MEASURE and INTEGRATION D-MATH, ETH Z¨Urich

MEASURE AND INTEGRATION D-MATH, ETH Z¨urich

Francesca Da Lio1

August 30, 2021

1Department of Mathematics, ETH Zürich, Rämistrasse 101, 8092 Zürich, Switzerland. Abstract

The aim of this course is to provide notions of abstract measure and integration which are more general and robust than the classical notion of Jordan measure and Riemann integral. For a presentation of Jordan measure and Riemann integral we suggest to look at the lecture notes of Analysis I & II by Michael Struwe (https://people.math.ethz.ch/ struwe/Skripten). ∼ The concept of measure of a general set in Rn originated from the classical notion of volume of a cube in Rn as product of the length of its sides. Starting from this, by a covering process one can assign to any subset a nonnegative number which “quantiﬁes its volume”. Such an association leads to the introduction of a set function called “Lebesgue measure” which is deﬁned over all subsets of Rn. Although the Lebesgue measure was initially developed in Euclidean spaces, the theory is independent of the geometry of the background space and applies to abstract spaces as well. Measure theory is now applied to most areas of mathematics such as functional analysis, geometry, probability, dynamical systems and other domains of mathematics. All results presented here are classical and their presentation follows the one in Struwe’s lecture notes. The proofs are mostly inspired by those in the books by Evans & Gariepy, and by Cannarsa & D’Aprile. Some References

Robert G. Bartle. The elements of Integration and Lebesgue • Measure, Wiley Classics Library, 1995. Piermarco Cannarsa and Teresa D’Aprile. Introduction to • measure theory and functional analysis, Vol. 89, Springer, 2015. Lawrence Craig Evans and Ronald F. Gariepy. Measure theory • and ﬁne properties of functions, revised edition, CRC Press, 2015. Gerald B. Folland. Modern Techniques and Their Applications, • second edition, John Wiley & Sons, INC, 1999. Lecture Notes by Prof. Michael Struwe, FS 13, • https://people.math.ethz.ch/ struwe/Skripten/AnalysisIII- ∼ FS2013-12-9-13.pdf

2 Contents

1 Measure Spaces 1 1.1 Algebras and σ-Algebras of Sets ...... 1 1.1.1 Notation and Preliminaries ...... 1 1.1.2 Algebras and σ-Algebras ...... 3 1.2 Measures ...... 6 1.2.1 Additive and σ-Additive Functions ...... 6 1.2.2 Measures and Measure Spaces ...... 8 1.2.3 Construction of Measures ...... 16 1.3 Lebesgue Measure ...... 21 1.4 Comparison between Lebesgue and Jordan Measure .. 27 1.5 An example of a non-measurable set ...... 29 1.6 An uncountable n-null set: the Cantor Triadic Set .. 32 L 1.7 Lebesgue-Stieltjes Measure ...... 34 1.8 Hausdorﬀ Measures ...... 41 1.8.1 Examples of non-integer Hausdorﬀ-dimension sets 48 1.9 Radon Measure ...... 52

2 Measurable Functions 55 2.1 Inverse Image of a Function ...... 55 2.2 Deﬁnition and Basic Properties ...... 55 2.3 Lusin’s and Egoroﬀ’s Theorems ...... 61 2.4 Convergence in Measure ...... 66

3 Integration 69 3.1 Deﬁnitions and Basic Properties ...... 69 3.2 Comparison between Lebesgue and Riemann-Integral . 82 3.2.1 Review of Riemann Integral ...... 83 3.3 Convergence Results ...... 86

3 3.4 Application: Diﬀerentiation under integral sign .... 91 3.5 Absolute Continuity of Integrals ...... 92 3.6 Vitali’s Theorem ...... 94 3.7 The Lp(Ω, µ) spaces, 1 p ...... 100 ≤ ≤∞ 4 Product Measures and Multiple Integrals 118 4.1 Fubini’s Theorem ...... 118 4.2 Applications ...... 129 4.3 Convolution ...... 131 4.4 Application:Solution of the Laplace Equation ...... 139

5 Differentiation of Measures 142 5.1 Differentiability of Lebesgue Integrals ...... 142 5.2 Differentiation of Radon measures ...... 147 5.3 Differentiation of absolutely continuousfunctions . . . . 151

4 Chapter 1

Measure Spaces

In this chapter we develop measure theory from an abstract point of view in order to construct a large variety of measures on a generic space X. In the particular case of X = Rn, a special role is played by Radon measures which have important regularity properties.

1.1 Algebras and σ-Algebras of Sets

1.1.1 Notation and Preliminaries We denote by X a nonempty set, by (X)or2X the set of all subsets P of X and the empty set. For any subset A of X we denote by Ac its ∅ complement, i.e. Ac = x X x / A . (1.1.1) { ∈ | ∈ } For any A,B (X) we set ∈ P A B = A Bc. (1.1.2) \ ∩ Let (A ) be a sequence in (X). The De Morgan’s identity holds n n P ∞ c ∞ c An = An . (1.1.3) n=1 n=1 [ \

1 We deﬁne1

∞ ∞ lim sup An = Am n →∞ n=1 m=n \ [ (1.1.4) ∞ ∞ lim inf An = Am . n →∞ n=1 m=n [ \

If L = limsupn An = liminfn An, then we set L = limn An →∞ →∞ →∞ and we say that (An)n converges to L. Remark 1.1.1. Note that:

1. liminf An lim sup An. n ⊆ n →∞ →∞ 2. If (A ) is increasing (A A ,n N), then n n n ⊆ n+1 ∈ ∞ lim An = An . (1.1.5) n →∞ n=1 [ Proof. Since An is increasing we have

∞ ∞ A A , n N . m ⊆ m ∀ ∈ m=1 m=n [ [ Therefore

∞ ∞ ∞ lim sup An = Am = Am. n n=1 m=n m=1 →∞ \ [ [ ∞ On the other hand for every n N we have Am = An and ∈ m=n therefore T ∞ lim inf An = Am. n →∞ m=1 [ 1We can “think” about the limit supremum and inﬁmum for sets in the following way:

lim sup An = x An, for inﬁnitely many n n→∞ { ∈ }

lim inf An = x An, for all but ﬁnitely many n n→∞ { ∈ }

2 3. If (A ) is decreasing (A A ,n N), then n n n ⊇ n+1 ∈ ∞ lim An = An . (1.1.6) n →∞ n=1 \

1.1.2 Algebras and σ-Algebras

Deﬁnition 1.1.2. Let (X), is called an algebra in X A ⊆ P A if a) X . ∈ A b) A,B = A B . ∈ A ⇒ ∪ ∈ A c) A = Ac . ∈ A ⇒ ∈ A

Remark 1.1.3. If is an algebra and A, B , then A B and A ∈ A ∩ A B belong to . Therefore, the symmetric diﬀerence A ∆ B = \ A (A B) (B A) belongs to as well. is also stable under ﬁnite \ ∪ \ A A unions and intersections

A1 An A1,...,An = ∪···∪ ∈ A ∈ A ⇒ ( A1 An . ∩···∩ ∈ A

Deﬁnition 1.1.4. An algebra (X) is called a σ-algebra if E ⊆ P for any sequence (A ) of elements in , we have ∞ A . n n E n=1 n ∈ E S

Remark 1.1.5. If isa σ-algebra in X and (A ) , then ∞ A E n n ⊆ E n=1 n ∈ by De Morgan’s identity (1.1.3). Moreover one can show (easy E T exercise!) that:

lim inf An , lim sup An . (1.1.7) n ∈ E n ∈ E →∞ →∞

3 Exercise 1.1.6.

1. (X) and = , X are σ-algebras. (X) is the largest P A {∅ } P σ-algebra in X and the smallest. A

2. In X = [0, 1), the class consisting of and of all ﬁnite A0 ∅ unions of half-closed intervals of the form

n A = [a ,b ), 0 a

Ac = [0, a ) [b , a ) [b , 1) , 1 ∪ 1 2 ∪···∪ n ∈ A0 moreover, is stable under ﬁnite unions. It suﬃces to A0 observe that the union of two (not necessarily disjoint) intervals of the form [a, b) and [c, d) belongs to . A0

3. In an infinite set X, consider the class . A = A (X) A is finite, or Ac is finite , (1.1.8) A { ∈ P | } is an algebra. The only point that needs to be checked A is that is stable under a finite union. Let A,B . If A ∈ A A,B are both finite, then so is A B. In all other cases, ∪ (A B)c is finite. but not a σ-algebra. ∪ A

4. In an uncountable set X, consider the class

= A (X) A is countable or Ac is countable . E { ∈ P | } (1.1.9) Then is a σ-algebra which is strictly smaller than (X) E P (here “countable” stands for “at most countable”).

4 Deﬁnition 1.1.7. Let (X). The intersection of all σ- K ⊆ P algebras including K is called the σ-algebra generated by K and will be denoted by σ( ). K

Exercise 1.1.8. Let (X). Verify that the intersection of σ- K ⊆ P algebras containing is actually a σ-algebra and it is the smallest K σ-algebra including . K

Example 1.1.9. 1. Let (X,d) be a metric space. The σ-algebra generated by all open sets of X is called the Borel σ-algebra of X and is denoted by (X). The elements of (X) are called Borel sets. B B 2. Let X = R and be the set of all half-closed intervals [a, b) I with a b. We show that σ( )= (R). ≤ I B “σ( ) (R)” : this follows from the fact that I ⊆B ∞ 1 [a, b)= a , b . − n n=1 \ “ (R) σ( )”: Let V be an open set in R, then it is a countable B ⊆ I union of some family of open intervals. Indeed, let (qk) be the sequence including all rational numbers of V and denote by Ik the largest open interval contained in V and containing q . We clearly have ∞ I k ∪k=0 k ⊂ V , but also the opposite inclusion holds: it suﬃces to consider, for any x V , r > 0 such that (x r, x + r) V, and k such that ∈ − ⊂ q (x r, x + r) to obtain that (x r, x + r) I , by the maximality k ∈ − − ⊂ k of I and then x I . k ∈ k Any open interval (a, b) can be represented as

∞ 1 1 (a, b)= a + ,b

5 1.2 Measures

1.2.1 Additive and σ-Additive Functions

Deﬁnition 1.2.1. Let (X) be an algebra and µ: A ⊆ P A → [0, + ] with µ( )=0. ∞ ∅ i) We say that µ is additive if for any ﬁnite family A ,...,A of mutually disjoint sets, we have 1 n ∈ A n n µ Ak = µ(Ak). (1.2.1) k=1 k=1 [ P ii) We say that µ is σ-additive if, for any sequence (A ) n n ⊆ A of mutually disjoint sets such that ∞ A , we have k=1 k ∈ A S ∞ ∞ µ Ak = µ(Ak). (1.2.2) k=1 k=1 [ P

6 Remark 1.2.2.

1. If µ is σ-additive function on then it is also additive. A 2. If µ is additive on , it is monotone with respect to the A inclusion: indeed, if A,B and B A, then µ(A) = ∈ A ⊆ µ(B)+ µ(A B). Therefore, µ(A) µ(B). \ ≥ 3. Let µ be additive on and let (A ) be mutually A n n ⊆ A disjoint sets such that ∞ A . Then k=1 k ∈ A n ∞ S n µ A µ A = µ(A ) n N, (1.2.3) k ≥ k k ∀ ∈ k=1 k=1 k=1 [ [ P therefore ∞ µ A ∞ µ(A ). (1.2.4) k ≥ k k=1 k=1 [ P 4. Any σ-additive function µ on is σ-subadditive, that is, A for any sequence (A ) in such that ∞ A n n A k=1 k ∈ A ∞ S µ A ∞ µ(A ). k ≤ k k=1 k=1 [ P

Indeed, let (An)n be such that n∞=1 An and deﬁne ⊆ A n 1 ∈ A B = A and B = A − A . Then B are mutually 1 1 n n \ k=1 k ∈ AS n disjoint, µ(B ) µ(A ) for all n N and n ≤ n S ∈ ∞ ∞ Bi = Ai. i=1 i=1 [ [ Therefore

∞ ∞ µ A = µ B = ∞ µ(B ) ∞ µ(A ). (1.2.5) i i i ≤ i i=1 i=1 i=1 i=1 [ [ P P

5. In view of points 3 and 4 an additive function is σ-additive if and only if it is σ-subadditive.

7 Definition 1.2.3. A σ-additive function µ on an algebra A ⊆ (X) is said to be P finite, if µ(X) < + ; • ∞ σ-finite, if there exists a sequence (A ) such that • n n ⊆ A ∞ A = X and µ(A ) < + n N. n n ∞ ∀ ∈ n=1 [

Exercise 1.2.4. In X = N consider the algebra

= A (X) A is finite or Ac is finite . (1.2.6) A ∈ P | The function µ: [0, + ] defined as • A→ ∞ # A if A is finite µ(A)= (1.2.7) ( if Ac is finite ∞ is σ-additive (# A = number of elements of A). The function ν: A [0, + ] defined as ν( )=0 and • → ∞ ∅ 1 n if A is finite ν(A)= n A 2 (1.2.8)  ∈ P if Ac is finite  ∞ is additive but not σ-additive.

1.2.2 Measures and Measure Spaces Let X be a set.

8 Deﬁnition 1.2.5. A mapping µ: (X) [0, + ] is called a P → ∞ measure on X if i) µ( )=0 and ∅ ∞ ∞ ii) µ(A) µ(Ak) whenever A Ak. ≤ k=1 ⊆ k=1 P S

Warning: Most texts call such a mapping µ an outer measure reserving the name measure for µ restricted to the collection of µ-measurable subsets of X (see below).

Remark 1.2.6. The condition ii) in Deﬁnition 1.2.5 implies the subadditivity and the monotonicity.

Deﬁnition 1.2.7. Let µ be a measure on X and A X, then µ ⊆ restricted to A written µ A is the measure deﬁned by

(µ A)(B)= µ(A B) B X. (1.2.9) ∩ ∀ ⊆

Deﬁnition 1.2.8. [Carath´eodory criterion of measurability] A X is called µ-measurable if for all ⊆ B X ⊆ µ(B)= µ(B A)+ µ(B A). (1.2.10) ∩ \

Remark 1.2.9. 1) If µ(A) = 0, then A is µ-measurable. Indeed, let B X. By the subadditivity, we have ⊆ µ(B) µ(B A)+ µ(B A). (1.2.11) ≤ ∩ \ 9 On the other hand, we have

µ(B A)+ µ(B A) µ(A)+ µ(B)= µ(B), (1.2.12) ∩ \ ≤ which yields the claim. 2) By the subadditivity the condition (1.2.10) is equivalent to

µ(B) µ(B A)+ µ (B A). ≥ ∩ \

Theorem 1.2.10. Let µ: (X) [0, + ] be a measure. P → ∞ Then the set

Σ= A X: A is µ-measurable { ⊆ } is a σ-algebra.

Proof. We proceed by steps.

i) X Σ: B X, it holds ∈ ∀ ⊆ µ(B X)+ µ(B X)= µ(B)+ µ( )= µ(B). (1.2.13) ∩ \ ∅

ii) A Σ= Ac Σ: B X it holds ∈ ⇒ ∈ ∀ ∈ B Ac = B A, B Ac = B A. (1.2.14) ∩ \ \ ∩ Therefore

µ(B Ac)+ µ(B Ac)= µ(B A)+ µ(B A)= µ(B). (1.2.15) ∩ \ \ ∩

iii) Let A ,...,A Σ. By induction on m we show that m A 1 m ∈ k=1 k ∈ Σ. S m = 1 is trivial. m 1 − (induction step) m 1= m: Let A = Ak. − ⇒ k=1 S

10 By inductive hypothesis, we assume A is µ-measurable. Thus, for B X, we have ∀ ⊆ µ(B)= µ(B A)+ µ(B A). (1.2.16) ∩ \

Since Am is also µ-measurable, we have

µ(B A)= µ (B A) Am + µ (B A) Am \ \ ∩ \ \ (1.2.17) = µ(B A) A + µB (A A ) . \ ∩ m \ ∪ m Now we observe that

B (A A )=(B A) [(B A) A ]. ∩ ∪ m ∩ ∪ \ ∩ m Thus by (1.2.17)

µ(B)= µ(B A)+ µ(B A) ∩ \ = µ(B A)+ µ (B A) A + µ (B A) A ∩ \ ∩ m \ \ m µ B (A A ) + µ B (A A ) ≥ ∩ ∪ m \ ∪ m (1.2.18) m From Remark 1.2.9 it follows that A Am = Ak Σ. ∪ k=1 ∈ S iv) Let (Ak)k N Σ. We show that A = k∞=1 Ak is µ-measurable. ∈ ⊂ We may suppose that A A = , k = ℓ, otherwise we consider k ∩ ℓ ∅ 6 S A = A , A = A A , A = A (A A ) 1 1 2 2 \ 1 3 3 \ 1 ∪ 2 n 1 − (1.2.19) Ae = A e A . e n n \ k k[=1 We cane note that ∞ ∞ Ak = Ak. k[=1 k[=1 e Since each Am is µ-measurable, we have

11 m m m µ B A = µ B A A + µ B A A ∩ k ∩ k ∩ m ∩ k \ m k=1 k=1 k=1 [ m 1 [ [ − = µ(B A )+ µ B A ∩ m ∩ k k=1 [m 1 m 1 − − = µ(B Am)+ µ B Ak Am 1 + µ B Ak Am 1 ∩ ∩ ∩ − ∩ \ − k=1 k=1 [ m 2 [ − = µ(B Am)+ µ(B Am 1)+ µ B Ak ∩ ∩ − ∩ k=1 ... [ (1.2.20) m = µ(B Ak). k=1 ∩ P By using the monotonicity of µ, we get

m m µ(B)= µ B A + µ B A ∩ k \ k k=1 k=1 (1.2.21) m [ [ µ(B Ak)+ µ(B A) ≥ k=1 ∩ \ P We let m + : → ∞ ∞ µ(B) µ(B Ak)+ µ(B A) ≥ k=1 ∩ \ µP(B A)+ µ(B A) ≥ ∩ \ by the σ-subadditivity. Therefore A Σ, by the Remark 1.2.9 ∈

Deﬁnition 1.2.11. If µ: (X) [0, + ] is a measure on X, P → ∞ Σ is the σ-algebra of µ-measurable sets, then (X, Σ, µ) is called a Measure Space.

12 Exercise 1.2.12. 1. Let x X. Define, for every A (X) ∈ ∈ P 1 x A δx(A)= ∈ (1.2.22) 0 x / A. ∈ Then δ is a measure on (X), called the Dirac measure x P at x. 2. For every A (X) we define ∈ P #A if A is finite µ(A)= (1.2.23) + otherwise ; ∞ µ is a measure on (X) and it is called the counting P measure. Every A is µ-measurable. 3. µ: (X) [0, + ], µ( ) = 0, µ(A) = 1, A = . In this P → ∞ ∅ 6 ∅ case, A X is µ-measurable A = or A = X. ⊆ ⇐⇒ ∅

13 Theorem 1.2.13. Let (X, Σ, µ) be a Measure Space and let A Σ, k N. Then the following conditions hold: k ∈ ∈ ∞ i) σ-additivity: Aj Aℓ = (j = ℓ) = µ Ak = ∩ ∅ 6 ⇒ k=1 ∞ S µ(Ak). k=1 P ii) Continuity from below: A A A 1 ⊆ 2 ⊆ ··· ⊆ k ⊂ ∞ Ak+1 ... = µ Ak = lim µ(Ak). ⊆ ⇒ k + k=1 → ∞ S iii) Continuity from above: A A A 1 ⊇ 2 ⊇ ··· ⊇ k ⊇ A ..., µ(A ) < + . Then k+1 ⊇ 1 ∞ ∞ µ Ak = lim µ(Ak). k + → ∞ k\=1

Proof. i) In the proof of Theorem 1.2.10 we showed that B X: ∀ ⊆ m m µ B A = µ(B A ) m. (1.2.24) ∩ k ∩ k ∀ k=1 k=1 [ P By taking B = X, we get in particular

m m µ Ak = µ(Ak). (1.2.25) k=1 k=1 [ P By using the monotonicity and the σ-subadditivity, we get m ∞ µ Ak lim µ Ak ≥ m + k=1 → ∞ k=1 [ m [ = lim µ(Ak) (1.2.26) m + → ∞ k=1 ∞ P ∞ = µ(Ak) µ Ak k=1 ≥ k=1 P S 14 ii) Consider the pairwise disjoint family of sets Ak = Ak Ak 1, \ − A1 = A1: e e A3

e A1

Fig. 1.1 e

By using i), we get m ∞ µ( k∞=1Ak)= µ( k∞=1Ak)= µ(Ak)= lim µ(Ak) m + ∪ ∪ k=1 → ∞ k=1 m P P = lim e µ(Ak) µe(Ak 1)+ µ(A1) e m + − → ∞ k=2 − = lim µP(Am). m + → ∞ (1.2.27) iii) One considers A = A A ,k N. We have k 1 \ k ∈ = A A .... (1.2.28) e ∅ 1 ⊆ 2 ⊆ We observe that k N ∀ ∈ e e µ(A1)= µ(Ak)+ µ(Ak). Thus e µ(A1) lim µ(Ak)= lim µ(Ak) − k + k + → ∞ → ∞ ∞ e ∞ = µ Ak = µ A1 Ak ii) \ (1.2.29) k[=1 k\=1 e ∞ = µ(A ) µ A . 1 − k k\=1 15 We can conclude.

Remark 1.2.14. If µ(A )=+ , then iii) in Theorem 1.2.13 1 ∞ may be false. Consider #A if A is ﬁnite µ(A)= + otherwise . ∞ Take A := m N : m n . We have µ(A )=+ and n { ∈ ≥ } 1 ∞ ∞ µ Ak =0 = lim µ(Ak). 6 k + → ∞ k\=1

1.2.3 Construction of Measures Let X = . 6 ∅ Deﬁnition 1.2.15. Let (X). is called a covering of K ⊆ P K X if i) . ∅∈K ∞ ii) (Kj)j N : X = Kj. ∃ ∈ ⊆K j=1 S

Example 1.2.16. The open intervals I = n (a ,b ), a b R k=1 k k k ≤ k ∈ are a covering of X = Rn. Each algebra (X) is a covering QA ⊆ P for X, since , X . ∅ ∈ A

16 Theorem 1.2.17. Let be a covering for X, λ: K K → [0, + ] with λ( )=0. Then ∞ ∅ ∞ ∞ µ(A) = inf λ(Kj): Kj , A Kj (1.2.30) j=1 ∈K ⊆ j=1 n P S o is a measure on X.

Proof of Theorem 1.2.17. We observe that µ 0, µ( ) = 0. We show ≥ ∅ that it is σ-subadditive. Let A ∞ A . The inequality µ(A) ⊆ k=1 k ≤ ∞ µ(A ) is trivial if the right-hand side is infinite. Therefore k=1 k S assume that ∞ µ(A ) is finite. For any k N and any ε > 0 P k=1 k ∈ there exists (K ) such that Pj,k j ⊆K ∞ ε A K and ∞ λ(K ) µ(A )+ . k ⊆ j,k j,k ≤ k 2k j=1 j=1 [ P Since A K . By definition of µ, we have ⊂ j,k j,k

S ∞ µ(A) λ(Kj,k) µ(Ak)+ ε. (1.2.31) ≤ j,k ≤ k=1 P P We let ε 0 and the σ-subadditivity follows. →

Exercise 1.2.18. X = , = , X , λ: [0, + ], λ( )= 0, 6 ∅ K {∅ } K → ∞ ∅ λ(X) = 1. The measure on X induced by λ is exactly the measure µ deﬁned by µ( )=0 and µ(A)=1if A = . ∅ 6 ∅

If = (X) is an algebra and λ: [0, + ] is additive, K A ⊆ P A → ∞ a natural question is: is the measure µ deﬁned in Theorem 1.2.17 a σ-additive extension of λ? We will give a suﬃcient condition in order for this to hold.

17 Deﬁnition 1.2.19. Let (X) be an algebra. A mapping λ: A ⊆ P [0, + ] is called a pre-measure if A→ ∞ i) λ( )=0. ∅ ii) λ(A) = ∞ λ(A ) for every A such that A = k=1 k ∈ A ∞ A , A mutually disjoint. k=1 k Pk ∈ A

λ isS called σ-ﬁnite if there exists a covering X = k∞=1 Sk, S and λ(S ) < + , k (we may assume without loss k ∈ A k ∞ ∀ S of generality the Sk’s are mutually disjoint).

Given a pre-measure λ, we obtain as above a measure µ deﬁned by

∞ ∞ µ(A) = inf λ(Ak): A Ak, Ak . (1.2.32) k=1 ⊆ k=1 ∈ A n P S o

Theorem 1.2.20. (Carath´eodory-Hahn extension) Let λ: [0, + ] be a pre-measure on X, µ be deﬁned as in A → ∞ (1.2.32). Then it holds: i) µ: (X) [0, + ] is a measure. P → ∞ ii) µ(A)= λ(A) A . ∀ ∈ A iii) A is µ-measurable. ∀ ∈ A

Proof of Theorem 1.2.20. i) It follows from Theorem 1.2.17. ii) Let A . Clearly it holds µ(A) λ(A). ∈ A ≤ For the opposite inequality, let A ∞ A , A mutually disjoint, ⊆ k=1 k k A . We set A = A A . Clearly, A are mutually k ∈ A k k ∩ ∈S A k disjoint and A = k∞=1 Ak. Since λ is a pre-measure, it follows e e S ∞ ∞ λ(A)=e λ(Ak) λ(Ak). (1.2.33) k=1 ≤ k=1 P P 18 e Thus ∞ ∞ λ(A) inf λ(Ak) A Ak, Ak ≤ k=1 ⊆ k=1 ∈ A (1.2.34) n o = µ(A).P S iii) Let A and B X be arbitrary. For every ε > 0 choose ∈ A ⊆ B , k N with B ∞ B and k ∈ A ∈ ⊆ k=1 k ∞ S λ(Bk) µ(B)+ ε. k=1 ≤ P Since A,B , we have k ∈ A λ(B )= λ(B A)+ λ(B A). (1.2.35) k k ∩ k \ Moreover, we have

∞ B A B A ∩ ⊆ k ∩ k=1 [ (1.2.36) ∞ B A (B A). \ ⊆ k \ k[=1 Thus, by deﬁnition of µ, we have

µ(B A)+ µ(B A) ∞ λ(B A)+ ∞ λ(B A) ∩ \ ≤ k ∩ k \ k=1 k=1 (1.2.37) P∞ P = λ(Bk) µ(B)+ ε k=1 ≤ P We let ε 0 and we conclude the proof. →

Theorem 1.2.21. (Uniqueness of Charathéodory- Hahn extension) Let λ: [0, + ] be as in Theorem 1.2.20 and σ- A → ∞ finite, µ be the measure induced by λ, Σ be the σ-algebra of the µ-measurable sets and let µ: (X) [0, + ] be P → ∞ another measure with µ = λ. Then µ Σ = µ, namely the |A | Charathéodory-Hahn extension is unique.e e e

19 Proof. i). Let A ∞ A with A , k N. From the σ-sub- ⊆ ∪k=1 k k ∈ A ∈ additivity ofµ ˜ it follows

µ˜(A) Σ∞ µ˜(A )=Σ∞ λ(A ). ≤ k=1 k k=1 k

By taking the inﬁmum over all (Ak)k we obtain µ˜(A) µ(A). (1.2.38) ≤ ii) We show that if A Σ then also the inverse inequality holds. To ∈ this purpose we consider ﬁrst of all the case there is S such that ∈ A A S and λ(S) < + . ⊆ ∞ By i) we get

µ˜(S A) µ(S A) µ(S)= λ(S) < . (1.2.39) \ ≤ \ ≤ ∞ Since A Σ, S Σ it follows by the sub-additivity ofµ ˜ that ∈ ∈A⊂ µ˜(A)+˜µ(S A) µ(A)+ µ(S A)= µ(S) \ ≤ \ = λ(S)=˜µ(S) µ˜(A)+˜µ(S A).(1.2.40) ≤ \ From (1.2.39) and (1.2.40) it follows that µ(A)=˜µ(A).

Suppose now that X = k∞=1Sk where (Sk)k N are mutually disjoint, ∪ ∈ S , λ(S ) < , for all k N. We consider the disjoint union k ∈ A k ∞ ∈ A = ∞ A , A = A S , k N. For every m it holds ∪k=1 k k ∩ k ∈ µ˜( m A )= µ( m A ). ∪k=1 k ∪k=1 k By taking the limit m + we get → ∞ m m µ˜(A) lim µ˜( k=1Ak)= lim µ( k=1Ak)= µ(A). m m ≥ →∞ ∪ →∞ ∪ We can conclude the proof.

Remark 1.2.22. i) From Theorem 1.2.21 it is not clear if every A Σ is also µ˜-measurable. ∈ ii) In general the statement of Theorem 1.2.21 cannot be improved as the following example shows.

20 Example 1.2.23. Let X = [0, 1], = , X and λ: A {∅ } A → [0, + ], µ as in the exercise 1.2.18 . We have Σ = , X . ∞ {∅ } We will see in the next section that Lebesgue measure 1 is also L an extension of λ but 1([0, 1/2])= 1/2 = µ([0, 1/2])= 1. L 6

1.3 Lebesgue Measure

Lebesgue measure is the extension of “volume” of the so-called elementary sets.

Deﬁnition 1.3.1. i) For a =(a ,...,a ), b =(b ,...,b ) Rn we set 1 n 1 n ∈ n (ai,bi) if ai

vol (a, b) = vol (a, b] = vol [a, b) = vol [a, b] n (bi ai) + if ai

iii) An elementary set is the ﬁnite disjoint union of intervals with volume m m vol Ik = vol(Ik). k=1 k=1 [ P

Remark 1.3.2.

21 m n i) Let I = Ik = Jj where Ik,Jj are intervals with Ik Iℓ = k=1 j=1 ∩ ∅ if k = ℓ andS J SJ = , if h = i, then 6 h ∩ i ∅ 6 m n vol(Ik)= vol(Jj). k=1 j=1 P P ii) The class of all elementary sets is an algebra, the vol function is a pre-measure. In particular, we have the following property:

if I is an elementary set such that I = k∞=1 Ik, Ik elementary mutually disjoint sets, then one can see (easy exercise) that S ∞ vol(I)= vol(Ik). k=1 P In what follows, we shall use the notion of dyadic cubes to obtain a basic decomposition of open sets in Rn. For every k N let be the ∈ Dk collection of half open cubes n a a +1 = i , i , a Z . Dk 2k 2k i ∈ i=1 n Y h o In other words, is the collection of cubes with edge length 1 and D0 vertices at points with integer coordinates. Bisecting each edge of a cube in , we obtain from it 2n subcubes of edge length 1. If we D0 2 continue bisecting, we obtain ﬁner and ﬁner collections k of cubes k D n such that each cube in has edge length 2− and is the union of 2 Dk disjoint cubes in . Dk+1 The cubes of the collection Q : Q , k =0, 1, 2,... ∈ Dk are called dyadic cubes . Remark 1.3.3. Dyadic cubes have the following properties:

n a) For every k 0, R = Q k Q with disjoint union. ≥ ∈D b) If Q and P Swith h k, then Q P or P Q = . ∈ Dk ∈ Dh ≤ ⊂ ∩ ∅ kn c) If Q , then vol(Q)=2− . ∈ Dk 22 Lemma 1.3.4. Every open set in Rn can be written as a countable union of disjoint dyadic cubes.

Proof. Let = V be an open set in Rn. Let be the collection of all ∅ 6 S0 cubes in which lie entirely in V . Let be the set of cubes in D0 S1 D1 which lie in V but which are not subcubes of any cube in . More S0 generally, for k 1, let be the cubes in which lie in V but which ≥ Sk Dk are not subcubes of any cube in 0, 1,..., k 1 (see Fig. 1.2). S S S − If = , then is countable since each is countable and S ∪kSk S Dk the cubes in are non-overlapping by construction. Moreover, since S V is open and the cubes in become arbitrarily small as k + , Dk → ∞ then by Remark 1.3.3 a) each point of V will eventually lie in a cube of some k. Hence V = Q Q and the proof is complete. S ∈S S

Fig: 1.2

Definition 1.3.5. The Lebesgue measure n is the Carathéodory- L Hahn extension of the volume defined on the algebra of elementary sets.

Deﬁnition 1.3.6. A measure µ on Rn is called Borel, if every Borel set is µ-measurable.

23 As a consequence of Lemma 1.3.4 we get that n is a Borel measure L and the σ-algebra of the n-measurable sets contains the σ-algebra of L the Borel sets. In the sequel we are going to list some properties of Lebesgue measure.

Theorem 1.3.7. For every A Rn it holds ⊆ n(A)= inf n(G), G open. (1.3.2) L A G L ⊆

Proof. “ ”: This inequality follows from the monotonicity: if A ≤ ⊆ G = n(A) n(G). ⇒ L ≤ L “ ”: We may suppose without loss of generality that n(A) < + ≥ L ∞ (otherwise the inequality is trivial). For ε> 0, choose intervals (Iℓ)ℓ N ∈ with A ∞ I and ⊆ ℓ=1 ℓ S ∞ n vol(Iℓ) (A)+ ε. ℓ=1 ≤ L P Since we are assuming that n(A) < + we have vol(I ) < + L ∞ ℓ ∞ for every ℓ> 0. For every ℓ we choose an open bounded interval Iℓ Iℓ such that ε ⊃ vol(I ) vol(I )+ ℓ , ℓ N. We set G = ∞ I , G is an open set ℓ ≤ ℓ 2 ∈ ℓ=1 ℓ containing A. Moreover, e e S e n ∞ ∞ ∞ ε (G) vol(Iℓ) vol(Iℓ)+ ℓ L ≤ ℓ=1 ≤ ℓ=1 ℓ=1 2 (1.3.3) P P P n(A)+2e ε. ≤ L We let ε 0 and get the result. →

24 Theorem 1.3.8. Let A Rn. The following conditions are ⊆ equivalent:

i) A is n-measurable. L ii) ε> 0 G A open: n(G A) <ε. ∀ ∃ ⊇ L \

Proof of Theorem 1.3.8. i) = ii): ⇒ Assume ﬁrst that n(A) < . For every ε > 0, choose G A, • L ∞ ⊇ G open with n(G) n(A)+ ε. L ≤ L Since A is n-measurable, by choosing B = G in the criterion of L measurability, we get n(G)= n(G A)+ n(G A) L L ∩ L \ (1.3.4) = n(A)+ n(G A). L L \ Therefore, n(G A)= n(G) n(A) ε. (1.3.5) L \ L − L ≤ n n If (A)= , we set A = A [ k,k] , k N, with A = ∞ A . • L ∞ k ∩ − ∈ ℓ=1 ℓ For every ε> 0 choose G A , G open with k ⊇ k k S ε n(G A ) < , k N. L k \ k 2k ∈

Then G = ∞ G is open, A G and ℓ=1 k ⊆ S n ∞ n (G A) (Gk Ak) < ε, since L \ ≤ k=1 L \ P ∞ ∞ G A = (G A) (G A ). \ k \ ⊆ k \ k k[=1 k[=1 ii) = i): For ε> 0, choose A G, G open with ⇒ ⊆ n(G A) ε. L \ ≤ 25 Since G is n-measurable, we have for every B Rn: L ⊆ n(B)= n(B G)+ n(B G) L L ∩ L \ n(B A)+ n(B A) n(G A) (1.3.6) ≥ L ∩ L \ − L \ n(B A)+ n(B A) ε. ≥ L ∩ L \ − In the second inequality, we have used the relation B A (B G) \ ⊆ \ ∪ (G A). We let ε 0 and we get the result. \ →

Corollary 1.3.9. Let A Rn. Then A is n-measurable if ⊆ L and only if it can be “approximated” from inside and outside: ε> 0 F A G, F closed, G open such that ∀ ∃ ⊆ ⊆ n(G A)+ n(A F ) ε. (1.3.7) L \ L \ ≤

Proof. If (1.3.7) holds, then the condition ii) of Theorem 1.3.8 holds and therefore A is n-measurable. Conversely, for ε > 0 choose G L ⊇ Ac open such that

n(G Ac) <ε (Theorem 1.3.8) . L \ Set F = Gc A. Then F is closed with ⊆ n(A F )= n(A G)= n(G Ac) <ε L \ L ∩ L \ and (1.3.7) follows.

Corollary 1.3.10. A Rn is n-measurable iﬀ it holds ⊆ L ε> 0 F A G, F closed, G open: n(G F ) < ε. ∀ ∃ ⊆ ⊆ L \ (1.3.8)

26 Proof. We observe that if F A G then ⊆ ⊆ G F =(G A) (A F ). (1.3.9) \ \ ∪ \ (1.3.7) = (1.3.8): From (1.3.9) it follows that ⇒ n(G F ) n(G A)+ n(A F ). L \ ≤ L \ L \ (1.3.8) = (1.3.7): We observe that ⇒ G A G F, A F G F. \ ⊆ \ \ ⊆ \ Therefore n(G A)+ n(A F ) 2 n(G F ). L \ L \ ≤ L \

1.4 Comparison between Lebesgue and Jordan Measure

To conclude, we compare Lebesgue measure with Jordan measure. We recall that A Rn bounded is Jordan-measurable if µ(A)= ⊆ µ(A), where

µ(A)= χ dµ := sup vol(E): E A, E elementary set A ⊆ } Z Rn µ(A)= χ dµ := inf vol(E): A E,E elementary set ; A ⊆ } Z R In this case we denote the common value by µ(A).

χA denotes the characteristic function of A: 1 x A χA(x)= ∈ 0 x / A. ∈

Theorem 1.4.1. Let A Rn be bounded, then ⊆ i) µ(A) n(A) µ(A). ≤ L ≤ ii) If A is Jordan-measurable, then A is n-measurable and L n(A)= µ(A). L

27 Proof. i) It holds

n ∞ ∞ (A) = inf vol(Ik), A Ik,Ik intervals L k=1 ⊆ k=1 n Pm S m o inf vol(Ik) A E = Ik,Ik intervals ≤ k=1 ⊆ k=1 n o = µ(A).P S (1.4.10)

Moreover, for every E = m I A, I mutually disjoint, it k=1 k ⊆ k holds S vol(E)= n(E) n(A) (1.4.11) L ≤ L and therefore µ(A) n(A). ≤ L ii) If A is Jordan-measurable, then µ(A) = µ(A) and from i), we have µ(A)= µ(A)= µ(A)= n(A). (1.4.12) L Now we show that A is n-measurable. Since A is bounded we L have n(A) < + . Since A is Jordan-measurable, ε > 0 I , L ∞ ∀ ∃ ε Iε elementary sets such that Iε A I and ⊇ ⊇ ε vol(Iε) ε < µ(A) < vol(I )+ ε. − ε We may assume that G = Iε is open. It follows that n n ε ε (G A) (I Iε) = vol(I Iε) L \ ≤ L \ \ (1.4.13) = vol(Iε) vol(I ) < 2ε. − ε Therefore A is n-measurable according to Theorem 1.3.8. L

Theorem 1.4.2. Lebesgue measure is invariant under isometries namely under Φ: Rn Rn, x x +Rx, where → 7→ 0 R is a rotation (RtR = Id).

28 Proof. Exercise! We present a further property of Lebesgue measure.

Deﬁnition 1.4.3. A Borel measure µ on Rn is called Borel regular if for every A Rn there exists B A Borel set such ⊆ ⊇ that µ(A)= µ(B).

Corollary 1.4.4. n is Borel regular. L

Proof. Without loss of generality let n(A) < (otherwise B = Rn). L ∞ Choose G open such that A G and k ⊆ k 1 n(G ) < n(A)+ , k N. L k L k ∈ In particular n(G ) < n(A)+1 < + . Without loss of generality L 1 L ∞ we may assume G G ,k N. Set B = ∞ G . Then B is Borel k+1 ⊆ k ∈ k=1 k and n n Tn (B)= lim (Gk)= (A). L k + L L → ∞

1.5 An example of a non-measurable set

In this section we will construct a subset of R which is not 1 measurable. L We will need the axiom of the choice in the following form:

Zermelo’s Axiom: For any family of arbitrary nonempty disjoint sets indexed by a set A, E , α A , there exists a set consisting of exactly one { α ∈ } element from each E , α A. α ∈ For x, y [0, 1) we deﬁne ∈ x + y, if x + y < 1 x y = (1.5.1) ⊕ ( x + y 1, if x + y 1. − ≥ 29 We observe that if E [0, 1)isa 1-measurable set, then E x [0, 1) ⊆ L ⊕ ⊆ is also 1-measurable and 1(E x)= 1(E) x [0, 1). Indeed, we L L ⊕ L ∀ ∈ have E x = E E (1.5.2) ⊕ 1 ∪ 2 where

E1 := E [0, 1 x) x = E [0, 1 x)+ x ∩ − ⊕ ∩ − (1.5.3) E := E [1 x, 1) x = E [1 x, 1)+(x 1). 2 ∩ − ⊕ ∩ − − E ,E are 1-measurable with 1 2 L 1(E )= 1 E [0, 1 x) L 1 L ∩ − 1(E )= 1E [1 x, 1) L 2 L ∩ − and E E = . Therefore, E x is measurable and 1 ∩ 2 ∅ ⊕ 1(E x)= 1(E )+ 1(E ) L ⊕ L 1 L 2 = 1 E [0, 1 x) + 1 E [1 x, 1) L ∩ − L ∩ − = 1(E). L In [0, 1) we deﬁne the following equivalence relation: x, y [0, 1), ∈ x y if x y Q. By the Axiom of Choice, there exists a set ∼ − ∈ P [0, 1) such that P consists of exactly one representative point from ⊆ each equivalence class. Let Q [0, 1) = rj j N be an enumeration of ∩ { } ∈ Q [0, 1) with r = 0, and for j N we deﬁne ∩ 0 ∈ P = P r . (1.5.4) j ⊕ j Observe the following.

1. (Pj) is a disjoint family. Let x P P , then x = p r = p r , r , r Q. ∈ n ∩ m n ⊕ n m ⊕ m n m ∈ Therefore p p Q, namely p ,p are in the same equivalence n − m ∈ n m class. By construction p = p and r = r = P P . n m n m ⇒ n ≡ m + ∞ 2. [0, 1) = Pj. j=0 S 30 Let x [0, 1) and [x] be its equivalence class. There exists a ∈ unique p P such that x p. If x = p, then x P = P ; if ∈ ∼ ∈ 0 x>p, then x = p + r = p r for some r = x P , if x

1 ∞ 1 ∞ 1 1= [0, 1) = (Pj)= (P ). L i=1 L i=1 L P P This is impossible since the right-hand side is = + or = 0. ∞ Since P is not 1-measurable, there exists B R such that L ⊆ 1(B) < 1(B P )+ 1(B P ). L L ∩ L \ B P and B P are two disjoint sets for which the additivity ∩ \ of the measure does not hold. We observe that 1(P ) > 0 (if L 1(P )=0= P is 1-measurable). If E P is 1-measurable, L 1 ⇒ L ⊆ L 1 then (E) = 0. Indeed, set Ei = E ri, Ei is -measurable L1 1 ⊕ L 1 and (E ) = (E). Moreover, ∞ E = F [0, 1), F , - L i L i=0 i ⊂ L measurable with S 1 1 ∞ 1 1= ([0, 1]) (F )= (Ei) L ≥ L i=0 L P = ∞ 1(E) (1.5.6) i=0 L = 1(E)=0 P ⇒ L Exercise 1.5.1. For every A R with 1(A) > 0, there exists B A ⊆ L ⊂ such that B is not 1-measurable. L Proof. Suppose for instance that A (0, 1). Set for i 0, Bi = A Pi. 1 1 ⊆ ≥ ∩ If Bi were -measurable, then (Bi) = 0 (since Bi Pi), therefore 1 L L ⊆ ∞ (B ) = 0. Since A = ∞ B , we would get a contradiction. i=0 L i i=0 i P S 31 Exercise 1.5.2. Show that every countable subset of R has Lebesgue measure zero. Proof. Let α R. Then α [α ε, α + ε] ε> 0. Therefore ∈ { }⊆ − ∀ 1 α vol[α ε, α + ε]=2ε ε> 0 (1.5.7) L { }≤ − ∀ and this gives 1 α = 0. L { } ∞ If A = α1, α2,... = αn is a countable set, then { } n=1{ } S1 1 (A) αn =0. (1.5.8) L ≤ n L { } P

1.6 An uncountable n-null set: the Cantor Triadic L Set

Let X = [0, 1]. For any ﬁxed integer b 2 and any x [0, 1] one ≥ ∈ can expand x in base b: there exists a sequence of “digits” d (x) i ∈ 0,...,b 1 such that { − } ∞ i x = di(x) b− i=1 P which is also denoted x =0,d1,d2 ... . This expansion is not entirely unique (for instance, in base 10, we have 0.1=0.099 = 0.09). However, the set of those x X for which there exists more than one expansion ∈ is countable. The Cantor set is deﬁned as follows: take b = 3, so the digits are in 0, 1, 2 and let { } C = x [0, 1] d (x) 0, 2 i ∈ | i ∈{ } ∀ be the set of those x which have no digit 1 in their 3-expansion.

Claim 1.6.1. 1(C)=0, but C is not countable. L Proof of Claim 1.6.1.

32 We observe that C = ∞ C , where • n=1 n C = xT [0, 1] d (x) =1 i n . n ∈ | i 6 ∀ ≤

Each Cn is a Borel subset of [0, 1]; indeed, we have for instance

C = 0, 1 2, 1 1 3 ∪ 3 h i h i (1.6.1) C = 0, 1 2, 1 2, 7 8, 1 . 2 9 ∪ 9 3 ∪ 3 9 ∪ 9 h i h i h i h i

More generally, Cn is a Borel subset of [0, 1] which is a disjoint n n union of 2 intervals of length 3− . Thus by additivity n 1 n n 2 (C )=2 3− = L n · 3 and since C C , 1(C ) < + , we have n+1 ⊆ n L 1 ∞ 1 1 1 (C)= Cn = lim (Cn)=0. (1.6.2) L L n + L n → ∞ \ C is uncountable. • Indeed, the function

f ∞ di = ∞ di 1 (1.6.3) 3i 2 2i i=1 i=1 P P maps C onto [0, 1], (see Exercise 1.6.3 below).

33 Remark 1.6.2. Fix some 0 <β < 1 and let I1 = [0, 1]. For each n N let I I be the collection of intervals obtained ∈ n+1 ⊆ n by excluding from every interval in In, say with length ℓ, its centered subinterval of length βℓ. We deﬁne by Cβ = n∞=1In the ∩ 1 fat Cantor set corresponding to β. If the parameter β < 3 then Cβ is not Jordan measurable. Actually the inner Jordan measure µ(Cβ)=0 since Cβ has empty interior, whereas the outer Jordan β measure µ(Cβ)=1 1 2β > 0. The Lebesgue measure of Cβ is 1 β − − (Cβ)=1 1 2β . Actually for all n N, In In+1 has length Ln 1 n − − ∈ \ 2 − β and thus I C has length 1 \ β

∞ n 1 n β 2 − β = . 1 2β n=1 X −

Exercise 1.6.3. Show that f in (1.6.3) is surjective but not injective, (f(x) coincides at the opposing ends of one of the middle third removed. For instance: 7 =0.202, 8 =0.2200, so 9 9 f 7 =0.101= f 8 =0.11. ) 9 9

1.7 Lebesgue-Stieltjes Measure

Let F : R R be nondecreasing and continuous from the left: → F (x0)= lim F (x) x0 R. x x− ∀ ∈ → 0 For a, b R we set ∈ F (b) F (a) if a

Let ΛF be the induced measure of λF :

∞ ∞ ΛF (A) = inf λF [ak,bk), A [ak,bk) . k=1 ⊆ k=1 n P S o 34 ΛF is called the Lebesgue-Stieltjes Measure generated by F . Is ΛF Borel or even Borel regular? The following criterion is useful.

Deﬁnition 1.7.1. A measure µ on Rn is called metric if A,B Rn with ∀ ⊆ dist(A,B) = inf a b , a A, b B > 0, {| − | ∈ ∈ } we have µ(A B)= µ(A)+ µ(B). (1.7.2) ∪

Theorem 1.7.2. (Carath´eodory’s criterion for Borel- measures) A metric measure µ on Rn is Borel.

Proof of Theorem 1.7.2. It is enough to show that every closed set F of Rn is µ-measurable, namely B Rn we have ∀ ⊆ µ(B) µ(B F )+ µ(B F ). (1.7.3) ≥ ∩ \ If µ(B)= then (1.7.3) is obvious. ∞ Assume instead µ(B) < . For k 0, deﬁne ∞ ≥ 1 F = x Rn : d(x,F ) . (1.7.4) k ∈ ≤ k n o Since µ is metric and 1 dist(B F , B F ) > 0, \ k ∩ ≥ k we have by the monotonicity of µ

µ(B F )+ µ(B Fk)= µ (B F ) (B Fk) ∩ \ ∩ ∪ \ (1.7.5) µ(B ). ≤

35 Claim 1.7.3. µ(B F ) µ(B F ) as k + . \ k → \ → ∞ Proof of Claim 1.7.3. For k N we set ∈ 1 1 R =(F F ) B = x B :

∞ ∞ B F = B F = (B F ) \ \ ℓ \ ℓ ℓ\=1 ℓ[=1 ∞ =(B F ) (B F ) (B F ) \ 1 ∪ \ ℓ+1 \ \ ℓ ℓ[=1 ∞ =(B F ) (B F ) (B F ) \ k ∪ \ ℓ+1 \ \ ℓ ℓ[=k ∞ =(B F ) (F F ) B \ k ∪ ℓ \ ℓ+1 ∩ ℓ[=k ∞ =(B F ) R . \ k ∪ ℓ ℓ[=k Thus

∞ µ(B Fk) µ(B F ) µ(B Fk)+ µ(Rℓ). (1.7.7) \ ≤ \ ≤ \ ℓ=k P We show that ∞ lim µ(Rℓ)=0. (1.7.8) k + → ∞ ℓ=k P To this purpose we prove that

∞ µ(Rℓ) < + . (1.7.9) ℓ=1 ∞ P We observe that dist(R , R ) > 0, if i j 2. Since µ is metric, it i j | − | ≥

36 follows by induction that

m m µ(R2k)= µ R2k µ(B) k=1 ≤ k=1 P [ (1.7.10) m m µ(R )= µ R µ(B). 2k+1 2k+1 ≤ k=1 k=1 P [ By adding (1.7.9) and (1.7.10) and letting m + , we prove that → ∞ ∞ µ(R ) < + . k=1 k ∞ P Theorem 1.7.4. The Lebesgue-Stieltjes measure is Borel regular.

Proof. Step 1. We ﬁrst show that ΛF is Borel. According to Theorem 1.7.2 it is enough to show that ΛF is metric. Let A,B R with δ = dist(A,B) > 0. For ε > 0, choose a ,b R ⊆ k k ∈ with

∞ ∞ A B [a ,b ) and λ [a ,b ) < Λ (A B)+ ε. ∪ ⊆ k k F k k F ∪ k[=1 Xk=1 Up to a further subdivision of the intervals [ak,bk), we can suppose that k 0 one has b a <δ. ∀ ≥ | k − k| Therefore, for every k N, we have either A [a ,b ) = or ∈ ∩ k k ∅ B [a ,b )= . The covering ([a ,b )) is decomposed in a covering ∩ k k ∅ k k k of A and a covering of B. It follows A B Λ (A B) Λ (A) + Λ (B) F ∪ ≤ F F ΛF [ak,bk) + ΛF [ak,bk) ≤ [ak,bk) [ak,bk) (1.7.11) P∈A P∈B λF [ak,bk) ΛF (A B)+ ε. ≤ k ≤ ∪ P We let ε 0 and we get the claim. → 37 Step 2. We show that Λ is Borel regular. Let A R with F ⊆ ΛF (A) < + (otherwise we choose B = R). For j N we choose the ∞j j ∈ sequences (ak)k,(bk)k with

∞ A [aj ,bj ) =: B ⊆ k k j k[=1 and ∞ j j 1 λF [ak,bk) ΛF (A)+ . k=1 ≤ j P We set B = ∞ B , B is Borel with A B and B B . It follows j=1 j ⊆ ⊆ j that T Λ (A) Λ (B) Λ (B ) F ≤ F ≤ F j (1.7.12) ∞ j j 1 λF [ak,bk) ΛF (A)+ . ≤ k=1 ≤ j P We let j + and get Λ (A) = Λ (B)= Λ is Borel regular. → ∞ F F ⇒ F

The set of half-open intervals is not an algebra. Nevertheless, it holds the following result:

Theorem 1.7.5. For a

Λ [a, b) = λ [a, b) = F (b) F (a) (1.7.13) F F −

Proof. Let’s consider a, b R with a

Λ [a, b) λ [a, b) F ≤ F 2. Now we show that

Λ [a, b) λ [a, b) F ≥ F

38 Let ([a ,b ) be a covering of [a, b), namely [a, b) [a ,b ). Since k k k ⊆ ∪k k k F is left-continuous for every ε> 0 there are δ > 0,δk > 0 such that F (b) F (b δ) ε (1.7.14) − − ≤ k F (a ) F (a δ ) 2− ε, k 0 (1.7.15) k − k − k ≤ ∀ ≥ We observe that ∞ [a, b δ] (a δ ,b ). − ⊆ k − k k k[=0 Since [a, b δ] is compact we can extract a ﬁnite sub-covering of the − open intervals ((a δ ,b )) , i.e. k − k k k m [a, b δ] (a δ ,b ). − ⊆ k − k k k[=0 By discarding any unnecessary (a δ ,b ) and reindexing the rest, we k − k k may assume without restriction that ak δk

λF ([a, b)) = F (b) F (a)= F (b) F (b δ)+ F (b δ) F (a) m− − − − − ε + F (b ) F (a δ ) ≤ k − k − k k=0 Xm m = ε + F (b ) F (a )+ F (a ) F (a δ ) k − k k − k − k Xk=0 Xk=0 ∞ ∞ k F (b ) F (a )+ ε + 2− ε ≤ k − k Xk=0 Xk=0 ∞ = F (b ) F (a )+3ε. (1.7.16) k − k Xk=0 By taking in (1.7.16) the inﬁmum over all covering of [a, b) and then letting ε 0 we get → λ ([a, b)) Λ ([a, b)), F ≤ F 39 and we conclude the proof of Theorem 1.7.5.

Exercise 1.7.6. i) For F (x)= x, Λ = 1. F L ii) For F (x) = 1, x > 0 and F (x) = 0, x 0, it follows that ≤ ΛF = δ0.

40 1.8 Hausdorﬀ Measures

Hausdorff measures are among the most important measures. They generalize Lebesgue measure and permit to measure the subsets of Rn of dimension less than n, such as the submanifolds (surfaces and curves in R3) and the so-called fractal sets (a fractal is a never ending pattern that repeats itself at different scales). They allow us to define the dimension of every set of Rn even with complicated geometry.

Deﬁnition 1.8.1. For s 0, δ > 0 and = A Rn we set ≥ ∅ 6 ⊆ s s δ(A) = inf rk, A B(xk,rk), 0

In the deﬁnition of s(A) we require that r > 0 since we want to Hδ k avoid the case 00. In the case s> 0 we get the same value s(A) with Hδ or without such a condition. s is a measure on Rn. It is deﬁned in terms of the radius of the balls Hδ B(x ,r ) := y Rn: y x 0

Deﬁnition 1.8.2. s is called the s-dimensional Hausdorﬀ H measure on Rn.

Theorem 1.8.3. For s 0, s is a Borel regular measure ≥ H on Rn.

Proof of Theorem 1.8.3.

41 i) s is a measure. H Let s 0. Clearly, s( )=0. ≥ H ∅ n s Let A ∞ A R . Since for every δ > 0 is σ-subadditive, ⊆ k k ⊆ Hδ we have the following inequalities S s s s δ(A) δ(Ak) (Ak). (1.8.4) H ≤ k H ≤ k H P P We let δ 0 and get the σ-sub-additivity of s. → H ii) s is metric and therefore Borel. H Let A,B Rn with δ = dist(A,B). We select 0 <δ< δ0 . ⊆ 0 4 Suppose A B B(x ,r ), r <δ. Set ∪ ⊆ k k k k =SB(xk,rk), B(xk,rk) A = A ∩ 6 ∅ (1.8.5) = B(x ,r ), B(x ,r ) B = . B k k k k ∩ 6 ∅ Then A B(x ,r ) ⊆ k k B(xk,rk) [ ∈A (1.8.6) B B(x ,r ) . ⊆ k k B(xk,rk) [ ∈B

We observe that if B(x ,r ) , then B(x ,r ) B = . Indeed, k k ∈ A k k ∩ ∅ suppose by contradiction that there exist x A and y B such ∈ ∈ that x x

s s s s δ(A)+ δ(B) rk + rk H H ≤ B(xk,rk) B(xk,rk) ∈A ∈B (1.8.8) Ps P = rk. k P 42 Taking the inﬁmum over B(x ,r ) covering A B, we get k k k ∪ s(A SB) s(A)+ s(B) (1.8.9) Hδ ∪ ≥Hδ Hδ

provided δ < δ0 . Let δ 0 and get 4 → s(A B) s(A)+ s(B). (1.8.10) H ∪ ≥H H It follows that s(A B)= s(A)+ s(B) and we conclude. H ∪ H H iii) s is Borel regular. H Let A Rn, suppose s(A) < + = s(A) < + δ > 0. ⊆ H ∞ ⇒Hδ ∞ ∀ For δ = 1, choose B(x ,r ) A with r < 1 and ℓ k k,ℓ k,ℓ ⊇ k,ℓ ℓ S s s 1 rk,ℓ 1 (A)+ ℓ . k ≤H ℓ P

Set Aℓ = k B(xk,ℓ,rk,ℓ) and B = ℓ Aℓ. B is a Borel set with A B. ⊆ S T For each ℓ, we have

s s s s 1 (A) 1 (B) 1 (Aℓ) rk,ℓ H ℓ ≤ H ℓ ≤H ℓ ≤ k s 1 P 1 (A)+ . (1.8.11) ≤ H ℓ ℓ

We let ℓ + and get → ∞ s(B)= s(A). (1.8.12) H H

43 Remark 1.8.4. 1) We observe that 0 is the so-called counting measure. H Actually let k be a positive integer and E be a set with k elements and let δ0 > 0 be the minimal distance between the elements of E. Then every covering of balls of radius 0 <δ<δ0 consists of at least k sets. On the other hand if E = x ,...,x then { 1 k} k B(x ,δ) is a covering of E. It follows that 0(E) = k for ∪i=1 i Hδ every 0 <δ<δ . It follows that 0(E) = k as well. If E is 0 H inﬁnite than by monotonicity we have 0(E) k for all k and H ≥ therefore 0(E)=+ . Finally 0( ) = 0 by deﬁnition. H s∞ H ∅ 2) In general Hδ are not Borel measures and therefore they are not metric.

In order to deﬁne the dimension of an arbitrary set, we need the following

Lemma 1.8.5. Let A Rn and 0 s 0= s(A)=+ . H ⇒H ∞

Proof.

i) Let s(A) < + . For δ > 0, A B(x ,r ) with r < δ, it H ∞ ⊆ k k k k holds S t t t s s t s s δ(A) rk = rk− rk δ − rk. (1.8.13) H ≤ k k ≤ k P P P By considering the inﬁmum with respect to k B(xk,rk), we get t t s s (A) δ − (A). S (1.8.14) Hδ ≤ Hδ Let δ 0 and we get t(A)=0. → H ii) It follows from i).

44 Example 1.8.6. Let Q =[ 1, 1]n Rn. Then it holds − ⊆ n n n n n n 2− (Q) (Q) 2− n 2 (Q). (1.8.15) L ≤H ≤ L

k 1 Proof. Let δ > 0 and k > 0 be such that r = 2− − √n<δ. We k (k+1)n decompose Q in subcubes Q of edge length 2− , 1 ℓ 2 ℓ ≤ ≤

(0, 1)

k =1 n =2 ℓ = 16 (1, 0) ( 1, 0) − 1 2 | {z }

(0, 1)

r n k 2 2− 1 (k+1) r = = n 2 2− si=1 2 P 

Fig. 1.3

Each Qℓ is included in a ball with the same center and radius r

45 2(k+1)n 1 (k+1) Therefore Q B(x ,r). Since r = n 2 2− <δ, we have ⊆ ℓ=1 ℓ n n n (k+1)n δ (QS) r = r 2 H ≤ 1 ℓ 2(k+1)n n (k+1)n (k+1)n ≤ ≤P = n 2 2− 2 n n n n (1.8.16) = n 2 =2 2− n 2 n n n = (Q) n 2 2− . L

Conversely, if (B(xk,rk))k with rk <δ is a covering of the ball B(0, 1), then the following estimate holds

n n wn = B(0, 1) (B(xk,rk) L ≤ k L P n = wn rk k P where n/2 + π ∞ t 1 x w = , Γ(t)= x − e− dx t> 0, n Γ(1 + n) 2 Z0 Γ is the Euler function. Therefore rn 1 and k k ≥ n(Q)P n B(0, 1) Hδ ≥Hδ n = inf rk : B(0, 1) B(xk,rk), rk <δ (1.8.17) k ⊆ k n n n o 1=2P− (Q) S ≥ L

Let δ 0 in(1.8.16) and in (1.8.17) and we get the result. →

Remark 1.8.7.

1. Actually, n(A)= w n(A) for all n-measurable sets A Rn. L n H L ⊆ 2. For A Rn, if s>n, then it holds s(A) = 0. Actually we can ⊆ H write Rn = Q where Q are dydic cubes. For Example 1.8.6 ∪ℓ ℓ ℓ and Lemma 1.8.5 it follows that s(Q ) = 0 for every ℓ N. H ℓ ∈ Therefore s(Rn)=0. H 46 Definition 1.8.8. The Hausdorff dimension of a set A Rn ⊆ is defined to be

dim (A) := inf s 0: s(A)=0 . H { ≥ H }

47 Figure 1.1: Example of fractal in everyday life: the cauliﬂower

Remark 1.8.9. 1. From Lemma 1.8.5 it follows that

dim (A) = sup s 0: s(A)=+ . H { ≥ H ∞} 2. Observe that dim (A) n. Let s = dim (A). Then t(A)=0 H ≤ H H for all t > s and t(A)=+ for all t < s; s(A) may be H ∞ H any number between 0 and , inclusive. Furthermore, dim (A) ∞ H need not be an integer. Even if dim (A) = k is an integer and H 0 < k(A) < , Aneed not bea “k-dimensional surface” in any H ∞ sense. 3. If A Rn is open, then dim (A) = n. Indeed, each open set ⊆ H contains a ball B such that n(B) > 0 = n(A) > 0 = L ⇒ H ⇒ dim (A) n = dim (A)= n. H ≥ ⇒ H Sets with non-integer Hausdorﬀ dimension are important and most sets in nature are fractals with non-integer dimension. A fractal set is a rough or fragmented geometric set that can be splitted into parts, each of which is a reduced-size copy of the whole.

1.8.1 Examples of non-integer Hausdorﬀ-dimension sets 1. Triadic Cantor Set

48 Figure 1.2: Approximation of the coastline of Great Britain by polygons

C = lim Ck = k∞=0 Ck, Ck closed sets with C0 = [0, 1], Ck is k + → ∞ obtained from Ck 1 by removing the open middle third from each −T connected component of Ck 1. In this case: −

ℓn(2) (d+1) d d dim (C)= =: d 2− < (C) < 2− , H ℓn(3) H

where ℓn(x) = loge(x).

Heuristics:

s> 0, λ> 0, A Rn • ∀ ∀ ⊆ s(λA)= λs s(A). H H If there exists d such that d(C) (0, + ) then, since C is • HC ∈ ∞ the union of two copies of 3 , we have

d(C)=2 d C = 2 d(C)= 2 =1= H H 3 3d H ⇒ 3d ⇒ d = log 2= ℓn 2 . 3 ℓn 3

49 2. Cantor Dust

2 A0 = [0, 1] A = 1 2 3 4 1 ∪ ∪ ∪ y = x 1 − 2 = union of squares through which the two lines y = x 1, y = x + 1 pass − 2 − 2 y = x + 1 − 2 Fig. 1.4

2 2 A A is the union of 4 squares of length 4− obtained by 2 ⊆ 1 repeating the same construction in every sub-squares of A1.

Finally one considers A = k∞=1 Ak. One has dimT (A)=1 (1.8.18) H

Proof. We show that 1 1(A) √2. 2 ≤H ≤ 2 k k k Each set Ak consists of 4 squares of length 4− . The 4 balls of k √2 radius 4− 2 cover Ak and thus A. Thus

1 1 √2 k √2 (A) (A ) k k : r =4− 0 < δ. (1.8.19) Hδ ≤Hδ k ≤ 2 ∀ ≥ 0 k0 2

Let Π: R2 R be the projection Π(x, y)= x. We have Π(A)= → [0, 1]. Let A ∞ B(z ,r ) with z =(x , y ) and r < δ. We ⊆ k=1 k k k k k k get S ∞ [0, 1] = Π(A) Π B(x ,r ) ⊆ k k k=1 [ (1.8.20) ∞ = (x r ,x + r ). k − k k k k[=1

50 Thus

1 ∞ 1 ∞ ∞ 1 1= [0, 1] (Ik)=2 rk = rk 2. (1.8.21) L ≤ k=1 L k=1 ⇒ k=1 ≥ P P P where I =(x r ,x + r ). k k − k k k 3. The Koch Curve Description:

Fig. 1.5

Begin with a straight line. Divide it into 3 equal segments and replace the middle segment by the two sides of an equilateral triangle of the same length. Now repeat, taking each of the four resulting segments dividing them into three equal parts and replacing each of the middle segments by two sides of an equilateral triangle. Continue with the construction.

The Koch Curve is the limiting curve obtained by applying this construction on inﬁnite number of times.

The length of the intermediate curve at the nth-iteration of the 4 n construction is (3) . Therefore the length of the Koch curve is infinite. Moreover, the length of the Koch curve between any two points of the curve is also infinite since there is a copy of the Koch ℓn 4 curve. In this case, the Hausdorff dimension is d = ℓn 3 > 1 (the 1 curve is made of 4 copies of itself rescaled by the factor 3).

51 1.9 Radon Measure

Deﬁnition 1.9.1. A measure µ on Rn is called a Radon measure if µ is Borel regular and µ(K) < , for every compact K Rn. ∞ ⊆

Example 1.9.2. i) n is a Radon measure on Rn. L ii) s for s

For general Radon measures an analogous of Theorem 1.3.7 holds.

Theorem 1.9.3. (Approximation by open and compact sets) Let µ be a Radon measure on Rn. i) For every A Rn it holds ⊆ µ(A) = inf µ(G): A G, G open . (1.9.2) ⊆ ii) For every A Rn µ-measurable it holds ⊆ µ(A) = sup µ(F ): F A, F compact . ⊆ (1.9.3)

To prove Theorem 1.9.3 we need the following lemma:

52 Lemma 1.9.4. For every Borel set B Rn it holds the ⊆ following: ε> 0 G B, G open such that µ(G B) <ε. ∀ ∃ ⊇ \ (No proof)

Proof of Theorem 1.9.3. i) If µ(A)=+ then (1.9.2) is obvious. Let us suppose µ(A) < ∞ + . ∞ Assume ﬁrst that A is a Borel set and ﬁx ε > 0. Then by Lemma 1.9.4 there exists G open such that G A and µ(G A) < ⊇ \ ε. Since µ(G) = µ(G A)+ µ(G A), we get µ(G) < µ(A)+ ε ∩ \ and i) follows. Now let A be an arbitrary set. Since µ is Borel regular, there exists a Borel set B A such that µ(A)= µ(B). Then ⊇ µ(A)= µ(B) = inf µ(U), B U open { ⊆ } (1.9.4) inf µ(U), A U open . ≥ { ⊆ } ii) Let A be a µ-measurable.

Case 1: µ(A) < + . • ∞ Set ν = µ A. ν is a Radon measure. We apply i): ε > 0 ⌊ ∀ there exists G open such that G Rn A and ⊇ \ ν(G) ν(Rn A)+ ε = ε. ≤ \ Let C = Rn G, C is closed and C A. Moreover, \ ⊆ µ(A C)= µ A (Rn C) = ν(Rn C) \ ∩ \ \ (1.9.5) = ν(G) < ε. Thus µ(A) µ(C)+ ε. (1.9.6) ≤ (µ(A)= µ(A C)+µ(A C), since C is measurable). Therefore ∩ \ µ(A) = sup µ(C): C A, C closed . (1.9.7) { ⊆ } 53 Case 2: µ(A)=+ . • ∞ Deﬁne D = x : k 1 x

If C is closed, then C B is compact and ∩ m µ(C)= lim µ(C Bm). m + → ∞ ∩ Hence, for each µ-measurable set A: sup µ(K): K A compact = sup µ(C): C A closed . { ⊆ } { ⊆ } We conclude the proof.

Theorem 1.9.5. Let µ be a Radon measure on Rn and let A Rn. The following two conditions are equivalent: ⊆ i) A is µ-measurable, ii) ε> 0 G A, G open such that µ(G A) <ε. ∀ ∃ ⊃ \

The proof is the same of that of Theorem 1.3.8 by using Theorem 1.9.3 instead of Theorem 1.3.7.

54 Chapter 2

Measurable Functions

2.1 Inverse Image of a Function

Let X,Y be nonempty sets. For any map ϕ: X Y and A (Y ), we set → ∈ P 1 ϕ− (A)= x X : ϕ(x) A = ϕ A , (2.1.1) { ∈ ∈ } { ∈ } 1 ϕ− (A) is called the inverse image or preimage of A.

Some Properties

1 c 1 c i) ϕ− (A )= ϕ− (A) A (Y ). ∀ ∈ P ii) If A,B (Y ): ∈ P 1 1 1 ϕ− (A B)= ϕ− (A) ϕ− (B). ∩ ∩ iii) If A (Y ): { k} ⊆ P

1 ∞ ∞ 1 ϕ− Ak = ϕ− (Ak). k[=1 k[=1 1 1 Consequently, if (Y, , µ)isa measure space, then ϕ− ( )= ϕ− (A): F F { A is a σ-algebra in X. ∈ F} 2.2 Deﬁnition and Basic Properties

Let µ be a measure on Rn,Ω Rn µ-measurable. ⊆ 55 Deﬁnition 2.2.1. f: Ω [ , ] is called µ-measurable if → −∞ ∞ 1 1 i) f − + , f − are µ-measurable. { ∞} {−∞} 1 ii) f − (U) is µ-measurable for every U R open. ⊆ Remark 2.2.2. The following two conditions are equivalent to ii) in Deﬁnition 2.2.1. 1 iii) f − (B) is µ-measurable for each Borel set B R. 1 ⊆ iv) f − (( , a)) is µ-measurable, a R. −∞ ∀ ∈ Remark 2.2.3. We consider R =[ , + ] with the topology generated −∞ ∞ by the open sets of R and the neighborhoods [ , a),(a, + ], a R −∞ ∞ ∈ of . ±∞ Then f:Ω R is µ-measurable if and only if 1 → v) f − (U) is µ-measurable, U R open ∀ ⊆ or if and only if 1 vi) f − ([ , a)) is µ-measurable, a R. −∞ ∀ ∈ Proof of vi) = i) and vi) = iv): ⇒ ⇒ vi) = i): We use the representation ⇒ 1 1 f − ( )= f − [ ,k) {−∞} −∞ k Z \∈ 1 1 f − ( + )=Ω f − [ ,k) . { ∞} \ −∞ k Z [∈

1 1 1 vi) = iv): f − (( , a)) = f − ([ , a)) f − ( ). ⇒ −∞ −∞ \ {−∞}

Exercise 2.2.4. Let f: Ω R be µ-measurable and g: R R be → → continuous. Then g f is µ-measurable. ◦

56 Theorem 2.2.5. (Properties of Measurable Functions)

i) Let f,g: Ω R be µ-measurable functions. Then: → f + g, f g, f , sign(f) , max(f,g), min(f,g) and (if · | | g(x) =0) f are µ-measurable, where 6 g f(x) f(x) if f(x) =0 (signf)(x)= | | 6 ( 0 if f(x)=0

ii) If f : Ω R are µ-measurable. Then: k → inf fk, sup fk lim inf fk, lim sup fk are µ-measurable. k k k k →∞ →∞

Proof.

1 i) In view of Remark 2.2.2, f:Ω R is µ-measurable iﬀ f − ( , a) → −∞ is µ-measurable a R. ∀ ∈ Let f, g:Ω R be µ-measurable, then → 1 1 1 (f + g)− ( , a)= f − ( ,r) g− ( , s) . (2.2.1) −∞ r+s

1 1 g− , 0 a< 0 a 1 1  − ( , a) = 1  g− ( , 0) a =0 g −∞  −∞  g 1( , 0) g 1 1 , + a> 0 − − a  −∞ ∪ ∞  (2.2.3)  57 1 f Therefore and are µ-measurable. g g + + Finally set s = max s, 0 , s− = max s, 0 . The maps s s , { } {− } → s s− are continuous. Therefore, by Exercise 2.2.4 we have → + + f ,f −, f = f + f − | | sup(f,g)= f +(g f)+ (2.2.4) − inf(f,g)= f (g f)− − − are µ-measurable. The function x sign(x) is continuous except at the origin. If 7→ 1 U R is open, sign− (U) is either open or of the form V 0 ⊆ ∪{ } where V is open, so sign(x) is Borel measurable function (in the sense that the pre-image of a Borel subset of R is a Borel set of R ). Therefore signf is µ-measurable. ii) Let f :Ω [ , + ] be µ-measurable. Then k → −∞ ∞ 1 − ∞ 1 inf fk [ , a)= fk− [ , a) k 1 −∞ −∞ ≥ k=1 [ (2.2.5) 1 − ∞ ∞ 1 1 sup fk [ , a)= fk− , a k 1 −∞ −∞ − ℓ ≥ ℓ[=1 k\=1 h = inf f , sup f are µ-measurable. ⇒ k k k k We complete the proof by noting

lim inf fk = sup inf fk k m 1 k m →∞ ≥ ≥

lim sup fk = inf sup fk. k m 1 k m →∞ ≥ ≥

We now discuss the functions that are building blocks for the theory of integration. Given A Rn, the function χ : Rn R deﬁned by ⊆ A → 1 if x A χA(x)= ∈ (2.2.6) ( 0 if x / A ∈ 58 is called the characteristic function of the set A. It is easily checked that the characteristic function χA of A is µ- measurable if and only if A is measurable. A simple function is a function of the form

+ ∞ n f(x)= di χAi (x), di R, Ai R ,Ai mutually disjoint. i=1 ∈ ⊆ P If Ai are µ-measurable, then f is called a µ-measurable simple function. Equivalently, f : Rn R isa µ- measurable simple function → if and only if f is µ- measurable and the range of f is an at most countable subset of R. The following theorem provides a useful way to decompose of a nonnegative µ-measurable function.

Theorem 2.2.6. Let f: Ω [0, + ] be µ-measurable.Then → ∞ there are µ-measurable sets A Ω such that k ⊆ ∞ 1 f = χAK . k=1 k P

Proof of Theorem 2.2.6: We set

1 A = x Ω: f(x) 1 = f − [1, + ], A is µ-measurable. 1 { ∈ ≥ } ∞ 1 We deﬁne by induction the following sets:

k 1 1 − 1 Ak = x Ω: f(x) + χAj k =2, 3,.... (2.2.7) ∈ ≥ k j=1 j n P o

Claim 2.2.7. f(x)= ∞ 1 χ (x). k Ak k=1 P Proof of Claim 2.2.7.

59 ∞ 1 “f(x) χAk (x)”: ≥ k=1 k This inequalityP follows directly from the deﬁnition of A if sup k : x k { ∈ A = + . Otherwise, consider k = max k: x A , and we use k} ∞ 0 { ∈ k} the fact that x A . We get ∈ k0 k 0 1 f(x) χ (x) ≥ k Ak Xk=1 and since x / A for k k +1 we can also write ∈ k ≥ 0 ∞ 1 f(x) χ (x). ≥ k Ak Xk=1

∞ 1 “f(x) χAk (x)”: ≤ k=1 k P We consider diﬀerent cases.

∞ 1 ∞ 1 i) f(x)=+ = x Ak, k and χAk (x) = = ∞ ⇒ ∈ ∀ k=1 k k=1 k + = f(x). P P ∞ ∞ 1 ii) f(x)=0= x / Ak, k = χAk (x)=0. ⇒ ∈ ∀ ⇒ k=1 k P iii) 0 < f(x) < + = k0 > 0 : x / k k Ak, namely k0 > ∞ ⇒ ∀ ∈ ≥ 0 ∀ 0 : k = k(k ) k such that x / A . (Indeed, if k and ∃ 0 ≥ 0 ∈ Tk ∃ 0 x k k Ak then χAk (x)=1 k k0 and ∈ ≥ 0 ∀ ≥ T ∞ 1 + = χAk (x) f(x) < + . ∞ k=k0 k ≤ ∞ P Hence for inﬁnitely many k, we have x / A and thus ∈ k k 1 1 0 f(x) χAn (x) < . (2.2.8) ≤ − n=1 n k By letting k + , we get P → ∞ ∞ 1 f(x) χAn (x). (2.2.9) ≤ n=1 n P 60 Remark 2.2.8.

k 1 1. Set fk = j=1 j χAj (x). Then fk is an increasing sequence that converges to f. P If f is bounded, then the convergence is uniform:

sup f(x) fk(x) 0 as k + . x R | − |→ → ∞ ∈ + 2. Through the approximation of f and f − we can approximate every µ-measurable function f:Ω [ , + ] by step-functions. → −∞ ∞

Proposition 2.2.9. Let f: Ω R be continuous, µ be a Borel → measure. Then f is µ-measurable.

1 n Proof. Let U R be an open set. Then f − (U)= O Ω where O R ⊆ ∩ ⊂ open. Since µ is a Borel measure (any open set in Rn is measurable), 1 f − (U) is µ-measurable.

Notation: The expression “µ-a.e.” means “almost everywhere with respect of the measure µ”, that is except possibly on a set A with µ(A)=0.

2.3 Lusin’s and Egoroﬀ’s Theorems

Let µ be a Radon measure on Rn.

61 Theorem 2.3.1. (Egoroﬀ) Let Ω Rn be µ-measurable with µ(Ω) < + . Let f : ⊆ ∞ k Ω R be µ-measurable, k N, f: Ω R be µ-measurable → ∀ ∈ → and f(x) ﬁnite µ-a.e.. Moreover f (x) f(x) as k + k → → ∞ for µ-a.e. x Ω. Then it holds: δ > 0 A Ω, A µ- ∈ ∀ ∃ ⊆ measurable with µ(Ω A) <δ and \

sup fk(x) f(x) 0 as k + , (2.3.1) x A | − |→ → ∞ ∈

namely (fk)k converges uniformly to f on A. Since µ is a Radon measure and µ(Ω) < + it also holds: δ > 0 F ∞ ∀ ∃ ⊆ Ω, F compact with µ(Ω F ) <δ and \

sup fk(x) f(x) 0 as k + , (2.3.2) x F | − |→ → ∞ ∈

namely (fk)k converges uniformly to f on F .

Exercise 2.3.2. Show that the conclusion of Egoroﬀ’s theorem can fail if we drop the assumption that the domain has ﬁnite measure.

Solution: 1 Let µ = be the Lebesgue measure and fk(x)= χ[ k,k](x), fk(x) 1 L − → pointwise as k + . For every δ > 0 there does not exists a µ- → ∞ measurable set A such that µ(R A) δ and f (x) uniformly on \ ≤ k A. Suppose there is δ > 0 and a µ-measurable set A R, such ⊆ that sup f (x) 1 0, and µ(R A) δ. Then N > 0 such A | k − | → \ ≤ ∃ that [ N,N]c Ac which is a contradiction, since µ(R A) δ and − ⊂ \ ≤ µ([ N,N]c)=+ . − ∞ Proof of Theorem 2.3.1: Let δ > 0. Deﬁne

∞ 1 C = x Ω: f (x) f(x) > (i, j =1, 2,... ). (2.3.3) ij ∈ | k − | 2i k[=j n o

62 Then C C i, j. Since f (x) f(x) for µ-a.e x and µ(Ω) < i,j+1 ⊆ i,j ∀ k → , we have ∞ ∞ lim µ(Cij)= µ Cij =0. (2.3.4) j + → ∞ j=1 \ Hence for every i, N(i) > 0 such that ∃ δ µ(C ) < , i,N(i) 2i+1 we set

∞ ∞ δ δ A =Ω C , µ(Ω A)= µ C < ∞ = . (2.3.5) \ i,N(i) \ i,N(i) 2i+1 2 i=1 i=1 i=1 [ [ P Moreover, x A, i, k N(i) ∀ ∈ ∀ ∀ ≥ 1 f (x) f(x) . (2.3.6) | k − |≤ 2i Thus f f is uniformly on A. k → Now let F A be compact such that µ(A F ) < δ . From (2.3.5) ⊆ \ 2 it follows, µ(Ω F ) µ(Ω A)+ µ(A F ) δ/2+ δ/2. (2.3.7) \ ≤ \ \ ≤ We can conclude the proof. The type of convergence involved in the conclusion of Egoroﬀ’s theorem is sometimes called almost uniform convergence. Any continuous function is µ-measurable. The next result gives a continuity property that characterizes measurable functions.

Theorem 2.3.3. (Lusin’s Theorem) We assume that Ω Rn is µ-measurable and µ(Ω) < + . ⊆ ∞ Let f: Ω R be µ-measurable, ﬁnite µ-a.e.. Then ε > → ∀ 0 K Ω compact such that µ(Ω K) < ε and f is ∃ ⊆ \ |K continuous.

63 Remark 2.3.4. 1) By f we mean the restriction of f to the set K. |K The conclusion of the theorem states that if f is viewed as a function defined only on K, then f is continuous. However, the theorem does not make the stronger assertion that the function f defined on Ω is continuous at the points of K. 2) If we drop the condition µ(Ω) < + , in general we cannot ∞ find a compact set F satisfying the conditions of Theorem 2.3.3. Nevertheless for every ε> 0 C Ω closed such that µ(Ω C) < ε ∀ ∃ ⊆ \ and f is continuous. |C

Example 2.3.5.

Ω=[0, 1] R ⊆ f = χ[0,1] Q \ f: [0, 1] R is not continuous but f [0,1] Q: [0, 1] Q R is → | \ \ → continuous.

Example 2.3.5 shows also that we cannot take ε = 0 in Lusin’s theorem.

Proof of Theorem 2.3.3: For each positive integer i, let B R be { ij}⊆ disjoint Borel sets such that R = j∞=1 Bij and diam Bij = sup x y , 1 {| − | x, y Bij < . S ∈ } i 1 Deﬁne Aij = f − (Bij), Aij is µ-measurable and let

∞ 1 Ω= A (Ω = Ω f − ). ij ∪ {±∞} j=1 [ e e Since µ is a Radon measure, there exists K A compact such that ij ⊆ ij ε µ(A K ) < . ij \ ij 2i+j

64 Then ∞ ∞ ∞ µ Ω K = µ A K \ ij ij \ ij j=1 j=1 j=1 [ [ [ e ∞ µ (A K ) ≤ ij \ ij j=1 [ (2.3.8) ∞ µ(Aij Kij) ≤ j=1 \ P ∞ ε ε < i+j = i . j=1 2 2 Since P n ∞ ε lim µ Ω Kij = µ Ω Kij < , n + \ \ 2i → ∞ j=1 j=1 [ [ N(i) > 0 such thate e ∃ N(i) ε µ Ω K < . \ ij 2i j=1 [ e Set D = N(i) K , D is compact. For each i and j we ﬁx b B i j=1 ij i ij ∈ ij and deﬁne g : D R, g (x) = b if x K for all j N(i). We S i i → i ij ∈ ij ≤ observe that gi is continuous since it is locally constant (the sets Kij for j N(i) are compact, disjoint sets and so they are at positive ≤ distance apart). Furthermore, f(x) g (x) < 1 x D (diam B < 1). Set | − i | i ∀ ∈ i ij i K = i∞=1 Di, K is compact and T ∞ µ(Ω K)= µ (Ω D ) ∞ µ(Ω D ) < ε. \ \ i ≤ \ i i=1 i=1 [ P Note that e e e 1 Ω K = Ω K f − K \ \ ∪ {±∞}\ 1 µ(Ω K) µ(Ω K)+ µ(f − ( K) \ ≤ e \ ±∞}\ ε +0 . ≤ e 65 Since f(x) g (x) < 1 x D , we see that g (x) f(x) is uniformly | − i | i ∀ ∈ i i → on K as i + . Thus f is continuous. → ∞ |K 2.4 Convergence in Measure

Let µ be an arbitrary measure on Rn,Ω Rn µ-measurable and let ⊆ f,f :Ω R be µ-measurable and f(x) < + µ-a.e.. k → | | ∞

Deﬁnition 2.4.1. The sequence (fk)k converges in measure µ to µ f, in short: f f as k if ε> 0 k → →∞ ∀

lim µ( x Ω: f(x) fk(x) >ε )=0. k + { ∈ | − | } → ∞ Question: Which is the relation between convergence in measure, pointwise convergence and uniform convergence?

Theorem 2.4.2. Let µ(Ω) < + . If fk f µ-a.e. as µ ∞ → k + then f f as k + . → ∞ k → → ∞

Proof. From Theorem 2.3.1 it follows that for every δ > 0 F Ω ∃ δ ⊆ µ-measurable with µ(Ω F ) < δ (if µ is a Radon measure we can \ δ actually assume that F is compact) and ε> 0 n > 0 such that δ ∀ ∃ ε sup fn(x) f(x) <ε n nε. x Fδ | − | ∀ ≥ ∈ Therefore, for n n ≥ ε x Ω f (x) f(x) >ε Ω F . (2.4.1) { ∈ | n − | }⊆ \ δ Hence µ x Ω: f (x) f(x) >ε µ(Ω F ) <δ { ∈ | n − | }≤ \ δ Since δ > 0 is arbitrary, we can conclude.

66 Remark 2.4.3.

1) Theorem 2.4.2 does not hold in general if µ(Ω) = + . Take 1 ∞ fk(x) = χ( k,k)(x), Ω = R, µ = . Then fk(x) f 1 in R µ − L → ≡ but f 9 1: ε> 0 k ∀ 1( x R : f (x) 1 ε )= 1 ( k,k)c = . (2.4.2) L { ∈ | k − |≥ } L − ∞ 2) The converse of Theorem 2.4.2 does not hold. Take Ω = [0, 1) ⊆ R, µ = 1 L fk(x)= χ k−2n k+1−2n (x), k N. [ 2n , 2n ) ∈ For every k 1, the index n is chosen so that 2n k < 2n+1. ≥ ≤ 1 2 We observe that µ( f (x) > 0 ) = < 0 as k + . { k } 2n k → → ∞ µ Thus f f 0 as k + . But for every x Ω, n N, k → ≡ → ∞ ∈ ∈ k N, we have ∈ 1 if k = [2nx]+2n fk(x)= ( 0 for the remaining k 2n +1,..., 2n+1 1 , ∈{ − }

where [s] = sup n N, n s . Thus f cannot converge to { ∈ ≤ } k f 0! ≡ However, it holds the following

µ Theorem 2.4.4. Let f f. Then there exists a k → subsequence f which converges to f µ-a.e. as n + . { kn } → ∞

µ Proof. Since f f, there exists k 1 such that k → n ≥ n n µ( x Ω: f (x) f(x) > 2− ) < 2− k k . { ∈ | k − | } ∀ ≥ n We deﬁne n A = x Ω: f (x) f(x) > 2− (2.4.3) n { ∈ | kn − | }

67 and for h 1 ﬁxed ≥ Eh = An. n h [≥ By the σ-subadditivity of the measure, we have

∞ ∞ 1 h+1 µ(Eh) µ(An) n =2− . (2.4.4) ≤ n=h ≤ n=h 2 P P Let x Ω E , then x / A n h. Thus h N, we have f (x) ∈ \ h ∈ n ∀ ≥ ∀ ∈ kn converges to f(x)on Ω E . \ h Set E = ∞ E , since µ(E ) < + and E is decreasing, we have h=1 h 1 ∞ h

T µ(E)= lim µ(Eh)=0. (2.4.5) h →∞ Since f (x) converges to f(x) x Ω E , h, then f (x) converges kn ∀ ∈ \ h ∀ kn to f(x), x Ω E, and we conclude. ∀ ∈ \

68 Chapter 3

Integration

In this chapter we assume that µ is a Radon measure on Rn and Ω Rn is µ-measurable. ⊆ 3.1 Deﬁnitions and Basic Properties

Deﬁnition 3.1.1. A function g: Ω R is called a simple → function if the image of g is at most countable.

Notation:

+ f = max(f, 0), f − = max( f, 0) − + + We recall that f = f f −, f = f + f −. − | | Deﬁnition 3.1.2. (Integral of simple functions) If g :Ω [0, + ] is a nonnegative, simple, µ-measurable → ∞ function, we deﬁne

1 g dµ y µ(g− y ). (3.1.1) Ω ≡ 0 y { } Z ≤ ≤∞ P We will use the convention that 0 =0. ·∞

69 Deﬁnition 3.1.3. If g :Ω [ , + ] is a simple, µ- → −∞+ ∞ measurable function, and either g dµ < or g−dµ < , Ω ∞ Ω ∞ we call g a µ-integrable simple function and deﬁne R R + g dµ g dµ g− dµ. (3.1.2) ≡ − Z Z Z Therefore if g is µ-integrable simple function:

1 g dµ = yµ(g− y ). (3.1.3) y { } Z −∞≤ ≤∞ P

Deﬁnition 3.1.4. Let f: Ω [ , ]. We deﬁne the upper integral → −∞ ∞

f dµ = inf g dµ : g a µ-integrable simple function, Ω Ω Z n Z g f µ-a.e. R¯ (3.1.4) ≥ ∈ o and the lower integral

f dµ = sup g dµ : g a µ-integrable simple function, Ω Ω Z n Z g f µ-a.e. R¯. (3.1.5) ≤ ∈ o

Deﬁnition 3.1.5. A µ-measurable function f: Ω R is called → µ-integrable if Ω f dµ = Ω f dµ, in which case we write R R

f dµ f dµ = f dµ. (3.1.6) ≡ ZΩ ZΩ ZΩ

70 Exercise 3.1.6. Show that

f dµ f dµ. (3.1.7) ≤ ZΩ ZΩ

Warning: For us a function is “integrable” provided it has an integral, even if this integral equals + or ∞ −∞ (see Evans-Gariepy).

Proposition 3.1.7. Let f: Ω [0, + ] µ-measurable. → ∞ Then f is µ-integrable.

Proof. We may suppose that fdµ< + . Ω ∞ R (Indeed, if f dµ =+ , then f dµ =+ as well). Ω ∞ Ω ∞ R R In particular, we have f(x) < + µ-a.e. ∞ Case 1: Assume µ(Ω) < + . For ε> 0 given, we set • ∞ A = x Ω: kε f(x) < (k + 1) ε , k 0. (3.1.8) k { ∈ ≤ } ≥ 1 We set Ω′ := k 0Ak = k 0f − ([kε, (k + 1)ε)). We have µ(Ω ∪ ≥ ∪ ≥ \ Ω′)=0. We deﬁne the simple functions

∞ e(x)= ε k χAk (x) and k=0 P (3.1.9) ∞ g(x)= ε (k + 1) χAk (x). k=0 P Clearly, we have e(x) f(x) S 0 there are simple functions e , g :Ω Q R with ℓ ℓ ∩ ℓ → e f g in Ω =Ω Q ℓ ≤ ≤ ℓ ℓ ∩ ℓ and ε e dµ g dµ e dµ + . (3.1.11) ℓ ≤ ℓ ≤ ℓ 2ℓ ZΩℓ ZΩℓ ZΩℓ Observe that ℓ∞=1 eℓ χΩℓ and ℓ∞=1 gℓ χΩℓ are simple functions with P ∞ P ∞ eℓ χΩℓ f(x) gℓ χΩℓ . ℓ=1 ≤ ≤ ℓ=1 We sum up in (3.1.11P ) over ℓ and weP get

∞ ∞ eℓ dµ f dµ f dµ eℓ dµ + ε. ℓ=1 Ω Qℓ ≤ Ω ≤ Ω ≤ ℓ=1 Ω Qℓ Z ∩ Z Z Z ∩ P P We let ε 0 and conclude the proof. →

Proposition 3.1.8. (Monotonicity) Let f ,f : Ω R be µ-integrable with f f µ-a.e.. Then 1 2 → 1 ≥ 2 f dµ f dµ. (3.1.12) 1 ≥ 2 ZΩ ZΩ

72 Proof. If g f and f f µ-a.e., then g f µ-a.e.. Therefore ≥ 1 1 ≥ 2 ≥ 2

f1 dµ = f1 = inf g dµ f2 dµ g ≥ ZΩ ZΩ ZΩ ZΩ (3.1.13)

= f2 dµ, ZΩ where the inﬁmum is taken over all simple µ-integrable function g f ≥ 1 µ-a.e..

Deﬁnition 3.1.9. i) A function f: Ω R is µ-summable if f is µ-measurable → and f dµ < + . | | ∞ ZΩ ii) A function f: Ω R is locally µ-summable in Ω, if f → |K is µ-summable for each compact set K Ω. ⊆

Proposition 3.1.10. i) If f is µ-summable, then it is µ-integrable.

ii) If f =0 µ-a.e., then f is µ-integrable and Ω f dµ =0. R Proof of Proposition 3.1.10:

i) We have f = f+ f . Since 0 f f the functions f are − − ≤ ± ≤ | | ± µ-integrable and f dµ < + . For ε > 0 simple functions Ω ± ∞ ∃ 0 e f g µ-a.e. with ≤ ± ≤ ± ≤ ±R 0 e dµ f dµ g dµ ≤ ± ≤ ± ≤ ± ZΩ ZΩ ZΩ e dµ + ε< + ≤ ± ∞ ZΩ

73 (see the proof of Proposition 3.1.7).

Thus e := e+ g , g := g+ e are simple functions satisfying − − − − e f g µ-a.e. ≤ ≤ We may assume without restriction that e = g = 0 on x ± ± { ∈ Ω: f ±(x)=0 . It holds }

e dµ f dµ f dµ g dµ ≤ ≤ ≤ ZΩ ZΩ ZΩ ZΩ e dµ +2ε. ≤ ZΩ We let ε 0 and get the result. → ii) It is trivial (choose e = g = 0). In this case we get f dµ 0 Ω ≥ R and f dµ 0 and therefore Ω ≤ R f dµ = f dµ =0. ZΩ ZΩ

We conclude the proof.

Proposition 3.1.11. Let f: Ω [0, + ] be µ-measurable. → ∞ i) If f dµ =0= f(x)=0 µ-a.e. Ω ⇒ ii) If R fdµ< + = f(x) < + µ-a.e. Ω ∞ ⇒ ∞ R

Proof of Proposition 3.1.11:

i) Assume f is not zero µ-a.e.. Consider the measurable sets 1 A = x : f(x) , k 1. k ≥ k ≥ n o 74 Note that A A and k ⊆ k+1 A = x Ω: f(x) > 0 . k { ∈ } [k We have

0 < µ x Ω: f(x) > 0 = lim µ(Ak). (3.1.14) { ∈ } k + → ∞ Therefore there exists k 1 such that µ(A ) > 0. The function ≥ k s(x) = 1 χ (x) is a simple function such that s(x) f and k Ak ≤ s(x) dµ = 1 µ(A ) > 0= fdµ> 0, which is a contradiction. Ω k k ⇒ Ω ii)R Suppose that f(x)=+ xR A with µ(A) > 0. Then the ∞ ∀ ∈ simple function s(x) = + χ (x) satisfy 0 s f, with ∞ A ≤ ≤ s(x) dµ =+ . Hence f dµ =+ which is a contradiction. Ω ∞ Ω ∞ R R

Corollary 3.1.12. Let f ,f : Ω R integrable with f = f 1 2 → 1 2 µ-a.e. Then

f1 dµ = f2 dµ. (3.1.15) ZΩ ZΩ

Proof. It follows directly from Proposition 3.1.8.

Theorem 3.1.13. (Tchebychev Inequality) Let f: Ω R be µ-summable. Then for every a> 0 it holds → 1 µ( x Ω: f(x) > a ) f dµ. (3.1.16) { ∈ | | } ≤ a | | ZΩ

Proof. Choose in Proposition 3.1.8 f1 = f and f2 = a χ x Ω: f(x) >a . | | { ∈ | | }

75 Corollary 3.1.14. Let f,f : Ω R be µ-integrable with k →

lim fk f dµ =0. k + | − | → ∞ ZΩ µ Then f f as k + and f f µ-a.e. for a subsequence k → → ∞ kn → fkn .

Proof. From Theorem 3.1.13 it follows that for ε> 0 ∀ 1 µ x Ω: f f >ε < f f 0 as k + . (3.1.17) { ∈ | k − | } ε | k − |→ → ∞ ZΩ µ Therefore, f f as k + and the second part is a consequence k → → ∞ of Theorem 2.4.4.

Theorem 3.1.15. Let f,g: Ω R be µ-summable, λ R. → ∈ Then f + g, λf are µ-summable and

(f + g) dµ = f dµ + g dµ Ω Ω Ω Z Z Z (3.1.18) λf dµ = λ f dµ. ZΩ ZΩ

Proof.

i) The two relations hold for µ-integrable simple functions of the form

∞ ∞ f(x)= ak χAk and g(x)= bk χBk (3.1.19) k=1 k=1 P P if both satisfy either f +dµ < + and g+dµ < + or Ω ∞ Ω ∞ f −dµ < + and g−dµ < + . Such conditions are satisﬁed Ω ∞ Ω R ∞ R for instance if we assume that both f and g are µ-summable. R R

76 We observe that in the representation of f and g in (3.1.19) can always suppose that A = B k N. Otherwise we can consider k k ∀ ∈ (Ak Bℓ)k,ℓ N (we leave all the statements in this part of the proof ∩ ∈ as an exercise). ii) We know that f + g is µ-measurable. For ε> 0 we choose simple ε ε µ-integrable functions fε,f ,gε,g such that

f f f ε and g g gε ε ≤ ≤ ε ≤ ≤

f εdµ fdµ<ε and f dµ f <ε − − ε ZΩ ZΩ ZΩ ZΩ

gεdµ gdµ<ε and g dµ g <ε − − ε ZΩ ZΩ ZΩ ZΩ ε ε Thus fε + gε, f + g are µ-measurable simple functions such that

f + g f + g f ε + gε (3.1.20) ε ε ≤ ≤ Moreover since we are assuming that f,g are both µ-summable ε ε we have in particular (f )−dµ < + and (g )−dµ < + , ∞ ∞ ZΩ ZΩ (f )+dµ < + (f ε)+dµ < + . It follows that f + g , ε ∞ ∞ ε ε ZΩ ZΩ ε ε ε ε f + g are µ-integrable since (f + g )− dµ < + and (fε + Ω ∞ Ω g )+ dµ < + and therefore Z Z ε ∞ (f ε + gε) dµ = f εdµ + gεdµ (3.1.21) ZΩ ZΩ ZΩ (fε + gε) dµ = fεdµ + gεdµ. (3.1.22) ZΩ ZΩ ZΩ

77 By combining (3.1.20) and (3.1.21) and (3.1.22) we get

f + g dµ (f ε + gε) dµ = f εdµ + gεdµ ≤ ZΩ ZΩ ZΩ ZΩ f dµ + g dµ +2ε ≤ ZΩ ZΩ (f + g) dµ f dµ + g dµ ≥ ε ε ZΩ ZΩ ZΩ f dµ + g dµ 2ε. ≥ − ZΩ ZΩ Since ε> 0 is arbitrary, it follows that f + g is µ-integrable with

(f + g) dµ = f dµ + g dµ. ZΩ ZΩ ZΩ iii) We apply the result to f and g and we get | | | | f + g dµ ( f + g ) dµ | | ≤ | | | | ZΩ ZΩ (3.1.23) = f dµ + g fµ < + . | | | | ∞ ZΩ ZΩ Thus f + g is µ-summable and a) is proved. iv) Let ε> 0, choose simple functions f f f ε such that ε ≤ ≤ f εdµ fdµ<ε − ZΩ ZΩ f dµ f dµ < ε. − ε ZΩ ZΩ ε Thus λfε, λf are simple functions with

λf λf λf ε if λ 0 ε ≤ ≤ ≥ and λf λf λf ε if λ< 0. ε ≥ ≥

78 We consider λ> 0 (the case λ = 0 is trivial and λ < 0 is similar to the case λ> 0). It follows

λf dµ λf εdµ = λ f εdµ ≤ ZΩ ZΩ ZΩ (3.1.24) λ f dµ + λε ≤ ZΩ and

(λf) dµ λ fε dµ Ω ≥ Ω Z Z (3.1.25) λ f dµ λε. ≥ − ZΩ Thus λf is µ-integrable with

(λf) dµ = λ f dµ. (3.1.26) ZΩ ZΩ Moreover,

Thus λf is µ-summable as well and b) is proved.

Corollary 3.1.16. Let f :Ω R be µ-summable. Then → f dµ f dµ. (3.1.28) Ω ≤ Ω | | Z Z

79 Lemma 3.1.17. Let f: Ω R be µ-summable and Ω Ω be → 1 ⊆ µ-measurable. Then f = f and fχ are µ-summable on Ω 1 |Ω1 Ω1 1 and Ω respectively and

f1 dµ = f χΩ1 dµ. ZΩ1 ZΩ

Proof. We ﬁrst note that f1 and fχΩ1 are µ-measurable functions. Step 1: We show the result for simple µ-summable functions. Let g(x)= ∞ a χ with A A = if i = i and A = Ω. In this i=1 i Ai i1 ∩ i2 ∅ 1 6 2 ∪i i case we have: P ∞ g1(x)= g Ω (x)= aiχAi Ω | 1 ∩ 1 i=1 X ∞ g dµ = a µ(A Ω ) 1 i i ∩ 1 Ω1 i=1 Z X ∞ ∞ gχΩ (x)= aiχAi χΩ = aiχAi Ω 1 1 ∩ 1 i=1 i=1 X X ∞ gχ dµ = a µ(A Ω ). Ω1 i i ∩ 1 Ω i=1 Z X We also observe that both g and gχ are µ-summable because gχ 1 Ω1 | Ω1 |≤ g . | | Step 2: Let ε > 0 and choose simple µ-integrable functions g, h such that g f h µa.e.x Ω ≤ ≤ ∈ and

fdµ gdµ + ε ≤ ZΩ ZΩ hdµ fdµ + ε. ≤ ZΩ ZΩ Claim 1: f1 is µ-integrable.

80 Proof of Claim 1. We have

0 f dµ f dµ ≤ − ZΩ ZΩ h dµ g dµ = hχ dµ gχ dµ ≤ 1 − 1 Ω1 − Ω1 ZΩ1 ZΩ1 ZΩ ZΩ = (h g)χ dµ (h g)dµ = hdµ gdµ 2ε. − Ω1 ≤ − − ≤ ZΩ ZΩ ZΩ ZΩ We let ε 0 and get f is µ-integrable. → 1 Claim 2: fχΩ1 is µ-integrable. Proof of Claim 2.

0 fχ dµ fχ dµ ≤ Ω1 − Ω1 ZΩ ZΩ hχ dµ gχ dµ ≤ Ω1 − Ω1 ZΩ ZΩ = (h g)χ dµ (h g)dµ = hdµ gdµ 2ε. − Ω1 ≤ − − ≤ ZΩ ZΩ ZΩ ZΩ We let ε 0 and get fχ is µ-integrable. → Ω1 Claim 3: fdµ = fχ . Ω1 Ω Ω1 Proof of Claim 3. R R f dµ fχ dµ h dµ gχ dµ 1 − Ω1 ≤ 1 − Ω1 ZΩ1 ZΩ ZΩ1 ZΩ = hχ dµ gχ dµ Ω1 − Ω1 ZΩ ZΩ = (h g)χ dµ 2ε. − Ω1 ≤ ZΩ On another hand:

f dµ fχ dµ g dµ hχ dµ 1 − Ω1 ≥ 1 − Ω1 ZΩ1 ZΩ ZΩ1 ZΩ = gχ dµ hχ dµ Ω1 − Ω1 ZΩ ZΩ = (g h)χ dµ 2ε. − Ω1 ≥− ZΩ 81 Therefore f dµ fχ dµ 2ε. 1 − Ω1 ≤ ZΩ1 ZΩ

By letting ε 0 we get the result and we conclude the proof. →

Corollary 3.1.18. Let f: Ω R be a µ-summable function and → Ω Ω with µ(Ω )=0. Then 1 ⊆ 1 f dµ =0. (3.1.30) ZΩ1

Proof. We observe that f χΩ1 = 0 µ-a.e.. Therefore Ω f χΩ1 dµ = 0 and the result follows from Lemma 3.1.17. R

Proposition 3.1.19. Let f: Ω R be µ-summable, Ω = → Ω Ω , Ω , Ω µ-measurable sets, Ω Ω = . Then 1 ∪ 2 1 2 1 ∩ 2 ∅ f dµ = f dµ + f dµ. (3.1.31) ZΩ ZΩ1 ZΩ2

Proof. We have f = f χΩ1 + f χΩ2 . The result follows from Lemma 3.1.17 and Corollary 3.1.18.

3.2 Comparison between Lebesgue and Riemann- Integral

82 3.2.1 Review of Riemann Integral

Let I =[a, b] be an interval of R and let P = a = x ,x ,...,x = b { 0 1 n } be a partition of I. If f: I R is a bounded function and P is a partition of I, we → deﬁne the upper and the lower Riemann sums respectively by

n S(P,f)= (xi xi 1) sup f(x) − i=1 − x (xi− ,xi] ∈ 1 Pn = (xi xi 1) Mi − − i=1 (3.2.1) Pn s(P,f)= (xi xi 1) inf f(x) − − x (xi−1,xi] i=1 ∈ Pn = (xi xi 1) mi. i=1 − − P Let be the set of all partitions of I. We deﬁne P

b f(x) dx = inf S(P,f), P R a { ∈ P} Z (3.2.2) b f(x) dx = sup s(P,f), P . R { ∈ P} Za We say that a bounded function f is Riemann integrable if

b b b f(x) dx = f(x) dx =: f(x) dx. R R Za Za Za If n

ϕ(x)= ci χ(xi−1,xi] (3.2.3) i=1 X where x partition of I =[a, b] and c ,...,c R, its Lebesgue { i}i=0,...,n 1 n ∈ integral is deﬁned by

n ϕ(x) dx = ci(xi xi 1). (3.2.4) − − Z[a,b] i=1 P 83 Thus we can write

S(P,f)= ϕ(x) dx [a,b] Z (3.2.5) s(P,f)= ϕ(x) dx, Z[a,b] where n

ϕ(x)= Mi χ(xi−1,xi] (3.2.6) i=1 Pn

ϕ(x)= mi χ(xi−1,xi] (3.2.7) i=1 P It is clear that ϕ(x) f(x) ϕ(x). In view of the above considerations ≤ ≤ we can write

b f(x) dx = inf ϕ(x) dx, ϕ as in (3.2.3), ϕ f in I R a [a,b] ≥ Z n Z o b f(x) dx = sup ϕ dx, ϕ as in (3.2.3), ϕ f in I . R a [a,b] ≤ Z n Z o (3.2.8)

84 Figure 3.1: The Riemann integral considers the area under a curve as made out of vertical rectangles, the Lebesgue integral considers horizontal slabs that are not necessarily just rectangles.

Example 3.2.1. We know that the Dirichlet function χQ [0,1] =: f ∩ is not Riemann integrable. Indeed, for every partition P of [0, 1], we have S(P,f)=1 and s(P,f) = 0. Therefore

1 1 f(x) dx =1, f(x) dx =0. (3.2.9) R R Z0 Z0 Moreover, if we consider the sequence

fn(x)= χ r ,...,rn (x), { 1 }

where rn is an enumeration of Q [0, 1], we have fn(x) is - { } 1 ∩ R integrable and fn(x) dx = 0 and fn χ[0,1] Q as n + . R 0 ↑ ∩ → ∞ It makes no sense to wonder if R 1 1 lim fn(x) dx = χQ [0,1](x) dx. n + R R ∩ → ∞ Z0 Z0 Unlike the Riemann integral the Lebesgue integral is such that the characteristic functions of 1-measurable sets are 1-integrable and L L it behaves well under “passage to the limit” as we will see in the sequel.

85 Proposition 3.2.2. Let f: [a, b] R be a bounded R- → integrable function. Then it is 1-integrable and L b f(x) dx = fd 1. R L Za Z[a,b]

Proof. We observe that every simple function of the type ϕ(x) = n 1 i=1 ci χ(xi−1,xi] is a -integrable simple function since (xi 1,xi] are L − 1-measurable. Moreover the following holds: LP inf ϕ(x) dx, ϕ as in (3.2.3), ϕ f in I [a,b] ≥ n Z o inf gd 1 : g a 1-integrable simple function, g f µ-a.e. in I ≥ [a,b] L L ≥ n Z o

sup gd 1 : g a 1-integrable simple function, g f µ-a.e. in I [a,b] L L ≤ n Z o sup ϕ dx, ϕ as in (3.2.3), ϕ f in I ≥ [a,b] ≤ n Z o It follows that

b b f(x) dx fd 1 fd 1 f dx. R ≤ L ≤ L ≤ R Za Z[a,b] Z[a,b] Za

3.3 Convergence Results

Let µ be a Radon measure on Rn and Ω be µ-measurable.

86 Theorem 3.3.1. (Fatou’s Lemma) Let f : Ω [0, + ] be µ-measurable, for all k N. Then k → ∞ ∈ it holds

lim inf fk dµ lim inf fk dµ. (3.3.1) k + ≤ k + ZΩ → ∞ → ∞ ZΩ

Proof. Note that all fk’s are µ-integrable and hence f := liminfk fk →∞ as well. It is suﬃcient to show that

g dµ lim inf fk dµ ≤ k ZΩ →∞ ZΩ for every simple function g = ∞ a χ with g f. j=0 j Aj ≤ Without loss of generality weP can assume that g 0, a = 0, a > ≥ 0 j 0 j 1, A A = (for i = j). ∀ ≥ i ∩ j ∅ 6 ∞ We ﬁx 0 t a for ℓ k . j,k { ∈ j ℓ j ≥ }

Clearly: B B and j,k ⊆ j,k+1 lim µ(Bj,k)= µ(Aj). (3.3.2) k →∞ Fix J, k N: ∈ J J f dµ f dµ f dµ k ≥ k ≥ k ZΩ j=1 ZAj j=1 ZBj,k P J P t aj µ(Bj,k). ≥ · j=1 P We let ﬁrst k + : • → ∞ (3.3.2) J lim inf fkdµ t aj µ(Aj). k ≥ · →∞ ZΩ j=1 P 87 Then we let J + : • → ∞ ∞ lim inf fkdµ t aj µ(Aj)= t g dµ. k ≥ · →∞ ZΩ j=1 ZΩ P Finally, we let t 1 and we conclude the proof. • →

Example 3.3.2. The condition fk 0 in Theorem 3.3.1 is n n ≥ 1 necessary. Take µ = ,Ω= R , f = n χ . Then: L k −k Bk(0)

lim inf fk, dµ = w1, k − →∞ ZΩ n as fk(x) dx = B(0, 1) = w1, n −L − ZR but liminf fk = f 0. k + ≡ → ∞

Theorem 3.3.3. (Monotone Convergence Theorem / Beppo Levi’s Theorem) Let f : Ω [0, + ] be µ-measurable for all k 1 and be k → ∞ ≥ such that f f f ... . Then it holds 1 ≤···≤ k ≤ k+1 ≤

lim fk dµ = lim fk dµ. (3.3.3) k k ZΩ →∞ →∞ ZΩ

Proof. Since (fk)k is a non decreasing sequence, for every j it holds fj dµ limk fk dµ (j =1,... ) and hence: Ω ≤ Ω →∞ R R lim fk dµ lim fk dµ. k ≤ k →∞ ZΩ ZΩ →∞

88 The opposite inequality follows from Fatou’s Lemma:

lim fk dµ = lim inf fk dµ k k ZΩ →∞ ZΩ →∞ Fatou′s Lemma lim inf fk dµ = lim fk dµ. ≤ k k →∞ ZΩ →∞ ZΩ

Remark 3.3.4. We can also start with Beppo Levi’s Theorem and deduce Fatou’s Lemma by setting for every k 1: gk := infℓ k fℓ We ≥ ≥ have 0 g1 gk 1 gk fℓ (ℓ k 1) and ≤ ≤ − ≤ ≤ ≥ ≥

lim gk = liminf fk, gk inf fℓ dµ. k + k + ≤ ℓ k → ∞ → ∞ ZΩ ≥ ZΩ From Beppo Levi’s Theorem it holds

lim inf fk dµ = lim gk dµ = lim gk dµ k k + k + ZΩ →∞ ZΩ → ∞ → ∞ ZΩ

lim inf fk dµ. ≤ k + → ∞ Z

Theorem 3.3.5. (Dominated Convergence / Lebesgue Theorem) Let g :Ω [0, ] be µ-summable and f :Ω R¯, → ∞ → f :Ω R¯ be µ-measurable. Suppose f g and f f { k}k → | k|≤ k → µ-a.e. as k + . Then: → ∞

lim fk f dµ =0. (3.3.4) k + | − | → ∞ ZΩ Moreover:

lim fk dµ = f dµ. (3.3.5) k + → ∞ ZΩ ZΩ

89 Proof. We observe that

f = lim fk g µ-a.e. . | | k | |≤ →∞ Thus: f f f + f 2g (3.3.6) | k − | ≤ | k| | |≤ and hence f f is µ-summable. Then: | k − |

2g dµ = lim inf(2g fk f ) dµ k − | − | ZΩ ZΩ →∞ Fatou′sLemma lim inf (2g fk f ) dµ ≤ k Ω − | − | →∞ Z (3.3.7)

= 2g dµ lim sup fk f dµ Ω − k Ω | − | Z →∞ Z

= lim fk f dµ =0. ⇒ k | − | →∞ ZΩ

The equation (3.3.5) follows from the fact that

fk dµ f dµ fk f dµ. Ω − Ω ≤ Ω | − | Z Z Z

90 3.4 Application: Diﬀerentiation under integral sign

Theorem 3.4.1. Let f: R [0, 1] R be a given function × → satisfying the following conditions:

i) x: y f(x, y) is 1-summable on [0, 1]. ∀ 7→ L ii) ∂f exists and it is bounded in R [0, 1]. ∂x × Then the map y ∂f (x, y) is, for every ﬁxed x R, 1- 7→ ∂x ∈ L summable on [0, 1] and

d 1 1 ∂f f(x, y) dy = (x, y) dy. dx 0 0 ∂x Z Z

Proof. For x R, h = 0, we have ∈ 6 1 1 1 1 1 f(x+h, y) dy f(x, y) dy = f(x+h, y) f(x, y) dy . h − h − Z0 Z0 Z0 (3.4.1) Given h 0 as k + , it holds k → → ∞ 1 ∂f gk(y)= f(x + hk, y) f(x, y) (x, y), as k + . hk − → ∂x → ∞ (3.4.2) Moreover by the mean value theorem and the assumption ii) we have ∂f gk(y) sup (x, y) C. (3.4.3) | |≤ R [0,1] ∂x ≤ ×

∂f (x, ) is 1-measurable with respect to y [0, 1] and 1-summable ∂x · L ∈ L since it is limit of 1-measurable function and bounded. L By Dominating Convergence Theorem we have: 1 1 ∂f g (y) dy (x, y) dy, as k + . k → ∂x → ∞ Z0 Z0 91 This holds sequence h 0. Hence: ∀ k → d 1 1 1 1 f(x, y) dy = lim f(x + h, y) dy f(x, y) dy dx 0 h 0 h 0 − 0 Z → Z Z 1 1 1 ∂f = lim f(x + h, y) f(x, y) dy = (x, y) dy. h 0 h − dx → Z0 Z0

3.5 Absolute Continuity of Integrals

Let f:Ω R be µ-summable. For A Ω µ-measurable, we set → ⊆ ν(A)= f dµ. ZA According to Corollary 3.1.18 we have

µ(A)=0= ν(A)=0. (3.5.1) ⇒

Exercise 3.5.1. ν is σ-additive, ν is a Radon measure in the case f 0 µ-a.e. . ≥ Notation:

ν = µ f.

In particular we have

ν = µ χA = µ A.

Let Σµ and Σν be the σ-algebra respectively of the µ and ν-measurable sets A Rn. ⊆ Deﬁnition 3.5.2. A measure ν such that Σ Σ and with the µ ⊆ ν property (3.5.1) is called “absolutely continuous” with respect to µ and we write ν µ. ≪

92 Theorem 3.5.3. Let f: Ω R be µ-summable. Then ε> → ∀ 0 δ > 0 such that ∃ A Ω µ-measurable with µ(A) <δ = f dµ < ε. ∀ ⊆ ⇒ | | ZA

Proof. We assume by contradiction that there exist ε > 0, and a k sequence of µ-measurable subsets A Ω with µ(A ) < 2− and k ⊆ k f dµ ε k N. | | ≥ ∀ ∈ ZAk For ℓ 1 we deﬁne ≥ ∞ Bℓ := Ak. k[=ℓ The following properties hold B B . • ℓ+1 ⊆ ℓ ∞ 1 ℓ ℓ: µ(Bℓ) µ(Ak) < 2 − . • ∀ ≤ k=ℓ µ(B ) < + P. • 1 ∞ f dµ f dµ ε k ℓ. • | | ≥ | | ≥ ∀ ≥ ZBℓ ZAk ∞ For A = Bℓ: µ(A)= lim µ(Bℓ)=0. • ℓ + ℓ=1 → ∞ Now observe thatT g := f χ f χ as ℓ + and g f ℓ | | · Bℓ → | | · A → ∞ | ℓ| ≤ | | µ-summable, ℓ. ∀ By Dominated Convergence Theorem it follows

ε lim f dµ = lim gℓ dµ ≤ ℓ + | | ℓ + → ∞ ZBℓ → ∞ ZΩ

= lim gℓ dµ = f dµ =0. ℓ | | ZΩ →∞ ZA We get a contradiction and we can conclude.

93 3.6 Vitali’s Theorem

We formulate in this section a necessary and suﬃcient condition to pass to the limit under integral sign. Let f, f :Ω R be µ-summable, k N. k → ∈ Deﬁnition 3.6.1. The family f is called uniformly µ- { k} summable if ε > 0 δ > 0: k N, A Ω µ-measurable ∀ ∃ ∀ ∈ ∀ ⊆ with µ(A) <δ we have f dµ<ε. | k| ZA

Theorem 3.6.2. (Vitali’s Theorem) If µ(Ω) < + , the following conditions are equivalent: ∞ µ i) f f (as k + ) and f is uniformly µ- k → → ∞ { k} summable.

ii) lim fk f dµ =0. k + | − | → ∞ ZΩ

Remark 3.6.3. The condition µ(Ω) < + is in general necessary: n n 1 ∞ Ω= R , µ = , f = n χ . We have: L k k B(0,k) µ f is uniformly µ-summable, f 0 as k + . (3.6.1) { k} k → → ∞ n But n f dµ = w = (B(0, 1)) > 0 and n f dµ = 0. R | k| n L R ProofR of Vitali’s Theorem: R ii) = i): ⇒ µ If f f dµ 0 as k + , then f f by Corollary 3.1.14. Ω | k − | → → ∞ k → Now we show that f is uniformly µ-summable. R { k} For ε> 0, let k0 = k0(ε) > 0 be such that

f f dµ < ε, for all k k . | k − | ≥ 0 ZΩ 94 Let δ > 0 be such that A Ω µ-measurable with µ(A) <δ, we have ∀ ⊆ f dµ<ε (3.6.2) | | ZA and

max fk dµ<ε (3.6.3) 1 k k0 | | ≤ ≤ ZA (see Theorem 3.5.3). Thus for k k and if µ(A) <δ, we have ≥ 0 f dµ f dµ + f f dµ | k| ≤ | | | k − | ZA ZA ZA (3.6.4) 2ε. ≤ Thus, we have the desired property. i) = ii): ⇒ µ µ Let f f as k + and be uniformly µ-summable. Since f f, k → → ∞ k → there exists a sub-sequence f f µ-a.e. as n + (see Theorem kn → → ∞ 2.4.4). Given ε > 0, we choose δ > 0 such that for A Ω, µ(A) < δ, we ⊆ have f dµ<ε and f dµ<ε k 1. (3.6.5) | | | k| ∀ ≥ ZA ZA Correspondent to such a δ > 0 we can ﬁnd by Egoroﬀ-Theorem a closed set F Ω such that µ(Ω F ) <δ and ⊆ \

sup fkn (x) f(x) 0 as n + . x F | − |→ → ∞ ∈ Let n be such that ε sup fkn (x) f(x) < , n n. x F | − | µ(Ω) ∀ ≥ ∈

95 Thus for n n: ≥

fkn (x) f(x) dµ = fkn f dµ + fkn f dµ Ω | − | Ω F | − | F | − | Z Z \ Z

fkn + f dµ + sup fkn f dµ ≤ Ω F | | | | F F | − | Z \ ε Z < 2ε + µ(F ) < 3ε. µ(Ω) (3.6.6)

Since ε> 0 is arbitrary, we have

lim fk (x) f(x) dµ =0. (3.6.7) n + | n − | → ∞ ZΩ

Claim 3.6.4. The whole sequence fk satisﬁes (3.6.7).

Proof of Claim 3.6.4: Suppose that

lim sup fk(x) f(x) dµ > 0. k + Ω | − | → ∞ Z

Then there exists a subsequence fkn such that

lim fkn (x) f(x) dµ = limsup fk(x) f(x) dµ > 0. n + Ω | − | k + Ω | − | → ∞ Z → ∞ Z

At this point we can repeat the same arguments as above to fkn and get that f ′ subsequence of f such that ∃{ kn } kn

lim fk′ f dµ =0, n + | n − | → ∞ ZΩ which is a contradiction. Vitali’s Theorem permits to improve Lebesgue’s Theorem.

96 Theorem 3.6.5. Let µ(Ω) < + and fn be a sequence of µ- ∞ µ { } measurable functions such that f f as n . Suppose n → → ∞ that n N f (x) g (x) µ-a.e., where g is a sequence of ∀ ∈ | n | ≤ | n | { n} µ-summable functions such that

lim gn(x) g(x) dµ =0 n + | − | → ∞ ZΩ for a µ-summable functions g. Then

lim fn(x) f(x) dµ =0. (3.6.8) n + | − | → ∞ ZΩ

Proof. Since limn + gn(x) g(x) dµ = 0, by Vitali’s Theorem, → ∞ Ω | − | the sequence g is uniformly µ-summable, namely ε > 0 δ > 0 { n} R ∀ ∃ such that if µ(A) <δ, then

sup gn(x) dµ ε. (3.6.9) n N A | | ≤ ∈ Z Hence, we have

f (x) dµ g (x) dµ<ε n N, if µ(A) < δ. (3.6.10) | n | ≤ | n | ∀ ∈ ZA ZA

Thus the sequence (fn)n is also uniformly µ-summable and from Vitali’s Theorem, we get the result and we conclude.

97 Theorem 3.6.6. Let µ(Ω) < + and u ,u: Ω R¯ be ∞ k → µ-measurable functions such that

u pdµ < + , u pdµ < + | | ∞ | k| ∞ ZΩ ZΩ u u pdµ 0 as k + , | k − | → → ∞ ZΩ where 1 p< + . Then ≤ ∞ u dµ u dµ as k + . (3.6.11) k → → ∞ ZΩ ZΩ

Proof. The case p = 1 is trivial. Suppose p> 1.

1) Tchebychev Inequality yields

µ x : u (x) u(x) >ε = µ x : u (x) u(x) p >εp { | k − | } { | k − | } p p ε− u (x) u(x) dµ 0 ≤ | k − | → ZΩ as k + . → ∞ (3.6.12)

µ Hence u u as k + . k → → ∞ ii) u are uniformly µ-summable. { k}k Indeed for ε> 0 there exists δ > 0, k > 0 such that

µ(A) <δ = u pdµ<ε and u u pdµ<ε k k. ⇒ | | | k − | ∀ ≥ ZA ZΩ We have

p p p 1 p p u ( u + u u ) 2 − ( u + u u ). (3.6.13) | k| ≤ | | | k − | ≤ | | | k − |

98 Thus, for every A Ω µ-measurable we have ⊆ u dµ (1 + u p) dµ | k| ≤ | k| ZA ZA p 1 p p (3.6.14) µ(A)+2 − u + u u dµ ≤ | | | k − | ZA δ +2pε (1+2p) ε ≤ ≤ if µ(A) <δ<ε and k k. Thus u u dµ 0 as k + by ≥ Ω | k − | → → ∞ Vitali’s Theorem and thus R u dµ u dµ as k + . (3.6.15) k → → ∞ ZΩ ZΩ

Exercise 3.6.7.

1. By applying Lebesgue’s Theorem to the counting measure on N, verify that 2 i lim n ∞ sin − =1. n + → ∞ i=1 n P 2. Let f:[a, + ) R be a locally bounded function and locally ∞ → Riemann integrable. Then f is 1-summable iff f is absolutely L + ∞ Riemann integrable in the generalized sense (namely a f(x) dµ = j | | limj + f(x) dµ exists and it is finite) and in this case → ∞ a | | R R j f(x) d 1 = lim f(x) dµ. [a,+ ) L j + a Z ∞ → ∞ Z Proof. 1. Let g := f χ . g are 1-summable since they are j | | [a,j] j L Riemann-integrable. Moreover, gj is an increasing sequence which converges to f pointwise. | | Beppo-Levi’s Theorem says that that f is 1-summable iff | | L 1 lim gj d < + . j + L ∞ → ∞ ZR

99 Since j g d 1 = f(x) dx j 0 j L | | ∀ ≥ ZR Za this is equivalent to say that f is Riemann integrable in a generalized | | sense. 2. Now set f = fχ . f are -integrable = f are 1- j [a,j] j R ⇒ j L integrable and f f in R as j + , f f . We have just j → → ∞ | j| ≤ | | said that f is 1-summable iﬀ it is absolutely -integrable in the L R generalized sense. By Lebesgue Theorem, we have

j 1 1 1 fd = lim fj d = lim f(x) d . (3.6.16) [a,+ ) L j R L j + a L Z ∞ →∞ Z → ∞ Z It follows that if f is not absolutely -integrable in the generalized 1 R sin x 1 sense, then it is not -summable. For instance, x is not - L j sin x L summable but there exists limj + dx < + . → ∞ a x ∞ In the same way we can prove that if f is locally -integrable on R R X = [a, b) (or X = (a, b], X = (a, b) ), then f is 1-summable on X L iﬀ it is -absolutely integrable in the generalized sense (namely there R b ε exists ﬁnite limε 0 − f(x) dµ) and in this case: → a | | b ε R − fd 1 = lim f(x) dx. L ε 0 Z[a,b) → Za

3.7 The Lp(Ω,µ) spaces, 1 p ≤ ≤ ∞ We suppose as usual that µ is a Radon-measure and Ω is a µ-measurable set.

100 Deﬁnition 3.7.1. For f: Ω R, 1 p< , let → ≤ ∞ 1 p p f Lp(Ω,µ) := f dµ + k k Ω | | ≤ ∞ Z and

f L∞(Ω,µ) := µ ess sup f(x) k k − x Ω | | = inf C ∈ [0, + ]: f C µ-a.e. + . { ∈ ∞ | |≤ }≤ ∞ For 1 p we set ≤ ≤∞ p (Ω, µ)= f :Ω R : f µ-measurable, f p < + . L { → k kL (Ω,µ) ∞}

Remark 3.7.2. For f ∞(Ω, µ), we have f(x) f ∞ for µ-a.e. ∈ L | |≤k kL x.

Proof. Consider a sequence c with c f ∞ as k . { k} k ↓k kL →∞ Set A = x Ω: f(x) > c , k N. By definition µ(A )=0 k k { ∈ | | k} ∈ k ∀ and thus A = A has measure zero as well. Therefore f(x) k k | | ≤ f ∞ x Ω A. k kL ∀ ∈ \S Remark 3.7.3. Consider the Lebesgue measure L1 on [0, 1] and the space p((0, 1], L1). Let us set for every α R L ∈ ϕ (x)= xα x (0, 1]. (3.7.1) α ∀ ∈ Then ϕ p((0, 1]) iff αp +1 > 0. α ∈ L Solution: ϕ p((0, 1]) iff there exists finite the following limit α ∈ L 1 xpα+1 1 lim xpα dx = lim δ 0 δ δ 0 1+ pα δ → Z → (3.7.2) 1 δ1+ pα = lim − . δ 0 pα +1 → The last limit is finite iff αp +1 > 0. Inh this case,i we have 1 1 xpα dL1 = lim xpα dL1 = . (3.7.3) δ 0 1+ αp Z[0,1] → Zδ

101 Exercise 3.7.4. Show that p((0, 1], L1) is not an algebra. L We observe that in general is not a norm since ϕ = 0 iﬀ k·kp k kp ϕ(x)=0 for µ-a.e.

Recall: Let Y be a vector space. A norm on Y is a mapping : k·k Y [0, + ), y y such that → ∞ 7→ k k i) y =0 y = 0. k k ⇐⇒ ii) αy = α y α R, y Y . k k | |k k ∀ ∈ ∀ ∈ iii) y + y y + y y , y Y . k 1 2k≤k 1k k 2k ∀ 1 2 ∈

The space Y endowed with the norm is called a normed space. k·k It is also a metric space with the distance given by d(y , y )= y y 1 2 k 1 − 2k y , y Y . 1 2 ∈ In order to construct a vector space on which is a norm, let k·kp us consider the following equivalence relation on p(Ω, µ): L ϕ ψ ϕ = ψ µ-a.e. ∼ ⇐⇒ Let us denote by Lp(Ω, µ) the quotient space p(Ω, µ)/ = [ϕ]: ϕ L ∼ { ∈ p(Ω, µ) . One can verify that Lp(Ω, µ) is a vector space. L } The precise definition of addition of two elements [ϕ1],[ϕ2] p ∈ L (Ω, µ) is the following: let f1,f2 be two “representative” of [ϕ1] and [ϕ2], respectively such that f1,f2 are finite everywhere (such representatives exist: exercise). Then [ϕ1]+[ϕ2]=[f1 + f2]. We set [ϕ] = ϕ [ϕ] Lp(Ω, µ). k kp k kp ∀ ∈ This definition is independent on the particular element ϕ [ϕ]. ∈ Then, since the zero element of Lp(Ω, µ) is the class consisting of all functions vanishing µ-a.e., it is clear that [ϕ] = 0 [ϕ] = 0. k kp ⇐⇒ To simplify the notation we will hereafter identify [ϕ] with ϕ and we will talk about “Functions in Lp(Ω, µ)” when there is no danger of confusion with the understanding that we regard equivalent functions (i.e. functions differing only on a set of measure zero) as identical elements of the space Lp(Ω, µ).

102 Theorem 3.7.5. The space Lp(Ω, µ) is a complete normed space for 1 p . ≤ ≤∞

Proof. Step 1. We ﬁrst show that Lp(Ω, µ) is a normed space and in particular that p is a norm. k·kL i) The positive homogeneity is clear:

λf ∞ = λ f ∞ k kL | |k kL 1 1 p p p p λf Lp = λf dµ = λ f dµ . k k Ω k k | | Ω | | Z Z 

ii) Triangular Inequality We need some preliminary results.

Lemma 3.7.6. (Young Inequality) Let 1

ap Proof of Lemma 3.7.6: We suppose b > 0. The function a ab p 1 7→ − with a 0 has the maximum at a∗ = b p−1 . Thus for all a 0 ≥ ≥ p p a p b p−1 ab b p−1 b − p ≤ − p (3.7.5) q 1 1 b = b p−1 1 = . − p q 1 1 The numbers p,q such that p + q = 1 are called conjugate.

103 Corollary 3.7.7. (H¨older Inequality) Let 1 p,q be conjugate, f Lp(Ω, µ), g Lq(Ω, µ). Then ≤ ≤∞ ∈ ∈ it holds f g L1(Ω, µ) and · ∈

fg 1 f p g q . (3.7.6) k kL ≤k kL k kL

Proof of Corollary 3.7.7: We suppose p q without loss of generality ≤ Case p = 1, q = • ∞ fg f g ∞ µ-a.e. | | ≤ | |k kL

Thus by monotonicity of the integral, we have

0 and g q > 0. k kL k kL f(x) g(x) | | | | We set f(x) = f p and g(x) = g q . We apply Lemma 3.7.6 to k kL k kL f, g: e e f p gq f g + . (3.7.8) e e ≤ p q Integrating over Ω, we get e e ee 1 1 f g dµ f p dµ + gq dµ. | | ≤ p q ZΩ ZΩ ZΩ Thus ee e e f(x) g(x) dµ | | ZΩ 1. (3.7.9) f p g q ≤ k kL k kL 104 and we conclude.

Corollary 3.7.8.

i) Let f Lp, g Lq with 1 = 1 + 1 1. Then fg Lr and ∈ ∈ r p q ≤ ∈ fg r f p g q . k kL ≤k kL k kL ii) If µ(Ω) < , then ∞ Ls(Ω) Lr(Ω), 1 r < s + . (3.7.10) ⊆ ≤ ≤ ∞

Proof. i) The property is clear if p = q = r =+ . If r< , the property r p∞ r q∞ 1 1 follows from Corollary 3.7.7: f L r , g L r and p + q = 1. | | ∈ | | ∈ r r rs 1 1 s r ii) If µ(Ω) < + . Set g 1 L s−r . Observe that = + − and ∞ ≡ ∈ r s rs apply i).

105 Exercise 3.7.9. Show that In general the inclusion in ii) is strict in the sense that there are functions in Lr that does not belong to Ls with r < s. Hint: Take 1 Ω= 0, 1 , f(x)= , f Lr, 2 1 ( log x)2 x r ∈ − but f / Ls, with s>r. Another example is Ω = (0, 1), f(x) = 1 ∈ p log( ) L p 1, but f / L∞. x ∈ ∀ ≥ ∈

Corollary 3.7.10. (Minkowski Inequality) Let 1 p and f,g Lp(Ω, µ). Then f + g Lp and ≤ ≤∞ ∈ ∈ f + g p f p + g p . k kL ≤k kL k kL

Proof.

p =1: This case follows from the pointwise estimate • f + g f + g . | | ≤ | | | | p = : This case follows from the estimate • ∞ (f + g)(x) f ∞ + g ∞ µ-a.e. x Ω. | |≤k kL k kL ∈ 1

p p 1 p p (f + g)(x) 2 − f(x) + g(x) x Ω | | ≤ | | | | ∀ ∈

p p p 1 it follows that f + g L and f + g − L p−1 . By applying the ∈ | | ∈

106 H¨older inequality, we get

p p 1 f + g p = f + g f + g − dµ k kL | | | | ZΩ p 1 p 1 f f + g − dµ + g f + g − dµ ≤ Ω | | · | | Ω | | | | Z 1 Z p−1 p p p 1 p p f f + g − · p−1 (3.7.11) ≤ Ω | | Ω | | Z 1Z p−1 p p p 1 p p + g f + g − · p−1 Ω | | Ω | | Z Z p 1 =( f + g ) f + g − . k kp k kp k kp p 1 By dividing both sides by f + g − , we get k kp f + g p f p + g p . (3.7.12) k kL ≤k kL k kL

Exercise 3.7.11. (Interpolation Inequality) Let 1 p

Lemma 3.7.13. (Lp(Ω, µ) is complete) p Let (fk)k N be a Cauchy sequence in L (Ω, µ). Then there exists p ∈ f L such that f f p 0 as k + . ∈ k k − kL → → ∞

107 Proof. p = : ε> 0 k N such that ∞ ∀ ∃ ε ∈ f f ∞ <ε ℓ, m k . k ℓ − mkL ∀ ≥ ε For ℓ, m N we set ∈ E = x Ω: f (x) f (x) > f f ∞ . (3.7.13) ℓ,m { ∈ | ℓ − m | k ℓ − mkL }

By deﬁnition of ess-sup, Eℓ,m has measure zero . We deﬁne

E0 = Eℓ,m. (3.7.14) ℓ,m[=1 We have µ(E ) = 0. Now let x Ω E , then x / E ℓ, m. Thus 0 ∈ \ 0 ∈ ℓ,m ∀ if ℓ, m k , we have ≥ ε

fℓ(x) fx(x) fℓ fm < ε. (3.7.15) | − |≤k − k∞ This implies that (f (x)) is a Cauchy sequence in R ( x Ω E ) k k ∀ ∈ \ 0 and since R is complete, it converges to a real number f(x). Thus if (fk(x))k is a Cauchy sequence in L∞(Ω), then it converges µ-a.e. to a function f(x). We consider the inequality

f (x) f (x) ε ℓ, m k , x Ω E . | ℓ − m |≤ ≥ ε ∈ \ 0 Take the limit as m + and we get → ∞ f (x) f(x) ε ℓ k x Ω E . (3.7.16) | ℓ − |≤ ∀ ≥ ε ∀ ∈ \ 0

Since µ(E0) = 0, we have

f (x) f(x) ∞ ε ℓ k (3.7.17) k ℓ − kL ≤ ∀ ≥ ε and

f ∞ f f ∞ + f ∞ c if ℓ k . (3.7.18) k kL ≤k ℓ − kL k ℓkL ≤ ≥ ε

From (3.7.18) it follows that f L∞(Ω). ∈ 108 p< : ∞

We are going to construct a convergent subsequence (fkn )n N. One ∈ chooses kn so that 1 f f p < ℓ, m k . k ℓ − mkL 2n ∀ ≥ n

1 p (One proceeds as follows: choose k1 such that fℓ fm L < 2 ℓ, m k −1 k ∀ ≥ p k1, then choose k2 > k1 such that fℓ fm L < 22 ℓ, m k2, etc.). pk − k ∀ ≥ We claim that fkn converges in L . In order to simplify the notation we write fn instead of fkn , so that 1 f f p n 1. (3.7.19) k n+1 − nkL ≤ 2n ∀ ≥

n Let g (x)= f (x) f (x) so that g p 1. n k=1 | k+1 − k | k nkL ≤ As a consequenceP of the monotone convergence theorem, gn(x) converges to g(x) µ-a.e. in Ω where g(x) = ∞ f (x) f (x) k=1 | k+1 − k |∈ Lp. On the other hand for m ℓ 2, we have ≥ ≥ P fm(x) fℓ(x) fm(x) fm 1(x) + + fℓ+1(x) fℓ(x) | − | ≤ | − − | · · · | − | = gm(x) gℓ 1(x) − − g(x) gℓ 1(x). ≤ − − (3.7.20)

It follows that µ-a.e. on Ω, fn(x) is a Cauchy sequence and converges to a limit f(x). We have µ-a.e. on Ω

f(x) f (x) 2g(x) ℓ 2 (3.7.21) | − ℓ |≤ ∀ ≥ and in particular f Lp. ∈ Finally we conclude by dominated convergence that f f p 0 k n − kL → since f (x) f p 0 and | n − | → f f p 2p g p L1. | n − | ≤ | | ∈

We can easily see that the entire sequence fn converges to f .

109 Indeed, for all n N and k>k , we have ∈ n f f p f f p + f f p k k − kL ≤k k − kn kL k kn − kL 1 (3.7.22) + f f p . ≤ 2n k kn − kL Thus 1 p p lim sup fk f L lim n + fkn f L =0. (3.7.23) k + k − k ≤ n + 2 k − k → ∞ → ∞

The proof of Theorem 3.7.5 is concluded as well.

A Banach space X is called separable if it contains a countable dense subset.

Example 3.7.14. 1. (R, ). The countable dense subset is Q. | · | 2. (C0([a, b], R), ). The countable subset is the set of k·k∞ the polynomials with rational coeﬃcients (Stone-Weierstrass Theorem).

Theorem 3.7.15. Let Ω Rn be open and 1 p < . ⊆ ≤ ∞ Then it holds

i) Lp(Ω, µ) is separable.

0 p ii) Cc (Ω) is dense in L (Ω, µ).

We recall that C0(Ω) denotes the space of continuous function f:Ω c → R with supp(f)= x Ω: f(x) =0 Ω and supp f is compact. { ∈ 6 }⊆

Remark 3.7.16. For p = the conditions of Theorem 3.7.15 do ∞ not hold in general.

110 1. LetΩ=[0, 1], µ = 1. For0 0 is more than countable. Suppose there is a countable set (ek)k dense in L∞((0, 1]) with

1 inf ft ek L∞ < t> 0. k k − k 2 ∀ We would have

L∞([0, 1]) B(e , 1/2) ⊆∪n n

in contrast with the fact no pair of functions of the family (ft)t>0 belongs to the same ball. 2. If Ω Rn is compact and µ = n then C(Ω) is a closed, proper ⊂ L0 subset of L∞(Ω) and therefore Cc (Ω) L∞(Ω).

Proof of Theorem 3.7.15: i) We suppose Ω = Rn. Observe that if we prove that Lp(Rn) is separable then Lp(Ω) is separable as well for every Ω Rn. There is a canonical isometry from Lp(Ω) into Lp(Rn) ⊆ that to every f Lp(Ω) associate f˜ Lp(Rn) where f˜(x) = f(x) if ∈ ∈ x Ω and f˜(x)=0is x Ωc. Therefore Lp(Ω) can be identiﬁed with ∈ ∈ a subset of Lp(Rn) and therefore it is separable. Let E be the countable set deﬁned by • N n E = ak χQk , ak Q,Qk R is a cube of a dyadic k=1 ∈ ⊆ (3.7.24) n P ℓ decomposition of length 2− , ℓ 0 . ≥ o

Claim 3.7.17. E is dense in Lp.

Proof of the Claim 3.7.17: Let f 0. We know that f is the monotone ≥ limit of a sequence of characteristic functions

k f = 1 χ . k j Aj j=1 P 111 By Lebesgue Theorem f f p 0. Therefore it is enough to show k k − kL → the following property:

Claim 3.7.18. χ E for every µ-measurable A Rn with A ∈ ⊆ µ(A) < . ∞

Proof of the Claim 3.7.18: Since µ is a Radon measure, we have µ(A)= inf µ(G) G open. (3.7.25) A G ⊆ Let us choose G A open G G with µ(G ) µ(A)+ 1 , k N. k ⊇ k+1 ⊆ k k ≤ k ∈ Since χ χ 0, 1 , it follows that Gk − A ∈{ } p p 1 χGk χA dµ = χGk A dµ = µ(Gk A) . (3.7.26) | − | | \ | \ ≤ k ZR ZR Thus we may suppose A open. Every open set can be expressed as countable disjoint union of dyadic cubes Iℓ: A = ℓ∞=1 Iℓ and µ(A)= ∞ µ(I ) (Lemma 1.3.4). ℓ=1 ℓ S P In particular: n n χ = χIℓ χA as n + . Iℓ ℓ=1 → → ∞ ℓ=1 S P Since µ(A) < (= χ Lp), we have by Lebesgue Theorem ∞ ⇒ A ∈ n p χA χIℓ L 0 (3.7.27) k − ℓ=1 k → as n + and clearly P → ∞ n

χIℓ E n. (3.7.28) ℓ=i ∈ ∀ P This concludes the proof of Claim 3.7.18 and 3.7.17. If f is a general function, then we apply the above arguments to + f and f −. C0(Rn) is dense in Lp(Rn). • c Notation:

112 Truncation operator

p n Claim 3.7.19. Given f L and ε> 0 g L∞(R) and K R ∈ ∃ ∈ ⊆ compact such that g 0 on Kc and ≡ ε f g < . k − kp 2

Proof of Claim 3.7.19: Let f = χ T (f), n 1. n B(0,n) n ≥ We have f f and by Lebesgue Theorem f f p 0 as n → k n − kL → n + . Let us choose g = f , n large enough. → ∞ n

n Claim 3.7.20. Let g L∞(R ) with compact support and δ > 0. ∈ Them there exists g C0(Rn) such that 1 ∈ c

g g 1 < δ. k − 1kL

Proof of Claim 3.7.20: We assume g 0= g is the pointwise limit ≥ ⇒ of simple functions gn and we can pass to the limit under integral sign by Lebesgue Theorem, namely g g 1 0 as n + . k n − kL → → ∞ Therefore we can prove Claim 3.7.20 when g = χA, A measurable and bounded. We know that F closed and G open such that ∃ F A G and µ(G F ) <δ ⊆ ⊆ \ |

(we may assume that G is compact). By Urysohn Lemma there is g1: Rn R continuous such that → 113 a) 0 g (x) 1, ≤ 1 ≤ c b) g1(x)=0on G ,

c) g1(x)=1on F ; and

χA g1 dµ µ(G F ) < δ. n | − | ≤ \ ZR We may always suppose without loss of generality that the function g in Claim 3.7.20 satisﬁes g ∞ g ∞ (otherwise we take g (x)= 1 k 1kL ≤k kL 1 T g with n = g ∞ ). n 1 k kL We observe that e

1 p−1 p p g g p g g 1 2 g ∞ . k − 1kL ≤k − 1kL k kL p−1 ε ∞ p 1/p If δ > 0 is chosen small enough so that (2 g L ) δ < 2, then ε k k g g p < . Now we have k − 1kL 2 ε ε f g p f g p + g g p + = ε (3.7.29) k − 1kL ≤k − kL k − 1kL ≤ 2 2 and we can conclude. Let Ω Rn be open. Then C0(Ω) is dense in Lp(Ω). • ⊆ c Let f Lp(Ω) and f˜ be deﬁned as f˜(x) = f(x) if x Ω and ∈ ∈ f˜(x)=0is x Ωc. We have f˜ Lp(Rn). For every ε > 0 there is n ∈ ∈ gε Cc(R ) such that ∈ ε f˜ g p < . (3.7.30) k − εkL 2 n One considers a sequence of open sets On of R such that O¯n is compact, O¯ O and O = Ω. We set for every n 1, n ⊆ n+1 ∪n n ≥ c dOn+1 (x) ψn(x)= gε(x) . c dOn+1(x) + dOn (x) We have ψ = 0 in Oc and ψ C (Ω). Moreover ψ = g in O , n n+1 n ∈ c n ε n ψ g , and since O Ω as n + we have ψ g in Ω and | n| ≤ | ε| n → → ∞ n → ε ψ g p 0 as n + . Therefore for all ε > 0 there is n 1 k n − εkL → → ∞ ε ≥ such that ε ψ g p < . (3.7.31) k nε − εkL (Ω) 2 114 The following estimate holds:

p ˜ p p f ψnε L (Ω) f gε L (Ω) + ψnε gε L (Ω) k − k ≤ k − k k − k ε ε f˜ g p n + ψ g p + = ε ≤ k − εkL (R ) k nε − εkL (Ω) ≤ 2 2 We can conclude the proof.

Theorem 3.7.21. Let f,f Lp(Ω, µ), k N, 1 p< . k ∈ ∈ ≤ ∞ Then following two conditions are equivalent:

i) f f p 0 k . k k − kL → →∞ µ ii) f f as k + and f p f p as k + . k → → ∞ k kkL →k kL → ∞

Proof. i) = ii): ⇒ µ The fact that f f as k + is a consequence of Tchebychev k → → ∞ inequality as we have already observed several times. From Minkowski inequality it follows that

f p f p f f p . |k kL −k kkL |≤k k − kL ii) = i): We observe that ⇒ p 1 p p 1 p p 2 − f +2 − f f f 0. | k| | | − | k − | ≥ Let (f ) be a subsequence of (f ) converging to f as n + µ-a.e. kn k → ∞

115 By Fatou’s Lemma, we get

p p p 2 f dµ lim sup fkn f dµ = Ω | | − n + Ω | − | Z → ∞ Z p 1 p p 1 p p lim inf 2 − fk +2 − f fk f dµ n | n | | | − | n − | ≥ →∞ Z p 1 p p 1 p p p p lim inf (2 − fk +2 − f fk f ) dµ =2 f . n | n | | | − | n − | | | ZΩ →∞ ZΩ p = lim sup fkn f dµ =0. ⇒ n + Ω | − | → ∞ Z (3.7.32) It follows that p lim fk f dµ =0. n + | n − | → ∞ ZΩ

Claim 3.7.22. f f p 0 as k . k k − kL → →∞

Proof of Claim 3.7.22. Suppose by contradiction that

p lim sup fk f dµ > 0. k + Ω | − | → ∞ Z Then there is a subsequence fkn such that

p p lim fkn (x) f(x) dµ = limsup fk(x) f(x) dµ > 0. n + Ω | − | k + Ω | − | → ∞ Z → ∞ Z µ Since f f as n + , f ′ f µ-a.e. as n′ + . Then by kn → → ∞ ∃ kn → → ∞ the prevoius arguments we get that f f p 0 which contradicts k kn′ − kL → that p lim fk f dµ > 0. n + | n − | → ∞ ZΩ

116 Remark 3.7.23. For p 1, the function x x p is convex, therefore ≥ 7→ | | x + y p 1 1 x p + y p 2 ≤ 2 | | 2 | | (3.7.33) p p 1 p p 1 p x + y 2 − x +2 − y x, y R. | | ≤ | | | | ∀ ∈ One can directly prove that a, b R ∀ ∈ p p 1 p p 1 p a + b 2 − a 2 − b 0. | | − | | − | | ≤ p p 1 p 1 p It is enough to consider f(x) = 1+ x 2 − 2 − x and show | | − − | | that f(x) 0. ≥

117 Chapter 4

Product Measures and Multiple Integrals

4.1 Fubini’s Theorem

Let X,Y be sets.

Deﬁnition 4.1.1. Let µ be a measure on X and ν be a measure on Y . We deﬁne the measure µ ν: (X Y ) [0, + ] by × P × → ∞ setting for each S X Y ⊆ × ∞ (µ ν)(S) = inf µ(Ai) ν(Bi) × i=1 n SP ∞ A B ⊆ i=1 i × i (4.1.1) A X µ-measurable i ⊆S B Y µ-measurable, i N . i ⊆ ∈ o The map µ ν is called the Product Measure. ×

Remark 4.1.2. If we define for every A X µ-measurable, B Y ⊆ ⊆ ν-measurable λ(A B)= µ(A) ν(B). (4.1.2) × Then λ is a pre-measure on the algebra of finite disjoint unions. Thus µ ν is the Caracthéodory extension of λ. × Exercise 4.1.3. X = Rk, Y = Rℓ and µ = k, ν = ℓ = µ ν = n L L ⇒ × L n = k + ℓ.

118 Digression We consider for simplicity R2. Let f: R2 R be a 2-summable → L function, namely a measurable function such that

f(x, y) d 2 < + . 2 | | L ∞ ZR We wonder if (by analogy with Riemann-integral for continuous functions) if the above integral can be splited in two integrals on R, namely

i) y R: x f(x, y) 1-measurable. ∀ ∈ 7−→ | | L ii) y f(x, y) d 1(x) is 1-measurable and summable. 7−→ | | L L ZR

Example 4.1.4. Let P be a non-measurable subset of R which is contained in [0, 1] and let E = P Q. Then E is measurable in R2. × Indeed, E [0, 1] Q which is a rectangular of null-measure ⊆ × 2([0, 1] Q) = 1[0, 1] 1(Q)=0 L × L × L = E is 2-measurable. ⇒ L Let us consider f(x, y)= χ (x, y)= χ (x) χ (y), f is 2-measurable E P Q L and its integral is zero, since f 0 2-a.e. Let y R be ﬁxed. Then ≡ L ∈ f(x, y)= χ (x) if y Q and = 0 otherwise. P ∈ Thus if y Q, f(x, y) is not measurable and it has no sense to write ∈ f(x, y) d 1(x). This means that we cannot apparently split the R L integral of f over R2 into two integrals. Actually the function f(x, y) R is measurable for 1-a.e. y. Therefore we can deﬁne for 1-a.e. y R L L ∈

119 the function y f(x, y) d 1(x) which is identically zero. We 7−→ R | | L can also re-deﬁne theZ function when y Q (the value of its integral ∈ does not change since 1(Q) = 0) in order to obtain a 1-measurable L L function which also 1-summable. This example shows that even L if the function obtained by freezing one of the two variables is not measurable, the values of x or y for which we get a non-measurable function have measure zero and therefore they can be neglected when we integrate.

Theorem 4.1.5. (Fubini) Let µ, ν be Radon measures on X = Rk, Y = Rℓ respectively, µ ν be the product measure on X Y = Rn, n = ℓ + k. × × Then it holds

i) For every A X µ-measurable, B Y ν-measurable, ⊆ ⊆ we have A B is µ ν-measurable and × × (µ ν)(A B)= µ(A) ν(B). (4.1.3) × ×

120 ii) Let S X Y be (µ ν)-measurable with (µ ν)(S) < ⊆ × × × + . Then the set S = x :(x, y) S is µ-measurable ∞ y { ∈ } for ν a.e. y and the mapping

y µ(S )= χ (x) dµ 7−→ y Sy ZX is ν- summable with

(µ ν)(S)= µ(S ) dy = χ (x) dµdν. × y Sy ZY ZY ZX An analogous property holds for S = y :(x, y) S . x { ∈ } iii) µ ν is a Radon measure. × iv) If f: X Y R is µ ν-summable, then the mapping × → × x f(x, y) and y f(x, y) are µ-summable for 7−→ 7−→ ν-a.e. y and ν-summable for µ-a.e. x and the mappings

y f(x, y) dµ(x) 7−→ ZX x f(x, y) dν(y) 7−→ ZY are respectively ν- and µ-summable and

fd(µ ν) = f(x, y) dµ(x) dν(x) X Y × Y X Z × Z Z = f(x, y) dµ(y) dµ(x). X Y Z Z 

Proof. We deﬁne the set = S X Y : S satisﬁes 1 and 2 (4.1.4) F ⊆ × where n o

121 1 x χ (x, y) is µ-measurable µ-a.e. y 7−→ S and

2 y χ (x, y) dµ(x) = µ(S ) is ν-measurable. 7−→ S y ZX For S we set ∈F

ρ(S)= χS(x, y) dµ(x) dν(y) . (4.1.5) Y X Z Z 

Goal: To show that every S X Y µ ν-measurable belongs to and ⊆ × × F f(S)=(µ ν)(S), in particular for S = A B. We deﬁne the families × × = A B, A µ-measurable, B Y ν-measurable P0 { × ⊆ ⊆ } ∞ = S , S P1 j j ∈ P0 j=1 (4.1.6) n [ o ∞ = R , R . P2 j j ∈ P1 j=1 n \ o

Claim 4.1.6. , , . P0 P1 P2 ⊆F

Proof of the Claim 4.1.6: i) Clearly, . Let S = A B. We have P0 ∈F × A if y B Sy = ∈ (4.1.7) ( if y / B. ∅ ∈ Thus Sy is clearly µ-measurable. Moreover, µ(Sy)= µ(A) χB. Thus y µ(S ) (4.1.8) 7−→ y is ν-measurable as well and ρ(S)= µ(A) µ(B).

122 ii) . We ﬁrst observe that for A B , A B it holds P1 ⊆F 1 × 1 2 × 2 ∈ P0 (A B ) (A B )=(A A ) (B B ) (4.1.9) 1 × 1 ∩ 2 × 2 1 ∩ 2 × 1 ∩ 2 ∈ P0 and

(A B ) (A B )= (A A ) B (A A ) B B . (4.1.10) 1× 1 \ 2× 2 1\ 2 × 1 ∪ 1∩ 2 × 1\ 2 A B 2 × 2

(A B ) (A B ) 1 × 1 \ 2 × 2 Fig. 4.1

Let now S = j∞=1 Sj. We deﬁne

n 1 S − S = S , S = S S , S = S S . (4.1.11) 1 1 2 2 \ 1 n n \ k k[=1 e e e Therefore each member of is a countable disjoint union of members P1 of . If S = ∞ S , S mutually disjoint, then the map P0 j=1 j j ∈ P0 S k ∞ x χS(x, y)= χSj (x, y)= lim χSj (x, y) 7−→ k + 0 → ∞ j P P is measurable since it is the limit of µ-measurable functions. We have A : y χ (x, y) dµ(x) is ν-measurable. Thus J 7−→ X Sj R k

y χS(x, y) dµ(x)= lim χSj (x, y) dµ(x) 7−→ k + ZX ZX → ∞ 0 k P (4.1.12) = lim χSj (x, y) dµ(x) k ւ →∞ 0 ZX by Beppo-Levi Theorem P

123 is ν-measurable. Moreover,

ρ(S)= χS(x, y) dµ dν Y X Z Z ∞ = χSj (x, y) dµdν Y j X Z Z (4.1.13) ∞ P = χSj (x, y) dµ dν j=0 Y X Z Z P∞ = ρ(Sj). j=0 P Above we have applied two times Beppo Levi-s Theorem.

Claim 4.1.7. For R = ∞ R , R and ρ(R ) < + j, j=0 j j ∈ P1 j ∞ ∀ then T k ρ(R)= lim ρ Rj . (4.1.14) k + → ∞ 0 T 

Proof of Claim 4.1.7: Let R = ∞ R , R , ρ(R ) < + j. j=0 j j ∈ P1 j ∞ ∀ We may assume that R R (otherwise we consider the family j+1 ⊆ Tj (Rj)j:

R1 = R1, R2 = R1 R2 e ∩ (4.1.15) Rj = Rj+1 Rj. e ∩e e Since χ (x, y) χ (x,e y) we applye twice the Lebesgue Theorem and Rj ≤ Rj

124 get

ρ(R)= χR(x, y) dµ dν Y X Z Z 

= lim χRj (x, y) dµ dν Y X y Z Z (4.1.16) = lim χRj (x, y) dµ dν Y j + X Z → ∞ Z 

= lim χRj (x, y) dµ) dν j + Y X → ∞ Z Z = lim ρ(Rj). j + → ∞ Therefore (4.1.14) holds.

Claim 4.1.8. For S X Y , we have ⊆ × (µ ν)(S) = inf ρ(R) S R, R . (4.1.17) × | ⊆ ∈ P1 

Proof of Claim 4.1.8: First we observe that if S R = ∞ A B , ⊆ i=1 i × i then n S ∞ ρ(R) ρ(Ai Bi)= µ(Ai) µ(Bi) ≤ i=1 × i=1 (4.1.18) ↓ P P Claim 4.1.6 Thus inf ρ(R): S R (µ ν)(S1). (4.1.19) { ⊆ ∈ P1}≤ × Moreover, there exists a disjoint sequence A′ B′ in such that { j × j} P0 ∞ R = (A′ B′ ). j × j j=0 [ Thus

∞ ∞ ρ(R)= ρ(Aj′ Bj′ )= µ(Ai′) ν(Bj′ ) j=0 × j=0 (4.1.20) P (Pµ ν)(S). ≥ × 125 We prove i) of Theorem 4.1.5. Let A B . By deﬁnition of µ ν and that ρ(A B)= µ(A) ν(B), × ∈ P0 × × we have

(µ ν)(A B) µ(A) ν(B)= ρ(A B) ρ(R) × × ≤ × ≤ R with A B R. Claim 4.1.8 implies that ∀ ∈ P1 × ⊆ (µ ν)(A B)= ρ(A B)= µ(A) ν(B). × × ×

Claim 4.1.9. A B is µ ν-measurable. × ×

Proof of Claim 4.1.9. Take T X Y , R with T R. Then ⊆ × ∈ P1 ⊆ R (A B), R (A B) are disjoint and belong to . Consequently, \ × ∩ × P1 (µ ν) T (A B) +(µ ν) T (A B) × \ × × ∩ × ≤ ρ(R (A + B)+ ρ R (A B) = ρ(R). \ ∩ × According to Claim 4.1.8, we get

(µ ν) T (A B) +(µ ν) T (A B) (µ ν)(T ). (4.1.21) × \ × × ∩ × ≤ × = A B is µ ν-measurable. ⇒ × ×

Claim 4.1.10. For each S X Y there is a set R such ⊆ × ⊆ P2 that S R and ⊆ ρ(R)=(µ ν)(S). ×

Proof of Claim 4.1.10: If(µ ν)(S)=+ , then take R = X Y . × ∞ × Otherwise, for j N, we choose S R with (µ ν)(S) ∈ ⊆ j ∈ P1 × ≤ ρ(R ) (µ ν)(S)+ 1. j ≤ × j We may suppose R R . We set R = R S. Since by j+1 ⊂ j j j ⊇ Claim 4.1.7 limj ρ(Rj)= ρ(R), we get the property, since R 2. →∞ T ∈ P We prove ii) of Theorem 4.1.5.

126 Let S X Y be µ ν-measurable and (µ ν)(S) < + . We ⊆ × × × ∞ choose R according to Claim 4.1.10 with S R and ∈ P2 ⊆ (µ ν)(S)= ρ(R). × We have two cases.

a) (µ ν)(S)=0= ρ(R)=0= µ(Ry) dν =0= µ(Ry) = × ⇒ ⇒ Y ⇒ 0 ν-a.s. y. Z Now we have 0 X (x, y) X (x, y), ≤ S ≤ R and in particular 0 χ (x) χ (x). ≤ Sy ≤ Ry Since µ(R )=0= µ(S ) = 0. Thus S is µ-measurable. y ⇒ y y Moreover, y µ(S ) dν =0 (4.1.22) 7−→ y Zy and thus it is ν-measurable. In particular, ρ(S)=0.

b) (µ ν)(S)= ρ(R) > 0. Consider R S: × \ (µ ν)(R S)=(µ ν)(R) (µ ν)(S) × \ × − × = ρ(R) (µ ν)(S)=0. − × We remark that every member of is µ ν-measurable and if P2 × R ,(µ ν)(R) < , then (µ ν)(R)= ρ(R). ∈ P2 × ∞ × Since (µ ν)(R S)= 0, then by a) ρ(R S) = 0. This means that × \ \ µ(R S) = µ(R S )=0. \ y y \ y Since R S is µ-measurable (having null measure), then S is µ- y − y y measurable as well and µ(Ry)= µ(Sy),

(µ ν)(S)= ρ(R)= µ(R ) dy × y ZY

= µ(Sy) dy = ρ(S). ZY 127 iii) No proof.

Proof of iv) of Theorem 4.1.5. If f = χ < + , then iv) follows from S ∞ ii). If f 0 we write ≥ ∞ 1 f = χAj (x, y), (4.1.23) j=1 j P with A (µ ν)-measurable. j × Because of ii), the linearity of integrals, the property holds for the partial sums k 1 fk = χAj . (4.1.24) j=1 j By passing to the limit as k + Pand applying Beppo-Levi Theorem, → ∞ we get the property for f 1 f(x, y) dµ dν = ∞ X (x) dµ(x) dν(y) j j ZY ZX ZY ZX j=1 P 1 = ∞ χ dµ(x) dν(y) j Aj j=1 ZY ZX P ∞ 1 = (µ ν)(Aj)= f(x, y) dµ ν. j=1 j × X Y × Z × P (4.1.25)

+ For general f, consider f = f f −. − Remark 4.1.11. A set S X Y that satisﬁes 1 and 2 is ⊆ × not necessarily (µ ν)-measurable: take X = Y = R, µ = ν = 1, × L A [0, 1] not measurable and set ⊆ 1 S = (x, y): x χ (y) < ; y [0, 1] . (4.1.26) { | − A | 2 ∈ }

S is not (µ ν)-measurable, but y [0, 1], S is µ-measurable × ∀ ∈ y and µ(Sy)=1. If f: X Y R+ is µ ν-measurable, then f is µ ν-summable if • × → × × one of the two iterated integrals exists ﬁnite (Tonelli ’s Theorem).

128 It may happen that the two iterated integrals exist without f • being µ ν-measurable (take f = χ , S as above) or without f × S being µ ν-summable (take f(x, y)= sin(y) with (x, y) (0, 1] × x ∈ × [0, 2π]. In this case we have 2π f(x, y)dy = 0 for all x (0, 1] 0 ∈ but the function is not summable). R 4.2 Applications

x2 y2 1. Show that f(x, y) = − is not Lebesgue summable in [0, 1] (x2+y2)2 × [0, 1], namely x2 y2 2 − 2 2 dxdy =+ . [0,1] [0,1] (x + y ) ∞ Z ×

1st Proof. We apply Fubini’s Theorem: if f L1(X Y ) then for ∈ × µ a.e. x X, f(x, ) L1(Y ) and f(x, y) dν L1(X), for ν-a.e. ∈ · ∈ Y ∈ y Y , f( , y) L1(X) and f(x, y) dµ L1(Y ). Moreover, ∈ · ∈ X R ∈ f(x, y) d(µ νR)= f(x, y) dν dµ X Y × X Y Z × Z Z (4.2.1) = f(x, y) dµ dν . Y X Z Z If f L1([0, 1] [0, 1]) then the two iterated integrals should be equal. ∈ × We compute separately the two iterated integrals (we observe that we can treated the two iterated integral as Riemann improper integrals): 1 x2 y2 1 x2 + y2 2y2 − dy = − dy (x2 + y2)2 (x2 + y2)2 Z0 Z0 1 1 1 2y2 = dy + − dy x2 + y2 (x2 + y2)2 Z0 Z0 1 1 y 1 1 1 = 2 2 dy + 2 2 2 2 dy 0 x + y x + y 0 − 0 x + y Z Z 1 = x2 +1 1 1 1 π = 2 dx = arctan x = . ⇒ 0 x +1 0 4 Z

129 On the other hand: 1 1 x2 y2 π − dxdy = . (x2 + y2)2 − 4 Z0 Z0

2nd Proof. One computes directly x2 y2 1 y y2 x2 1 x2 y2 2 − 2 2 d(µ ν)= 2 − 2 2 dx + 2 − 2 2 dx dy [0,1]2 (x + y ) × 0 0 (x + y ) y (x + y ) Z Z h Z Z i 1 2

y y y y2 x2 | 1 {z } 2x2 − dx = dx | {z dx } (x2 + y2)2 (x2 + y2) − (x2 + y2)2 Z0 Z0 Z0 y 1 x y y 1 = 2 2 dx + 2 2 2 2 2 dx 0 (x + y ) y + x 0 − 0 (x + y ) Z Z y 1 = = 2y2 2y 1 x2 y2 1 1 1 2x2 − dx = dx + dx (x2 + y2)2 − (x2 + y2) (x2 + y2) Zy Zy Zy 1 1 x 1 1 1 = 2 2 dx 2 2 + 2 2 dx − y (x + y ) − x + y y y x + y Z Z 1 y 1 1 = + = + . 1+ y2 2y2 −1+ y2 2y Therefore 1 y y2 x2 1 x2 y2 1 1 1 1 2 − 2 2 dx + 2 − 2 2 dy = dy + 2 dy 0 0 (x + y ) y (x + y ) 0 y 0 −1+ y Z hZ Z i Z Z 1 1 = log y arctan y =+ . | | 0 − 0 ∞

2. Gaussian Integral We want to compute + ∞ x2 e− dx. Z−∞ 130 x2 There is no elementary indeﬁnite integral for e− dx but the deﬁnite + x2 integral ∞ e− dx can be evaluated: −∞ R

R (x2+y2) (x2+y2) e− dxdy = e− dx dy 2 ZR ZR ZR + + ∞ x2 ∞ y2 = e− dx e− dy Z−∞ Z−∞ + 2 ∞ x2 = e− dx . Z−∞ On the other hand: 2π + (x2+y2) ∞ r2 e− dxdy = e− rdrdθ 2 ZR Z0 Z0 + ∞ r2 ∞ s =2π re− dr = π e− ds = π 0 ↓ 0 s= r2 Z − Z + x2 Therefore ∞ e− dx = √π. −∞ R

4.3 Convolution

We suppose µ = n is the n-dimensional Lebesgue measure. For L Ω Rn, 1 p we will write for simplicity Lp(Ω) or Lp instead ⊆ ≤ ≤ ∞ of Lp(Ω, n). L

Lemma 4.3.1. Let f: Rn R be n-measurable. Then the → L function F (x, y)= f(x y): Rn Rn R − × → is 2n-measurable. L

131 Proof. With the substitution z := x y we obtain − 1 F − ([ , a) = (x, y): f(x y) < a = (x,x z): f(z) < a −∞ { − } { − } = T (Rn z : f(z) < a ) ×{ } where T : R2n R2n (x, z) (x,x z). → 7→ − T is Lipschitz continuous and Rn z : f(z) < a is 2n-measurable. ×{ } L The claim follows from the following lemma:

Lemma 4.3.2. (no proof) Let T : Rn Rn be Lipschitz continuous with → T (x) T (y) L x y x, y Rn | − |≤ | − | ∀ ∈ and let A Rn be n-measurable. Then T (A) is n- ⊆ L L measurable.

Deﬁnition 4.3.3. Let f,g : Rn R be n-measurable functions → L such that for a.e x Rn the function y f(x y)g(y) is n- ∈ 7→ − L integrable. We deﬁne the convolution product of f and g by

(f⋆g)(x)= f(x y) g(y) dy for a.e x Rn. (4.3.1) n − ∈ ZR

132 Remark 4.3.4.

1. If f,g : Rn [0, + ] are n-measurable functions, → ∞ L then, since the function f(x y) g(y) is positive and − n-measurable, (f ⋆g): Rn [0, + ] is well deﬁned L → ∞ for every x Rn. ∈ 2. By making in (4.3.1) the change of variable z = x y − and using the translation invariance of the Lebesgue measure we obtain that the function f(x y)g(y) is n- − L integrable iﬀ the function f(z)g(x z) is n-integrable − L and (f ⋆g)(x) = (g⋆f)(x). This proves that the convolution is commutative.

Theorem 4.3.5. Let f,g L1(Rn). Then for a.e x Rn the ∈ ∈ function y f(x y)g(y) is n-summable and f⋆g L1(Rn) 7→ − L ∈ with f⋆g 1 f 1 g 1 . k kL ≤k kL k kL

Proof of Theorem 4.3.5. By Lemma 4.3.1 the function F (x, y) = f(x y) g(y) is 2n-measurable. − L i) We suppose f,g 0. Then F is 2n-integrable. By Tonelli’s ≥ L Theorem, F is 2n-summable if and only if one of the iterated L

133 integrals exist ﬁnite

f⋆g L1 = f(x y) g(y) dy dx k k Rn Rn − Z Z = f(x y) g(y) dx dy Rn Rn − Z Z = g(y) f(x y) dx dy Rn Rn − Z Z = f 1 g 1 < + . k kL k kL ∞ ii) For general f,g it follows from i) that

F (x, y) = f(x y) g(y) L1(R2n). | | | − | | |∈ Thus by Fubini’s Theorem f ⋆ g L1(Rn). We have | | | |∈

f⋆g L1 = f(x y) g(y) dy dx k k Rn Rn − ≤ Z Z

f(x y) g(y) dy dx = (4.3.2) n n | − | | | ZR ZR

f ⋆ g (x) dx = f ⋆ g L1 = f L1 g L1. n | | | | k | | | |k k k k k ZR In particular it follows that for a.e x Rn the function y ∈ 7→ f(x y)g(y) is in L1(Rn). −

134 Corollary 4.3.6. Let 1 p + , 1 q + . ≤ ≤ ∞ ≤ ≤ ∞ i) f L1, g Lp Then for a.e x Rn the function y ∈ ∈ ∈ 7→ f(x y)g(y) is n-summable and f⋆g Lp with f ⋆ − L ∈ k g p f g p . kL ≤k kL1 k kL ii) f Lp, g Lq, 1 1 + 1 = 1+ 1. Then for a.e ∈ ∈ ≤ p q r x Rn the function y f(x y)g(y) is n-summable ∈ 7→ − L and f⋆g Lr with ∈ f⋆g r f p g q . k kL ≤k kL k kL

Proof of Corollary 4.3.6.

i) The case p = 1 follows from Theorem 4.3.5. For 1

1 1 p f(x y) q f(x y) p g(y) dy dx. ≤ Rn Rn | − | | − | | | Z Z From the case p = 1, it follows that for a.e. x Rn, y ∈ 7−→ f(x y) g(y) p is µ-summable and thus | − | | | 1 y f(x y) p g(y) Lp(Rn). (4.3.3) 7−→ | − | | |∈

1 Moreover, y f(x y) q Lq. From H¨older inequality it 7−→ | − | ∈ follows

1 1 1 n f(x y) g(y) = f(x y) q f(x y) p g(y) L (R ) for a.e x Rn. | − | | | | − | | − | | |∈ ∈ (4.3.4)

135 We have p f⋆g pdx = f(x y) g(y) dy dx Rn | | Rn Rn − Z Z Z p 1 1 f(x y) q f(x y) p g(y) dy dx. ≤ n n | − | | − | | | ZR ZR p p q p Rn f⋆g dx f(x y) dy f(x y) g(y) dx | | ≤ n n | − | n | − | | | ZR ZR ZR R p q p f 1 f ⋆ g 1 ≤ k kL k | | | | kL

p q p f f 1 g 1 ≤ k kL1 k kL k| | kL

p q +1 p f 1 g p ≤ k kL k kL

= f⋆g Lp f L1 g Lp . ⇒ k k ≤k k k k (4.3.5)

The case p = follows form H¨older inequality. ∞ ii) We assume without loss of generality that p q. Since 1+ 1 = ≤ r 1 + 1 1+ 1, we get p q r. p q ≤ q ≤ ≤

The case r = follows from H¨older inequality. • ∞

The case p =1 follows from i) (q = r). •

Thus 1

136 We have

We apply the generalized H¨older inequality (see Corollary 3.7.8) in (4.3.6) and get that for a.e x Rn, the function y f(x y)g(y) ∈ 7→ − is in L1(Rn). Moreover the following estimate holds:

r 1 p r 1 q r p q f⋆g f − r s g − r t f ⋆ g | | ≤ k | | kL k | | kL | | | | r−p r−q r ( rp ) r( rq ) = f pdx g qdx f p ⋆ g q (4.3.9) | | | | | | | | Z p q Z r(1 r ) r(1 r ) p q = f p − g q − f ⋆ g k kL k kL | | | | Therefore:

1 p q r r(1 r ) p r(1 r ) q r − − p q f⋆g L f Lp f Lp g Lq g Lq = f L g L . k k ≤ k k k k k k k k k k k k 

Other properties of convolutions

Deﬁnition 4.3.7. Let 1 p . We say that a function f: p ≤ ≤ ∞ Rn R belongs to L (Rn) if f χ Lp(Rn) for every compact → loc K ∈ K Rn. ⊆ Note that if f Lp (Rn) then f L1 (Rn). ∈ loc ∈ loc

137 Proposition 4.3.8. Let f C0(Rn) and g L1 (Rn). Then ∈ c ∈ loc (f⋆g)(x) is well deﬁned for every x Rn and moreover f⋆g ∈ ∈ C0(Rn).

Proof. Note that for every x Rn the function y f(x y) g(y) ∈ 7−→ − is n-summable on Rn and therefore (f ⋆g)(x) is deﬁned for every L x Rn. ∈ Let x x and let K be a ﬁxed compact set in Rn such that n → x supp f K n. Therefore, we have f(x y)=0 n and n − ⊆ ∀ n − ∀ y / K (y / K = y / x supp f = y x / supp f and ∈ ∈ ⇒ ∈ n − ⇒ − n ∈ − y x / supp f ). − ∈− We deduce that f(x y) f(x y) w ( x x ) χ (y) | n − − − |≤ f | n − | K where wf is the uniform modulus of continuity of f: (f⋆g)(x ) (f⋆g)(y) | n − |

Proposition 4.3.9. (no proof) Let f Ck(Rn) k 1 and g L1 (Rn). Then f⋆g Ck(Rn) ∈ c ≥ ∈ loc ∈ and Dα(f⋆g)= Dαf⋆g α k. ∀| |≤ n 1 n In particular, if f Cc∞(R ) and g Lloc(R ), then f ⋆g n ∈ ∈ ∈ C∞(R ).

138 4.4 Application: Solution of the Laplace Equation

For u C2(Rn) let ∈ n ∂2u ∆u = 2 = div( u). i=1 ∂xi ∇ n P n Given f C∞(R ) we look for a solution u C∞(R ) of the Laplace ∈ c ∈ equation ∆u = f in Rn. − n We consider ﬁrst the problem to determine a function u C∞(R ) ∈ c from ∆u. We recall the Gauss Theorem.

Theorem 4.4.1. Let Ω Rn be a smooth bounded open set, ⊆ ϕ, ψ C2(Ω). Then it holds ∈ ∂ψ ∂ϕ ϕ∆ψ ψ∆ϕ = ϕ ψ dσ (4.4.1) − ∂n − ∂n ZΩ Z∂Ω where ~n is the outward normal vector to ∂Ω and ∂ϕ = ϕ ~n. ∂n ∇ ·

n We choose ψ = u C∞(R ) and Ω open set containing supp(u) = ∈ c x : u(x) =0 . For every ϕ C2(Rn) it follows from Theorem 4.4.1 { 6 } ∈ u ∆ϕdx = ϕ ∆udx. − − ZΩ ZΩ If we succeed in ﬁnding ϕ that solves in a suitable sense the equation: ∆ϕ = δ − 0 n (δ is the Delta-Dirac distribution with δ (u) = u(0) u C∞(R )), 0 0 ∀ ∈ c then we expect a representation of the form u(n)= ϕ⋆ ( ∆u) − = ϕ(x y)( ∆u)(y) dy. n − − ZR 139 1 n =1 ϕ(x)= x x R. −2 | | ∀ ∈ In this case, we have

∆ϕ(x)= ϕ′′(x)=0, x =0. 6 Moreover, + ∞ ϕ⋆ ( ∆u)(x )= u′′(x) ϕ(x x) dx − 0 − 0 − Z∞ x ε + 0− ∞ = lim u′′(x) ϕ(x0 x) dx + u′′(x) ϕ(x0 x) dx − ε 0 − x ,+ε − → h Z−∞ Z 0 i x0 ε + = lim (u′(x)ϕ(x0 x)) − +(u′(x)ϕ(x0 x)) ∞ ε 0 x0+ε − → − −∞ − x ε + 0− ∞ lim u′(x) ϕ′(x0 x) du + u′(x) ϕ′(x0 x) dx − ε 0 − x +ε − → h Z−∞ Z 0 i x ε + 0− ∞ = lim u(x) ϕ′(x0 x) + u(x) ϕ′(x0 x) − ε 0 − − x0+ε → h −∞ x0 ε + − ∞ lim( u(x)ϕ′′(x0 x) dx + u(x)ϕ′′(x0 x) dx) − ε 0 − x +ε − → Z−∞ Z 0

= lim u(x0 ε) ϕ′(ε) u(x0 + ε) ϕ′( ε) ε 0 − → − − − = u(x0) lim [ϕ′( ε) ϕ′(ε)] ε 0 → − −

= u(x0). (4.4.2) 1 1 n =2 ϕ(x)= log . 2π x | | n =3

1 2 n ϕ(x)= n 2 := Cn n − (n 2) αn 1 n − | | − − | | n 1 αn 1 = vol(S − ). ϕ is called the fundamental solution of the Laplace − equation.

140 n Theorem 4.4.2. Let f Cc∞(R ). Then u = ϕ⋆f n ∈ ∈ C∞(R ) and solves ∆u = f. −

Proof. See Proof of Theorem 4.2.3 in Struwe’s Notes.

141 Chapter 5

Diﬀerentiation of Measures

5.1 Diﬀerentiability of Lebesgue Integrals

We start with the following example.

Example 5.1.1. x i) Let f: R R be continuous, x0 R and let F (x) := f(t) dt → ∈ x0 be a primitive of f. Then it holds F C1(R) and ∈ R 1 x+h F ′(x) = lim f(t) dt = f(x). h 0 h → Zx ii) In general, for f C(Rn), it holds ∈ 1 f(x) = lim f(y) dy. (5.1.1) r 0 n(B(x,r)) → L ZB(x,r)

Question: Do similar formulas hold when f L1 (Rn) or if we ∈ loc replace n with a general Radon measure µ? L

In this section we suppose µ = n. L Notation: We denote the average of f over the set E by 1 f dµ := f dµ − µ(E) ZE ZE 142 provided 0 < µ(E) < + and the integral on the right-hand side is ∞ deﬁned.

Theorem 5.1.2. (Lebesgue Diﬀerentiation Theorem) Let f L1 (Rn). Then for µ- a.e. x Rn ∈ loc ∈ lim f dµ = f(x). (5.1.2) r 0 − → ZB(x,r)

Remark 5.1.3. In Theorem 5.1.2 we can replace balls with cubes. To prove Theorem 5.1.2 we need to associate to f L1 (Rn) the function ∈ loc

f ∗(x) = sup f dµ. (5.1.3) − | | r>0 ZB(x,r)

Deﬁnition 5.1.4. f ∗ is called Hardy-Littlewood Maximal function.

Remark 5.1.5. We showed that if f L1(Rn) then ε > 0 δ > 0: ∈ ∀ ∃ A Rn, µ(A) <δ, A µ-measurable and we have f dµ<ε. ∀ ⊆ A | | It follows that for r> 0 ﬁxed the mapping R

x f(y) dµ 7−→ − | | ZB(x,r) is continuous.

Thus f ∗ is lower semicontinuous and thus measurable. However, 1 n f ∗ L (R ) only is the trivial case f = 0. Indeed, if f is not zero ∈ up to a set of measure zero, then R, C > 0 such that f ∗(x) n ∃ ≥ c x − x R. (Exercise). | | ∀ | |≥ We next show that it is “almost in L1”.

143 Proposition 5.1.6. Let f L1(Rn). It holds for all a> 0 ∈ 5n µ x : f ∗(x) > a f 1. (5.1.4) { }≤ a k kL

Proof of Theorem 5.1.2. Suppose f L1(Rn) (otherwise we consider ∈ for a given x Rn the function g(x) = f(x) χ L1(R)). Then ∈ B1(x) ∈ f C0(Rn) such that f f in L1(Rn) as k . We have ∃ k ∈ c k → →∞

lim sup −B(x,r)[f(y) f(x)] dy lim sup −B(x,r) f(y) f(x) dy r 0 − ≤ r 0 | − | → → R R

lim sup −B(x,r) fk(y) f(y) dy + limsup −B(x,r) fk(y) fk(x) dy + fk(x) f(x) ≤ r 0 | − | r 0 | − | | − | → → R R (f fk)∗(x) +limsup fk(y) fk(x) dy + fk(x) f(x) . ≤ − r 0 −B(x,r) | − | | − | 1 → Z 3 2 | {z } | {z (5.1.5)} | {z } Since f are continuous k 1, we have 2 =0. • { k} ∀ ≥ For ε> 0 we set • Aε := x : limsup f(y) f(x) dy > 2ε . (5.1.6) r 0 −B(x,r) | − | → Z We have

A x :(f f )∗(x) >ε x :(f (x) f (x) >ε . (5.1.7) ε ⊆{ − k }∪{ k − k | } This inclusion holds for all ε > 0 and for all k 1. Thus from ≥ Proposition 5.1.6 and Tchebychev Inequality, we get

µ(A ) µ x :(f f )∗(x) >ε + µ x : f(x) f (x) >ε ε ≤ { − k } { | − k | } C (5.1.8) f f 1 as k + . ≤ ε k k − kL → → ∞ We have

∞ A := x : limsup f(y) f(x) dy > 0 = A 1 . (5.1.9) n r 0 −B(x,r) − | → Z n=1 [ 144 Thus µ(A) = 0. It follows that for µ-a.e x Rn we have ∈ f(x) lim f(y) dy lim f(x) f(y) dy =0 | − r 0 − |≤ r 0 − | − | → ZB(x,r) → ZB(x,r) for µ a.e. x (namely x / A). ∀ ∈ Actually, we have proved the following strong property:

Corollary 5.1.7. Let f L1 (Rn). Then for µ-a.e. x Rn, we ∈ loc ∈ have lim f(y) f(x) dy =0 . (5.1.10) r 0 − | − | → ZB(x,r)

Deﬁnition 5.1.8. A point x Rn which satisﬁes (5.1.10) is ∈ called a Lebesgue point of f.

In order to prove Proposition 5.1.6, we need the following result.

Theorem 5.1.9. (Vitali’s covering theorem) Let be a family of nondegenerate closed balls B = B(x,r) Rn F ⊆ with d = sup diam B,B < . For each B, we consider 0 { ∈ F} ∞ B = B(x, 5r) the concentric ball with radius 5 times the radius of B. Then there is a countable family of mutually disjoint G⊆F ballsb such that B B. ⊆ B B [∈F [∈G b Proof of Theorem 5.1.9.

1. We write d = sup diam B B . Set = B : d0 < 0 { | ∈ F} Fj { ∈ F 2j diam B d0 . We deﬁne inductively as follows: ≤ 2j−1 } Gj ⊆Fj a) is any maximal disjoint collection of balls in . G1 F1

145 b) Assuming 1, 2,..., k 1 have been selected, we choose k G G G − G to be any maximal disjoint subcollection of

k 1 − B B B′ = B′ . ∈Fk | ∩ ∅ ∀ ∈ Gj j=1 n [ o

Finally deﬁne ∞ . Clearly, is a collection of disjoint G ≡ j=1 Gj G balls and . G⊆F S

2. Claim. For each B , there exists a ball B′ so that ∈ F ∈ G B B′ = and B B . ∩ 6 ∅ ⊆ ′ Proof of the Claim. Fix B c . Then there exists an index j such that ∈F j B j. By the maximality of j, there exists a ball B′ k=1 k, ∈ F Gd0 d0 ∈ G with B B′ = . But diam B′ > and diam B = diam B ∩ 6 ∅ 2j ≤ 2j−1 ⇒ S ≤ 2 diam B′. Therefore B B . ⊆ ′ (B = B(x ,r ), B′ = B(x ,r ): r 2r . If y B = x y < 1 1 c2 2 1 ≤ 2 ∈ ⇒ | 1 − | r1 = y x2 y x1 + x1 x2 r1 + r1 + r2 2r1 + r2 5r2). ⇒ | − | ≤ | − | | − |≤ ≤ ≤

Proof of Proposition 5.1.6. Let a > 0. Set A = x : f ∗(x) > a . { } For x A, choose r = r(x) > 0 with f(y) dy a n B(x,r) . ∈ B(x,r) | | ≥ L We set w := n(B(0, 1)). It follows that n L R aw rn = a n B(0,r) f dy f . n L ≤ | | ≤k kL1 ZB(x,r) We obtain in particular the uniform condition

1 f L1 n r(x) r0 = k k x A. ≤ awn ∀ ∈ We consider the family = B x,r(x) ; x A . F ∈ It holds A B B. We choose a sub-family of disjoint balls ⊆ ∈F according to Theorem 5.1.9 with S A B B. ⊆ ⊆ B B [∈F [∈G b 146 We have

n(A) n(B)=5n n(B) L ≤ B L B L ∈G ∈G 5Pn b P 5n 5n f dy = f dy f L1 ≤ a B B | | a B | | ≤ a k k ∈G Z ZB∈G P S and we can conclude.

5.2 Diﬀerentiation of Radon measures

We would like to extend the result of Theorem 5.1.2 to other measures on Rn. We first must find a replacement for Vitali’s Theorem (Theorem 5.1.9). This theorem relies heavily on the fact that expanding a ball concentrically by a factor only enlarge its Lebesgue measure proportionally whereas no such relation may hold for general measures. In order to bypass this difficulty, we shall present a covering lemma that is purely geometric in nature, that is, which makes no mention of measure.

Theorem 5.2.1. (Besicovitch’s Covering Theorem)

There exists a constant Nn depending only on n with the following property: If is any collection of nondegenerate closed balls in F Rn with sup diam B B < + { | ∈ F} ∞ and if A is the set of centers of balls in , then there exists F ,..., such that is the countable collection of disjoint G1 GNn ∈F Gi balls in and F Nn A B . ⊆ i=1 B i [ [∈G

Note that the number of overlapping of the balls belonging to diﬀerent is controlled by N . Gj n Let µ and ν be Radon measures on Rn.

147 Deﬁnition 5.2.2. For each point x Rn, deﬁne ∈ ν B(x,r) lim sup if µ B(x,r) > 0 r > 0 r 0 µ B(x,r) ∀ Dµ ν(x) =  →   + , if µ B(x,r) =0 for some r> 0 ∞   ν B(x,r) lim inf if µ B(x,r) > 0 r> 0 r 0 ∀ D ν(x) =  → µB(x,r) µ   + , if µ B(x,r) =0 for some r> 0. ∞  

Deﬁnition 5.2.3. If D ν(x) = D ν(x) < + , we say ν is µ µ ∞ diﬀerentiable with respect to µ at x and write

D ν(x) D ν(x)= D ν(x). µ ≡ µ µ

Dµ ν is the derivative of ν with respect to µ. We also call Dµ ν the density of ν with respect to µ.

Our goals are to study a) when Dµ ν exists and b) when ν can be recovered by integrating Dµ ν.

Theorem 5.2.4. Let µ and ν be Radon measures on Rn. Then Dµ ν exists and is ﬁnite µ-a.e. Furthermore, Dµ ν is µ- measurable.

Deﬁnition 5.2.5. The measure ν is absolutely continuous with respect to µ, written ν µ, ≪ provided µ(A)=0 implies ν(A)=0 A Rn. ∀ ⊆

148 Deﬁnition 5.2.6. The measure ν and µ are mutually singular, written ν µ ⊥ if there exists a Borel subset B Rn such that µ(Rn B) = ⊆ \ ν(B)=0.

Theorem 5.2.7. (Diﬀerentiation theorem for Radon measure) Let µ, ν be Radon measures on Rn with ν µ. Then ≪

ν(A)= Dµ ν dµ ZA for all µ-measurable sets A Rn. ⊆

Remark 5.2.8. This is a version of the Radon-Nikodym Theorem. Observe, we prove not only that ν has a density with respect to µ, but also that this density Dµ ν can be computed by “diﬀerentiating” ν with respect to µ. These assertions comprise in eﬀect the fundamental Theorem of Calculus for Radon measures on Rn.

Next, we generalize Theorem 5.1.2 to general Radon measures by applying Theorem 5.2.7.

Theorem 5.2.9. Let µ be a Radon measure on Rn, f L1(Rn). ∈ Then for µ-a.e. x Rn it holds ∈ f(x) = lim f dµ. r 0 − → ZB(x,r)

Proof of Theorem 5.2.9. We suppose without restriction that f ∈ L1(Rn). We write + f = f f −. − 149 For B Rn µ-measurable, we set ⊆

ν±(B)= f ±dµ ZB and for a general A Rn, we set ⊆ ν±(A) := inf ν±(B); A B, Bµ-measurable . { ⊆ } By Theorem 3.5.3 (Satz 3.3.1 in Struwe’s Notes) and Theorem 1.4, ν± deﬁne Radon measures with ν± µ. From Theorem 5.2.4 it follows ≪

ν±(A) := D ν± dµ = f ± dµ A µ-measurable. µ ∀ ZA ZA This implies that Dµ ν± = f ± µ-a.e.. Consequently,

+ ν B(x,r) ν− B(x,r) lim f(y) dy = lim − r 0 − r 0 µ B(x,r) → ZB(x,r) → + = D ν (x) D ν−(x) µ − µ + = f (x) f −(x) − = f(x) for µ-a.e..

Corollary 5.2.10. Let µ be a Radon measure on Rn, 1 p< . ≤ ∞ Then lim f(x) f(y) p dµ =0 (5.2.11) r 0 − | − | → ZB(x,r) for µ-a.e. x

Deﬁnition 5.2.11. A point x for which (5.2.11) holds is called a Lebesgue point of f with respect to µ.

150 Proof of Corollary 5.2.10. Let r ∞ be a countable dense subset of { i}i=1 R. By Theorem 5.2.9 we have:

p p lim f ri dµ = f(x) ri r 0 | − | | − | → ZB(x,r) for µ-a.e. x and i =1, 2,... . Thus there exists a set A Rn such that ⊆ µ(A)=0 and x Rn A implies ∈ \ p p lim f ri dµ = f(x) ri i. r 0 | − | | − | ∀ → ZB(x,r) Fix x Rn A and ε> 0. Let us choose r such that f(x) r p < ε . ∈ \ i | − i| 2p Then

p p 1 p lim sup f f(x) dµ 2 − lim sup f ri dµ r 0 B(x,r) | − | ≤ r 0 B(x,r) | − | → Z h → Z p + limsup f(x) ri dµ r 0 B(x,r) | − | → Z i p 1 p p =2 − f(x) r + f(x) r | − i| | − i| =2p f(x) r p < ε. | − i|

5.3 Diﬀerentiation of absolutely continuous functions

Let F : R R be non-decreasing and continuous from the left → F (x0) = lim F (x) x0 R. x x0 ∀ ∈ ↑ For a, b R we set ∈ F (b) F (a) if a b λF (a, b)= − ≤ ( 0 otherwise .

Consider the measure ΛF induce by λF

∞ ∞ Λ (A) = inf λ (a ,b ); A [a ,b ] . F F k k ⊆ k k h Xk=1 k[=1 i 151 ΛF is the Lebesgue-Stiltjes measure associated to F . It is a Borel regular measure. Moreover,

Λ (a, b) = F (b) F (a)= λ [a, b] F − F if a

Deﬁnition 5.3.1. F is called absolutely continuous if it holds:

ε> 0 δ > 0: (ak,bk)k N, ak bk ak+1, ∀ ∃ ∀ ∈ ≤ ≤ ∞ ∞ bk ak <δ = F (bk) F (ak) < ε. k=1 | − | ⇒ k=1 | − | P P

Remark 5.3.2.

i) F absolutely continuous = F uniformly continuous. ⇒ ii) F Lipschitz continuous = F absolutely continuous. ⇒ iii) A monotone, continuous function is not necessary absolutely continuous (see for instance Cantor-Lebesgue Function)

Theorem 5.3.3. The following two conditions are equivalent: i) Λ is absolutely continuous with respect to 1. F L ii) F is absolutely continuous in R.

Proof of Theorem 5.3.3. i) = ii): Since Λ is absolutely continuous with respect to 1, we ⇒ F L have the representation

Λ (A)= fd 1 F L ZA

152 1 1 1 for every A -measurable with f = D 1 ΛF Lloc(R, ). It follows −L L ∈ L that for (a ,b ) k N with a b a , we have { k k } ∈ k ≤ k ≤ k+1 ∞ ∞ F (bk) F (ak) = ΛF [ak,bk) 0 if k=1 | − | k=1 → P1 ∞ S [ak,bk) = bk ak 0. L k=1 | − |→ S P ii) = i): Let A Rn be a Lebesgue null set ( 1(A) = 0). We recall ⇒ ⊆ L that 1(A) = inf 1(F ): F A open . L {L ⊇ } Thus for δ > 0 let B A open such that 1(B) <δ. B can be written ⊆ L B = k∞=1 Ik, Ik = [ak,bk) mutually disjoint. We may suppose that a 0 is arbitrary, it followsP that Λ (A) = 0, namely Λ 1. F F ≪ L From Theorem 5.2.7 and 5.3.3 it follows:

Theorem 5.3.4. (Rademacher) Let F be nondecreasing and absolutely continuous. Then F is diﬀerentiable 1-a.e. and L 1 0 F ′(x)= D 1 ΛF (x) := f(x) L (R) ≤ L ∈ and for all x , x R it holds: 0 ∈ x F (x) F (x ) = Λ [x ,x) = f(y) dy. − 0 F 0 Zx0

153 Proof. We estimate F (x + r) F (x) 1 x+r lim sup − f(x) = limsup f(y) f(x) dy r 0 r − r 0 r x − → → Z 1 x+r lim sup f(y) f(x) dy ≤ r 0 r x | − | → Z 1 x+r 2limsup f(x) f(x) dy =0 ≤ r 0 2r x r | − | → Z − for 1-a.e. x. L Remark 5.3.5. Conversely, if f L1(R), f 0, then the function ∈ ≥ x F (x)= f(y) dy (x0 is given) Zx0 is absolutely continuous and non-decreasing. Thus by Theorem 5.3.4 1 it is diﬀerentiable -a.e. with F ′ = f. L The Cantor Function Consider the standard triadic Cantor set [0, 1]. We recall that C ∈ k = ∞ C , where C is the disjoint union of 2 closed intervals. C k=0 k k If D = [0, 1] C , then D consists of 2k 1 intervals Ik (ordered T k \ k k − j from left to right) removed in the k-step of the construction of the Cantor set.

Let fk be the continuous function on [0, 1] which satisﬁes

1. fk(0) = 0.

2. fk(1) = 1.

k k k 1 3. fk(x)= j2− on Ij , j =1,... 2 − .

4. fk is linear on each interval of Ck.

For instance: k =1 , f (0) = 0, f (1) = 1, f (x) = 1 if 1 x 2 1 1 1 2 3 ≤ ≤ 3 and f1 is linear on C1.

154 k =2 : 0 x =0  1 1 2  x  4 9 ≤ ≤ 9    1 1 2 f2(x)=  x  2 3 ≤ ≤ 3  3 7 8 x  4 9 ≤ ≤ 9    1 x =1 .    k By construction each fk is monotone increasing, fk+1 = fk on Ij , k k 1 j = 1, 2,... , 2 1 and f f 2− − . Hence (f f ) − | k − k+1| ≤ k k − k+1 converges uniformly on [0, 1]. P Let f = limk fk. Then f is non-decreasing, continuous, f(0) = →∞ 1 1 0, f(1) = 1, f is constant -a.e., in particular f ′(x) 0, -a.e. But L ≥ L f is not absolutely continuous.

155