤ τ we can find a nondecreasing family of partitions σ1 , . . . , σn with σ1 = σ , σn = τ and σi+1 \ σi containing just one point. Therefore, in the proof of the monotonicity of σ  → Iσ ( f ) we need only to show that Iσ ( f ) ≤ Iσ ∪{t} ( f ) whenever t ∈ (0, ∞) \ σ . Let σ = {t0 , . . . , t N } and let i be the last index such that ti < t. If i < N we use the inequality (ti+1 − ti ) f (ti+1 ) = (ti+1 − t) f (ti+1 ) + (t − ti ) f (ti+1 ) ≤ (ti+1 − t) f (ti+1 ) + (t − ti ) f (t)  adding to both sides j=i (t j+1 −t j ) f (t j+1 ) we obtain Iσ ( f ) ≤ Iσ ∪{t} ( f ). If i = N the argument is even easier, because the difference Iσ ∪{t} ( f ) − Iσ ( f ) is given by (t − t N ) f (t). Now, let f, g : (0, +∞) → [0, +∞)  ∞ be given; since  ∞ Iσ ( f + g) = Iσ ( f ) + Iσ (g) we get Iσ ( f + g) ≤ 0 f (t) dt + 0 g(t) dt. Since σ ∈  is arbitrary, this proves that  ∞  ∞  ∞ f (t) + g(t) dt ≤ f (t) dt + g(t) dt. 0

0

0

∞ In order to prove the converse inequality, fix L < 0 f (t) dt, M < ∞ 0 g(t) dt and find σ, η ∈  with Iσ ( f ) > L and Iη (g) > M; then  ∞ f (t) + g(t) dt ≥ Iσ ∪η ( f + g) = Iσ ∪η ( f ) + Iσ ∪η (g) 0

≥ Iσ ( f ) + Iη (g) > L + M. ∞ Letting L ↑ 0 f (t) dt and M ↑ 0 g(t) dt the inequality is proved. Exercise 2.6. We will prove that f ∗ is lower semicontinuous, the proof of the upper semicontinuity of f ∗ being analogous. Let (xn ) ⊂ R be converging to x and use the definition of f ∗ (xn ) to find yn ∈ R such that ∞

|xn − yn | <

1 n

and

1 f (yn ) ≤ f ∗ (xn ) + . n

158 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Then (yn ) still converges to x, so that f ∗ (x) ≤ lim inf f (yn ) ≤ lim inf f ∗ (xn ) + n→∞

n→∞

1 = lim inf f ∗ (xn ). n→∞ n

Exercise 2.7. Let t ∈ R and let (xn ) ⊂ { f ∗ ≤ t} be convergent to x. Then, the lower semicontinuity of f ∗ gives f ∗ (x) ≤ lim inf f ∗ (xn ) ≤ t. n→∞

This proves that x ∈ { f ∗ ≤ t}, so that { f ∗ ≤ t} is closed. The proof for f ∗ is similar. Since the Borel σ –algebra is generated by halflines, it follows that f ∗ and f ∗ are Borel, and the same is true for the set { f ∗ = f ∗ }, that coincides with . Exercise 2.8. Set ϕ0 := ϕ, A0 := {ϕ0 ≥ a0 } and ϕ1 := ϕ − a0 1 A0 ≥ 0. Then, set A1 := {ϕ1 ≥ a1 } and ϕ2 := ϕ1 − a1 1 A1 and so on. If ϕ(x) = +∞ then ϕn (x) = +∞ for all n, so that x belongs to all sets n ai 1 Ai (x) = +∞. We then assume that ϕ(x) < +∞ in the Ai and i=0 following. By construction we have that 0 ≤ ϕi+1 ≤ ϕi ≤ · · · ≤ ϕ0 = ϕ, hence n n   (ϕi − ϕi+1 ) = ϕn+1 + ai 1 Ai . ϕ = ϕn+1 + i=0

i=0



This proves that ϕ ≥ i ai 1 Ai . If the inequality were strict for some x ∈ X with ϕ(x) < +∞, we could find ε > 0 such that ϕi (x) ≥ ε for all i ∈ N, and since ai < ε for i large  enough, we would get x ∈ Ai for i large enough. But since the series i ai is not convergent, we would get  a 1 (x) = ∞, a contradiction. i i Ai Exercise 2.9. Assume by contradiction that the absolute continuity prop−i erty fails. Then, for some ε > 0 we can find Ai with μ(Ai ) < 2 and Ai |ϕ| dμ ≥ ε. It follows that the set B := lim supi Ai is μ–negligible, and

Ai \ B ↓ ∅. Bn := 



i≥n

Since Bn |ϕ| dμ ≥ An |ϕ| dμ ≥ ε we find a contradiction with the dominated convergence theorem applied to the functions 1 Bn |ϕ|, pointwise converging to 0.  Exercise 2.10. Let ε > 0 be given and let δ > 0 be such that A |ϕ| dμ < ε/2 whenever A ∈ E and μ(A) < δ. The triangle inequality gives, with the same choice of A, A |ϕn | dμ < ε for n > n 0 , provided ϕn − ϕ1 < ε/2 for n > n 0 . Since ϕ1 , . . . , ϕn0 are integrable, we can find δi > 0 such that A |ϕi | dμ < ε whenever A ∈ E and μ(A) < δi . If

159 Introduction to Measure Theory and Integration

 δ0 = min{δ, mini δi }, we have A |ϕn | dμ < ε/2 whenever n ∈ N, A ∈ E and μ(A) < δ. A possible example for the second question is = [0, 1], μ = λ the n Lebesgue measure, and ϕn = 2n 1[2−n ,21−n ) . The uniform integrability is a direct consequence of the convergence of ϕn to 0 in L 1 . If ϕn ≤ g, then ∞ ∞   ϕn = ϕn ≤ g  ∞

∞

n=1

n=1

but n=1 ϕn = 1 1/n = +∞. Exercise 2.11. (a) For any y ∈ X we have gλ (x) ≤ g(y) + λd(x, y) ≤ g(y) + λd(x  , y) + λd(x, x  ). Since y is arbitrary we get gλ (x) ≤ gλ (x  ) + λd(x, x  ). Reversing the roles of x and x  the inequality is achieved. (b) Clearly the family (gλ ) is monotone with respect to λ, and since we can always choose y = x in the minimization problem we have gλ (x) ≤ g(x). Assume that supλ gλ (x) is finite (otherwise the statement is trivial) and let xλ such that gλ (x) + λ−1 ≥ g(xλ ) + λd(x, xλ ). This inequality implies that xλ → x as λ → ∞ and, now neglecting the term λd(x, xλ ), that 1 gλ (x) + ≥ g(xλ ). λ Passing to the limit in this inequality as λ → ∞ and using the lower semicontinuity of g we get supλ gλ (x) ≥ g(x). Exercise 2.12. Let us first assume that f is bounded. For ε > 0 we consider the functions  x+ε 1 f (x  , y) dx  . f ε (x, y) := 2ε x−ε Since x  → f (x, y) is continuous, we can apply the mean value theorem to obtain that f ε (x, y) → f (x, y) as ε ↓ 0. So, in order to show that f is a Borel function, we need only to show that f ε are Borel. We will prove indeed that f ε are continuous: let xn → x and yn → y; since f (x  , yn ) → f (x  , y) for all x  ∈ R, we have 1[xn −ε,xn +ε] (x  ) f (x  , yn ) → 1[x−ε,x+ε] (x  ) f (x  , y)

for all x  ∈ R \ {x − ε, x + ε}. Therefore, since f is bounded, the dominated convergence theorem yields  1 1 (x  ) f (x  , y) dx  f ε (x, y) = 2ε R [x−ε,x+ε]  1 = 1 (x  ) f (x  , yn ) dx  = lim f ε (xn , yn ). lim n→∞ 2ε n→∞ R [xn −ε,xn +ε]

160 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

In the general case when f is not bounded we approximate it by the bounded functions f h (x) := max{−h, min{ f (x), h}}, with h ∈ N, that are still separately continuous, and therefore Borel.

Chapter 3 Exercise 3.1. On the real line, endowed with the Lebesgue measure, the function (1 + |x|)−1 belongs to L 2 , but not to L 1 , and the function |x|−1/2 1(0,1) (x) belongs to L 1 , but not to L 2 . Turning back to the general case, if ϕ ∈ L p1 ∩ L p2 with p1 ≤ p2 , from the inequality |ϕ| p ≤ max{|ϕ| p1 , |ϕ| p2 } ≤ |ϕ| p1 + |ϕ| p2

∀ p ∈ [ p1 , p2 ]

(that can be verified considering separately the cases |ϕ| ≤ 1 and |ϕ| > 1) we get that ϕ ∈ L p for all p ∈ [ p1 , p2 ]. Exercise 3.2. The statement is trivial if  f q = 0, so we assume that  f q > 0. For > 0 the set X := {| f | > } has finite μ–measure, by the Markov inequality, hence the inclusion between L r spaces for finite measures gives that | f |1 X ∈ L p (X, E , μ). Since the dominated convergence theorem gives   | f |q dμ = 0 lim | f − f 1 X |q dμ = lim ↓0

↓0

X

X\X

we can choose f˜ = f 1 X for > 0 small enough. Exercise 3.3. By homogeneity we can assume that ϕ p = 1 and ψq = 1. Since    p |ϕ| |ψ|q ϕ p ψq + − |ϕ||ψ| dμ = + −1=0 p q p q X and the function among parentheses is nonnegative, it follows that if vanishes μ–a.e. In particular, for μ–a.e. x, |ϕ(x)| is a minimizer of y →

yq − |ψ(x)|y q

in [0, +∞). But this problem has a unique minimizer, given by |ψ(x)|q−1, and we conclude. Exercise 3.4. It suffices to apply H¨older’s inequality to the functions |ϕ|r and |ψ|r , with the dual exponents p/r and q/r, to obtain ϕψrr ≤ |ϕ|r  p/r |ψ|r q/r = ϕrp ψrq .

161 Introduction to Measure Theory and Integration

Exercise 3.5. The positive part and the negative part of ϕ − ϕn have the same integral, hence   |ϕ − ϕn | dμ = 2 (ϕ − ϕn )+ dμ. X

X

The condition lim infn ϕn ≥ ϕ ensures that (ϕ −ϕn )+ is pointwise convergent to 0; in addition, since ϕn are nonnegative, the functions are dominated by ϕ + . Therefore the dominated convergence theorem gives the result. Exercise 3.6. If ψn → ψ μ–a.e. we apply Fatou’s lemma to the functions ψn + ϕn to obtain   lim inf ψn + ϕn dμ ≥ ψ + ϕ dμ. n→∞

Therefore

X





 ψn dμ + lim inf

lim sup n→∞

X

X

n→∞



ϕn dμ ≥ X

ϕ dμ + X

ψ dμ. X

 Subtracting ψ dμ from both sides the statement is achieved. In the  general case, let n(k) be a subsequence such that limk ϕn(k) dμ = lim infn X ϕn , and let n(k(s)) be a further subsequence converging to ϕ μ–a.e. Then    lim inf ϕn dμ = lim ϕn(k(s)) dμ ≥ lim inf ϕn(k(s)) dμ n→∞ s→∞ X n X X  ≥ lim inf ϕn dμ. X n→∞

Exercise 3.7. We show only how (3.13) implies g(t x + (1 − t)y) ≤ tg(x) + (1 − t)g(y) for all x, y ∈ J and t ∈ [0, 1]. We prove first, by induction on m, that m  2 2m   1 1 xi ≤ g(xi ) g m 2 2m i=1 i=1 for all x1 , . . . , x2m ∈ J . The case m = 1 is (3.13) and the induction step can be achieved grouping the terms as follows:  m−1 m−1 2m 2  1 1 1 2 1 xi = xi + x2m−1 +i . 2m 2 i=1 2m−1 2m−1 i=1 i=1

162 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Now, considering the case when xi = x for 1 ≤ i ≤ k and xi = y otherwise, we get g(t x + (1 − t)y) ≤ tg(x) + (1 − t)g(y)

with t =

k . 2m

Since g is continuous, by approximation we get g(t x + (1 − t)y) ≤ tg(x) + (1 − t)g(y) for all x, y ∈ J and t ∈ [0, 1]. Exercise 3.8. Let us first show the existence of z 0 . Let A = g(R) and let u n = g(z n ) with u n ↓ inf A. Since u n is uniformly bounded from above, our assumption on g ensures that (z n ) is bounded. By the BolzanoWeierstrass theorem we can find a subsequence z n(k) convergent to z ∈ R. The continuity of g gives that u n(k) = g(z n(k) ) converge to g(z). It follows that inf A is finite and coincides with g(z). Now, by applying the convexity inequality of the previous exercise with x = z 2 , y = z 0 and t = (z 1 − z 0 )/(z 2 − z 0 ), we get g(z 2 ) − g(z 1 ) g(z 1 ) − g(z 0 ) ≥ ≥0 z2 − z1 z1 − z0 for z 0 < z 1 < z 2 , proving the monotonicity of g in [z 0 , +∞). The argument in (−∞, z 0 ] is analogous.   Exercise 3.9. Fatou’s lemma   gives lim infn ϕn dμ ≥ lim infn ϕn dμ ≥ ϕ dμ. Therefore tn := ϕn dμ → t := ϕ dμ; we can apply Exercise 3.5 to the functions ϕn /tn to obtain that ϕn /tn → ϕ/t in L 1 . From this, taking into account that tn → t, the convergence of ϕn to ϕ in L 1 follows. Exercise 3.10. Let (c) := (c)/c and notice that |ϕi | ≤ c(|ϕi |)/(c) = (|ϕi |)/(c) on {|ϕi | ≥ c}. Therefore    (|ϕi |) M |ϕi | dμ ≤ |ϕi | dμ ≤ dμ+ +cμ(A). (c) A A∩{|ϕi |≥c} (c) A∩{|ϕi | 0 such that cδ < ε/2. The inequality above yields A |ϕi | dμ < ε whenever μ(A) < δ. Exercise 3.11. Let ( f n ) ⊂ Cb (X) be converging in L 1 to f , and let f n(k) be a subsequence pointwise convergent μ–a.e. to f . Then, given any ε > 0, by Egorov theorem we can find a Borel set B ⊂ X with μ(B) < ε and f n(k) → f uniformly on B c . By the inner regularity of the measure we can find a closed set C ⊂ B c such that μ(X \ C) < ε. The function f restricted to C, being the uniform limit of bounded continuous functions, is bounded and continuous.

163 Introduction to Measure Theory and Integration

Chapter 4 Exercise 4.1. Notice that ·, · is obviously symmetric, that x, −y = −x, y = −x, y and that x, x = x2 ≥ 0, with equality only if x = 0. Notice that the parallelogram identity gives x + x  + 2y2 + x − x  2 = 2x + y2 + 2x  + y2 = 8x, y + 8x  , y − 2x − y2 − 2x  − y2 and x + x  − 2y2 + x − x  2 = 2x − y2 + 2x  − y2 = 8x, −y + 8x  , −y − 2x + y2 − 2x  + y2 . Subtracting and dividing by 4 we get x + x  , 2y = 4x, y + 4x  , y − 2x, y − 2x  , y. So, we proved that x + x  , 2y = 2x, y + 2x  , y. Using the relation u, 2v = 4u/2, v (due to the definition of ·, · and the homogeneity of  · ), we get & % x + x 1 1 , y = x, y + x  , y. 2 2 2 Setting x = t1 v, x  = t2 v, and defining the continuous function φ(t) = tv, y, we get   t1 + t2 1 1 φ = φ(t1 ) + φ(t2 ). 2 2 2 This means that φ and −φ are convex in R, so that φ is an affine function, and since φ(0) = 0 we get φ(t) = tφ(0), i.e. tu, y = tu, y. Coming back to the identity above, we get x + x  , y = x, y + x  , y. Exercise 4.2. Assume that y = π K (x). For all z ∈ K and t ∈ [0, 1] we have y + t (z − y) belongs to K , so that y + t (z − y) − x2 ≥ y − z2 . Expanding the squares we get t 2 z − y2 + 2tz − y, y − x ≥ 0

∀t ∈ [0, 1].

This implies (either dividing by t > 0 and passing to the limit as t ↓ 0, or computing the right derivative at t = 0) that z − y, x − y ≤ 0. Conversely, if for some y ∈ K this condition holds for all z ∈ K , the

164 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

argument can be reversed to get y + t (z − y) − x ≥ y − x for all t ≥ 0. Choosing t = 1 we get z − x ≥ y − x, proving that y = π K (x). Exercise 4.3. Let Yk be the vector space spanned by { f 1 , . . . , f k } and let us prove by induction on k ≥ 1 that f i is orthogonal to f j whenever 1 ≤ i < j ≤ k. First we observe that if this property holds for some k, then Yk is k-dimensional and coincides with the vector space spanned by {v1 , . . . , vk } (being contained in it, and with the same dimension). The orthogonality of the vectors f i can be obtained just noticing that f k = vk −

k−1  vk , ei ei . i=1

So, f k = vk − πYk−1 (vk ) is orthogonal to all vectors in Yk−1 . It follows that ek , ei  = 0 for all i < k.  Exercise 4.4. Let y = x− k x, ek ek ; we know that the series converges in H by Bessel’s inequality. In order to show that k x, ek ek = π X (x) it suffices to prove that y is orthogonal to all vectors in X. But since any vector v ∈ X can be represented as a series, it suffices to show that v, ei  = 0 for all i. The continuity and linearity of the scalar product give y, ei  = x, ei  −

∞  x, ek x, ei  = x, ei  − x, ei  = 0. k=0

Exercise 4.5 Since X and its scalar product coincide with L 2 ([0, 1], P ([0, 1]), μ), where μ is the counting measure in [0, 1], we obtain that X is an Hilbert space. Let us prove by contradiction that X is not separable. If S = { f n }n≥1 were a dense subset, it could be possible to find a countable set D ⊂ [0, 1] such that f n (x) = 0 for all n and all x ∈ [0, 1] \ D. Since [0, 1] is not countable we can find x0 ∈ [0, 1] \ D and define g0 (x) equal to 1 if x = x0 and equal to 0 if x  = x0 . We claim that g0 does not belong to the closure of S. If this property fails, we can find a sequence ( f n(k) ) ⊂ S convergent to g0 μ–a.e. in [0, 1]; but, convergence μ–a.e. corresponds to pointwise convergence and since g0 (x0 )  = 0, while f n(k) (x0 ) = 0 for all k, we obtain a contradiction. Exercise 4.6. By Parseval identity we know that x  → (x, ei )) is a linear isometry from H to 2 . As a consequence, taking the parallelogram identity into account, the scalar product is preserved. Exercise 4.7. We consider the class of orthonormal systems {ei }i∈I of H , ordered by inclusion. Zorn’s lemma ensures the existence of a maximal

165 Introduction to Measure Theory and Integration

system {ei }i∈I . Let V be the subspace spanned by ei , let Y be its closure (still a subspace) and let us prove that Y = H . Indeed, if Y were a proper subspace of H , we would be able to find, thanks to Corollary 4.5, a unit vector e orthogonal to all vectors in Y , and in particular to all vectors ei . Adding e to the family {ei }i∈I the maximality of the family would be violated. Now, by the just proved density of V in H , given any x ∈ H we can find a sequence of vectors (vn ), finite combinations of vectors ei , such that x − vn  → 0. If we denote by Jn ⊂ I the set of indexes used to build the vectors {v1 , . . . , vn }, and by Hn the vector space spanned by {ei }i∈Jn , we know by Proposition 4.6 that  x − x, ei ei  ≤ x − vn  → 0. i∈Jn

As a consequence, setting J = ∪n Jn , we have x =



i∈J x, ei ei .

Chapter 5 Exercise 5.1. The functions sin mx cos lx are odd, therefore their integral on (−π, π) vanishes. To show that sin mx is orthogonal to sin lx when l = m, we integrate twice by parts to get  π  m π sin mx sin lx dx = cos mx cos lx dx l −π −π  m2 π = 2 sin mx sin lx dx. l −π The integrals of products cos mx cos lx can be handled analogously. Exercise 5.2. Since for N < M we have 

N  n=0

xn −

M  n=0

xn  ≤

M  i=N +1

xi  ≤

∞ 

xi 

i=N +1

 we obtain that ( 0N xi ) is a Cauchy sequence in E. Therefore the completeness of E provides the convergence series. Passing to the  of the limit as N → ∞ in the inequality  0N xi  ≤ 0N xi  and using the continuity of the norm we obtain (5.15). √ Exercise 5.3. We consider only the first system gk = 2/π sin kx, the proof for the second one being analogous. The fact that (gk ) is orthonormal can be easily checked noticing that gk are restrictions to (0, π) of odd functions, and using the orthogonality of sin kx in L 2 (−π, π). Analogously, if f ∈ L 2 (0, π) let us consider its extension f˜ to (−π, π) as an

166 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

odd function and its Fourier series, which obviously contains no cosinus. In (0, π) we have N 

bk sin kx =

k=1

N   f, gk gk , k=1

where the scalar products are understood in L 2 (0, π). Therefore, from the convergence of the Fourier series in L 2 (−π, π) to f˜, which implies convergence in L 2 (0, π) to f , the completeness follows. Exercise 5.4. Clearly ek , ek  = 1, while *π )  π 1 ikx −ilx i(k−l)x e e dx = dx =0 whenever k  = l. e i(k − l) −π −π As a consequence (ek ) is an orthonormal system. N  f, ek ek of f depends linearly on Since the Fourier series SN f = −N f , in order to show completeness we need only to show S N f → f when f is real-valued and when f is imaginary-valued (i.e. i f is real-valued). We consider only the first case, the second one being analogous. Setting ck =  f, ek , we have  π 1 ck = √ f (x) cos kx − i f (x) sin kx d x. 2π π √ As a consequence, for k ≥ 1 we have 2/πck = ak − ibk , where ak and bk are√the coefficients of the real Fourier series of f , and for √ k ≤ −1 we have 2/πck = a−k + ib−k . For k = 0, instead, we have 2/πc0 = a0 . Taking into account these relations and setting b0 = 0, we have

N 1  eikx = ck √ (cos kx + i sin kx)(ak − ibk ) 2 k=1 2π k=−N N 

+

−1 

(cos kx + i sin kx)(a−k

 − ib−k )

k=−N

  N a0 (cos kx + i sin kx)(ak − ibk ) = + Re 2 k=1 =

N a0  ak cos kx + bk sin kx, + 2 k=1

and the convergence of S N f to f follows by the convergence in the realvalued case.

167 Introduction to Measure Theory and Integration

Exercise 5.5. It suffices to note that 1 2π



π

f (x)e

−π

−ikx

2 dx

= ( f, ek )2 ,

where (ek ) is the orthonormal system of Exercise 5.4 and to use its completeness. 2N ikz e = (ei(2N +1)z − 1)/(ei z − 1), Exercise 5.6. From the identity i=0 we get N 

eikz = e−i N z

k=−N

=

e

2N 

eikz = e−i N z

k=0 i(N +1/2)z

ei(2N +1)z − 1 = ei z − 1

− e−i(N +1/2)z sin((N + 1/2)z) = (A.15) i z/2 −i z/2 e −e sin(z/2)

and we call this term G N (z). Hence  N   1  π f (y)e−iky dy eikx S N f (x) = 2π −π k=−N  π N  1 = f (y)eik(x−y) dy 2π −π k=−N  π 1 f (y)G N (x − y) dy. = 2π −π Using the fact that sin((N + 1/2)z)/ sin(z/2) has, still because of (A.15), mean value 1 on (−π, π), we get f (x) − S N f (x) =

1 2π



π

−π

( f (x) − f (y))G N (x − y) dy.

Exercise 5.7. We apply the Parseval identity to the function f (x) = x 2 , whose Fouries series contains no sinus. It is simple to check, by integration by parts, that a0 = 2π 2 /3 and that ak = 4k −2 cos kx for k ≥ 1. We have then 1 π



π

−π

∞ ∞  a2  16 2 4 x 4 dx = π 4 = 0 + ak2 = π 4 + . 5 2 18 k4 k=1 k=1

Rearranging terms, we get

∞ 1

k −4 = π 4 /90.

168 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Exercise 5.8. The polynomials Pn are given by Q n /Q n 2 , where Q n are recursively defined by Q 0 = 1 and Q n (x) := x n −

n−1 n−1   x n , Q k  x n , Pk Pk (x) ∀n ≥ 1. Q k (x) = x n − Q , Q  k k k=0 k=0

√ (a) Since Q 0 = 1, P0 = 1/ 2 and Q 1 = √ x − x, P0 P0 = x, because x, P0  = 0. As a consequence P1 (x) = 3/2x. Since x 2 , P1  = 0, we have also Q 2 (x) = x 2 − x 2 , P0 P0 − x 2 , P1 P1 = x 2 −

1 3

√ and this leads, with simple calculations, to P2 (x) = 45/8(x 2 − 1/3). (b) Let H be the closure of the vector space spanned by Cn . This space contains all monomials x n , and therefore all polynomials. Since the polynomials are dense in C([a, b]), for the sup norm, they are also dense in L 2 (a, b). It follows that H = L 2 (a, b). By Proposition 4.13 we conclude that (Cn ) is complete. (c) Set ( n 2n + 1 1 ˜n (x) := z n d (x 2 − 1)n , P z n := 2 2n n! dn x Clearly the polynomial P˜n has degree n. So, in order to show that P˜n = Pn , we have to show that P˜n is orthogonal to all monomials x k , k = 0, . . . , n −1, and that  P˜n 2 = 1. Since P˜n has zeros at ±1 with multiplicity n, all its derivatives at ±1 with order less than n are zero. Therefore, for k < n we have 

) *1  1 d n−1 d n−1 x k−1 n−1 (x 2 − 1)n dx  P˜n , x k  = z n x k n−1 (x 2 − 1)n − k d x d x −1 −1 = ··· ) n−k *1 d k 2 n = 0. = (−1) k!z n n−k (x − 1) d x −1 In order to prove that  P˜n 2 = 1, still integrating by parts we have 

1

n+1 d n−1 2 n d − 1) (x (x 2 − 1)n dx = · · · n−1 x n+1 x d d −1 (A.16)  1 2n 2 2 n d 2 n = zn (1 − x ) 2n (x − 1) dx. d x −1

 P˜n , P˜n  = −z n2

169 Introduction to Measure Theory and Integration

On the other hand  1  1 2 n (1 − x ) dx = 2n (1 − x 2 )n−1 x 2 dx −1

−1



= −2n so that 

1

−1

 (1 − x ) dx + 2n 2 n

1

−1

(1 − x 2 )n−1 dx,

 1 2n (1 − x ) dx = (1 − x 2 )n−1 dx = · · · 2n + 1 −1 −1  1 (2n)!! 2(2n)!! = (1 − x 2 )0 dx = . (2n + 1)!! −1 (2n + 1)!! 1

2 n

Taking into account that d 2n 2 (x − 1)n = (2n)! = (2n)!!(2n − 1)!! = 2n n!(2n − 1)!! d 2n x from (A.16) we get  P˜n , P˜n  =

1 2(2n)!! n 2n + 1 2 n!(2n − 1)!! = 1. 2 22n (n!)2 (2n + 1)!!

Exercise 5.9. Recall that 1 ck = 2π



π

−π

f (x)e−ikx dx.

Integrating by parts once and using that f (−π) = f (π) we get  π 1 1 ck = f  (x)e−ikx dx. ik 2π −π Continuing in this way, in m steps we get  π 1 1 f (m) (x)e−ikx dx. ck = (ik)m 2π −π

Chapter 6 Exercise 6.1. Let us prove the inclusion (F 1 × F 2 ) × F 3 ⊂ F 1 × (F 2 × F 3 ),

170 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

the proof of the converse one being analogous. We have to show that all products A × B, with A ∈ F 1 × F 2 and B ∈ F 3 belong to F 1 × (F 2 × F 3 ). Keeping B fixed, the class of sets A for which this property holds is a σ –algebra that contains the π–system of measurable rectangles A = A1 × A2 (because A × B = A1 ×(A2 × B) and A2 × B ∈ F 2 ×F 3 ), and therefore the whole product σ –algebra F 1 × F 2 . For all A in the product σ –algebra we have  μ3 (A x1 x2 ) dμ1 × μ2 (x1 , x2 ) (μ1 × μ2 ) × μ3 (A) =  X 1 ×X 2 = μ3 (A x1 x2 ) dμ2 (x2 ) dμ1 (x1 ) X1 X2  = μ2 × μ3 (A x1 ) dμ1 (x1 ) = μ1 × (μ2 × μ3 )(A). X1

×

n

Exercise 6.2. Obviously the cubes belong to 1 B (R), and thanks to Lemma 6.9 the same is true for the open sets. It follows that B (Rn ) is n contained in 1 B (R). Let us consider the class   M := B ⊂ R : B × R × · · · × R ∈ B (Rn ) .

×

This class contains the open sets (because the product of open sets is open) and it is a σ –algebra, so it contains B (R). We have thus proved that all rectangles B1 × R × · · · × R, with B1 Borel belong to B (Rn ). By a similar argument we can show that all rectangles R × · · · × R × Bi × R × · · · × R are Borel. Intersecting rectangles in these families we obtain that all rectangles with Borel sides belong to B (Rn ) and we conclude. Exercise 6.3. Assume that A, B ∈ L 1 ; then there exist Borel sets A , B  and Borel Lebesgue negligible sets N A , N B with AA ⊂ N A and BB  ⊂ N B . Since A × B  ∈ B (R2 ), by the previous exercise, (A × B)(A × B  ) ⊂ (N A × R) ∪ (R × N B ) and N A × R and R × N B are L 2 negligible, we obtain that A × B ∈ L 2 . This proves that L 2 contains the generators of L 1 × L 1 , and therefore the whole σ –algebra. In order to show the strict inclusion, we consider the set E = F × {0}, where F ⊂ R is not Lebesgue measurable. Since E is L 2 –negligible we have E ∈ L 2 . On the other hand, since the 0

171 Introduction to Measure Theory and Integration

section E 0 coincides with F, and therefore it does not belong to L 1 , the set E can’t belong to the product of the two σ –algebras. Exercise 6.4. Let A be the σ –algebra generated by these sets; since these sets are obviously cylindrical, A is contained in the product σ –algebra. n The class of sets B ⊂ 1 X i such that B × X n+1 × X n+2 × · · · ∈ A is a σ –algebra containing the measurable rectangles A1 × · · · × An , and n therefore contains the product σ –algebra 1 F i . Therefore A contains the cylindrical sets and, by definition, the whole product σ –algebra. Exercise 6.5. The#sections Ty := {(x, z) : (x, y, z) ∈ T } are squares with length side 2 r 2 − |y|2 for 0 ≤ |y| ≤ r, hence  r  r 1 16 3 2 L (Ty ) dy = 8 (r 2 − y 2 ) dy = 8(r 3 − r 3 ) = r 3 . L (T ) = 3 3 −r 0

×

×

Exercise 6.6. For x ∈ Rn (with n ≥ 3) let   r := (x12 + x22 )1/2 , Ar := (x3 , . . . , xn ) : (x32 + · · · + xn2 ) < 1 − r 2 . Then, using polar coordinates we get   1 ωn = L n−2 (Ar ) dx1 dx2 = 2πωn−2 r(1 − r 2 )(n−2)/2 dr {r<1}

0

2π ωn−2 . = n Therefore ω2k =

2k−1 π k−1 πk ω2 = 2k(2k − 2) · · · 4 k!

and an analogous argument gives ω2k+1 = 2k+1 π k /(2k + 1)!!. Exercise 6.7. In order to show that ωn =

π n/2 n ( +1) 2

we show that the right

hand side satisfies the same recursion formula of the previous exercise. √ Since (thanks to the identities (1) = 1, (1/2) = π) the formula holds when n = 1, 2, this will prove that the identity holds for all n. For n ≥ 2 we have π n/2 π · π (n−2)/2 2π π (n−2)/2 . = = n ( n2 + 1) ( n2 ) n ( (n−2) + 1) 2 2 Exercise 6.8. We know, by Exercise 2.4, that there exist a λ–negligible set N ∈ F ×G and a F ×G –measurable function F˜ : X ×Y → [0, +∞] ˜ is contained in N . By applying the Fubini–Tonelli such that {F  = F}

172 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

theorem to 1 N we obtain that N x is ν–negligible in Y for μ–a.e. x ∈ X. ˜ Since {F(x, ·)  = F(x, ·)} ⊂ N x , still Exercise 2.4 gives that F(x, ·) is ν–measurable for μ–a.e. x ∈ X. This proves statement (i). Since, still for μ–a.e. x ∈ X, the integral on Y (with respect to ν) of F(x, ·) coincides ˜ with the integral of F(x, ·), statements (ii) and (iii) follow by applying ˜ the Fubini–Tonelli theorem to F. Exercise 6.9. Indeed, μ(D y ) = μ({y}) = 0 for all y ∈ Y , so that Y μ(D y ) dν(y)  = 0. On the other hand, ν(Dx ) = ν({x}) = 1 for all x ∈ X, so that X ν(Dx ) dμ(x) = 1.  Exercise 6.10. Let (h(k)) be a subsequence such that k  f h(k) − f 1 is convergent. Then the Fubini–Tonelli theorem gives   ∞  X

k=0

 | f h(k) (x, y) − f (x, y)| dν(y) dμ(x)

Y

=

∞   k=0

| f h(k) (x, y) − f (x, y)| dμ × ν < ∞. X×Y

 It follows that k  f h(k) (x, ·) − f (x, ·) L 1 (ν) is finite for μ–a.e. x ∈ X, and for any such x the functions f h(k) (x, ·) converge to f in L 1 (ν). Choosing Y = { y¯ } and ν = δ y¯ , to provide a counterexample it is sufficient to consider any example (see Remark 3.7) of a sequence converging in L 1 but not μ–almost everywhere.  Exercise 6.11. It suffices to apply (6.15) to |h| to show that |h| d f μ is  finite if and only if |h| f dμ is finite. Exercise 6.12. We prove the property for the sup, the property for the inf being analogous. If A = B1 ∪ B2 with B1 ∈ F and B2 ∈ F disjoint, we have   f dμ + g dμ f μ(B1 ) + gμ(B2 ) = B1 B2   ≤ f ∨ g dμ + f ∨ g dμ B1 B2  f ∨ g dμ. = A

The arbitrariness of this decomposition, proves that [( f μ) ∨ (gμ)](A) ≤ ( f ∨g)μ(A). The converse inequality can be obtained noticing that, in the chain of equalities-inequality above, the inequality becomes an equality if we choose B1 = A ∩ { f ≥ g} and B2 = A ∩ { f > g}. Exercise 6.13. It is easy to check that μ ≤ μi (respectively, μ ≥ μi ) for all i ∈ I , and that any measure ν with this property is less than μ (resp.

173 Introduction to Measure Theory and Integration

  greater k ν(Bk ) ≤ k μi(k) (Bk ) (resp.  than μ): just write ν(B) = μ (B ). So, it remains to show that μ and μ are σ -additive. ≥ i(k) k k For any map i : N → I , A1 , A2 ∈ F disjoint and any countable F – measurable partition of A1 ∪ A2 we have ∞ 

μi(k) (Bk ) =

k=0

∞ 

μi(k) (Bk ∩ A1 ) +

k=0

∞ 

μi(k) (Bk ∩ A2 ).

k=0

Estimating the right hand side from below with μ(A1 ) + μ(A2 ) we get (because (Bk ) is arbitrary) that μ is superadditive, i.e. μ(A1 ∪ A2 ) ≥ μ(A1 ) + μ(A2 ). With a similar argument one can prove not only that μ is subadditive, but also that μ is σ –subadditive (it suffices to consider a countable F –measurable family, instead of 2 sets). Now, let us prove that μ is subadditive and μ is superadditive. Let A1 , A2 ∈ F be disjoint and let Bk1 , Bk2 be countable F –measurable partitions of A1 and A2 respectively. If i 1 , i 2 : N → I we define i(2k) = i 1 (k), B2k = Bk1 and i(2n + 1) = i 2 (n), B2k+1 = Bk2 , so that μ(A1 ∪ A2 ) ≤

∞ 

μi(k) (Bk ) =

k=0

∞ 

μi1 (k) (Bk1 ) +

k=0

∞ 

μi2 (k) (Bk2 ).

k=0

By the arbitrariness of Bk1 , Bk2 , i 1 and i 2 we conclude that μ(A1 ∪ A2 ) ≤ μ(A1 ) + μ(A2 ). With a similar argument one can prove that μ is even σ –subadditive (one has to use a bijection between N × N and N) and that μ is superadditive. Exercise 6.14. If for all ε > 0 there exists δ > 0 satisfying A ∈ F , μ(A) < δ

⇒

ν(A) < ε

then ν # μ: indeed, if μ(A) = 0 the implication above holds for all ε > 0, hence ν(A) = 0. If ν is finite, to prove the converse we argue by contradiction. Assume that, for some ε0 , we can find sets An ∈ F with μ(An ) < 2−n and ν(An ) ≥ ε0 . Then, by the Borel–Cantelli lemma the set A := lim supn An is μ–negligible. On the other hand, we have ν

∞ 

Am ≥ ν(An ) ≥ ε0

m=n

and therefore (here we use the assumption that ν is finite) ν(A) ≥ ε0 , contradicting the absolute continuity of ν with respect to μ. Exercise 6.15. Let B ∈ F be a μ–negligible set where ν is concentrated. Then ν(E) = ν(E ∩ B) for all E ∈ F . But, by the absolute continuity

174 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

of ν with respect to μ, we have ν(E ∩ B) = 0 because E ∩ B ⊂ B is μ–negligible. Exercise 6.16. Let B ∈ F be a ν–negligible set where σ is concentrated. Then σ (E) = σ (E ∩ B) ≤ μ(E ∩ B) + ν(E ∩ B) ≤ μ(E)

∀E ∈ F ,

where we used the fact that ν(E ∩ B) = 0 because E ∩ B ⊂ B is ν– negligible. Exercise 6.17. It is easy to check that the class of functions f satisfying f μ ≤ ν is a lattice. Hence, given a maximizing sequence ( f h ) in (6.20), possibly replacing f h by maxi≤n f i , we can assume that f h ↑ f . The monotone convergence theorem gives that f is a maximizer. In order to show that ν = f μ we set σ = ν − f μ ≥ 0 and notice that σ satisfies the following property: ⇒ μ(B) = 0. (A.17) t > 0, B ∈ F, t 1 B μ ≤ σ   Indeed, the integrals X ( f + t 1 B ) dμ and X f dμ have to coincide, because ( f + tχ B )μ ≤ ν. Exercise 6.18. We have to prove that any measure σ satisfying (A.17) is concentrated on a μ-negligible set. To this aim, let us consider the problem inf {μ(A) : A ∈ F, σ is concentrated on A} . By taking the intersection of a minimizing sequence it is easy to check that also this problem has a solution A; we have to show that μ( A) = 0. By the minimality of A, the implication F  B ⊂ A, μ(B) > 0

⇒

σ (B) > 0

(A.18)

holds. Let us consider the numbers   ξh := sup μ(B) : F  B ⊂ A, χ B μ ≥ 2h 1 B σ and let us prove that ξh → 0 as h → ∞. Given maximizers Bh ⊂ A, whose existence is easy to check, we have μ(Bh ) ≥ 2h σ (Bh ) and in  particular h σ (Bh ) < ∞. Hence   σ lim sup Bh = 0 h→∞

175 Introduction to Measure Theory and Integration

and (A.18) tells us that necessarily   0 = μ lim sup Bh ≥ lim sup μ(Bh ). h→∞

h→∞

Let us show now that the maximality of Bh implies that μ(C) ≤ 2h σ (C) for any set C ⊂ A \ Bh , i.e. t 1 A\Bh μ ≤ σ . Indeed, if there is C0 ⊂ A \ Bh with μ(C0 ) > 2h σ (C0 ), the maximality of Bh provides a minimal integer h 1 ≥ 1 and C1 ⊂ C0 satisfying μ(C1 ) ≤ 2h σ (C1 ) − 1/ h 1 . Let us consider C0 \ C1 ; we still have μ(C0 \ C1 ) > 2h σ (C0 \ C1 ) and the maximality of Bh provide a minimal integer h 2 ≥ h 1 and C2 ⊂ C0 \ C1 satisfying μ(C2 ) ≤ 2h σ (C2 ) − 1/ h 2 . Continuing in this way we have a nondecreasing sequence (h i ) of integers and (Ci ) ⊂ F such that μ(Ci ) ≤ 2h σ (Ci ) − 1/ h i and Ci ⊂ C0 \ ∪i−1 j=1 C j for all i ≥ 2; moreover h i is the least integer for which there is such Ci . Now limi h i = ∞, since the Ci are pairwise disjoint. Setting C = C0 \ ∪∞ 1 Ci , for all F ∈ F i−1 contained in C, since F ⊂ C0 \ ∪1 C j for all i ≥ 2, we have μ(F) ≥ 2h σ (F)−1/(h i −1) (if h i ≥ 2) and then μ(F) ≥ 2h σ (F). Hence Bh ∪C is an admissible set for the maximum problem defining ξh , against the maximality of Bh . We choose h in such a way that ξh < μ(A) and set t = 2−h , B = A\ Bh in (A.17). From (A.17) we conclude that μ(B) = 0, contradicting the fact that μ(B) = μ(A) − ξh > 0. Exercise 6.19. Let ν = ν + − ν − and let ν + = νa+ + νs+ , ν − = νa− + νs− be the Lebesgue decompositions with respect to μ of ν + and ν − respectively. Then, νa := νa+ − νa− and νs := νs+ − νs− provide a decomposition ν = νa + νs with νa , νs signed, |νa | # μ and |νs | ⊥ μ. If μ is signed and A provides a Hahn decomposition of μ (i.e. μ+(E) = μ(E ∩ A) and μ− (E) = −μ(E ∩ Ac )), we repeat the decomposition above in A, relative to ν and μ+ , and in B = Ac , relative to ν and μ− . Denoting by νaA + νsA and νaB + νsB the two decompositions obtained, νa (E) := νaA (E ∩ A)+νaB (E ∩ B),

νs (E) := νsA (E ∩ A)+νsB (E ∩ B)

provides the desired decomposition ν = νa + νs with |νa | # |μ| and |νs | ⊥ |μ|. The uniqueness of these decompositions can be proved with the same argument used in the case of nonnegative measures. Exercise 6.20. Let B ∈ F and let (Bi ) be a F –measurable partition of B; since !   ∞ ∞ !! ∞    ! !≤ ! | f μ(Bi )| = f dμ | f | dμ = | f | dμ, ! ! i=0

i=0

Bi

i=0

Bi

B

176 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

we obtain that | f μ|(B) ≤ | f |μ(B). To prove the converse inequality fix ε > 0 and define Bi = B ∩ f −1 (Ii ), where Ii = ε[i, i + 1), i ∈ Z. Since the oscillation of | f − εi| and || f | − ε|i|| in f −i (Ii ) are less than , we get ! ! ! ! ! ! ! ! ! | f | dμ−ε|i|μ(Bi )! ≤ εμ(Bi ), ! f dμ−εiμ(Bi )!! ≤ εμ(Bi ), ! ! ! Bi

hence

Bi

!  ! ! ! | f | dμ − ! ! Bi

Bi

! !! ! f dμ !! ≤ 2εμ(Bi ).

It follows that !   !! ! !≥ ! | f μ(Bi )| = f dμ | f | dμ − 2εμ(Bi ) ! ! B i i∈Z i∈Z i∈Z Bi  | f | dμ − 2εμ(B). = B

Since ε is arbitrary the converse inequality follows. Exercise 6.21. If x < 0 or x ≥ 1 all repartition functions are respectively equal to 0 or 1, so we need to consider only the case x ∈ [0, 1). The repartition function of 1[0,1] L 1 obviously is equal to x, while μh ((−∞, x]) =

#{i ∈ [1, h] : i ≤ hx} [hx] = , h h

where [s] denotes the integer part of s. Using the inequalities s − 1 < [s] ≤ s with s = hx we obtain that μh ((−∞, x]) → x. Exercise 6.22. The argument is similar to the one used in the proof of Theorem 6.27: if y < x < y  and y, y  ∈ D we have F(y) = lim Fh (y) ≤ lim inf Fh (x) ≤ lim sup Fh (x) h→∞

h→∞



h→∞



≤ lim Fh (y ) = F(y ). h→∞

Letting y ↑ x and y  ↓ x, we conclude. Exercise 6.23. We define a−h 2 = μ((−∞, −h]) and, for −h 2 < i ≤ h 2 , ai = μ((i − 1)/ h, i/ h]). Let us denote by μh the measure obtained in this way. If x ∈ (−h, h] and i is the smallest integer in (−h 2 , h 2 ] such that x ≤ i/ h, we have  *  * i−1  1 i −1 μ −∞, x − ai ≤ μh ((−∞, x]). ≤μ −∞, = h h j=−h 2

177 Introduction to Measure Theory and Integration

If x is not an atom of μ, this proves that lim inf μh ((−∞, x]) ≥ μ((−∞, x]). h

Analogously *  *  i  i 1 ai ≥ μh ((−∞, x]). ≥μ −∞, = μ −∞, x + h h j=−h 2 If x is not an atom of μ, this proves that lim sup μh ((−∞, x]) ≤ μ((−∞, x]). h

Exercise 6.24. Let us assume that (6.31) holds. If Fi (x) → 1 as x → +∞ uniformly in i ∈ I , for any ε > 0 we can find x such that 1−Fi (x) < ε/2 for all i ∈ I . Analogously, we can find y < x such that Fi (y) < ε/2 for all i ∈ I . Then, the interval I = (y, x] satisfies μi (I ) > 1 − ε for all i ∈ I , because I c = (−∞, y] ∪ (x, +∞). Exercise 6.25. If μ is the weak limit and ε > 0 is given, let us choose an integer n ≥ 1 such that μ([1 − n, n − 1]) > 1 − ε and points x ∈ (−n, 1 − n) and y ∈ (n − 1, n) where the repartition functions of μh are converging to the repartition function of μ. Then, since μ((∞, x]) + 1 − μ((−∞, y]) = μ(R \ (x, y)) < ε, there exists n ε ∈ N such that supn≥nε μn ((∞, x]) + 1 − μn ((−∞, y]) < ε. Let now x  and y  be satisfying μn ((∞, x  ]) + 1 − μn ((−∞, y  ]) < ε

∀n = 0, . . . , n ε − 1.

Then, the interval I = [min{x, x  }, max{y, y  }] satisfies infn μn (I ) > 1 − ε. Exercise 6.26.   ∀g ∈ Cb (R) (that is, (6.32)); (a) limh R g dμh = R g dμ ∀g ∈ Cc (R); (b) limh R g dμh = R g dμ (c) Fh converge to F on all points where F is continuous; (d) Fh converge to F on a dense subset of R; (e) limh μh (R) = μ(R); (f) (μh ) is tight. 1

x2

We consider the functions ρh (x) := ρ(x +h), where ρ(x) = (2π)− 2 e− 2 is the Gaussian, and μh = ρh λ (λ being the Lebesgue measure), μ = 0.

178 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

In this case (c), (d), do not hold, because Fh (x) → 1  = 0 = F(x) for all x ∈ R, (e) does not hold and (b) holds. a ⇒ b, e. This is easy, because Cc (R) ⊂ Cb (R) and 1R ∈ Cb (R). a ⇒ c. This follows by second part of the proof of Theorem 6.28. d ⇔ c. This is Exercise 6.22. b∧e ⇒ c. This follows by the same argument used in the proof of second part of Theorem 6.28: the sequence (gk ) monotonically convergent to 1 A can be chosen in Cc (R), and this shows that lim infh μh (A) ≥ μ(A) for all A ⊂ R open. Using (e) and passing to the complementary sets, we obtain lim suph μh (C) ≤ μ(C) for all C ⊂ R closed. d ⇒ f . This follows by the same argument used in the solution of Exercise 6.25. d ∧ f ⇒ e. For all x ∈ D, with D dense, we have limh μh ((−∞, x]) = μ((−∞, x]). Since μh ((−∞, x]) → μh (R) as x → +∞ uniformly in h, we can pass to the limit as x ∈ D → +∞ to obtain limh μh (R) = limx→+∞ μ((−∞, x]) = μ(R). d ∧ f ⇒ a. This follows by the same argument used in the first part of the proof of Theorem 6.28, choosing the points ti in the partitions to be in the dense set where convergence occurs. Exercise 6.27. Set  1 2 2 eiξ x e−x /(2σ ) dx. g(ξ ) := √ 2 2πσ R Notice that g(0) = 1, and that differentiation theorems under the integral sign (2) and an integration by parts give  1 2 2 g  (ξ ) = √ ieiξ x (xe−x /(2σ ) ) dx 2 2πσ R  d σ2 2 2 i eiξ x e−x /(2σ ) dx = √ 2 2πσ R dx  ξσ2 2 2 eiξ x e−x /(2σ ) dx. = −√ 2 2πσ R Therefore g satisfies the linear differential equation g  (ξ ) = −σ 2 ξ g(ξ ), 2 2 whose general solution is g(ξ ) = ce−σ ξ /2 . Taking into account that g(0) = 1, c = 1. d iξ x e−x 2 /(2σ 2 ) | (2) In this case, the application of the theorem is justified by the fact that sup ξ ∈I | dξ e

is Lebesgue integrable for all bounded intervals I

179 Introduction to Measure Theory and Integration

Exercise 6.28. Let us approximate μ by μn = 1(−n,n) μ; using the inequality x, ξ, η ∈ R |eiξ x − eiηx | ≤ |x||ξ − η| we obtain that



|μˆ n (ξ ) − μˆ n (η)| ≤ |ξ − η|

R

|x| dμn (x) ≤ n|ξ − η|,

ˆ )| ≤ μ(R \ therefore μˆ n is uniformly continuous. Since |μˆ n (ξ ) − μ(ξ [−n, n]), we have that μˆ n → μˆ uniformly as n → ∞, therefore μˆ is uniformly continuous (indeed, given ε > 0, find n such that sup |μˆ n − μ| ˆ < ε/2 and δ = ε/(2n) to obtain |μˆ n (ξ ) − μˆ n (η)| ≤ ε/2 whenever |ξ − η| < δ, and then |μ(ξ ˆ ) − μ(η)| ˆ < ε). ˆ 0 ) = eiθ for Exercise 6.29. Obviously |μ(ξ ˆ 0 )| = 1, and we set c = μ(ξ some θ ∈ R. Since  |1 − ce ¯ i xξ0 |2 dμ(x) = 2 − cc ¯ − cc¯ = 0, R

we obtain that ei xξ0 = c for μ–a.e. x ∈ R. This implies that xξ0 − θ ∈ 2πZ for μ–a.e. x ∈ R, so that μ is concentrated on the set of points {(2nπ + θ)/ξ0 }n∈N , and it suffices to set x0 = θ/ξ0 to obtain the stated representation of μ as a sum of Dirac masses. Obviously |μ| ˆ ≡ 1 if μ is a Dirac mass. Conversely, if |μ| ˆ ≡ 1, we find x0 with μ({x0 }) > 0 and ξ0 , ξ0 ∈ R \ {0} with ξ0 /ξ0 ∈ / Q to obtain that μ is concentrated on the set {2nπ/ξ0 +x0 }n∈N and on the set {2nπ/ξ0 +x0 }n∈N . By our choice of ξ0 and ξ0 , the intersection of the two sets is the singleton {x0 }, and this proves that μ = δx0 .

Chapter 7 Exercise 7.1. Let C > 0 be such that |H (x) − H (y)| ≤ C|x − y| for all x, y ∈ R.  Let ε > 0 and let δ > 0 be such that i | f (bi )− f (ai )| < ε/C (b − a ) < δ. We have whenever i i i i |H ( f (bi )) − H ( f (ai ))| ≤   C i | f (bi ) − f (ai )| whenever i (bi − ai ) < δ. In particular, choosing f (t) = t, we see that Lipschitz functions are absolutely continuous. Exercise 7.2. We assume that both L 1 (E) > 0 and L 1 (R \ E) > 0. Let a ∈ R be such that L 1 ((a, ∞) ∩ E) > 0 and L 1 ((a, ∞) \ E) > 0, and define F(t) = L 1 (E ∩(a, t)). By our choice of a, F(t) and (t −a)− F(t) are not identically 0 in (a, +∞). If t > a is a rarefaction point of E, we have F+ (t) = lim h↓0

F(t + h) − F(t) L 1 ((t, t + h) ∩ E) = lim = 0. h↓0 h h

180 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Analogously, F− (t) = 0 and we find that F  is equal to 0 at all rarefaction points. A similar argument proves that F  = 1 at all density points. Let now t0 ∈ (a, ∞) where 0 < F(t0 ) < (t0 − a) and apply the mean value theorem to obtain t0 ∈ (a, t0 ) such that F(t0 ) = (t0 − a)F  (t0 ). By our choice of t0 it follows that F  (t0 ) ∈ (0, 1), a contradiction (because either t0 is a density point or a rarefaction point). 7.3. Assume first that ϕ is continuous and bounded. Let H (z):= Exercise z ϕ(y) dy. By the (classical) fundamental theorem of the integral calf (a) culus, H is differentiable and H  (z) = ϕ(z) for all z ∈ f (I ). By the chain rule and Exercise 7.1, the function  f (t) F(t) := ϕ(y) dy = H ( f (t)) f (a)

is absolutely continuous and it has derivative equal to H  ( f (t)) f  (t) = ϕ( f (t)) f  (t) at all points t where f is differentiable. On the other hand, still by the fundamental theorem of the integral calculus, the function  t ϕ( f (x)) f  (x) dx G(t) := a

has derivative equal to (ϕ ◦ f ) f  L 1 –a.e. in [a, b]. Since both F and G vanish at t = a, they coincide. By the dominated convergence theorem, the identity of the two functions persists if ϕ = 1 A , with A open (because 1 A is the pointwise limit of continuous functions). By applying Dynkin’s theorem to the class M of  f (t) t the sets E ∈ B ( f (I )) such that f (a) 1 E (y) dy = a 1 E ( f (x)) f  (x) dx we obtain that the formula holds for all ϕ = 1 E with E Borel. Eventually we obtain it for simple functions and, by uniform approximation, for bounded Borel functions. b Exercise 7.4. Choosing g = 1 N , by Exercise 7.3 we get a 1 f −1 (N ) f  dx = 0, because 1 N ◦ f = 1 f −1 (N ) . Let h + and h − be respectively the positive and negative part of f  1 f −1 (N ) . Since  a

b

h + dx −

 a

b

h − dx =

 a

b

f  1 f −1 (N ) dx = 0

for all intervals (a, b), it follows that h + = h − L 1 –a.e. in R. As a consequence, f  = 0 L 1 –a.e. in f −1 (N ).

181 Introduction to Measure Theory and Integration

Chapter 8 Exercise 8.1. Both are measures in (Z , H ). If B ∈ H then g◦ f # μ(B) = μ( f −1 (g −1 (B))), because (g ◦ f )−1 = f −1 ◦ g −1 . On the other hand, g# ( f # μ)(B) = f # μ(g −1 (B)) = μ( f −1 (g −1 (B))). Exercise 8.2. Let n ≥ 1 integer, 0 ≤ k < 2n and let us consider the interval I = [k/2n , (k + 1)/2n ). Then, f −1 (I ) is the cylindrical set of all binary sequences a0 a1 · · · such that a0 · · · an−1 is the binary expression of k. It follows that  ∞   −1 1 1 f (I ) = L 1 (I ). δ0 + δ1 2 2 i=0

×

because their common value is 2−n . On the other hand, f −1 ({1}) consists of a single point and therefore the identity above holds for I = {1}, the common value being 0. By additivity the identity holds for finite unions of sets of this type, a family stable under finite intersections. By the coincidence criterion the two measures coincide. Exercise 8.3. Let A ⊂ R be a dense open set whose complement C has strictly positive Lebesgue measure (Exercise 1.9), and let ϕ(t) := min {1, dist(t, C)}

t ∈ R.

By construction the function ϕ is continuous, nonnegative, bounded by 1, and vanishes precisely on C. Then, set ⎧ t ⎪ ⎪ ϕ(s) ds if t ≥ 0; ⎨ 0 F(t) := 0 ⎪ ⎪ ⎩− ϕ(s) ds if t < 0. t

We have F  = ϕ, so that F ∈ C 1 and its critical set C F = C has positive Lebesgue measure. It follows that F# L 1 is not absolutely continuous b with respect to L 1 . Finally, since a ϕ dt > 0 whenever a < b (because A ∩ (a, b)  = ∅) we obtain that F is strictly increasing. Exercise 8.4. Recall that F(C F ) is always Lebesgue negligible, regardless of any injectivity assumption on U . Hence, possibly replacing U by U \C F we can assume with no loss of generality that C F = ∅, i.e. D F(x) is nonsingular at any x ∈ U . Recall that, according to the local invertibility theorem, for any x ∈ U there exists a ball Br (x) contained in U such that the restriction to F is injective. Now, following the strategy of

182 Luigi Ambrosio, Giuseppe Da Prato and Andrea Mennucci

Lemma 6.9 we can cover U by a sequence of right open cubes {Q i }i∈I , pairwise disjoint, such that the restriction of F to a neighbourhood of Q i is injective (we keep dividing a cube until this property is achieved). n Let Q i = i=1 [ai , ai + δ); for bi < ai sufficiently close to ai and n ˜ Q˜ i = i=1 (bi , bi + δ) we have (by injectivity of F on Q i )

×

×

F# (1 Q˜ i L n ) =

1 1 ˜ Ln |JF | ◦ F −1 F( Q i )

and therefore we can pass to the limit to get F# (1 Q i L n ) =

1 1 L n. |JF | ◦ F −1 (y) F(Q i )

If we add both sides with respect to i ∈ I we get F# (1U L n ) =

 i∈I

 1 1 n 1 L = 1 F(U ) L n . F(Q ) i −1 |JF | ◦ F (y) |J |(x) F x∈F −1 (y)

References

[1] L. C ARLESON, On the convergence and growth of Fourier series, Acta Math. 116 (1966), 135–157. [2] W. F. E BERLEIN, Notes on Integration I: The Underlying Convergence Theorem, Comm. Pure Appl. Math. X (1957), 357–360. [3] H. F EDERER, “Geometric Measure Theory”, Springer, 1969. [4] F. R IESZ and B. NAGY, “Functional Analysis”, Dover, 1990. [5] W. RUDIN, “Real and Complex Analysis”, McGraw-Hill, 1987. [6] S. WAGON, “The Banach-Tarski Paradox”, Cambridge University Press, 1985. [7] K. YOSIDA, “Functional Analysis”, Springer, 1980.

LECTURE NOTES

This series publishes polished notes dealing with topics of current research and originating from lectures and seminars held at the Scuola Normale Superiore in Pisa.

Published volumes 1. M. T OSI , P. V IGNOLO, Statistical Mechanics and the Physics of Fluids, 2005 (second edition). ISBN 978-88-7642-144-0 2. M. G IAQUINTA , L. M ARTINAZZI , An Introduction to the Regularity Theory for Elliptic Systems, Harmonic Maps and Minimal Graphs, 2005. ISBN 978-88-7642-168-8 3. G. D ELLA S ALA , A. S ARACCO , A. S IMIONIUC , G. T OMASSINI , Lectures on Complex Analysis and Analytic Geometry, 2006. ISBN 978-88-7642-199-8

4. M. P OLINI , M. T OSI , Many-Body Physics in Condensed Matter Systems, 2006. ISBN 978-88-7642-192-0 P. A ZZURRI, Problemi di Meccanica, 2007. ISBN 978-88-7642-223-2 5. R. BARBIERI, Lectures on the ElectroWeak Interactions, 2007. ISBN 978-88-7642-311-6

6. G. DA P RATO, Introduction to Stochastic Analysis and Malliavin Calculus, 2007. ISBN 978-88-7642-313-0 P. A ZZURRI, Problemi di meccanica, 2008 (second edition). ISBN 97888-7642-317-8

A. C. G. M ENNUCCI , S. K. M ITTER , Probabilit`a e informazione, 2008 (second edition). ISBN 978-88-7642-324-6 7. G. DA P RATO, Introduction to Stochastic Analysis and Malliavin Calculus, 2008 (second edition). ISBN 978-88-7642-337-6 8. U. Z ANNIER, Lecture Notes on Diophantine Analysis, 2009. ISBN 978-88-7642-341-3

9. A. L UNARDI, Interpolation Theory, 2009 (second edition). ISBN 978-88-7642-342-0

186 Lecture notes

10. L. A MBROSIO , G. DA P RATO , A. M ENNUCCI, Introduction to Measure Theory and Integration, 2011. ISBN 978-88-7642-385-7, e-ISBN: 978-88-7642-386-4

Volumes published earlier G. DA P RATO, Introduction to Differential Stochastic Equations, 1995 (second edition 1998). ISBN 978-88-7642-259-1 L. A MBROSIO, Corso introduttivo alla Teoria Geometrica della Misura ed alle Superfici Minime, 1996 (reprint 2000). E. V ESENTINI, Introduction to Continuous Semigroups, 1996 (second edition 2002). ISBN 978-88-7642-258-4 C. P ETRONIO, A Theorem of Eliashberg and Thurston on Foliations and Contact Structures, 1997. ISBN 978-88-7642-286-7 Quantum cohomology at the Mittag-Leffler Institute, a cura di Paolo Aluffi, 1998. ISBN 978-88-7642-257-7 G. B INI , C. DE C ONCINI , M. P OLITO , C. P ROCESI, On the Work of Givental Relative to Mirror Symmetry, 1998. ISBN 978-88-7642-240-9 H. P HAM, Imperfections de March´es et M´ethodes d’Evaluation et Couverture d’Options, 1998. ISBN 978-88-7642-291-1 H. C LEMENS, Introduction to Hodge Theory, 1998. ISBN 978-88-7642-268-3 Seminari di Geometria Algebrica 1998-1999, 1999. A. L UNARDI, Interpolation Theory, 1999. ISBN 978-88-7642-296-6 R. S COGNAMILLO, Rappresentazioni dei gruppi finiti e loro caratteri, 1999. S. RODRIGUEZ, Symmetry in Physics, 1999. ISBN 978-88-7642-254-6 F. S TROCCHI, Symmetry Breaking in Classical Systems, 1999 (2000). ISBN 978-88-7642-262-1

L. A MBROSIO , P. T ILLI, Selected Topics on “Analysis in Metric Spaces”, 2000. ISBN 978-88-7642-265-2 A. C. G. M ENNUCCI , S. K. M ITTER, Probabilit`a ed Informazione, 2000. S. V. B ULANOV, Lectures on Nonlinear Physics, 2000 (2001). ISBN 978-88-7642-267-6

Lectures on Analysis in Metric Spaces, a cura di Luigi Ambrosio e Francesco Serra Cassano, 2000 (2001). ISBN 978-88-7642-255-3 L. C IOTTI, Lectures Notes on Stellar Dynamics, 2000 (2001). ISBN 978-88-7642-266-9

S. RODRIGUEZ, The Scattering of Light by Matter, 2001. ISBN 978-88-7642-298-0

G. DA P RATO, An Introduction to Infinite Dimensional Analysis, 2001. ISBN 978-88-7642-309-3

S. S UCCI, An Introduction to Computational Physics: – Part I: Grid Methods, 2002. ISBN 978-88-7642-263-8 D. B UCUR , G. B UTTAZZO, Variational Methods in Some Shape Optimization Problems, 2002. ISBN 978-88-7642-297-3

187 Lecture notes

A. M INGUZZI , M. T OSI, Introduction to the Theory of Many-Body Systems, 2002. S. S UCCI, An Introduction to Computational Physics: – Part II: Particle Methods, 2003. ISBN 978-88-7642-264-5 A. M INGUZZI , S. S UCCI , F. T OSCHI , M. T OSI , P. V IGNOLO, Numerical Methods for Atomic Quantum Gases, 2004. ISBN 978-88-7642-130-0