Notes on Measure and Integration

Notes on Measure and integration

Elias Hernandis

November 25, 2019

2

0.1 AcknowledgementsThese notes are based on the lectures of Karma Dajani ([email protected]) duringFall of 2019 at Universiteit Utrecht. These lectures are in turn based on thebook Measures, Integrals and Martingales by René L. Schiling (2nd edition).

Readers are asked to report errata to [email protected].

0.2 Why these notes?As mentioned before, these notes are based on lectures which are themselvesbased on a book. In fact, the lectures are almost one-to-one to the book, andthere are handwritten notes available online.

However, the process of writing these, as opposed to simply taking notes inclass allows me to make sure I understand the connections between the material.In other courses, where the contents were not as well organised as in this one, Ihad to reorganise the contents of the corresponding version of the notes for thatcourse. Even though this is mostly unnecessary in this course, the handwrittenlecture notes allow for more formal proofs or verifications where there was notime to complete them. Also, the book suggest that the reader verify someassumptions made during the proofs of some remarks or lemmas. Hence thisnotes are most valuable as a complement to the book.

Another valuable thing about these notes are the solutions to the assign-ments that we were given during the course. These have been compiled fromcorrections given to me by the TAs1.

0.3 RecommendationsThe official course page is [?]. In addition, we were offered exercise sets andmore material through Blackboard. Grades were also published there.

Always trust the book when there is a discrepancy. Though there may beerrata (in fact, see [?] for a list of them), it is way more likely that I’ve made amistake.

The solutions to problems of the second edition are available online, see [?].All problems from the first edition are included in the second edition, albeitwith a different numbering. Order is preserved, though.

1If you want to see what a disaster it was when I started this subject, look at the scannedcopies of submitted exercises...

Contents

3

4 CONTENTS

Chapter 1

σ-algebras

1.1 What is a σ-algebra?Definition 1.1 (Sigma algebra). Let X be a set and A ⊂ ℘(X) a collection ofsubsets of X. We say that A is a sigma algebra or σ-algebra (on X) if it satisfiesthese three properties:

1. It contains the set, X ∈ A;

2. it is closed under set complement A ∈ A Ô⇒ X ∖A = Ac ∈ A, ∀A ⊂ X,and

3. it is closed under countable1 union (An)n∈N ∈ A Ô⇒ ⋃n∈NAn ∈ A.

s

Definition 1.2 (Measurable space). A measurable space is a pair (X,A) whereX is a set and A is a σ-algebra .

From now on we shall assume X is a set and A is a σ-algebra on X.

Remark 1.3.

• From properties 1 and 2 we have ∅ =Xc ∈ A.2

• From properties 2 and 3 we have that A is also closed under countableintersection. Let (An)n∈N ⊂ A. Then, using De Morgan’s laws we have

⋂n∈N

An = (⋃n∈N

Acn)

c

∈ A

because Acn ∈ A because of 2, the union is included because of 3 and its

complement is also included because of 2. When a set has this property,we sometimes say A is ∩-stable(see ??).

1In these notes we shall follow the convention from [?, bottom of p. 8] and use countableto describe any set whose cardinality is less than or equal to that of the set of the naturalnumbers, i.e. #A ≤#N. This means that countable does not necessarily mean infinite.

2In fact, sometimes we substitute property one for ∅ ∈ A. This is equivalent as long as stilltake complements with respect to the universe X, i.e. Ac = X ∖A.

5

6 CHAPTER 1. σ-ALGEBRAS

• Clearly, finite unions and intersections of sets in the σ-algebra end up inthe σ-algebra . For the union of M sets take An = ∅,∀n >M and apply 3(or the equivalent for the intersection).

Example 1.4. The power set ℘(X) is the largest σ-algebra on X.

Example 1.5. ∅,X is the smallest σ-algebra on X.

Example 1.6. Given a subset A ⊂ X, the smallest σ-algebra which containsinformation about A is ∅,X,A,Ac. More on this later.

LetX be an uncountable set. ThenA = A ⊂X ∣ A is countable or Ac is coutable is a σ-algebra .

Proof. We shall prove the three properties of a σ-algebra

1. X ∈ A because Xc = ∅ is countable.

2. If A ∈ A and is countable the Ac ∈ A because (Ac)c is countable. If A ∈ Abut A is not countable then Ac must be so, hence Ac ∈ A.

3. Let (An)n∈N ⊂ A. To see if the union is included we distinguish to cases:

(a) If An is countable ∀n ∈ N then ⋃n∈NAn is also countable and there-fore in A.3

(b) If there is any Ak that is uncountable, then the union is uncountablebut

(⋃n∈N

An)c

= ⋂n∈N

Acn ⊂ Ac

k

Recall that Ack must be countable because Ak ∈ A and therefore

(⋃n∈NAn)c must also be countable. Thus, ⋃n∈NAn ∈ A.

Example 1.7 (Trace σ-algebra). Let (X,A) be a measurable space and letE ⊂X. Then

AE = A ∩E ∣ A ∈ A (1.1)

is a σ-algebra on E.

Proof.

1. Clearly, E =X ∩E ∈ AE since X ∈ A.

2. For the second property we need to take care to take the complementrelative to E, as we want AE to be a σ-algebra on E. Here we reserve Ac

to denote the complement relative to X and just write E ∖A if we meanrelative to E.

3Recall that the countable union of countable sets is still a countable set.

1.1. WHAT IS A σ-ALGEBRA? 7

Let B ∈ AE . Then there exists an A ∈ A such that B = A ∩E. We havethat the complement of B relative to E is

E ∖B = E ∩Bc = E ∩ (A ∩E)c = E ∩ (Ac ∪Ec) = Ac ∩E ∈ AE ,

since Ac ∈ A.

3. Finally, let (Bn)n∈N ⊂ AE . Then there exists (An)n∈N ⊂ A such thatBn = An ∩E, ∀n ∈ N. Then,

⋃n∈N

Bn = ⋃n∈N(An ∩E) = E ∩ ⋃

n∈NAn ∈ AE ,

since A is closed under countable union.

Example 1.8 (Preimage σ-algebra ). Recall some properties of preimages (thatare not always true for images): Set difference and union (and therefore com-plement and intersection) are well defined and behave as expected. Moreover,union and intersection behave as expected even for countable arities.

Let X,X ′ be sets, let A′ be a σ-algebra on X ′ and let f ∶ X → X ′ be afunction between the two sets. We claim that4 A = f−1(A′) ∣ A′ ∈ A′ is aσ-algebra on X5

Proof.

1. X = f−1(X ′) ∈ A because X ′ ∈ A.

2. If A ∈ A then Ac =X ∖A = f−1(X ′) ∖ f−1(A′) = f−1(X ′ ∖A′) ∈ A becauseX ′ ∖A′ = A′c ∈ A′.

3. Let (An)n∈N ⊂ A. By definition there must be a collection (A′n)n∈N ⊂ A′for which An = f−1(A′n), ∀n ∈ N. Recall that

⋃n∈N

An = ⋃n∈N

f−1(A′n) = f−1(⋃n∈N

A′n) ∈ A

because A′ is a σ-algebra hence ⋃n∈NA′n ∈ A′.

Now we will explore a result which will allow us to generate more examplesfrom existing ones and clarify operations between σ-algebras .

Lemma 1.9. Given a set X. The arbitrary intersection ⋂α∈I Aα of σ-algebrasis again a σ-algebra .

Proof. We shall go over the three properties of σ-algebras .4Here f−1 denotes the preimage of a set, not the inverse function. Recall that the preimage

of a function f ∶ X → X′ is defined as f−1(A′) = x ∈ X ∣ f(x) ∈ A′ ⊂ X′5When we write f ∶ X → X′ we mean that f is surjective, namely that f−1(X′) = X.

Without this, the first property of the σ-algebra cannot be guaranteed. However, this is notusually a problem, as we can always redefine f as f ∶ X → f(X) and take the trace σ-algebraAf(X) (c.f. ??).


1. X ∈ Aα, ∀α ∈ I hence X ∈ ⋂α∈I Aα

2. If A ∈ ⋂α∈I Aα then in particular A ∈ Aα, ∀α ∈ I. Because each Aα is aσ-algebra , then Ac ∈ An, ∀α ∈ I Ô⇒ Ac

n ∈ ⋂α∈I Aα.

3. If (An)n∈N ⊂ ⋂α∈I Aα then, as before, we have that (An)n∈N ⊂ An, ∀α ∈ I.Hence, ⋃n∈NAn ∈ Aα Ô⇒ ⋃n∈NAn ∈ ⋂α∈I Aα.

Definition 1.10 (σ-algebra generated by a collection of subsets). Let X be aset and G a collection of subsets of X. We denote by σ(G) the smallest σ-algebrawhich contains G and we define it by

σ(G) ∶=⋂C where G ⊂ C ∧ C is a σ-algebra

We also say that G is a generator of σ(G) or that σ(G) is generated by G.

Lemma 1.11. For any G ⊂ ℘(X), σ(G) exists and is a the smallest σ-algebracontaining G.

Proof. Let A = σ(G). Since ℘(X) ⊃ G and ℘(X) is a σ-algebra , the intersectionA is non-empty and contains G. Because of the previous result, A itself is alsoa σ-algebra . (So we have existence). Furthermore, A is the smallest σ-algebracontaining G because if there were another σ-algebra A′ containing G it wouldbe included in the intersection hence A ⊆ A′.

Remark 1.12.

1. If G is a σ-algebra , then σ(G) = G.

2. For any A ∈X, σ(A) = ∅,X,A,Ac.

3. Let G,F ⊂ ℘(X). If G ⊆ F then σ(G) ⊆ σ(F).

Proof of 3. G ⊆ F ⊆ σ(F). So σ(F) is a σ-algebra containing G. But σ(G)is the smallest σ-algebra containing G therefore σ(G) ⊆ σ(F).

1.1.1 Final remarks and the good set principleThe definition of a σ-algebra is more axiomatic than constructive. This is notcoincidental, σ-algebras are normally really hard to describe constructively sincethey are huge sets. This last notion of generators will prove very helpful in thecoming chapters: most of the time we will be able to get theorems that onlydepend on a generator of the σ-algebra at hand.

In particular, part 3 of this last remark will be used extensively throughoutthe course. Also, it gives rise to a common proof technique in measure theory:the good set principle.

Remark 1.13 (Good set principle). The good set principle is a proof techniquethat comes in handy when we want to show that every member of a σ-algebraA satisfies some property P . In general, this proof technique works as follows(cf. [?]).

1.2. THE BOREL σ-ALGEBRA ON RN 9

Define

B ∶= A ∈ A ∣ A has property P (1.2)

We will show that

1. B is a σ-algebra

2. There is a collection G ⊂ B such that σ(G) = A

Then, by ?? we have that A = σ(G) ⊂ σ(B), but since B is already a σ-algebra , we have A ⊂ B. Additionally, by definition of B, we already have thatA ⊂ B, therefore A = B, i.e. all the members of A satisfy property P .

1.2 The Borel σ-algebra on Rn

In what follows we will use some basic topology concepts. Let us write

• On ∶= A ⊂ Rn ∣ A is open ,

• Cn ∶= A ⊂ Rn ∣ A is closed , and

• Kn ∶= A ⊂ Rn ∣ A is compact .

The collection On is a topology, meaning it satisfies the following properties:

1. ∅,Rn ∈ On

2. It is closed under finite intersections, i.e. V,W ∈ On Ô⇒ V ∩W ∈ On.

3. It is closed under arbitrary unions, i.e.

(Aα)α∈I ∈ On Ô⇒ ⋃α∈I

Aα ∈ On.

We shall call the pair (Rn,On) a topological space. We now consider thesmallest σ-algebra containing On.

Definition 1.14 (Borel σ-algebra ). The Borel σ-algebra on Rn is the smallestσ-algebra containing On. We denote it by σ(On) or by B(Rn).

Theorem 1.15 (Topological generators of the Borel σ-algebra ).

B(Rn) ∶= σ(On) = σ(Cn) = σ(Kn)

Proof. First, we prove the first equality, i.e. σ(On) = σ(Cn) by proving mutualinclusion. To show σ(Cn) ⊂ σ(On) it is enough to show that Cn ⊂ σ(On) (recallremark ??). Let C ∈ Cn be any closed set in Rn. By definition Cc is openhence Cc ∈ On ⊂ σ(On). Because σ(On) is a σ-algebra it must be true that(Cc)c = C ∈ σ(On). The same holds for the other inclusion.

Now we turn our attention to σ(Cn) = σ(Kn). The inclusion σ(Kn) ⊂ σ(Cn)is trivial because every compact set is closed in Rn (recall remark ??). For theother one, it is again enough to show that Cn ∈ σ(Kn). Let C ∈ Cn and define


Ck ∶= C ∩Bk(0) which is6 closed and bounded. By construction C = ⋃k∈NCk ∈Kn thus Cn ∈ σ(Kn).

We would now like to find smaller sets of generators for the Borel σ-algebraon Rn. Let us define the following collections (where ⨉ denotes de cartesianproduct of the intervals):

• The collection of open rectangles (or cubes or hypercubes)

J o,n = n

⨉i=1(ai, bi) ∣ ai, bi ∈ R

• The collection of (from the right) half-open rectangles

J n = n

⨉i=1[ai, bi) ∣ ai, bi ∈ R

Theorem 1.16 (Borel interval generators). We have

B(Rn) = σ(J nrat) = σ(J o,n

rat ) = σ(J n) = σ(J o,n)

Proof. Let’s begin by proving B(Rn) = σ(J o,nrat ). Recall that B(Rn) = σ(On),

so to prove the previous equality it suffices to prove the following two mutualinclusions:

• σ(J o,nrat ) ⊆ σ(On). From remark ?? we have that to prove this it suffices

to say that every open rectangle is an open set and thus J o,nrat ⊂ On Ô⇒

σ(J o,nrat ) ⊆ σ(On).

• σ(On) ⊆ σ(J o,nrat ). To prove this we make the following claim:

U ∈ On Ô⇒ U = ⋃I∈J o,n

rat , I⊆UI

Again we shall attack this by proving the mutual inclusion of the two sets:

– It is clear that ⋃I∈J o,nrat , I⊂U I ⊆ U because of the restriction on the

union.– For the reverse containment, we have that as U is open, for any x ∈ U

there is a ball Bε(x) ⊆ U . Because the rationals Qn are dense in thereals Rn we can chose a rectangle I ⊂ Bε(x) and hence U is containedin the union.It is clear that all the sets I ⊆ U are also in σ(J o,n

rat ). However, forthe union of them to be inside de σ-algebra we must ensure thatthe number of sets that participate is countable. Each rectangle Ican be fully determined by two of its corners, which in turn havecoordinates in Qn. Therefore, the number of sets intervening in theunion is #(Qn × Qn) = #N and thus the union is again within theσ-algebra .

6Here Br(c0) and Br(c0) denote the open and closed balls of radius r and centre c0,respectively. Clearly these are both bounded sets.

1.2. THE BOREL σ-ALGEBRA ON RN 11

Because J o,nrat ⊂ J o,n ⊂ On we get for free that σ(J o,n

rat ) ⊆ σ(J o,n) ⊆ σ(On).We therefore conclude that

σ(J o,nrat ) = σ(J o,n) = σ(On)

Now we would like to prove that half open sets also yield the same Borelσ-algebra for Rn.

• We begin by noticing that we can write open sets as infinite unions of halfopen ones

n

⨉i=1(ai, bi) = ⋃

n∈N

n

⨉i=1[ai +

1

n, bi)

for both rectangles with rational and real endpoints. Thus, we have

J o,nrat ⊆ σ(J n

rat) Ô⇒ σ(J o,nrat ) ⊆ σ(J n

rat)J o,n ⊆ σ(J n) Ô⇒ σ(J o,n) ⊆ σ(J n)

(remember that σ-algebras are closed under countable unions).

• For the reverse containment we must notice that we can write (right)half-open sets as intersections of open ones

n

⨉i=1[ai, bi] = ⋂

n∈N

n

⨉i=1(ai −

1

n, bi)

for rectangles with both rational and real endpoints. Similarly, we have

J nrat ⊆ σ(J o,n

rat ) Ô⇒ σ(J nrat) ⊆ σ(J o,n

rat )J n ⊆ σ(J o,n) Ô⇒ σ(J n) ⊆ σ(J o,n)

• We conclude that

σ(J o,nrat ) = σ(J n

rat)σ(J o,n) = σ(J n)

With the previous equalities the theorem has been proved.

To recap, there are a few important points on this proof:

• First, we would like to have a more tangible generator for the Borel σ-algebra on Rn.

• We choose rectangles because they are easier to work with and have directapplication on probability theory (we could also have chosen balls, forinstance).

• They key to proving that two σ-algebras are the equal is to prove thateach is contained in the other. To do this, we use the generators: if thegenerator of A is contained in σ(A′) and the generator of A′ is containedin σ(A), we are done.


• To prove this containments we have had to write sets from the generatoras unions of sets from the other σ-algebra . It is key to make sure thatthese unions only iterate over a countable number of elements.

Example 1.17 (Another characterisation of the Borel σ-algebra on Rn). In thisexample we shall see that B = Br(x) ∣ x ∈ Rn, r ∈ R+ is also a generator ofB(Rn).

Proof. We proceed as before, first defining an auxiliary collection where theradii are all rational and the centres have rational coordinates:

B′ = Br(x) ∣ x ∈ Qn, r ∈ Q+

It is trivial that B′ ⊂ B ⊂ On. Therefore we have that

σ(B′) ⊆ σ(B) ⊆ σ(On) = B(Rn)

For the reverse inclusions, we will focus on proving that On ⊆ σ(B′). For thiswe claim that any open set U ∈ On can be written as

U = ⋃B∈B′,B⊆U

B

We need to verify two things. That the previous equality is true and that thenumber of sets that intervene in the union is countable.

1. It is clear that any set B ⊂ B′ is also in U by the definition of the union. Forthe reverse inclusion, we shall choose a point q ∈ Qn such that ∥x−q∥ < r/3.This is possible because Qn is dense in Rn. Next we will choose a radiusr′ ∈ Q such that r′ < r. This is also possible for the same reason. Now weconsider B = Br′(q) ∈ B′ which is assured to contain x.

2. Moreover, each of these balls is fully determined by an (n+1) tuple of ra-tionals (namely the center coordinates and the radius). Hence, the numberof sets in the union is #(Qn ×Q) =#N.

Chapter 2

Measures

2.1 Definition. Properties.Definition 2.1 (Measure). Let (X,A) be a measurable space. A set functionµ ∶ A→ [0,∞) is called a measure (on X) if

1. µ(∅) = 0 and

2. (σ-aditivity) if (An)n∈N ⊂ A is a pairwise disjoint (i.e. Ai ∩ Aj = ∅ ifi ≠ j) sequence, then

µ(⊍n∈N

An) = ∑n∈N

µ(An)

And, analogously to σ-algebras

Definition 2.2 (Measure space). Let (X,A) be a measurable space and µ ameasure. The triple (X,A, µ) is called a measure space.

In the definition of measure (??), the symbol ⊍ indicates that the union isdisjoint.

Some special measures get cool names, for instance:

• If µ(X) <∞ we say that µ is finite and that (X,A, µ) is a finite measurespace.

• IF µ(X) = 1 then (X,A, µ) is a probability space and we usually denoteit by (Ω,A,P).

From the two conditions on ??, one can derive many properties, but before,let us introduce some notation.

Let (An)n∈N be a sequence of sets. We say that (An) is an increasing, resp.decreasing, sequence and denote it by An ↑ A, resp. An ↓ A according to thefollowing definition.

An ↑ A ⇐⇒ A1 ⊆ A2 ⊆ . . . and A = ⋃n∈N

An (2.1)

An ↓ A ⇐⇒ A1 ⊇ A2 ⊇ . . . and A = ⋂n∈N

An (2.2)

13

14 CHAPTER 2. MEASURES

We say that an increasing sequence An ↑ A is an exhausting sequence ifAn ⊆X and A =X. We write An ↑X for an exhausting sequence.

Definition 2.3 (σ-finite measure). A measure µ is called σ-finite if there existsan exhausting sequence (An)n∈N ⊂ A such that µ(An) <∞, ∀n ∈ N. A measurespace with this kind of measure is called a σ-finite measure space.

Remark 2.4. Finiteness is stronger than σ-finiteness1. Namely, any finite mea-sure is σ-finite since one can choose the sequence An = X, ∀n ∈ N which is anexhausting sequence An ↑ X and µ(An) < ∞, ∀n ∈ N. On the other hand, anexample of a σ-finite measure which is not finite is the Lebesuge measure (see??).

Theorem 2.5 (Properties of measures). Let (X,A, µ) be a measure space andA,B,An,Bn ∈ A, ∀n ∈ N. Then,

1. (finite additivity) A ∩B = ∅ Ô⇒ µ(A ⊍B) = µ(A) + µ(B),

2. (monotonicity) if A ⊆ B then µ(A) ≤ µ(B),

3. if A ⊂ B and µ(A) <∞ then µ(B ∖A) = µ(B) − µ(A),

4. (strong additivity) µ(A ∪B) + µ(A ∩B) = µ(A) + µ(B),

5. (finite subadditivity) µ(A ∪B) ≤ µ(A) + µ(B),

6. (continuity from below) if An ↑ A then µ(A) = µ(⋃n∈NAn) =supn∈N µ(An) = limn→∞ µ(An),

7. (continuity from above) if µ(A) <∞,∀A ∈ A and An ↓ A then µ(A) =µ(⋂n∈NAn) = infn∈N µ(An) = limn→∞ µ(An), and

8. (sigma subadditivity) µ(⋃n∈NAn) ≤ ∑n∈N µ(An).

Proof.

1. Let (An)n∈N ⊂ A with A1 = A, A2 = B and Ai = ∅ for i > 2. It is clearthat An are disjoint since A ∩B = ∅ and An ∩ ∅ = ∅, ∀n ∈ N.

2. Write B = (B ∖A) ⊍A. Then, because of σ-aditivity we have

µ(B) = µ(B ∖A) + µ(A) ≥ µ(A) since µ(B ∖A) ≥ 0

3. As previously write µ(B) = µ(B ∖A) + µ(A). Because µ(A) < ∞ we cansubtract it on both sides to get µ(B) − µ(A) = µ(B ∖A).

4. Write A ∪B = A ∖B ⊍B ∖A ⊍A ∩B hence µ(A ∪B) = µ(A ∖B) + µ(B ∖A) + µ(A ∩B). Add µ(A ∩B) on both sides and group terms:

µ(A ∪B) + µ(A ∩B) = µ(A ∖B) + µ(A ∩B)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

µ(A)

+µ(B ∖A) + µ(A ∩B)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

µ(B)1By σ-finiteness we mean that a measure satisfies ??, not that it is s-finite, which we wont

see in this course. See [?] for more details.

2.2. EXAMPLES 15

5. µ(A ∩B) ≤ µ(A ∩B) + µ(A ∪B) = µ(A) + µ(B)

6. Define the sequence (Bn)n∈N ⊂ A by B1 = A1, Bn = An ∖An−1 for n > 1.It is clear that Bn is pairwise disjoint and that ⊍n∈NBn = ⋃n∈NAn = A.Hence

µ(A) = µ(⋃n∈N

An) = µ(⊍n∈N

Bn) = ∑n∈N

µ(Bn) = limm→∞

m

∑n=1

µ(Bn)

= limm→∞

µ(m

⊍n=1

Bn) = limm→∞

µ(Am) = supn∈N

µ(An)

since µ(An) is an increasing sequence. The introduction of the limits inthe previous chain of equalities has to be done carefully, as we are buildingon the definition of limits for sequences of numbers. The equality betweenthe first and the second lines comes from σ-additivity.

7. Let Dn = A1 ∖An, ∀n ∈ N. Then DN is an increasing sequence with

⋃n∈N

Dn = ⋃n∈N(A1 ∖An) = ⋃

n∈N(A1 ∩Ac

n) = A1 ∩ ⋃n∈N

Acn

= B1 ∩ (⋂n∈N

An)c

= A ∖ ⋂n∈N

An

Thus,

µ(A ∖ ⋂n∈N

An)(3)= µ(A) − µ(⋂

n∈NAn) = µ(⋃

n∈NDn)

(4)= limn→∞

µ(Dn) = limn→∞

µ(A1∖An)(3)= lim

n→∞(µ(A)−µ(An)) = µ(A)− lim

n→∞µ(An)

Substracting µ(A) <∞ from both sides we have

µ(⋂n∈N

An) = limn→∞

µ(An)

8. Let (An)n∈N be any countable subcollection of A. Define

En =n

⋃m=1

Am ∈ A

Then En ↑ ⋃n∈NEn = ⋃n∈NAn. Thus, by (6) we have

µ(⋃n∈N

An) = µ(⋃n∈N

En) = limn→∞

µ(En) = limn→∞

µ(n

⋃m=1

Am)(5)≤ lim

n→∞

n

∑m=1

µ(Am)

2.2 ExamplesThroughout this section, let (X,A) be a measurable space.


Example 2.6 (Dirac measure). We define the Dirac measure for a givenx0 ∈X as follows:

δx0(A) =⎧⎪⎪⎨⎪⎪⎩

0 if x0 /∈ A1 if x0 ∈ A

, ∀A ∈ A (2.3)

Clearly δx0 is a measure since it satisfies the two properties. First, δx0(∅) = 0since ∀x0 ∈ X, x0 /∈ ∅. Second, for any pairwise disjoint collection of sets(An)n∈N ⊂ A we have two possibilities:

• If x0 /∈ ⊍n∈NAn then clearly x0 /∈ An, ∀n ∈ N so

0 = µ(⊍n∈N

An) = ∑n∈N

µ(An) = 0

• Otherwise, if x0 ∈ ⊍n∈NAn then there must be only one n0 ∈ N such thatx0 ∈ An0 since (An) is a pairwise disjoint collection. Thus,

1 = µ(⊍n∈N

An) = ∑n∈N

µ(An) = µ(An0) + ∑n≠n0

µ(An) = 1 + 0

Example 2.7 (Counting measure). Let X = R and choose A = A ⊂ R ∣A is countable or Ac is countable . We already saw on the Chapter 1 thatA is a σ-algebra . Now define µ ∶ A→ [0,∞) as

µ(A) =⎧⎪⎪⎨⎪⎪⎩

0 if #A ≤#N1 if #Ac ≤#N

(2.4)

We have that µ(∅) = 0 since #∅ is countable. As for σ-additivity we must recallfrom set theory that the union of countable sets is also countable, so:

• If #An ≤#N, ∀n ∈ N then

0 = µ(⊍n∈N

An) = ∑n∈N

µ(An) = 0

• If there exists an n0 ∈ N such that Acn0

is countable then

(⊍n∈N

An)c

= ⋂n∈N

Acn ⊆ Ac

n0

and hence (⊍n∈NAn)c is countable. Furthermore, since the collection ispairwise disjoint, ∀n ∈ N, n ≠ n0 we have Ac

n ⊆ An0 so #Acn, ∀n ≠ n0.

Thus

1 = µ(⊍n∈N

An) = ∑n∈N

µ(An) = µ(An0) + ∑n≠n0

µ(An) = 1 + 0

Example 2.8 (Discrete probability measure). Let (Ω,A,P) be a measure spacewhere Ω = ω1, ω2, . . . is a countable set, A = ℘(Ω) (which is of course a σ-algebra ) and let (p1, p2, . . . ) be a probability vector where ∑n∈N pn = 1 (and piis the probability of ωi). Define the measure P ∶ A→ [0,∞) where

P(A) = ∑ωi∈A

pi =∞∑i=1

piδωi(A)

2.2. EXAMPLES 17

Let’s verify that P is a measure. First, P(∅) = 0 since ∅ cannot contain anyωi. As for σ-additivity, we have

P(∞⊍n=1

An) =∞∑i=1

piδxi (∞⊍n=1

An) =∞∑i=1

pi∞∑n=1

δxi(An)

=∞∑n=1(∞∑i=1

piδxiAn) =∞∑n=1

P(An)

Nota that here we can exchange the summations because all the terms arenon-negative so the convergence problems (oscillating convergence, that is) areeliminated.

Example 2.9 (Lebesgue measure). For now we shall define this measure onlyon n-dimensional rectangles. The generalisation to arbitrary Borel sets willcome in chapter 4.

Consider the measure space (Rn,B(Rn), λn) where Rn and B(Rn) are theusual suspects and λn ∶ B(Rn)→ [0,∞) is defined as:

λn (n

⨉i=1[ai, bi)) =

n

∏i=1(bi − ai) (2.5)

or, more informally, the hypervolume of the rectangle in question.We are not ready to verify that λn is a measure over B(Rn) yet. In fact, we

will need to wait until the end of Chapter 4, where, once we have the generali-sation to arbitrary Borel sets, we shall prove that it is a measure.

We are however, in a position to prove that λn is σ-finite. Let Ai = [−i, i)nfor all n ∈ N. Clearly, (Ai)n∈N is an exhausting sequence since Ai ↑ Rn. Also,λn(Ai) = (2i)n <∞,∀n ∈ N. Therefore λn is σ-finite.

Note that λn is not finite, since λ(Rn) /<∞.

Chapter 3

Uniqueness of measures

In this chapter we shall introduce some technicalities to be able to extend theLebesgue measure to arbitrary Borel sets in B(Rn). The first tool is a gener-alized version of a σ-algebra , where union closure is only required for disjointunions. We will explore the properties of this new construct and try to drawsimilarities to what we know about σ-algebras . Finally we will give a resulton the conditions that a σ-algebra must meet to be able to uniquely define ameasure on it by defining the measure on the elements of a generator.

3.1 PreliminariesDefinition 3.1 (Dynkin system). A family D ⊂ ℘(X) is called a Dynkin systemif it satisfies these three properties:

1. X ∈ D

2. ∀A ∈ ℘(X), A ∈ D Ô⇒ Ac ∈ D

3. For any countable collection of pairwise disjoint sets (An)

(An)n∈N ⊂ D Ô⇒ ⊍n∈N

An ∈ D

Remark 3.2. Every σ-algebra is a Dynkin system but the converse need notbe true.

We have an analogue version to the generator theorem for σ-algebras in thecase of Dynkin systems:

Definition 3.3. Let G ⊂ ℘(X) be a collection of subsets of X. We define theDynkin system generated by G as

δ(G) = ⋂G⊂C,CDynkin

C

Lemma 3.4. Let G ⊂ ℘(X) be a collection of subsets of X. Then δ(G) is aDynkin system and it is the smallest. Moreover, G ⊆ δ(G) ⊆ σ(G).

19

20 CHAPTER 3. UNIQUENESS OF MEASURES

Proof.

• δ(G) is a Dynkin system because of the restriction on the intersection.Note that the intersection is non-empty because ℘(X) is a Dynkin systemand that the intersection of Dynkin systems is also a Dynkin system (theproof is the same as the one for σ-algebras but using disjoint unions, see??).

• Now let’s look at why it is the smallest. Suppose there is another Dynkinsystem D that contains the collection G. Then D is a Dynkin system andtherefore intervenes in the intersection. Therefore δ(G) ⊂ D.

• Finally, G ⊆ δ(G) follows from the restriction in the intersection and δ(G) ⊆σ(G) follows from the fact that every σ-algebra is a Dynkin system andhence δ(D) = D (by minimality) and δ(σ(G)) = σ(G). Therefore, G ⊆σ(G) Ô⇒ δ(G) ⊂ δ(σ(G)) = σ(G).

Also, as with σ-algebras , we have

Remark 3.5. If F ⊆ G then δ(F) ⊆ δ(G).

So, we already saw that every σ-algebra is a Dynkin system and that Dynkinsystems can be generated from a collection of subsets. A natural question to askis, when is a Dynkin system a σ-algebra ? Let us introduce some terminologyfirst.

Definition 3.6 (Stable under finite intersection). Let D be a collection of sets.We say that D is stable under finite intersection, or closed under finite inter-section or ∩-stable if

C,D ∈ D Ô⇒ C ∩D ∈ D

Some sources call such a collection a Π-system.

Lemma 3.7. Let D be a Dynkin system. D is ∩-stable⇐⇒ D is a σ-algebra .

Proof. It is clear that going from right to left is true, since all σ-algebras are∩-stable.

For the implication from left to right we proceed as follows. Assuming Dis a ∩-stableDynkin system, to prove that D is a σ-algebra we only need togeneralise the last property to countable unions of arbitrary collections of setsin D, not just disjoint ones. Let (An)n∈N ⊂ D be a sequence of sets in D. Wedefine a new sequence (En)n∈N by

E1 =D1 for n = 1, En =Dn ∖En−1 for n > 1.

Rewriting the set difference as an intersection we have that

En =Dn ∖En−1 =Dn ∖n−1⋃i=1

Di =Dn ∩ (n−1⋂i=1

Dci) ∈ D

3.1. PRELIMINARIES 21

since D is ∩-stable. It is clear that En are pairwise disjoint, i.e. Ei ∩Ej = ∅ fori ≠ j. We also have that

⊍n∈N

En = ⋃n∈N

Dn ∈ D

Hence, D is a σ-algebra .

Remark 3.8. As a consequence we have that if G ⊂ ℘(X) and δ(G) is ∩-stablethen δ(G) is a σ-algebra containing G therefore σ(G) ⊆ δ(G). Since δ(G) ⊆σ(G) always holds, we have that

if δ(G) is ∩-stable then δ(G) = σ(G)

This is nice, but it would even be nicer if we could just argue about thegenerators, since Dynkin systems and σ-algebras are hard to reason about. We’lldo just that.

Lemma 3.9. Let G be a collection of subsets of X. If G is ∩-stablethen δ(G) =σ(G).

Proof. We only need to show that if G is ∩-stablethen δ(G) also is. Then,because of the previous remark we have δ(G) = σ(G).

We shall proceed in steps.

1. Claim 1. For each E ∈ δ(G), the collection DE = F ⊆ X ∣ F ∩E ∈ δ(G)is a Dynkin system. We prove the three properties of a Dynkin system.

(a) Clearly ∅ = ∅ ∩E ∈ DE

(b) For any F ∈ DE we have F c =X ∖ F ⊆X and

F c ∩E =(F c ∪Ec) ∩E= (F ∩E)c ∩E= (F ∩E) ⊍Ec ∈ δ(G),

since F ∩E ∈ δ(G) by hypothesis and Ec ∈ δ(G) since δ(G) is a Dynkinsystem.

(c) Finally, for any collection of disjoint subsets (Fn)n∈N ⊂ DE we needto show that the disjoint union is still in DE . We have that for alln ∈ N, Fn ∩E ∈ δ(G) by hypothesis so

⊍n∈N

Fn ∩E = ⊍n∈N(Fn ∩E) ∈ δ(G).

2. Claim 2. G ⊂ DG, ∀G ∈ G. Let G′ ∈ G. Since G is ∩-stablewe have thatG′ ∩G ∈ G and hence G′ ∩G ∈ δ(G) Ô⇒ G′ ∈ DG.

As a consequence of these claims we have that, since G ⊂ DG then δ(G) ⊂δ(DG) = DG. Therefore, for any E ∈ δ(G) and any G ∈ G we have that E ∩G ∈δ(G). This also shows that δ(G) ⊂ DE , for any E ∈ δ(G). In other words wehave that for any E,F ∈ δ(G), E ∩ F ∈ δ(G). Thus, δ(G) is ∩-stableimplyingthat δ(G) is a σ-algebra and hence δ(G) = σ(G).


3.2 Uniqueness of measures

Now we move on to define the requirements needed to be able to guaranteeuniqueness of a measure over a σ-algebra given a definition of the measure on agenerator of that σ-algebra .

Theorem 3.10 (Uniqueness of measures). Let (X,A) be a measurable spacewhere A = σ(G) for some collection G of subsets of X where G satisfies thefollowing:

1. G is ∩-stable(so δ(G) = σ(G)), and

2. there exists an exhausting sequence (Gn)n∈N ⊂ G such that Gn ↑ X (soX = ⋃n∈NGn).

If µ, ν are measures on A = σ(G) such that µ(G) = ν(G) <∞, ∀G ∈ G thenµ = ν, i.e. µ(A) = ν(A), ∀A ∈ A.

Proof. The plan is to use the good set principle (??) to prove that a set wherethe measures coincide is equal to the σ-algebra A. Therefore we can concludethat µ = ν everywhere.

Define, for every n ∈ N (and hence for every Gn ∈ G)

DN = A ∈ A ∣ µ(Gn ∩A) = ν(Gn ∩A). (3.1)

Note that µ(Gn ∩A) ≤ µ(Gn) <∞ so Dn is well defined. The goal is to see thatDn = A but that is too hard. We will prove that Dn is a Dynkin system.

1. Clearly X ∈ Dn since µ(Gn ∩X) = µ(Gn) = ν(Gn) = ν(Gn ∩X).

2. Let A ∈ Dn. Then

µ(Gn ∩Ac) = µ(Gn ∩ (X ∖A))= µ((Gn ∩X) ∖ (Gn ∩A))??= µ(Gn) − µ(Gn ∩A)= ν(Gn) − ν(Gn ∩A)??= ν(Gn ∖ (Gn ∩A))= ν(Gn ∩ (X ∖A)) = ν(Gn ∩Ac).

3.2. UNIQUENESS OF MEASURES 23

3. Let (Aj)j∈N be a pairwise disjoint sequence in Dn. Then1,

µ⎛⎝Gn ∩ ⊍

j∈NAj

⎞⎠= µ⎛⎝⊍j∈N(Gn ∩Aj)

⎞⎠

= ∑j∈N

µ(Gn ∩Aj)

= ∑j∈N

ν(Gn ∩Aj)

= ν⎛⎝⊍j∈N(Gn ∩Aj)

⎞⎠= ν⎛⎝Gn ∩ ⊍

j∈NAj

⎞⎠.

Now we want to prove that Dn = A. By definition we already have thatDn ⊆ A. On the other hand, µ(Gn ∩G) = ν(Gn ∩G), ∀G ∈ G (by hypothesis,the measures agree on the generators) so G ⊆ Dn. By ??, the previous impliesthat δ(G) ⊆ δ(Dn) = Dn, but since G is ∩-stablethen δ(G) = σ(G) and thusσ(G) = A ⊆ Dn.

So that means that

∀A ∈ A, µ(Gn ∩A) = ν(Gn ∩A).

We use ?? to take the limit and

∀A ∈ A, µ(A) = µ(X ∩A) = limn→∞

µ(Gn ∩A)

= limn→∞

ν(Gn ∩A) = ν(X ∩A) = ν(A).

Suppose that we don’t have an exhausting sequence on the generator. Aneat trick (that doesn’t always work) is to extend the generator to G ∪ Xand define the trivial exhausting sequence Gn = X, ∀n ∈ N. If it holds thatµ(X) <∞, i.e., if µ is a finite measure, then we are in a position to apply thistheorem. See exercise 5.9 for an opportunity to apply this trick.

Lemma 3.11.

1. The n-dimensional Lebesgue measure λn is invariant under translations,i.e.

λn(x +B) = λn(B), ∀x ∈ R‘n,B ∈ B(Rn),

where x +B = x + y ∣ y ∈ B.

2. Every measure µ on (Rn,B(Rn)) which is invariant under translations andsatisfies κ = µ([0,1)n) <∞ is a multiple of the Lebesgue measure µ = κλn.

Proof.

1This is the reason why we prove that Dn is a Dynkin system and not a σ-algebra . Doingthe latter would be much complicated because we wouldn’t have σ-additivity.


1. First of all, let us check that λ(x + B) is well defined, i.e. that B ∈B(Rn) Ô⇒ x +B ∈ B(Rn). There is a clever way to do this. Define

Ax = B ∈ B(Rn) ∣ x +B ∈ B(Rn) ⊂ B(Rn).

It is clear that Ax is a σ-algebra on Rn and that J ⊂ Ax since translationsof half-open intervals are still half-open intervals and hence in J and thusin B(Rn). Therefore B(Rn) = σ(J ) ⊂ Ax ⊂ B(Rn). Now we can start withthe meat of the proof.Define ν(B) = λn(x+B) for anyB ∈ B(Rn) and some fixed x = (x1, . . . , xn) ∈Rn. ν is a measure on (Rn,B(R‘n)) since

ν(∅) = λn(x + ∅) = λn (n

⨉i=1[xi + a, xi + a)) =

n

∏i=1(xi + a − (xi + a)) = 0

and ν (∞⊍j=1

Aj) = λn (∞⊍j=1

x +Aj) =∞∑j=1

λn(x +Aj) =∞∑j=1

ν(Aj).

Now take I = ⨉ni=1[ai, bi] ∈ J and note that

ν(I) = λn(x + I) = λn (n

⨉i=1[ai + xi, bi + xi])

=n

∏i=1(bi + x − (ai + x)) =

n

∏i=1(bi − ai) = λn(I)

This means that, if we restrict ourselves to the generator J , we have thatν ∣J = λn ∣J .2 Recall that B(Rn) = σ(J ) and that the generator J admitsthe exhausting sequence [−k, k)k∈N ⊂ J with ν([−k, k)) = λn([−k, k)) =(2k)n < ∞. Hence, using ??, the measures ν and λn must coincide inevery A ⊂ B(Rn).

2. Similarly, take I ∈ J but this time with rational endpoints ai, bi ∈ Q. Thenthere is some M ∈ N and some k(I) ∈ N and points xi ∈ Rn such that

I =k(I)⊍i=1(xi + [0, 1

M)n) . (3.2)

What we did here is pave I with little squares of side length 1M

and lowerleft corner xi, where M could be the common denominator of ai and bi.Now, µ (by hypothesis) and λ (by part 1) we can write

µ(I) =k(I)∑i=1

µ(xi + [0, 1

M)n

) =k(I)∑i=1

µ([0, 1

M)n

) = k(I)µ([0, 1

M)n

)

λn(I) =k(I)∑i=1

λn (xi + [0, 1

M)n

) =k(I)∑i=1

λn ([0, 1

M)n

) = k(I)λn ([0, 1

M)n

) ,

and for I = [0,1)n,

µ([0,1)n) = k([0,1)n)µ([0, 1

M)n

) =Mnµ([0, 1

M)n

)

λn([0,1)n) = k([0,1)n)λn ([0, 1

M)n

) =Mnλn ([0, 1

M)n

) ,

2Where f ∣ A denotes the restriction of f ∶ X → Y to the new domain A ⊂ X.

3.2. UNIQUENESS OF MEASURES 25

since it takes k([0,1)n) =Mn rectangles of side 1M

to cover [0,1)n. Thus

µ(I)µ([0,1)n) =

k(I)Mn

Ô⇒ µ(I) = k(I)Mn

µ([0,1)n)

λn(I)λn([0,1)n) =

k(I)Mn

Ô⇒ λn(I) = k(I)Mn

λn([0,1)n)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

=1

= k(I)Mn

.

Thus, µ(I) = µ([0,1)n)λn(I) = κλn(I) and application of ?? finishes theproof.

Chapter 4

Existence of measures

4.1 PreliminariesIn this chapter we shall explore a new structure, the semi ring and a new func-tion, the premeasure. They are analogous to a σ-algebra and a measure, respec-tively, but weaker. Then we shall prove that under some conditions, premeasurescan be extended to measures and semirings to σ-algebras . The most importantimplication of this result, called Caratheodory’s theorem, is that the Lebesguemeasure that we have so far only defined in the open n-dimensional intervalscan be extended to any Borel set in B(Rn) and thus is a proper measure.

Definition 4.1 (Semi-ring). Let S ⊂ X be a collection of subsets of a set X.We say that S is a semi-ring if the following are satisfied:

1. ∅ ∈ S,

2. S,T ∈ S Ô⇒ S ∩ T ∈ S (or S is ∩-stable), and

3. if S,T ∈ S then there exists a finite collection of pairwise disjoint setsS1, . . . , SM ∈ S such that S ∖T = ⊍M

j=1 Sj (so S ∖T is the disjoint union ofa finite collection in S).

We will see that J and Jrat are semi-rings.

Definition 4.2 (Premeasure). Let X be a set, S a semiring of subsets of X,and µ ∶ S → [0,∞) be a function. We say that µ is a premeasure if the followingare satisfied:

1. µ(∅) = 0,

2. if (Sn)n∈N is a pairwise disjoint collection of sets in S then

µ(⊍j∈N

Sj) =n

∑j∈N

µ(Sj),

in other words, σ-additivity.

What is missing for a premeasure to become a measure is the fact that it isnot defined on a σ-algebra on X, but rather on a weaker structure, the semi-ringS.

27

28 CHAPTER 4. EXISTENCE OF MEASURES

4.2 The Caratheodory theorem

Theorem 4.3 (Caratheodory). Let X be a set, S a semi-ring on X and µ apremeasure defined on S. Then µ has an extension to a measure µ defined onσ(S). If S contains an exhausting sequence Sn ↑ X with µ(Sn) < ∞ then theextension is unique.

And that’s it. We could end the chapter here or make it much longer byproving Caratheodory’s theorem. We may do so, if I manage to find the timeto write it down, but otherwise look it up [?, p. 41]

What we will do is apply the theorem to the Lebesgue measure and give anoutline of the proof.

Remark 4.4. The n-dimensional Lebesgue measure satisfies the hypothesis forthe premeasure in Caratheodory’s theorem.

Proof. TODO

TODO: outline of the proof

Chapter 5

Measurable mappings

In mathematics mappings between sets are a central topic. Moreover, whensets have a specific structure, we want mappings that preserve the that samestructure between the two sets. For instance we have

• groups and homomorphisms, which hold the group operation: f(a ⋅ b) =f(a) ⋅ f(b);

• topological spaces and continuous functions, which hold the topology ofthe spaces in question: for any open set V and continuous function f ,f−1(V ) is also an open set;

• and naturally we wish that between measurable spaces, there are appro-priate mappings that preserve the measurable structure (σ-algebra ).

Let’s dive right into it.

5.1 Definition. Properties.Definition 5.1 (Measurable map). Let (X,A) and (X ′,A′) be measurablespaces. A map T ∶ X → X ′ is said to be A/A′-measurable (or measurableunless this is too ambiguous) if

T −1(A′) = x ∈X ∣ T (x) ∈ A′ ∈ A, ∀A′ ∈ A′,

i.e. if the preimage of every measurable set in A′ is a measurable set in A.In the next chapter we will particularise this to mappings T ∶ X → R and

measurable spaces (X,A) and (R,B(Rn)) to pave the way for integration.

Remark 5.2.

1. In probability theory, a measurable map is usually called a random vari-able.

2. Consider the collection T −1(A′) = T −1(A′) ∣ A′ ∈ A′. Recall ??, inwhich we proved that the preimage of a σ-algebra is a σ-algebra . We canrephrase the definition of measurability as

T is measurable ⇐⇒ T −1A′ ⊂ A

29

30 CHAPTER 5. MEASURABLE MAPPINGS

3. If we write T ∶ (X,A) → (X ′,A′) then we usually mean that T is A/A′-measurable.

4. If T ∶ (Rn,B(Rn))→ (R,B(R)), then we simply say T is Borel-measurable.

Lemma 5.3 (Measurability can be checked only on the generators). Let (X,A)and (X ′,A′) be two measurable spaces with A′ = σ(G′). A map T ∶ X → X ′ isA/A′-measurable ⇐⇒ T −1(G′) ∈ A, ∀G′ ∈ G′ ⇐⇒ T −1(G′) ⊆ A.

Proof. If T is A/A′ measurable then we have that T −1(A′) ∈ A, ∀A′ ∈ A so, inparticular, T −1(G′) ∈ A, ∀G′ ∈ G ⊂ A′.

For the converse let us define

Σ′ ∶= A′ ⊂X ′ ∣ T −1(A′) ∈ A

As usual we will check that Σ′ is itself a σ-algebra on X ′ and therefore

A′ = σ(G′) ⊆ σ(Σ′) = Σ′ Ô⇒ T −1(A′) ∈ A, ∀A′ ∈ A′.

Let’s verify that Σ′ is indeed a σ-algebra on X ′.

1. X ′ ∈ Σ′ since T −1(X ′) =X ∈ A

2. For any A′ ∈ Σ′ we have that T −1(A′c) = T −1(X ′ ∖ A′) = T −1(X ′) ∖T −1(A′) =X ∖ T −1(A′) ∈ A since T −1(A′) ∈ A by hypothesis.

3. For any collection (A′n)n∈N ⊂ Σ′ we have that

T −1 (⋃n∈N

A′n) = ⋃n∈N

T −1(A′n) ∈ A,

since T −1(A′n) ∈ A by hypothesis.

So it is enough to check measurability on the generators only. This leads usto the following remark.

Remark 5.4. Any continuous function f ∶ Rn → Rm is a measurable mapping.

Proof. Recall that f is continuous if and only if for any open set V ⊂ Rm thenf−1(V ) ∈ Rn is also open. Recall that B(Rm) is also generated by the open setsOm hence if f is continuous, f−1(V ) ∈ On = B(Rn) Ô⇒ f is a measurablemap.

Keep in mind that not every measurable map is continuous.

Example 5.5 (A non-continuous measurable map.). Let f ∶ (R,B(R))→ (R,B(R))be a map defined by f(x) = 1A for some A ⊂ R. This map is clearly not contin-uous as 1A takes only two values, 0 and 1. However, it is indeed measurable.Take any a ∈ R. Then,

1A−1((a,∞)) = x ∈ R ∣ 1A(x) > a =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∅ if a ≥ 1A if 0 ≤ a < 1R if a < 0

∈ B(R)

By ?? we have that f is Borel measurable.

5.2. σ-ALGEBRAS IN RELATION TO MEASURABLE MAPS. IMAGE MEASURES.31

Before we move on, lets prove that the composition of measurable maps isanother measurable map, as we would do with any other structure-preservingmap in any other field of mathematics.

Lemma 5.6. Let (Xi,Ai) be measurable spaces for i = 1,2,3 and T ∶ X1 →X2, S ∶ X2 → X3 be two A1/A2- resp. A2/A3-measurable maps. Then S T ∶X1 →X3 is A1/A4-measurable.

Proof. We need to show that (S T )−1(A) ∈ A1 for anyA ∈ A3. Now (S T )−1(A) =T −1(S−1(A)) with S−1(A) ∈ A2 and hence T −1(S−1(A)) ∈ A1.

5.2 σ-algebras in relation to measurable maps.Image measures.

When dealing with measurable maps, we often have a σ-algebra on the codomainbut none is given for the domain. Naturally, we want to know which σ-algebrason X render a map T ∶ X → X ′ measurable when (X ′,A′) is the destinationmeasurable space.

It is clear that if we take A = ℘(X) to be the σ-algebra on X then any mapis measurable since

T −1(A′) ⊂X Ô⇒ T −1(A′) ∈ ℘(X), ∀A′ ∈ A′.

Also, from the examples on chapter 1 know that the preimage of a σ-algebra isanother σ-algebra . This would then be the smallest σ-algebra on X that makesT measurable but we cannot remove any sets from T −1(A′) without endangeringthe measurability of T .

What if we have many mappings to different measurable spaces that sharethe same origin set? We cannot guarantee that taking the union of all the σ-algebras which individually render each mapping measurable is again a σ-algebra. Let us formalise this.

Definition 5.7. Let (Ti)i∈I be arbitrarily many mappings Ti ∶X →X ′ from thesame set X into measurable spaces (Xi,Ai). The smallest σ-algebra on X thatmakes all Ti simultaneously measurable is

σ(Ti ∶ i ∈ I) ∶= σ (⋃i∈I

Ti−1(Ai)) .

We say that σ(Ti ∶ i ∈ I) is generated by the family Ti−1(Ai).

Lemma 5.8. The previous definition makes sense, i.e. A = σ(Ti ∣ i ∈ I) is indeeda σ-algebra which renders all Ti simultaneously A/Ai measurable.

Proof. Let Ai ∈ Ai for any i ∈ I. Then, clearly, Ti−1(Ai) ∈ T −1(Ai), so any Ti is

measurable. But is A a σ-algebra ? Well, that is the reason why we includedthe σ-hull in the above definition, to guarantee that the union does not breakthe σ-algebra structure.

To tie it back to measures we will give the following definition:


Definition 5.9 (Image measure). Let (X,A), (X ′,A′) be measurable spacesand T ∶ X → X ′ be an A/A′-measurable map. Then, for every measure µ on(X,A), we define the image measure of µ under T , denoted by T (µ) or µT −1,by

T (µ)(A′) = µ(T −1(A′)), ∀A′ ∈ A′

Lemma 5.10. The image measure is indeed a measure.

Proof. We just have to check the two properties of measures.

1. Clearly T (µ)(∅) = µ(T −1(∅)) = µ(∅) = 0.

2. For any pairwise disjoint collection (A′n)n∈N ⊂ A′ we have

T (µ)(⊍n∈N

A′n) = µ(T −1 (⊍n∈N

A′n))

= µ(⊍n∈N

T −1(A′n))

= ∑n∈N

µ(T −1(A′n)) = ∑n∈N

T (µ)(A′n)

5.3 ExercisesExample 5.11. This is a version of exercise 7.6 that I think is more preciselyformulated, partly inspired by [?].

Let (X,A) and (X ′,A′) be measurable spaces and T ∶ X → X ′ be a sur-jective, A/A′-measurable map. Then T (A) is a σ-algebra ⇐⇒ T −1 is A′/Ameasurable.

Exercise 5.3.1. Let T ∶ (X,A) → (X,A′) be a measurable map. Under whichcircumstances is the family of sets T (A) a σ-algebra ?.

Proof. We will see that the first and third properites of σ-algebras hold moreor less trivially. It is the closing under complements that gives us problems.

We claim that T (A) is a σ-algebra ⇐⇒ T −1 ∶X ′ →X is a measurable map.Let us assume that T is surjective, as when we define T −1 ∶ X ′ → X we areimplicitly denoting that. Otherwise, we can just redefine X ′ = T (X) and moveon.

For the reverse implication let’s verify the properties of a σ-algebra over X ′on T (A).

1. First, we need to show that X ′ ∈ T (A). This is clear from the fact thatX ∈ A and T is surjective.

2. For any A′ ∈ T (A) we need to show that A′c ∈ T (A). Since T is A/A′measurable, A = T −1(A′) ∈ A. Since A is a σ-algebra , we have that Ac isin A. Moreover, since T −1 is measurable then T −1(Ac) = T −1(T −1(A)c) =T −1(T −1(Ac)) = T (Ac) ∈ T (A).

5.3. EXERCISES 33

3. Finally, for any collection (A′n)n∈N ⊂ T (A) we need to show that ⋃n∈NA′n ∈

T (A). For each A′n ∈ T (A) we have a set An ∈ A such that A′n = T (An).Thus,

⋃n∈N

A′n = ⋃n∈N

T (An) = T (⋃n∈N

An) ∈ T (A),

since ⋃n∈NAn ∈ A.

Chapter 6

Measurable functions

In this chapter we restrict our attention to measurable maps whose domain isany measurable space (X,A) but whose domain is (R,B(R)). To distinguishthem from the general case, we shall use the same terminology as [?], and callthem measurable functions.

The following result is trivial but important, as the intervals we introducein the following lemma are easy to work with.

Lemma 6.1. Let (X,A) be a measurable space, and u ∶ X → R. Then, u isA/B(R)-measurable

⇐⇒ u−1((a,∞)) = x ∈X ∣ u(x) > a = u > a ∈ A⇐⇒ u−1([a,∞)) = x ∈X ∣ u(x) > a = u ≥ a ∈ A⇐⇒ u−1((−∞, a)) = x ∈X ∣ u(x) > a = u < a ∈ A⇐⇒ u−1((−∞, a]) = x ∈X ∣ u(x) > a = u ≤ a ∈ A

Notice that we have introduced some notation here, namely, for a functionu ∶X → R we denote by u > a the set x ∈X ∣ u(x) > a. We define analogousnotations for ≥,<,≤,=,≠, ∈, /∈ and more.

Proof. The proof follows immediately from ??. Recall that all the families ofintervals of the lemma are generators of the Borel σ-algebra on R.

6.1 The extended real line RThroughout this chapter, we will deal with the concepts of limn, lim supn, lim infn, supnand infn which will often be infinite. If we agree that −∞ < x < ∞, ∀x ∈ Rit makes sense to include the values ±∞ in R to build the extended spaceR = [−∞,+∞] = R ∪ ±∞. We would like R to inherit as much as possiblefrom the algebraic, topological and measurable structures of R.

6.1.1 Extension of the algebraic structureWe extend the algebraic structure by extending the addition and multiplicationtables as follows:

35

36 CHAPTER 6. MEASURABLE FUNCTIONS

+ x ∈ R +∞ −∞y ∈ R x + y +∞ −∞+∞ +∞ +∞ not defined−∞ −∞ not defined −∞

⋅ 0 x ∈ R ∖ 0 +∞ −∞0 0 0 0* 0*

y ∈ R ∖ 0 0 x ⋅ y sgn(y) ⋅ ∞ −sgn(y) ⋅ ∞+∞ 0* sgn(x) ⋅ ∞ +∞ −∞−∞ 0* −sgn(x) ⋅ ∞ −∞ +∞

Caution: here we understand ± as limits but 0 only as bona-fide 0 (i.e. notas a limit, which would cause convergence problems). Conventions are tricky.Expresions of the form

∞−∞ or ±∞±∞

should be avoided.

6.1.2 Extension of the topological structureDefinition 6.2 (Neighbourhoods in R). For some x ∈ R we say that a neigh-bourhood of x is a set of the form

(x − ε, x + ε) if x ∈ R(a,+∞] if x = +∞[−∞, a) if x = −∞

for some a, ε ∈ R.

Definition 6.3 (Open set in R). We say that a set U ⊆ R is open if, for everypoint x ∈ U there exists a neighbourhood B(x) of x such that x ∈ B(x) ⊆ R.

6.1.3 Extension of the measurable structureDefinition 6.4 (Borel σ-algebra on R). The Borel σ-algebra on R is defined by

B(R) ∶= B∗ = B ∪ S ∣ B ∈ B(R) ∧ S ∈ S,

where S = ∅,+∞,−∞,−∞,+∞.

The reason this extension is still called a Borel σ-algebra is justified by theabove definition of B(R), by the extension of the topological structure and bythe following lemma.

Lemma 6.5. B(R) is generated by sets of the form [a,∞] (or (a,∞] or [−∞, a]or [−∞, a)), where a ∈ R or a ∈ Q (and hence those intervals are subsets of R orQ, resp.)

Proof. It is analogous to the one given for ??.

6.2. SIMPLE FUNCTIONS 37

6.1.4 Final remarksDefinition 6.6 (Numerical function). A function u ∶ X → R that takes valueson R is called a numerical function.

Definition 6.7 (Set of measurable functions). Let (X,A) be a measurable space.We write

M ∶=M(A) ∶= u ∶X → R ∣ u is A/B(R)-measurable ,MR ∶=MR(A) ∶= u ∶X → R ∣ u is A/B(R)-measurable

for the families of real-values and numerical-valued measurable functions on X.

6.2 Simple functionsNow we will see some important (yet simple) examples. Throughout this section,(X,A) will be a measurable space.

Example 6.8 (Indicator functions). Let A ∈ A and define 1A ∶X → R by

1A(x) =⎧⎪⎪⎨⎪⎪⎩

0 if x ∈ A1 if x /∈ A

.

Then 1A is A/B(R)-measurable since

1A−1((−∞, a)) = 1A < a =

⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

∅ if a ≤ 0Ac if 0 < a ≤ 1R if a > 1

∈ A

In fact, from the proof we can see that 1A is A/B(R)-measurable if and onlyif A ∈ A. So measurability of 1A as a function is equivalent to the measurabilityof A as a set (recall that a set is measurable if it is part of the A which is thedomain of the measure at hand).

The following example is so important that we will give it as a definition.

Definition 6.9 (Simple function). Let A1, . . . ,AM ∈ A be pairwise disjointsubsets of X and y1, . . . , yM ∈ R. A function f ∶X → R of the form

f(x) =M

∑i=1

yi1Ai

is called a simple function.We denote by E(A) = E ⊆M the family of all simple functions f ∶ (X,A) →

(X,B(R)).

Note that

f(x) =⎧⎪⎪⎨⎪⎪⎩

yi if x ∈ Ai

0 if x /∈ ⊍Mi=1Ai


is an alternative definition. The i that makes x ∈ Ai hold is unique or non-existent since, (Ai) are pairwise disjoint, but do not necessarily cover the wholeX.

This small inconvenience (of the possibility that i does not exist) is easilyfixed by extending the collection of sets to be a partition by defining

A0 =X ∖M

⊍i=1

Ai and y0 = 0

and adding the set A0 to the sum in the definition of f .This leads to the following definition

Definition 6.10 (Standard representation of a simple function). Let f ∈ E bea simple function given by

f =n

∑n=0

an1An (6.1)

with Ai ∩Aj = ∅ and X = ⊍Mn=0An. Then ?? is called a standard representation

of the simple function f .

Remark 6.11. Note that this representation is not unique. However, we willsee that this does not matter.

Notice that a simple function takes only finitely many values. The converseis also true.

Lemma 6.12. Any measurable function g which takes only finitely many valuesy0, y1, . . . , yM can be expressed as a linear combination of simple functions.

Proof. Define Ai ∶= g = yi = x ∈ X ∣ g(x) = yi = g−1(yi). Ai ∈ A,∀i =0, . . . ,M since yi = (−∞, yi] ∖ (−∞, yi) ∈ B(R) and g is assumed to be mea-surable.

Since y0, . . . yM are all distinct, then Ai ∩ Aj = ∅ if i ≠ j. Furthermore, gtakes the value yi on Ai, and thus

g =M

∑i=0

yi1Ai ,

which is in fact the standard representation of g.

In general, the representation of a simple function as a linear combinationof simple functions is not unique. Consider the function

f(x) = 1[0,1](x) + 1[0, 23 ](x) + 1[0, 13 ](x) =

⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩

0 if x ∉ [0,1]3 if x ∈ [0, 1

3]

2 if x ∈ ( 13, 23]

1 if x ∈ ( 23,1]

which can also be given by its standard representation

f(x) = 0 ⋅ 1R∖[0,1] + 1 ⋅ 1( 23 ,1] + 2 ⋅ 1( 13 , 23 ] + 3 ⋅ 1[0, 13 ]Simple functions are the building blocks of all measurable functions (in the

sense that any measurable function is the limit of a sequence of simple functions).

6.3. SEQUENCES OF SIMPLE FUNCTIONS. THE SOMBRERO LEMMA.39

6.2.1 Properties of simple functions

Theorem 6.13 (Properties of simple functions). Let f, g ∈ E(A). Then thefollowing hold:

1. f ± g ∈ E(A) and f ⋅ g ∈ E(A)

2. f+ =max(f,0) and f− =max(−f,0) are in E(A)

3. ∣f ∣ ∈ E(A)

Proof. Let

f =M

∑i=0

ai1Ai , g =N

∑j=0

bj1Bj

be the standard representations of f and g, resp. Then,

1.

f ± g =M

∑i=0

N

∑j=0(ai ± bj)1Ai∩Bj ,

M

∑i=0

N

∑j=0(ai ⋅ bj)1Ai∩Bj

are the standard representations of f ± g and f ⋅ g, respectively and thusf ± g, f ⋅ g ∈ E(A).

2.

f+ = ∑i∣ai≥0

ai1Ai f− = ∑i∣ai≤0

−ai1Ai

are the standard representations of f+ and f−, resp. and hence f+, f− ∈E(A).

3. ∣f ∣ = f+ + f− ∈ E(A) by the first two properties.

6.3 Sequences of simple functions. The som-brero lemma.

Theorem 6.14 (Sombrero lemma). Let (X,A) be a measurable space and u ∶X → [0,∞] a non-negative A/B(R)-measurable function. Then, there exists anincreasing sequence (fn)n∈N ⊂ E(A) of non-negative simple functions such thatfor any x ∈X

u(x) = limn→∞

fn(x) = supn∈N

fn(x).

Note that in the previous statement, fn is a sequence of real-valued (asopposed to numerical) functions. Also, the limit is to be understood point-wise(as opposed to uniformly).


Proof. TODO

Corollary 6.15. Let (X,A) be a measurable space. Then, for any numericaland A/B(R)-measurable function u ∶ X → R there exists a sequence of simplefunctions (fn)n∈N ⊂ E(A) such that ∣fn∣ ≤ ∣u∣ and

u(x) = limn→∞

fn(x)

Moreover, if u is bounded then convergence is uniform (as opposed to point-wise).

Proof. Write u = u+−u−, where u+, u− are both non-negative functions. We firstshow that u+, u− are A/B(R)-measurable. We shall proceed by using ??, withgenerators of the form (a,∞]). For any a ∈ R we have

u+ > a =⎧⎪⎪⎨⎪⎪⎩

X if a < 0u ≥ a if a ≥ 0

∈ A

Similarly,

u− > a =⎧⎪⎪⎨⎪⎪⎩

X if a < 0−u ≥ a = u < a if a ≥ 0

∈ A.

Since u+, u− are non-negative measurable functions, ?? applies and thereexist two sequences (fn)n∈N, (gn)n∈N ⊂ E(A) such that fn ↑ u+ and gn ↑ u−.Hence

limn→∞

fn − limn→∞

gn = limn→∞(fn − gn) = lim

n→∞hn = u+ − u− = u.

Furthermore, ∣fn − gn∣ ≤ ∣fn∣ − ∣gn∣ = fn − gn ≤ u+ + u− = ∣u∣. Finally, if u isbounded then ∃N ∈ N such taht u(x) ≤ N,∀x ∈ X. Then from the proof of ??(applied to u+ and u−) we see that ∀n ≥ N and ∀x ∈X

∣fn(x) − gn(x) − u(x)∣ ≤ ∣fn − u+(x)∣ + ∣gn − u−(x)∣ ≤1

2n−1.

This implies uniform convergence.

Convention Given a sequence (un)n∈N ⊂MR(A), by supn∈N un, infn∈N un, lim supn∈N un

and lim infn∈N we mean the point-wise defined functions (supn∈N un)(x) = supn∈N un(x) =supun(x) ∣ n ∈ N and similarly for others.

Recall that

lim infn→∞

u(x) = supn≥1

infm≥n

um(x) (6.2)

and

lim supn→∞

u(x) = infn≥1

supm≥n

um(x). (6.3)

6.3. SEQUENCES OF SIMPLE FUNCTIONS. THE SOMBRERO LEMMA.41

Also,

vn(x) = infm≥n

um(x) ↑ lim infn→∞

u(x),

wm(x) = supm≥n

um(x) ↓ lim supn→∞

un(x)

and

infn∈N

un(x) ≤ lim infn→∞

un ≤ lim supn→∞

un(x) ≤ supn∈N

un(x).

If lim infn→∞ un(x) = lim supn→∞ un(x), then limn→∞ un(x) exists and equalsthe common value.

Corollary 6.16. Let (X,A) be a measurable space and (un)n∈N ⊂MR(A) a se-quence ofA/B(R)-measurable functions. Then, infn≥1 un, lim infn→∞ un, lim supn→∞ un, supn≥1are all A/B(R)-measurable functions.

Proof. First we prove that sup and inf are both A/B(R)-measurable.

supn≥1

un ≤ a =∞⋂n=1un ≤ a ∈ A

infn≥1

un ≥ a =∞⋂n=1un ≥ a ∈ A

Now we use these to prove the rest.lim infn→∞ un = supn≥1 infm≥n um. Set vn = infm≥n um as before and be-

cause of the above vn is A/B(R)-measurable. Since vn ↑ lim infn→∞ un we havelim infn→∞ = supn≥1 vn ∈ A. Hence lim infn→∞ un is also A/B(R)-measurable.

Similarly, write wn = supm≥n un ∈ A because of the above. Then, since wn ↓lim supn→∞ un we have lim supn→∞ un = infn≥1wn ∈ A and hence lim supn→∞ un

is A/B(R)-measurable.

In fact, we can deduce more.

Corollary 6.17. Let u, v ∈MR(A) be twoA/B(R)-measurable functions. Thenu ± v, u ∨ v =maxu, v and u ∧ v =minu, v are A/B(R)-measurable.

Proof. By ?? there exist two sequences of non-negative simple functions (fn)and (gn) such that fn ↑ u and gn ↑ v. Since fn and gn are simple functions, by?? we have that fn ± gn is also a simple function with limn→∞ fn + gn = u + v.Moreover, fn + gn is A/B(R)-measurable (since fn, gn are) and thus it is alsoA/B(R)-measurable and hence u + v is also A/B(R)-measurable.

Similarly, we can prove the corollary for u ∨ v and u ∧ v.

Remark 6.18. Applying the above to u+, u− we see that u isA/B(R)-measurableif and only if u+ and u− are A/B(R)-measurable.

Corollary 6.19. If u, v ∈MR(A), then the following sets are all measurable:

u ≤ v,u < v,u = v,u > v,u ≥ v


Proof. Not that u, v ∈ MR Ô⇒ u − v ∈ MR therefore we may rewrite theprevious sets as

u − v ≤ 0,u − v < 0,u − v = 0,u − v > 0,u − v ≥ 0.

With the exception of u − v = 0, all the others are generators of the Borelσ-algebra so, by ??, they are measurable. As for u − v = 0 we have that

u − v = 0 = (u − v)−1(0) = (u − v)−1 (∞⋂n=1[0, 1

n)) ∈ A

since ⋂∞n=1[0, 1n) ∈ B(R).

6.4 ExamplesExample 6.20. (This is actually exercise 9.8 in [?, p. 79]). Every functionu ∶ N→ R on (N,℘(N)) is ℘(N)/B(R)-measurable.

Proof. Since our σ-algebra is ℘(N), all subsets of N are measurable. Therefore

f ≤ α = k ∈ N ∣ f(k) ≤ α ⊂ N Ô⇒ f ≤ α ∈ ℘(N)

and we may apply ??.

Chapter 7

Integrals of non-negativefunctions

We are now ready to define the integral of a non-negative function in terms ofa measure.

7.1 Integral of a non-negative simple functionSince each u ∈M+(A) orM+

R(A) is a limit of an increasing sequence of simplefunctions (??), we concentrate first on the collection E+ = E+(A) of all A/B(R)-measurable non-negative simple functions.

Let f(x) = ∑Ni=0 ai1Ai be the standard representation of f ∈ E+. Recall that

this means that (Ai) is a collection of sets in A which define a partition of X(i.e. X = ⊍N

i=0Ai). Furthermore, on Ai, f takes the value ai.Let

Iµ(f) ∶=N

∑i=0

aiµ(Ai). (7.1)

We want to interpret Imu(f) as ∫ fdµ, but there might be a small problem,namely f might have more than one standard representation so we need tocheck that the definition of Iµ(f) is independent of the specific representation,i.e. they all give the same answer.

Lemma 7.1 (Integrals of simple functions are representation-invariant). If f =∑M

i=0 ai1Ai = ∑Nj=0 bj1Bj are two standard representations of a simple function

then ∑Mi=0 aiµ(Ai) = ∑n

j=0 bjµ(Bj).

Proof. Recall that since (Aj) and (Bj) are partitions we may write

Ai = Ai ∩M

⊍j=0

Bj =M

⊍j=0

Ai ∩Bj ,

Bj = Bj ∩N

⊍i=0

Ai =N

⊍i=0

Ai ∩Bj .

43

44 CHAPTER 7. INTEGRALS OF NON-NEGATIVE FUNCTIONS

Thus,

M

∑i=0

aiµ(Ai) =M

∑i=0

aiµ(M

⊍j=0

Ai ∩Bj) =M

∑i=0

aiN

∑j=0

µ(Ai ∩Bj)

=M

∑i=0

N

∑j=0

aiµ(Ai ∩Bj) =N

∑j=0

M

∑i=0

aiµ(Ai ∩Bj).

Now,

• if Ai ∩Bj = ∅ then aiµ(Ai ∩Bj) = 0 = bjµ(Ai ∩Bj), and

• if Ai∩Bj ≠ ∅ then ∃x ∈ Ai∩Bj so that f(x) = ai = bj . Thus aiµ(Ai∩Bj) =bjµ(Ai ∩Bj).

Therefore,

m

∑i=0

aiµ(Ai) =N

∑j=0

M

∑i=0

aiµ(Ai ∩Bj) =N

∑j=0

N

∑i=0

bjµ(Ai ∩Bj)

=N

∑j=0

bjM

∑i=0

µ(Ai ∩Bj) =N

∑j=0

bjµ(Bj)

So Iµ(f) is well-defined regardless of the chosen standard representation. Weare about to be ready to define the integral for arbitrary non-negative functions(once we verify that the properties of simple functions that have allowed us tostate ?? hold under integrals).

Theorem 7.2 (Properties of Iµ(f)). Let f, g ∈ E+(A) and λ ≥ 0, then

1. Iµ(1A) = µ(A), for any A ∈ A

2. Iµ(λf) = λIµ(f) (positive homogenous degree 1)

3. Iµ(f + g) = Iµ(f) + Iµ(f) (additive)

4. if f ≤ g then Iµ(f) ≤ Iµ(g) (monotone)

Proof. 1. Clearly in f = 1A there is only one term in the sum and a1 = 1hence Iµ(1A) = 1µ(A) = µ(A).

2.

Iµ(λf) = Iµ (λM

∑n=0

an1An)

= Iµ (M

∑n=0(λan)1An)

=M

∑n=0(λan)µ(An)

= λM

∑n=0

aNµ(An) = λIµ(f)

7.2. INTEGRAL OF A NON-NEGATIVE FUNCTION 45

3. Recall from ?? that

f + g =M

∑i=0

N

∑j=0(ai + bj)1Ai∩Bj .

Therefore,

Iµ(f + g) =M

∑i=0

N

∑j=0(ai + bj)µ(Ai ∩Bj)

=M

∑i=0

aiN

∑j=0

µ(Ai ∩Bj) +N

∑j=0

bjM

∑i=0

µ(Ai ∩Bj)

=M

∑i=0

aiµ(Ai) +N

∑j=0

bjµ(Bj)

= Iµ(f) + Iµ(g).

4. We write g = f + (g − f), where g − f ∈ E by ?? and g − f ≥ 0 since f ≤ g.Therefore, f − g ∈ E+ and

∫µ(f) ≤ ∫

µ(f) + ∫

µ(f − g)

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶≥0

= ∫µ(g).

7.2 Integral of a non-negative functionDefinition 7.3 ((µ)-integral of a non-negative function). Let u ∈M+

R(A), thenthe (µ)-integral of u is defined by

∫ udµ ∶= supIµ(g) ∣ g ≤ u, g ∈ E+(A) ∈ [0,∞] (7.2)

Some alternative notations are

∫ udµ = ∫Xudµ = ∫

Xu(x)µ(dx) = ∫

Xu(x)dµ(x).

Note that here we are taking the supremum over all simple functions lessthan or equal to µ. We will eventually show that it is enough to find a sequence(fn)n∈N ⊂ E+(A) such that fn ↑ u, since we will prove that

∫ udµ = limn→∞∫ fndµ = sup

n≥1∫ fndµ.

Before we move one, let’s make sure that we did not break anything. Namely,does the definition of the integral above extend the definition of Iµ for non-negative simple functions, i.e. Iµ(f) = ∫ fdµ if f is a non-negative simplefunction. The answer is yes!

Lemma 7.4. If f ∈ E+(A) then Iµ(f) = ∫ fdµ.


Proof. Since f ∈ E+(A) and f ≤ f , then by definition of the supremum, we haveIµ(f) ≤ ∫ fdµ. Now, by monotonicity of Iµ, for any g ∈ E+(A) such that g ≤ fone has Iµ(g) ≤ Iµ(f). Thus,

∫ fdµ = supIµ(g) ∣ g ≤ f, g ∈ E+(A) ≤ Iµ(f).

This shows that Iµ(f) = ∫ fdµ.

Remark 7.5. It is easy to see that µ-integrals are monotone1. Since if u ≤ v,then g ∈ E+(A) ∣ g ≤ u ⊂ h ∈ E+(A) ∣ h ≤ v. Hence,

∫ udµ = supIµ(g) ∣ g ∈ E+(A), g ≤ u ≤ supIµ(h) ∣ h ∈ E+(A), h ≤ v = ∫ vdµ

Now we come to the central theorem of this chapter.

Theorem 7.6 (Beppo-Lévi). Let (X,A, µ) be a measure space and let (un)n∈Nbe an increasing sequence inM+

R(A). Then,

u ∶= supn≥1

un = limn→∞

un ∈M+R(A),

and

∫ udµ = ∫ supn≥1

undµ = supn≥1∫ undµ, (7.3)

or, alternatively,

∫ udµ = ∫ limn→∞

undµ = limn→∞∫ undµ (7.4)

Notice that the assumption that un is an increasing sequence is necessary tobe able to interchange limits with suprema.

Proof. TODO. Important.

The following corollary is really just a more concise restatement of ??.

Corollary 7.7. Let u ∈M+R(A) be a non-negative, numerical, A/B(R)-measurable

function and (fn)n∈N ⊂ E+(A) be a sequence of non-negative simple functionssuch that fn ↑ u. Then,

∫ udµ = limn→∞∫ fndµ (7.5)

Proof. Clearly the hypothesis of ?? are satisfied. (Just take un = fn).

Now we extend ?? to the case for integrals of arbitrary non-negative func-tions.

1The reason we anticipate this property and not wait for ?? is that we need it for the proofof ??.

7.2. INTEGRAL OF A NON-NEGATIVE FUNCTION 47

Theorem 7.8 (Properties of integrals of non-negative functions). Let (X,A, µ)be a measure space. Let u, v ∈M+

R(A). Then,

1. ∫ 1Adµ = Iµ(A) = µ(A)

2. ∫ αudµ = α ∫ udµ, ∀α ≥ 0.a

3. ∫ (u + v)dµ = ∫ udµ + ∫ vdµ

4. if u ≤ v, then ∫ udµ ≤ ∫ vdµaHere we require α ≥ 0 to keep the functions non-negative. As soon as we generalise to

arbitrary functions this assumption is not needed.

This proof is left as exercise 9.3 in [?, p. 79].

Proof.

1. 1A is clearly a simple function so, by definition ∫ 1Adµ = Iµ(1A) = µ(A).

2. By ?? there exists a sequence of simple functions (fn)n∈N ⊂ E+ such thatfn ↑ u. Then,

∫ αudµ??= lim

n→∞∫ αfndµ??= lim

n→∞Iµ(αfn) ??= α lim

n→∞Iµ(fn) ??= α∫ udµ

3. Once again by ?? there exist sequences (fn)n∈N, (gn)n∈N ⊂ E+ such thatfn ↑ u and gn ↑ v. Then, (fn + gn) ↑ (u + v) and hence,

∫ (u + v)dµ??= lim

n→∞∫ (fn + gn)dµ??= lim

n→∞Iµ(fn + gn)

??= limn→∞

Iµ(fn) + limn→∞

Iµ(gn) ??= ∫ udµ + ∫ vdµ

4. Already proven in ??.

Corollary 7.9. Let (un)n∈N ⊂M+R(A). Then, ∑

∞n=1 un ∈M+

R(A) and

∫∞∑n=1

undµ =∞∑n=1∫ undµ (7.6)

This proof is left exercise 9.6 in [?, p. 79].

Proof. Let vm = ∑mn=1 un. Then vm ↑ ∑∞n=1 un and thus

∫∞∑n=1

undµ = ∫ limm→∞

vmdµ??= lim

m→∞∫ vmdµ = limm→∞∫

m

∑n=1

undµ

??= limm→∞

m

∑n=1∫ undµ =

∞∑n=1∫ undµ


Theorem 7.10 (Fatou’s lemma). Let (un)n∈N be any sequence in M+R(A).

Then,

u ∶= lim infn→∞

un ∈M+R(A) (7.7)

and

∫ lim infn→∞

undµ ≤ lim infn→∞ ∫ undµ (7.8)

The great thing about ?? is that it makes no assumptions of monotonicityand can be applied to any sequence of non-negative, numerical, measurablefunctions.

Proof. ?? follows from ??.As for ??, we have

∫ lim infn→∞

undµ = ∫ supn∈N

infj≥n

ujdµ

??= supn∈N∫ inf

j≥nujdµ

??≤ sup

n∈Ninfl≥n∫ uldµ

= lim infn→∞ ∫ undµ

where we used the fact that infj≥n ≤ ul for any l ≥ n.

We also have a similar result for lim sup, sometimes known as the reverseFatou’s lemma.

Corollary 7.11 (Reverse Fatou’s lemma). Let (un)n∈N ⊂M+(A) be a sequenceof non-negative numerical measurable functions. If un ≤ u for all n ∈ N and somev ∈M+(A) such that ∫ vdµ <∞, then

lim supn→∞

∫ undµ ≤ ∫ lim supn→∞

undµ (7.9)

Proof. Let wn = v−un ≥ 0, i.e. (wn)n∈N is a sequence of non-negative measurablefunctions. By ?? we get,

∫ lim infn→∞

wndµ ≤ lim inf ∫ wndµ

= lim infn→∞

(∫ vdµ − ∫ undµ)

= ∫ vdµ − lim supn→∞

∫ undµ.

7.3. EXAMPLES 49

(Recall that lim inf(−un) = − lim supun.) Thus,

∫ vdµ − lim supn→∞

∫ undµ ≥ ∫ lim infn→∞

wndµ

= ∫ lim infn→∞

(v −wn)dµ

= ∫ vdµ − ∫ lim supn→∞

undµ.

Since we assumed ∫ vdµ to be finite, we can subtract it from both sides and thelemma follows.

Note how this time we had to make sure that the sequence was bounded bya function v whose integral was finite. This was implicit in ??, only in this casethe bounding is from below and the bounding function is v = 0.

7.3 ExamplesExample 7.12. Consider the measure space (X,A, δy) where δy is the Diracmeasure (cf. ??) for some fixed y ∈X. We claim that

∫ udδy = ∫ u(x)δy(x) = u(y),∀u ∈M+R(A). (7.10)

Proof. First consider the case f ∈ E+ with standard representation f = ∑Mn=0 an1An

with Ai ∩Aj = ∅ if i ≠ j (cf. ??). We know that y lies in exactly one An, sayy ∈ An0 . Then

∫ fdδy = Iδy(f) =M

∑n=0

anδy(An) = an0 = f(y).

Now we take any sequence of simple functions fn ↑ u. By ?? we have

∫ udδy = limn→∞

fndδy = limn→∞

fn(y) = u(y)

Example 7.13. Consider the measure space (N,℘(N), µ) with

µ =∞∑n=1

anδn,

so that µ(n) = an, ∀n ∈ N. All the measurable functions u ∶ N→ R are of theform

u(x) =∞∑n=1

un1n(x), x ∈ N.


Then2,

∫ udµ = ∫∞∑n=1

un1ndµ

??=∞∑n=1∫ un1ndµ

=∞∑n=1

unµ(n)

=∞∑n=1

un ⋅ an.

2Note how we do not expand ∫ udµ = Iµ(u), since u is not a simple function (too manyterms, although as we see, it does not really matter).

Chapter 8

Integrals of measurablefunctions

In this chapter we will extend the notion of integral to the general case u ∈MR(A) (as opposed to just non-negative functions, which were discussed in theprevious chapter).

We have already seen that any function u ∈MR(A) may be written as thedifference of two non-negative functions u = u+−u−, hence we have the tendencyto define

∫ udµ = ∫ u+dµ − ∫ u−dµ

However, since ∫ u+dµ, ∫ u−dµ ∈ [0,∞] and therefore both can be +∞ wemust be careful with the definition of the integral for a general u.

Definition 8.1 (µ-integrable). Let u ∈MR(A), then u is said to be µ-integrableif

max∫ u+dµ,∫ u−dµ <∞, (8.1)

i.e. both are finite. In this case we define

∫ udµ ∶= ∫ u+dµ − ∫ u−dµ ∈ (−∞,+∞). (8.2)

Remark 8.2. If

min∫ u+dµ,∫ u−dµ <∞,

i.e. at most one of the integrals is +∞, then one can still define

∫ udµ ∶= ∫ u+dµ − ∫ u−dµ ∈ [−∞,+∞].

We will denote the set of all µ-integrable functions by

L1R(µ) ∶= u ∈MR(A) ∣ u is µ − integrable. (8.3)

Similarly, if we wish to restrict ourselves to real-valued functions we will write

L1(µ) ∶= u ∈MR(A) ∣ u is µ − integrable. (8.4)

51

52 CHAPTER 8. INTEGRALS OF MEASURABLE FUNCTIONS

Remark 8.3. Note that for u ≥ 0, ∫ udµ always exists, but it can have the value+∞. So for u ∈M+

R(A) we have

u ∈ L1R(µ) ⇐⇒ ∫ udµ <∞.

Some authors still call a positive function µ-integrable if it takes the value +∞.We will not use this convention. Take care to update your understanding bycomparing ?? and ??.

As before, if we need to stress the integration variable (the space where theintegral is defined), we write

∫ udµ = ∫Xu(x)dµ(x) = ∫

Xu(x)µ(dx).

The following theorem gives us for ways to check that a function u ∈MR(A)is µ-integrable.

Theorem 8.4 (Characterisation of µ-integrability). Let u ∈MR(A), then thefollowing are equivalent:

1. u ∈ L1R(µ);

2. u+, u− ∈ L1R(µ);

3. ∣u∣ ∈ L1R(µ);

4. ∃w ∈ L1R(µ) with w ≥ 0 and ∣u∣ ≤ w.

Proof.

• 1. ⇐⇒ 2. follows from the definition of µ-integrability.

• 2. ⇐⇒ 3. follows from ∣u∣ = u++u− and u+, u− ≤ ∣u∣ and the monotonicityfor non-negative measurable functions.

• 3. ⇐⇒ 4. follows from monotonicity of the integral of non-negativemeasurable functions.

53

Theorem 8.5 (Properties of the µ-integral). Let (X,A, µ) be a measure space,u, v ∈ L1

R(µ) and α ∈ R. Then, the following hold

1. αu ∈ L1R(µ) and ∫ αudµ = α ∫ udµ.

2. u + v ∈ L1R(µ) and ∫ (u + v)dµ = ∫ udµ + ∫ vdµ.

3. minu, v,maxu, v ∈ L1R(µ)

4. if u ≤ v, then ∫ udµ ≤ ∫ vdµ.

5.

∣∫ udµ∣ ≤ ∫ ∣u∣dµ.

Proof. For 1. and 2., we first use ?? to prove that αu and u + v are in L1R(µ).

Then, we rewrite each in terms of positive and negative parts to prove that theintegrals coincide.

1. We will prove that αu ∈ L1R(µ) ⇐⇒ ∣αu∣L

1R(µ). Now,

∫ ∣αu∣dµ = ∫ ∣α∣ ∣u∣dµ??= ∣α∣∫ ∣u∣dµ <∞.

Thus, by ?? we have that αu ∈ L1R(µ). To check ∫ αudµ = α ∫ udµ we

consider two cases:

• If α ≥ 0 then (αu)+ = αu+ and (αu)− = αu− hence

∫ αudµ = ∫ αu+dµ − ∫ αu−dµ = α∫ u+dµ − α∫ u−dµ

= α(∫ u+dµ − ∫ u−dµ) = α∫ udµ.

• If α < 0 then (αu)+ = −αu− and (αu)− = −αu+. Therefore,

∫ αudµ = ∫ ((αu)+ − (αu)−)dµ

= ∫ −α>0

u−dµ − ∫ −α>0

u+dµ

= (−α)∫ u−dµ − (−α)∫ u+dµ

= (−α) (∫ u−dµ − ∫ u+dµ)

= (−α) (−∫ udµ) = α∫ udµ

2. ∣u + v∣ ≤ ∣u∣ + ∣v∣ so by linearity for non-negative integrals we have

∫ ∣u + v∣dµ ≤ ∫ ∣u∣dµ + ∫ ∣v∣dµ <∞.


Hence, by ??, u + v ∈ L1R(µ). As before, (u + v) = (u + v)+ − (u + v)− =

u+ − u− + v+ − v− hence

(u + v)+ + u− + v− = (u + v)− = u+ + v+,

so

∫ [(u + v)+ + u− + v−]dµ = ∫ [(u + v)− = u+ + v+]dµ

where all functions are non negative. Applying ?? we have

∫ (u + v)+dµ + ∫ u−dµ + ∫ v−dµ = ∫ (u + v)−dµ + ∫ u+dµ + ∫ v+dµ.

Rewriting again (every term is finite since u, v, u+v ∈ L1R(µ) so subtraction

is not a problem), we have

∫ (u + v)+dµ − ∫ (u + v)−dµ = ∫ u+dµ − ∫ u−dµ + ∫ v+dµ − ∫ v−dµ,

or,

∫ (u + v)dµ = ∫ udµ + ∫ vdµ.

3. Observe that both max(u, v),min(u, v) ≤ ∣u∣ + ∣v∣ so

∫ max(u, v)dµ ≤ ∫ ∣u∣dµ + ∫ ∣v∣dµ <∞

∫ min(u, v)dµ ≤ ∫ ∣u∣dµ + ∫ ∣v∣dµ <∞

by ??. Thus, max(u, v),min(u, v) ∈ L1R(µ).

4. Suppose u ≤ v then u+ ≤ v+ and u− ≥ v−, so

∫ udµ = ∫ u+dµ − ∫ v−dµ ≤ ∫ v+dµ − ∫ v−dµ = ∫ vdµ.

5. Recall that one can defined ∣a∣ =max(a,−a). Then

∣∫ udµ∣ =max∫ udµ,−∫ udµ

=max∫ udµ,∫ −udµ

=max∫ ∣u∣dµ,∫ ∣u∣dµ = ∫ ∣u∣dµ.

Remark 8.6. Note that in property 2. we did had the assumption that (u+ v)does not take the values ±∞. Hence we cannot say that L1

R(µ) is a linear space.However, L1(µ), the set of all µ-integrable real-valued functions is a linear

space.

8.1. RESTRICTING THE DOMAIN 55

8.1 Restricting the domainLet (X,A, µ) be a measure space and u ∈ L1

R(µ). For any A ∈ A, the function1A ⋅u ∈MR(A) and ∣1A ⋅ u∣ ≤ ∣u∣. Hence, by ?? (4.) we have that 1A ⋅u ∈ L1

R(A).Similarly, if (X,A, µ) is a measure space and u ∈M+

R(A), for any A ∈ A wecan define the function 1A ⋅ u ∈M+

R(A) by ?? since 1A ⋅ u = min(1A, u) in thiscase (since u is non-negative).

When we wish to restrict the domain of integration, we write,

∫Audµ ∶= ∫ 1Audµ. (8.5)

Note that if u ∈ L1R(µ) ∫A udµ < ∞ but if u ∈M+

R(A) then ∫A udµ is welldefined but can take the value +∞.

We do we keep coming back to the case u ∈M+R(A)? Well, because in this

case we can associate a new measure on (X,A) as follows:

Lemma 8.7. Let (X,A, µ) be a measure space and u ∈M+R(A). Define ν ∶ A→

[0,∞] by

ν(A) ∶= ∫Audµ = ∫ 1Audµ. (8.6)

Then ν is a measure on (X,A). Moreover, if ∫ udµ <∞, i.e. u is, additionally,in L1

R(A), then ν is a finite measure.

Proof. As usual,

1.

ν(∅) = ∫∅ udµ = ∫ 0 ⋅ udµ = 0

2. Let (An)n∈N be a sequence of pairwise disjoint sets in A. Then

1⊍n∈N An = ∑n∈N

1An ,

and thus,

ν (⊍n∈N

An) = ∫⊍n∈N An

udµ

= ∫ 1⊍n∈N Anudµ

= ∫ (∑n∈N

1An)udµ

= ∫ (∑n∈N

1Anu)dµ

??= ∑n∈N∫ 1Anudµ = ∑

n∈Nν(An).


8.2 ExamplesExample 8.8. Let (X,A, δA) be a measure space where δA is the Dirac measurefor a fixed A ∈ A (see ??). Then for u ∶ A→ R, u ≥ 0 we have

∫ udδA = ∫ u(x)δA(dx) = u(A).

Then, if we generalise u ∈MR(A) we have

∫ udδA = ∫ (u+ − u−)dδA = ∫ u+dδA − ∫ u−dδA = u+(A) − u−(A) = u(A).

We conclude that u ∈ L1R(δA) ⇐⇒ ∣u(A)∣ <∞.

Example 8.9. Consider the measure space (N,℘(N), µ) where µ is the discreetmeasure given by

µ = ∑n∈N

αnδn.

We have seen that

∫ ∣u∣dµ = ∑n∈N

αn ∣u(n)∣ .

Thus, u ∈ L1R(µ) ⇐⇒ ∑n∈N αn ∣u(n)∣ <∞. Note that we can write

u(m) = ∑n∈N

u(n)1n(m),

so if α1 = α2 = ⋅ ⋅ ⋅ = 1 then u ∈ L1R(µ) ⇐⇒ ∑n∈N ∣u(n)∣ <∞. In particular, it is

necessary that u(n) ∈ R, ∀n ∈ N. We can in this case express

L1R(µ) = (xn)n∈N ∣ xn ∈ R ∧ ∑

n∈N∣xn∣ <∞ ∶= `1(N).

We call `1(N) the set of all summable sequences. This space is important infunctional analysis.

Example 8.10. If (X,A, µ) is a finite measure space, then any bounded func-tion u ∈ MR(A) is µ-integrable. That is, if for any given u ∈ MR(A) thereexists M ∈ R, M > 0 such that ∣u(x)∣ ≤M, ∀x ∈ A then u is µ-integrable.

∫ ∣u∣dµ = ∫ ∣u∣1Xdµ ≤ ∫ M ∫Xdµ =M ∫ 1Xdµ =Mµ(X) <∞.

Chapter 9

Null sets and the notationalmost everywhere

In this chapter we formally introduce the concept of a null set and the handynotation almost everywhere, which gives us a way to formalise ideas that we getintuitively.

Definition 9.1. Let (X,A, µ) be a measure space. The collection of µ-null setsis defined by

Nµ = A ∈ A ∣ µ(A) = 0 (9.1)

Note that Nµ is a σ-algebra , not necessarily on X, contained in A.

Definition 9.2. We say that a property Π = Π(x) holds µ almost everywhere,in short µ-a.e. or a.e., if there exits

N ∈ Nµ such that x ∈X ∣ Π(x) fails ⊂ N. (9.2)

Note that we do not require that we do not require that x ∈X ∣ Π(x) fails be a measurable set, it only has to be a subset of a measurable set. In practice,this set is indeed a µ-null set itself.

Example 9.3. Let Π be the property that u(x) = v(x) for u, v ∈MR(A). Thenboth x ∈ X ∣ u(x) = v(x) and x ∈ X ∣ u(x) ≠ v(x) are measurable (sinceu, v ∶ A → R). In this case, saying that Π holds a.e. means that x ∈ X ∣ u(x) ≠v(x) ∈ Nµ.

Remark 9.4. Note that the notation µ-a.e. is very tricky. For instance, thestatements

1. u has property Π almost everywhere

2. u is µ-a.e. equal to a function which has property Π everywhere

are very different.Consider the functions 1Q and 0 and the property Π(u) ≡ u is continuous.

Indeed, 0 is everywhere-continuous and 1Q is µ-a.e. equal to 0 since they onlydiffer in the rationals for which λ(Q) = 0 but 1Q is everywhere-discontinuous.

57

58CHAPTER 9. NULL SETS AND THE NOTATION ALMOST EVERYWHERE

Before moving on, we will prove a very handy inequality.

Theorem 9.5 (Markov inequality). Let u ∈ L1R(µ), A ∈ A and c > 0. Then,

µ(∣u∣ ≥ c ∩A) ≤ 1

c∫A∣u∣dµ. (9.3)

In particular if A =X then

µ(∣u∣ ≥ c) ≤ 1

c∫ ∣u∣dµ.

Informally, this means that an integrable function does not blow up (toomuch), i.e. an integrable function only attains large values on a set with a smallmeasure.

Proof.

µ(u ≥ c ∩A) = ∫ 1∣u∣≥c∩Adµ

= c

c∫ 1∣u∣≥c ⋅ 1Adµ

= 1

c∫Ac1∣u∣≥cdµ

≤ 1

c∫A∣u∣1∣u∣≥c´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

≥∣u∣

dµ ≤ 1

c∫A∣u∣dµ

Theorem 9.6. Let u ∈ L1R(µ) be a numerical integrable function on a measure

space (X,A, µ). Then,

1.

∫ ∣u∣dµ = 0 ⇐⇒ ∣u∣ = 0 µ-a.e. ⇐⇒ µ (x ∈X ∣ u(x) ≠ 0) = 0 (9.4)

2. For any N ∈ Nµ, we have

1Nu ∈ L1R(µ) and ∫N udµ = 0. (9.5)

Proof. Let us start with 2. Define fn = min∣u∣ , n. Then, fn ∈M+R(A) and

fn ↑ ∣u∣ and 1N ⋅ fn ↑ 1N ⋅ ∣u∣. By ??,

∫N∣u∣dµ = ∫ 1N ∣u∣dµ

= ∫ supn∈N

1Nfndµ

??= supn∈N∫ 1Nfndµ

≤ supn∈N∫ n1Ndµ = sup

n∈Nnµ(N) = sup

n∈N0 = 0.

59

Thus, ∫ 1N ∣u∣dµ = 0 Ô⇒ 1Nu ∈ L1R(µ). Now,

0 ≤ ∣∫Nudµ∣ ≤ ∫

N∣u∣dµ = 0,

thus ∫N udµ = 0.

Now we move on to 1. The second equivalence is true since u ∈ L1R(µ)

implies that u is measurable and hence u ≠ 0 is not just a subset of a µ-nullset but also a measurable set itself. For Ô⇒ in the first equivalence, assume∫ ∣u∣dµ = 0. Then,

µ(u ≠ 0) = µ(∣u∣ > 0)

= µ(∞⋃n=1∣u∣ ≥ 1

n)

??≤∞∑n=1

µ(∣u∣ ≥ 1

n)

??≤∞∑n=1

n∫ ∣u∣dµ

=∞∑n=1

0 = 0.

Thus ∣u∣ = 0 a.e.Conversely, assume ∣u∣ = 0 a.e. Then ∣u∣ ≠ 0 ∈ Nµ and

∫ ∣u∣dµ = ∫ (∣u∣1∣u∣=0 + ∣u∣1∣u∣=0)dµ

= ∫ ∣u∣=0

1∣u∣=0dµ + ∫ ∣u∣1∣u∣=0dµ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶=0 by part 2.

= 0

This theorem is very important and used widely as it allows us tochange the value of a function on a µ-null set without changing thevalue of the integral.

Corollary 9.7. Let u, v ∈MR(A) be such that u = v a.e. Then,

1. if u, v ≥ 0, then ∫ udµ = ∫ vdµ (though it is possible that both sides are+∞).

2. u ∈ L1R(A) ⇐⇒ v ∈ L1

R(A) and, also in this case ∫ udµ = ∫ vdµ.

Proof. Part 1. Assume u, v ∈M+R(A) with u = v a.e.


∫ udµ = ∫ (u1u=v + u1u≠v)dµ

= ∫ u1u=v´¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¶=v

dµ + ∫ u1u≠vdµ

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶0 by ??

= ∫ v1u=vdµ + ∫ v1u≠vdµ

´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶0 by ??

= ∫ (v1u=v + v1u≠v)dµ = ∫ vdµ.

Part 2 follows from part 1. Note that if u = v a.e. then u+ = v+ and u+ = v−a.e. So if u ∈ L1

R(µ), then by part 1,

max(∫ u+dµ,∫ u−dµ) =max(∫ v+dµ,∫ v−dµ) ,

which implies that v ∈ L1R(µ) and

∫ udµ = ∫ u+dµ − ∫ u−dµpart 1= ∫ v+dµ − ∫ v−dµ = ∫ vdµ.

Another consequence of ?? is that integrable functions take values in R µ-a.e. .

Corollary 9.8. Let u ∈ L1R(µ). Then u takes values in R µ-a.e. , i.e.

µ(x ∈X ∣ ∣u(x)∣ = +∞) = µ(∣u∣ =∞) = 0.In particular, we can find v ∈ L1

R(µ) such that u = v µ-a.e. and ∫ udµ =∫ vdµ.

Proof. We use Markov’s inequality to show that µ(N) = µ(∣u∣ =∞) = 0. SetN = ⋂

n∈N∣u∣ ≥ n.

where we have that

µ(∣u∣ ≥ 1) = ∫∣u∣≥1 1dµ ≤ ∫∣u∣≥1 ∣u∣dµ ≤ ∫ ∣u∣dµ <∞,

since u ∈ L1R(µ) Ô⇒ ∣u∣ ∈ L

1R(µ). Therefore, we may use continuity of measures

from above and Markov’s inequality to get

µ(N) = µ( limn→∞

n

⋂i=1u ≥ i)

= µ( limn→∞u ≥ n)

= limn→∞

µ(u ≥ n)

≤ limn→∞

1

n∫ ∣u∣dµ´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶<∞

<∞.

61

As a consequence of this corollary, we can now restrict our attention to L1(µ)without loss of generality. We do this since L1 is a much nicer set—it is a vectorspace and we need not take precautions when adding functions. The previousresults are easily adapted to this new space.

Finally, we give a result that is helpful for verifying that two functions areequal µ-a.e.

Corollary 9.9. Let B ⊆ A be a σ-algebra .

1. If u,w ∈ L1(B) (so u,w ∈M(B)) and ∫B udµ = ∫B wdµ for all B ∈ B, thenu = w µ-a.e.

2. If u,w ∈M+(B) and ∫B udµ = ∫B wdµ for all B ∈ B and µ ∣B is σ-finitethen u = w µ-a.e.

Proof. TODO

Chapter 10

Convergence theorems

The purpose of this chapter is to generalise the results that were discussedin chapter 7, namely ??, ?? and the reverse Fatou lemma. By generalisationwe mean that we will remove the restrictions about positive functions and/orincreasing sequences.

10.1 Convergence theoremsFirst, we extend ?? to negative functions.

Theorem 10.1 (Monotone convergence). Let (X,A, µ) be a measure space and(un)n∈N ⊂ L1(µ) be an increasing sequence of measurable functions such that

un ≤ un+1, ∀n ∈ N and define u = supn∈N

un = limn→∞

un ∈MR(A).

Then,

u ∈ L1(µ) ⇐⇒ supn∈N∫ undµ <∞, (10.1)

and in that case


undµ = ∫ supn∈N

un = supn∈N∫ undµ = lim

n→∞∫ undµ. (10.2)

Here we’re writting both the lim and the sup versions of the equalities to beextra clear at the expense of being too verbose. This is not the case in [?, p-88].

Proof. We want to use Beppo-Lévi but the functions may be negative, so wedefine a new sequence wn = un − u1 ≥ 0 since u1 ≤ un, ∀n ∈ N. Clearly, wn ∈L1(µ) ⊂M(A) and 0 ≤ w1 ≤ w2 ≤ . . . .

Firstly, by ??,

0 ≤ ∫ supn∈N

wndµ = supn∈N∫ wndµ.

63

64 CHAPTER 10. CONVERGENCE THEOREMS

Equivalently,

0 ≤ ∫ sup(un − u1)dµ

= supn∈N(∫ (un − u1)dµ)

= supn∈N(∫ undµ − ∫ u1dµ)

= supn∈N∫ undµ − ∫ u1dµ.

Note how we wait until sup is outside the integral to apply linearity, since wedon’t know if sup(un − u1) ∈ L1(µ) yet.

Assume u = supn un ∈ L1(µ). Then supn(un − u1) = supn un − u1 ∈ L1(µ),and the above gives

∫ supn

undµ − ∫ u1dµ = supn∫ undµ − ∫ u1dµ.

Since u1 ∈ L1(µ), its integral is finite and we can subtract it from both sides toget

supn∫ undµ = ∫ sup

nundµ <∞.

Conversely, assume supn ∫ undµ <∞. Then,

∫ supn(un − u1)dµ = sup

n∫ undµ − ∫ u1dµ <∞,

which implies that sup(un − u1) = supn un − u1 ∈ L1(µ). Since u1 ∈ L1(µ), wesee that supn un ∈ L1(µ) as required. Also, if supn un ∈ L1(µ) the above shows??.

Of course, we can state the same for decreasing sequences and infima, astaking un = −vn for some decreasing sequence (vn) is enough to fulfill the as-sumptions of the ??. Anyway, we can state the result.

Corollary 10.2. Let (un)n∈N ⊂ L1(µ) be a sequence of decreasing integrablefunctions such that

un ≥ un+1, ∀n ∈ N and define u = infn∈N

un = limn→∞

un ∈MR(A).

Then,

u ∈ L1(µ) ⇐⇒ infn∈N∫ undµ > −∞, (10.3)

in which case


undµ = ∫ infn∈N

undµ = infn∈N∫ undµ = lim

n→∞∫ undµ. (10.4)

Proof. Set vn = −un and apply the proof of ??.

10.2. APPLICATIONS TO PARAMETER DEPENDENT-INTEGRALS 65

Now we move on to a very important result. This time, not only we dropnon-negativity, but also, monotonicity. Of course, there is a price to pay forthis.

Theorem 10.3 (Lebesgue Dominated Convergence). Let (X,A, µ) be a mea-sure space and (un)n∈N ⊂ L1(µ) be a sequence of functions such that

1. there exists 0 ≤ w ∈ L1(µ) such that ∣un∣ ≤ w µ-a.e. for all n ∈ N, and

2. u = limn→∞ un exists in R µ-a.e. .

Then u ∈ L1(µ) and we have

limn→∞∫ ∣un − u∣dµ = 0, (10.5)

limn→∞∫ undµ = ∫ lim

n→∞undµ = ∫ udµ. (10.6)

Proof. TODO

Remark 10.4. We have proved the theorem for the case in which ∣un∣ ≤ weverywhere and limn→∞ un ∈ R everywhere, but they may both be relaxed toµ-a.e. .

10.2 Applications to parameter dependent-integrals

Theorem 10.5 (Continuity lemma). Let (X,A, µ) be a measure space and∅ ≠ (a, b) ⊂ R a non-degenerate open interval and u ∶ (a, b) × X → R be afunction satisfying

1. x↦ u(t, x) is in L1(µ) for every fixed t ∈ (a, b);

2. t↦ u(t, x) is continuous for every fixed x ∈X; and

3. ∣u(t, x)∣ ≤ w(x) for all (t, x) ∈ (a, b) ×X and some w ∈ L1+(µ).

Then the function v ∶ (a, b)→ R given by

t↦ v(t) ∶= ∫ u(t, x)µ(dx) (10.7)

is continuous.

66 CHAPTER 10. CONVERGENCE THEOREMS

Theorem 10.6 (Differentiability lemma). Let (X,A, µ) be a measure space,∅ ≠ (a, b) ⊂ R a non-degenerate open interval and u ∶ (a, b) × X → R be afunction satisfying

1. x↦ u(t, x) is in L1(µ) for every fixed t ∈ (a, b);

2. t↦ u(t, x) is differentiable for every fixed x ∈X; and

3. ∣∂tu(t, x)∣ ≤ w(x) for all (t, x) ∈ (a, b) ×X and some w ∈ L1+(µ).

Then the function v ∶ (a, b)→ R given by

t↦ v(t) ∶= ∫ u(t, x)µ(dx) (10.8)

is differentiable and its derivative is

∂tv(t) = ∫ ∂tu(t, x)µ(dx). (10.9)

10.3 Riemann integral vs. Lebesgue integral

Theorem 10.7. Let u ∶ [a, b] → R be a measurable and Riemann-integrablefunction. Then u ∈ L1(λ) and the Riemann integral and Lebesgue integralagree.

∫b

audx = ∫[a,b] udλ (10.10)

Remark 10.8. Note that the converse need not be true. For instance the Dirich-let function 1Q (see [?]) is Lebesgue integrable but not Riemann-integrable.

Proof. TODO

Theorem 10.9. Let u ∶ [a, b]→ R be a bounded and Riemann-integrable func-tion, then

x ∈X ∣ u is not continuous in x ⊆ N ∈ Nµ. (10.11)

Note that the additional subset N is not required if u is measurable.

10.3.1 Improper Riemann integralsCorollary 10.10. Let u ∶ [a, b] → R be a measurable function that is Riemannintegrable on [0,N] for all N ≥ 1. Then u ∈ L1([0,∞]) if, and only if,

limN→∞∫

N

0∣u∣dx <∞. (10.12)

In this case,

∫∞

0udx = ∫[0,∞] udλ.

Chapter 11

The function spaces Lp

11.1 A seminorm for Lp

We turn our attention to functions for which the p-th power of their absolutevalue is integrable. Throughout this chapter we will assume that (X,A, µ) issome measure space.

Definition 11.1 (Lp-space). Let 1 ≤ p ∈ R. We define

Lp(µ) = u ∶X → R ∣ u ∈M(A) ∧ ∫ ∣u∣pdµ <∞, p ∈ [1,∞) (11.1)

L∞(µ) = u ∶X → R ∣ u ∈M(A) ∧ ∃c > 0, µ(∣u∣ ≥ c) = 0, p =∞ (11.2)

As usual we might also refer to these sets by Lp = Lp(A) = Lp(X) if thechoice of measure is clear or we want to stress the underlying σ-algebras ordomains.

Definition 11.2 (p-seminorm). Let u ∶ X → R be a measurable function. Wedefine

∥u∥p = (∫ ∣u∣pdµ)

1p

, p ∈ [1,∞), (11.3)

∥u∥∞ = infc > 0 ∣ µ∣u∣ ≥ c = 0, p =∞. (11.4)

Remark 11.3. Of course u ∈ Lp(µ) ⇐⇒ u ∈M(A) ∧ ∥u∥p <∞.

It is not coincidental that we use the notation ∥⋅∥p which clearly resemblesa norm. Indeed, ∣⋅∣p is a semi-norm, i.e. a function with the properties of anorm except for the property that identifies ∥v∥p = 0 ⇐⇒ v = 0. What happenshere is that ∥u∥p = 0 ⇐⇒ u = 0 µ-a.e. and thus we don’t have a one-to-onecorrespondence between the 0 value of the norm and the vector space’s zeroelement (we will eventually fix this though).

Lemma 11.4. ∥⋅∥p is indeed a seminorm [?], i.e. it satisfies the following

1. (positive homogenous) ∥αv∥p = ∣α∣ ∥v∥p , ∀α ∈ R

2. (triangle inequality) ∀u, v ∈M(A), ∥u + v∥p ≤ ∥u∥p + ∥v∥p.

67

68 CHAPTER 11. THE FUNCTION SPACES LP

Proof.Let α ∈ R, u ∈M(A). We have

(∫ ∣αu∣pdµ)

1p

= (∫ αp ∣u∣p dµ)1p

= (∣α∣p ∫ ∣u∣pdµ)

1p

= ∣α∣ (∫ ∣u∣pdµ)

1p

It turns out that proving the triangle inequality is not so easy. We will dedicatethe rest of this section to it.

Definition 11.5 (Conjugate numbers). Let p, q ∈ R. We say that p and q areconjugate numbers if

1

p+ 1

q= 1, (11.5)

hence q = pp−1 .

We will not bother giving the previous definition too much formality, forexample, if p = 1 we can set 1

∞ = 0 and hence p = 1 and q = ∞ are conjugatenumbers. We will see that this spooky extension does not mess with the defi-nition of ∥⋅∥p (cf. ??). Also, note that 2 is the only number which has itself asits conjugate number.

Lemma 11.6 (Young’s inequality). Let p, q ∈ (1,∞) be conjugate numbers.Then

AB ≤ Ap

p+ Bq

q(11.6)

holds for all A,B ≥ 0. Equality occurs if, and only if, B = Ap−1.

Proof. TODO

Theorem 11.7 (Hölder’s inequality). Let u ∈ Lp and v ∈ Lq where p, q ∈ [1,∞]are conjugate numbers. Then uv ∈ L1(µ) and

∣∫ uvdµ∣ ≤ ∫ ∣uv∣dµ ≤ ∥u∥p ∥v∥q . (11.7)

Proof. For the first inequality, see ??.TODO

Corollary 11.8 (Cauchy-Schwarz inequality). Hölder’s inequality with p = q = 2is called the Cauchy-Schwarz inequality:

∫ ∣uv∣dµ ≤ ∥u∥2 ∥v∥2 . (11.8)

Theorem 11.9 (Minkowski’s inequality). Let u, v ∈ Lp, p ∈ [1,∞]. Then thesum u + v ∈ Lp and

∥u + v∥p ≤ ∥u∥p + ∥v∥p . (11.9)

11.1. A SEMINORM FOR LP 69

Note how we stated that u+ v ∈ Lp. We already had this for L1 (see ??) butnow we prove it for the more general case.

Proof. content...

We can further generalise this theorem to sequences of non-negative functionsusing Beppo-Lévi.

Corollary 11.10. Let (un)n∈N ⊂ Lp be any sequence of non-negative functionsin Lp with p ∈ [1,∞). Then

∥∞∑n=1

un∥p

≤∞∑n=1∥un∥p . (11.10)

Proof. By repeated applications of Minkowski’s inequality we have, for any N ∈N,

∥N

∑n=1

un∥p

≤N

∑n=1∥un∥p ≤

∞∑n=1∥un∥p .

Since the right hand side is independent of the choice of N , we can take thesupremum of the left side without breaking the inequality:

supN∈N∥

N

∑n=1

un∥p

≤ supN∈N

N

∑n=1∥un∥p ≤

∞∑n=1∥un∥p .

We now wish to see that this holds for N = ∞, for which we use Beppo-Lévi.Observe that (∑N

n=1 un)p is a sequence of increasing functions1. Thus,

supN∈N∥

N

∑n=1

un∥p

p

= supN∈N∫ (

N

∑n=1

un)p

dµ = ∫ (supN∈N

N

∑n=1

un))p

dµ

??= ∫ (∞∑n=1

un)p

= ∥∞∑n=1

un∥p

p

.

Taking the p-th root to get the norms and the proof follows.

Remark 11.11. We need the functions to be non-negative so that the functionwe are integrating, namely

∣N

∑n=1

un∣p

is an increasing sequence. At this point we don’t really care if it is non-negative or whatever, i.e. using Monotone Convergence instead of Beppo-Léviwould not help as the absolute value in the p-norm reduces the question to thenon-negative case.

As a counterexample consider a sequence of functions that oscilates around 0:−1,0,1,0,−1, . . . . The partial absolute value of the partial sums is 1,1,0,0,1, . . . ,which does not really converge. Another way of solving this problem would be toask for a sequence of arbitrary but increasing functions and applying MonotoneConvergence, but I feel its more useful to ask for any sequence of non-negativefunctions as we get the increasing part from taking the partial sums.

1p ≥ 1 so it does not mess with us here


11.2 A norm for Lp

Remark 11.12. Note that from ?? and ?? we can deduce

u, v ∈ Lp Ô⇒ αu + βv ∈ Lp, ∀α,β ∈ R. (11.11)

which shows that Lp is an R-vector space.

In addition, we already saw that ∣⋅∣p is a seminorm for Lp (cf. ??). The thingthat’s missing for ∣⋅∣p to be a norm is that ∣u∣p = 0 Ô⇒ u(x) = 0 for every x ∈X(as opposed to for almost every x). There is an easy, but rather technical wayto fix this.

• We introduce the equivalence relation for u, v ∈ Lp:

u ∼ v ⇐⇒ u ≠ v ∈ Nµ.

It is not hard to verify that ∼ is indeed an equivalence relation (see [?] fora definition).

• The space made up of the equivalence classes

[u]p = v ∈ Lp ∣ u ∼ v

is called the quotient space and denoted by Lp = Lp/ ∼.

• It is also not hard to see that Lp is a vector space by proving

[αu + βv]p = α[u]p + β[v]p.

• Moreover, it admits the norm2

∥[u]p∥p ∶= inf∥w∥p ∣ w ∈ Lp ∧w ∼ u,

and, fortunately, we have that ∥[u]p∥p = ∥u∥p so we will start abusingnotation identify [u] with u and ditch the distinction Lp vs Lp.

All the results in this chapter so far are still valid if u is µ-a.e. real valuedso there is no need to distinguish between the cases Lp

Rand Lp ∶= Lp

R.

11.3 Convergence and completenessDefinition 11.13 (Lp-convergence). A sequence of functions (un)n∈N ⊂ Lp inLp is said to be convergent in the space Lp with limit Lp − limn→∞un = u if, andonly if,

limn→∞

∥un − u∥p = 0. (11.12)

Remember, however that Lp-limits are only almost everywhere unique. Ifw1,w2 are both Lp-limits of the same sequence (un)n∈N we have

∥w1 −w2∥p??≤ lim

n→∞(∥w1 − un∥ + ∥un −w2∥p) = 0.

2To be honest, I still have not figured out why we require this infimum as the norms areall the same inside of an equivalence class...

11.3. CONVERGENCE AND COMPLETENESS 71

Remark 11.14. Pointwise convergence of a sequence (un)n∈N ⊂ Lp does notquarantee Lp-convergence.

Example needed...

We can however, give a weaker result by using Lebesgue’s dominated con-vergence theorem.

Lemma 11.15. Let (un)n∈N ⊂ Lp be a sequence of functions in Lp such thatlimn→∞ un(x) = u(x) for (almost) every x and ∣un∣ ≤ w for some function w ∈ Lp.Then, u ∈ Lp and Lp − limun = u.

Proof. We want to show that Lp − limun = u ⇐⇒ limn→∞ ∥un − u∥p = 0 ⇐⇒limn→∞ ∥un − u∥pp = 0. To do that we show that the sequence ∣un − u∣p satisfiesthe hypotheses of ??. We have

∣un − u∣p ≤ (∣un∣ + ∣u∣)p ≤ (2w)p = 2pwp

and therefore

limn→∞∫ ∣u − un∣p dµ = ∫ lim

n→∞∣un − u∣dµ = ∫ 0dµ = 0

which in turn implies that Lp − limun = u.

Definition 11.16 (Lp-Cauchy sequence). We call (un)n∈N an Lp-Cauchy se-quence if

∀ε > 0, ∃Nε ∶ ∀n1, n2 ≥ Nε, ∥un1 − un2∥p < ε. (11.13)

Remark 11.17. Any Lp-convergent sequence is an Lp-Cauchy sequence.

Proof. By Lp-convergence we have that for any ε/2 > 0, there exists an N suchthat for any n > N .

∥un − u∥p <ε

2.

We can rewrite that as

∥un1 − un2∥p ≤ ∥un1 − u∥p + ∥un2 − u∥p =ε

2+ ε

2= ε.

The converse is also true but much harder to prove.

Theorem 11.18 (Riesz-Fischer). The spaces Lp, p ∈ [1,∞] are complete, i.e.every Lp-Cauchy sequence (un)n∈N ⊂ Lp converges to some limit u ∈ Lp.

Proof. TODO

Theorem 11.19 (Riesz). Let (un)n∈N ⊂ Lp, p ∈ [1,∞) be a sequence such thatlimn→∞ un(x) = u(x) for almost every x ∈X and some u ∈ Lp. Then

limn→∞

∥un − u∥p = 0 ⇐⇒ limn→∞

∥un∥p = ∥u∥p . (11.14)


Proof. TODO

Remark 11.20. This theorem does not hold for p =∞.

Chapter 12

Product measures andFubini’s theorem

Remember how long it took us to be able to define the Lebesgue measure on(Rn,B(Rn)). It took three chapters. And we were not nearly done. It is truethat we proved that the n-dimensional Lebesgue measure existed and was welldefined on all B(Rn). However we still lack a lot of tools to be able to workwith it. In this chapter we will prove that for a set A ×B ∈ Rn ×Rm, the n ⋅m-dimensional Lebesgue measure is well behaved, i.e. λn⋅m(A×B) = λn(A)⋅λn(B).

Throughout this chapter we will assume that (X,A, µ) and (Y,B, ν) are twoσ-finite measure spaces.

12.1 Product σ-algebras and product measures

Lemma 12.1. Let A and B be two semi-rings, then A × B is a semi-ring.

You might want to consult what a semi-ring is in ??.

Proof. TODO

Definition 12.2 (Product σ-algebra). Let (X,A), (Y,B) be measurable spaces.Then the σ-algebra A ⊗ B = σ(A × B) is called a product σ-algebra and (X ×Y,A⊗ B) is called the product of measurable spaces.

Lemma 12.3. If A = σ(F) and B = σ(G) and if F and G contain exhaustingsequences (Fn)n∈N ⊂ F , Fn ↑X and (Gn)n∈N ⊂ G, Gn ↑ Y , then

σ(F × G) = σ(A × B) ∶= A⊗ B. (12.1)

73

74 CHAPTER 12. PRODUCT MEASURES AND FUBINI’S THEOREM

Theorem 12.4 (Uniqueness of product measures). Let (X,A, µ) and (Y,B, ν)be two measure spaces and assume that A = σ(F) and B = σ(G). If

1. F ,F are ∩-stable, and

2. F ,G contain exhausting sequences Fk ↑ X and Gn ↑ Y with µ(Fk) < ∞and ν(Gn) <∞ for all k,n ∈ N,

then there is at most one measure ρ on (X × Y,A⊗ B) satisfying

ρ(F ×G) = µ(F )ν(G), ∀F ∈ F , G ∈ G.

Theorem 12.5 (Existence of product measures). Let (X,A, µ) and (Y,B, ν)be σ-finite measure spaces. The set function

ρ ∶ A × B → [0,∞], ρ(A ×B) ∶= µ(A)ν(B)

extends uniquely to a σ-finite measure on (X × Y,A⊗ B) such that

ρ(E) ∶= ∫X×Y

1E(x, y)dρ(x) (12.2)

=∬ 1E(x, y)dµ(x)dν(y) =∬ 1E(x, y)dν(y)dµ(x) (12.3)

holds for all E ∈ A⊗ B. In particular, the functions

x↦ 1E(x, y), (12.4)y ↦ 1E(x, y), (12.5)

x↦ ∫ 1E(x, y)dν(y), (12.6)

y ↦ ∫ 1E(x, y)dµ(x) (12.7)

are A, resp. B-measurable for every fixed y ∈ Y , resp. x ∈X.

In the previous statement, when we write a double integral ∬ . . . dµdν weare implicitly writing brackets, i.e. ∬ . . . dµdν = ∫ (∫ dµ)dν.

Recall that we can only integrate measurable functions (??). That is thereason for the second part of the theorem, we need to make sure that whenwe use iterated integrals, whatever we’re integrating is measurable. Thus, thestructure of the proof will be to first verify that ?? – ?? are indeed measurableand then to prove the equality ??.

Proof. Uniqueness follows from ??.Existence. By σ-finiteness we have that there exist (An)n∈N ⊂ A and (Bn)n∈N ⊂

B with µ(An) <∞ and ν(Bn) <∞ for all n. Let En = An ×Bn ∈ A × B ⊆ A⊗ Band

Dn = D ∈ A⊗ B ∣ ?? − ?? hold for D ∩En. (12.8)

Here our E = D ∩En. We will show that A × B ⊂ Dn and that Dn is a Dynkinsystem (cf. ??, hence Dn = δ(Dn)). This, together with the facts that Dn ⊂

12.1. PRODUCT σ-ALGEBRAS AND PRODUCT MEASURES 75

A⊗ B and that A × B is ∩-stablewill imply that δ(A × B) = σ(A × B) = A⊗ B ⊂δ(Dn) =Dn ⊂ A⊗ B Ô⇒ Dn = A⊗ B.

1. A×B ⊂Dn. Let A×B ∈ A×B. Then (A×B)∩En = (A×B)∩(An ×Bn) =(A ∩An) × (B ×Bn). Therefore

1(A×B)∩En(x, y) = 1A∩An(x)1B∩Bn(y)

and thus

∬ 1(A×B)∩En(x, y)dµ(x)dν(y) =∬ 1A∩An(x)1B∩Bn(y)dµ(x)dν(y)

= µ(A ∩An)∫ 1B∩Bn(y)dν(y)

= µ(A ∩An)ν(B ∩Bn)

= ν(B ∩Bn)∫ 1A∩An(x)dµ(x)

=∬ 1A∩An(x)1B∩Bny(y)dν(y)dµ(x)

=∬ 1(A×B)∩En(x, y)dν(y)dµ(x).

2. Dn is a Dynkin system.

(a) X × Y ∈ A × B ⊂Dn by the previous claim.

(b) If D ∈Dn then 1Dc∩En = 1En − 1En∩D and

∬ 1Dc∩En(x, y)dµ(x)dν(y)

=∬ (1En(x, y) − 1En∩D(x, y))dµ(x)dν(y)

=∫ (∫ 1En(x, y)dµ(x) − ∫ 1En∩D(x, y)dµ(x))dν(y)

=∬ ∫ 1En(x, y)dµ(x)dν(y) −∬ 1En∩D(x, y)dµ(x)dν(y)

=∬ ∫ 1En(x, y)dν(y)dµ(x) −∬ 1En∩D(x, y)dν(y)dµ(x)

= ⋅ ⋅ ⋅ =∬ 1Dc∩En(x, y)dν(y)dµ(x),

where . . . means do the same calculations backwards. We concludethat Dc ∈Dn.

(c) Finally let (Dk)k∈N ⊂ Dn be a mutually disjoint collection of sets inDn. We will show that D = ⊍∞k=1Dk ∈ Dn. Therefore 1D = ∑∞k=1 1Dk


and 1D∩En = ∑∞k=1 1Dk∩En . Therefore,

∬ 1D∩En(x, y)dµ(x)dν(y)

=∬∞∑k=1

1Dk∩En(x, y)dµ(x)dν(y)

??= ∫ (∞∑k=1

1Dk∩En(x, y)dµ(x))dν(y)

??=∞∑k=1∬ 1Dk∩En(x, y)dµ(x)dν(y)

=∞∑k=1∬ 1Dk∩En(x, y)dν(y)dµ(x)

= ⋅ ⋅ ⋅ =∬ 1D∩En(x, y)dν(y)dµ(x).

And thus, D = ⊍∞k=1Dk ∈Dn.

We conclude that Dn is indeed a Dynkin system.

Finally, to finish the proof we let E ∈ A⊗B and we want to verify that ?? -?? hold for 1E . Recall that En = An ×Bn ↑X × Y so

1En ↑ 1X×Y and 1En∩E ↑ 1(X×Y )∩E = 1E .

We may write 1E = supn∈N supEn∩E . The supremum of measurable functionsis measurable so ?? and ?? hold for 1E . By Beppo-Lévi (??) we get that ??and ?? also hold for 1E . Finally, for ?? we repeat what we did for the thirdproperty of Dynkin systems:

∬ 1E(x, y)dµ(x)dν(y)

=∬ supn∈N

1En∩E(x, y)dµ(x)dν(y)

??= ∫ (supn∈N∫ 1En∩E(x, y)dµ(x))dν(y)

??= supn∈N∬ 1En∩E(x, y)dµ(x)dν(y)

= supn∈N∬ 1En∩E(x, y)dν(y)dµ(x)

= ⋅ ⋅ ⋅ =∬ 1E(x, y)dν(y)dµ(x).

Definition 12.6 (Product measure). The unique measure ρ constructed in ?? iscalled the product of the measures µ and ν, denoted by µ×ν. (X×Y,A⊗B, µ×ν)is called the product measure space.

Remark 12.7. The product measure ρ ∶ A×B → [0,∞], ρ(A×B) = µ×ν(A×B) =µ(A) ⋅ µ(B) is indeed very well defined. The product at the end does not failbecause we impose that both measurable spaces are σ-finite.

12.2. INTEGRALS WITH RESPECT TO PRODUCT MEASURES 77

Consider the example given in exercise 14.2 ([?, p. 149]). Let (X,A, µ), (Y,B, ν)be two σ-finite measure spaces and let A ∈ A, N ∈ B with ν(N) = 0. Thenµ × ν(A ×N) = 0.

Proof. We cannot just state µ×ν(A×N) = µ(A)⋅ν(N) = µ(A)⋅0 = 0 since we mayhave A =X and thus we would end up with∞⋅0 which is, in principle, undefined.But we have σ-finiteness (cf. ??) so there exist (Ak)k∈N ⊂ A, (Bk)k∈N ⊂ B withAk ↑X, Bk ↑ Y and µ(Ak), ν(Bk) <∞, ∀k ∈ N. Thus, we might use continuityfrom below to state

µ × ν(A ×N) = µ × ν((A ×N) ∩ (X × Y ))= µ × ν((A ×N) ∩ lim

k→∞(Ak ×Bk))

??= limk→∞

µ × ν((A ×N) ∩ (Ak ×Bk))

= limk→∞

µ × ν((A ∩Ak) × (N ∩Bk))

= limk→∞

µ(A ∩Ak)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

<∞

ν(N ∩Bk)´¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¹¶

=0

= limk→∞

µ(A ∩Ak) ⋅ 0 = 0

Now we show an alternative construction of the n-dimensional Lebesguemeasure to the one we did after ??.

Corollary 12.8. If n > d ≥ 1 then

(Rn,B(Rn), λn) = (Rd ×Rn−d,B(Rd)⊗ B(Rn−d), λd × λn−d).

12.2 Integrals with respect to product measuresIn this section we will basically repeat chapters ?? and ?? at lightning speed.

The following theorem is basically ?? but for non-negative measurable func-tions. It allows us to calculate the integral with respect to the product measurebuilding upon simple integrals.

Theorem 12.9 (Tonelli). Let (X,A, µ) and (Y,B, ν) be two σ-finite measurespaces. Let u ∶X × Y → [0,∞] be an A⊗ B measurable function. Then

1. x↦ u(x, y) and y ↦ u(x, y) are measurable for a fixed y, resp. x.

2. x↦ ∫Y u(x, y)dν(y), y ↦ ∫X u(x, y)dµ(x) are measurable

3.

∫X×Y

ud(µ × v) (12.9)

=∫Y∫Xu(x, y)dµ(x)dν(y) (12.10)

=∫X∫Yu(x, y)dν(y)dµ(x) ∈ [0,∞]. (12.11)


Proof. The proof is standard. Approximate u by a sequence of simple functions

fn(x, y) =N(n)∑j=0

αnj 1An

j(x, y).

By linearity of the integral and ?? we have that

x↦ fn(x, y), y ↦ fn(x, y), x↦ ∫Yfn(x, y)dµ(y), y ↦ ∫

Xfn(x, y)dµ(x)

are all measurable. The standard Beppo-Lévi argument shows that 1. and 2.are true. For 3. we do the same but applying Beppo-Lévi multiple times.

What Tonelli’s last statement implies is that even though the integrals maybe infinite, they still make sense, i.e. they are not undefined.

Next, we move on to another theorem which is basically the same as Tonelli’sbut which guarantees integrability and can be used on non non-negative func-tions.

Theorem 12.10 (Fubini). Let (X,A, µ) and (Y,B, ν) be two σ-finite measurespaces. Let u ∶ X × Y → R be an A⊗ B measurable function. If at least one ofthe three integrals

∫X×Y

ud(µ × v), ∫Y∫Xu(x, y)dµ(x)dν(y), ∫

X∫Yu(x, y)dν(y)dµ(x)

is finite then all three integrals are finite, u ∈ L1(µ × ν), and

1. x↦ u(x, y) is in L1(µ) for ν-a.e. y ∈ Y ,

2. y ↦ u(x, y) is in L1(ν) for µ-a.e. x ∈X,

3. y ↦ ∫X u(x, y)dµ(x) is in L1(ν),

4. x↦ ∫Y u(x, y)dν(y) is in L1(µ), and

5.

∫X×Y

ud(µ × v) = ∫Y∫Xu(x, y)dµ(x)dν(y) = ∫

X∫Yu(x, y)dν(y)dµ(x).

12.3 Application: distribution functionsLet (X,A, µ) be a σ-finite measure space and u ∈M+(A). Define F ∶ R→ [0,∞]by

F (t) = µ(u ≥ t) = µ(u−1([t,∞])). (12.12)

Lemma 12.11. If u ∈ L1 then F is left-continuous.

Proof. We want to show that if (tn)n∈N ⊂ (0,∞) is a sequence increasing towardst ∈ (0,∞) then limn→∞ F (tn) = F (t). Observe that F (tn) = u ≥ tn ↓ u ≥ t =F (t). Since u ∈ L1, using Markov’s inequality we have that

µ(u ≥ t1)??≤ 1

t1∫ ∣u∣dµ <∞.

12.3. APPLICATION: DISTRIBUTION FUNCTIONS 79

So we are in a position to apply continuity of measures. Then,

limn→∞

F (tn) = limn→∞

µ(u ≥ tn) ??= µ( limn→∞u ≥ tn) = µ(u ≥ t) = F (t).

So F is left-continuous.

The next theorem is very handy because it allows us to transform any µ-integral into a Lebesgue integral.

Theorem 12.12. Let (X,A, µ) be a σ-finite measure space and u ∶X → [0,∞])an A-measurable function. Then,

∫ udµ = ∫(0,∞) µ(u ≥ t)dλ(t) ∈ [0,∞]. (12.13)

I had a lot of trouble understanding this proof because it all seemed likehappy ideas to me. Below is a very step by step approach, for a more directone, but equal in essence, see [?, p. 146] or the handwritten lecture notes forchapter 14.

Proof. For this proof, we will use Tonelli’s theorem (??). The main idea is torewrite the integral in terms of another function but first we need to come upwith that other function and then to check that it fulfils the requirements ofTonelli’s theorem.

1. First observe that u(x) = u(x) − 0 = λ1([0, u(x)]). This gives us the ideato write

∫ u(x)dµ(x) = ∫ λ1([0, u(x)])dµ(x) =∬ 1[0,u(x)](t)dλ1(t)dµ(x).

2. Not the previous integral looks more like somewhere where we would applyTonelli’s theorem. We can continue transforming the integrand to get

1[0,u(x)](t) = 1t∈[0,∞)∣t≤u(x)(t) = 1x∈X ∣u(x)≥t(x).

3. In any case, the previous indicator function depends both on t ∈ [0,∞)and x ∈ X. For it to be a candidate for Tonelli’s theorem we need toverify that it is measurable. In particular, the integrand has to be A ⊗B([0,∞))/B(R2)-measurable, i.e. a A⊗ B([0,∞))-measurable function.

4. Since the integrand is an indicator function, it is sufficient to prove thatthe interval in question is A⊗B([0,∞))-measurable. For that, we rewritethe integrand as 1E(x, t) where

E ∶= (x, t) ∣ u(x) ≥ t,

and hence

1E(x, t) = 1t∈[0,∞)∣t≤u(x)(t) = 1x∈X ∣u(x)≥t(x).


To prove that E is A ⊗ B([0,∞))-measurable we define the function U ∶X × [0,∞] → R2 by U(x, t) ∶= (u(x), t). We can see that U is measurablefrom

U−1(A × I) = u−1(A) × I ∈ A × B([0,∞)),

where A ∈ A and I ∈ B([0∞)). So in particular, E = U−1(u(x) ≥ t) ∈A⊗ B([0,∞)), i.e. E is measurable and thus 1E(x, t) also is.

5. Finally, we are ready to apply Tonelli’s theorem

∫ u(x)dµ(x) = ∫ λ1([0, u(x)])dµ(x)

=∬ 1[0,u(x)](t)dλ1(t)dµ(x)

=∬ 1E(x, t)dλ1(t)dµ(x)??= ∬ 1E(x, t)dµ(x)dλ1(t)

=∬ 1x∈X ∣u(x)≥t(x)dµ(x)dλ1(t)

= ∫ µ(x ∈X ∣ u(x) ≥ t)dλ1(t)

= ∫ µ(u ≥ t)dλ1(t).

Chapter 13

Exercise sets

13.1 Exercise set 1Due September 20th, 2019.

Exercise 13.1.1. Let X be a nonempty set and A = A1,A2, . . . a collection ofdisjoint subsets of X such that X = ⋃∞n=1An. Show that each element A ∈ σ(A)is a union of at most a countable subcollection of elements of A. (3 pts)

The technique used here is similar to the good set principle (??), only thatit is not necessary to show that the good set is inside the σ-algebra .

Proof. The idea is to prove that σ(A) only contains countable unions of sets ofA.

Let us define

B = ⋃i∈I

Ai ∣ Ai ∈ A ∧ I ⊂ N .

We shall prove that B is a σ-algebra on A and thus σ(A) ⊆ B, i.e. σ(A) ismade up of unions of at most a countable subcollection of A.

1. Firstly, ∅ ∈ B since ∅ = ⋃i∈I Ai by choosing I = ∅ ⊂ N.

2. Secondly, for any set B = ⋃i∈I Ai ∈ B we have that

Bc = (⋃i∈I

Ai)c

=⋂i∈I

Aci =⋂

i∈I⋃j≠i

Aj = ⋃j/∈I

Aj = ⋃j∈Ic

Aj ∈ B,

since Ic = N ∖ I ⊂ N.

3. Finally, for any countable collection (Bn)n∈N ∈ B we need to show that theunion is also in B.

⋃n∈N

Bn = ⋃n∈N⋃i∈In

Ai = ⋃j∈J

Aj ∈ B,

since J = ⋃n∈N In ⊂ N.

81

82 CHAPTER 13. EXERCISE SETS

Since we have shown that B is a σ-algebra on A and thus σ(A) ⊂ B, we concludethat the elements of σ(A) must be all unions of at most countable subcollectionsof elements in A, which are the only elements of B.

Exercise 13.1.2. Let (X,D, µ) be a measure space, and let Dµ be the com-pletion of the σ-algebra D with respect to the measure µ (see exercise 4.15).We denote by µ the extension of the measure µ to the σ-algebra Dµ. Suposef ∶ X → X is a function such that f−1(B) ∈ D and µ(inf f(B)) = µ(B) for eachB ∈ D. Show that inf f(B) ∈ Dµ and µ(f−1(B)) = µ(B) for all B ∈ Dµ. (3 pts)

Proof. First we show that f−1(B) ∈ Dµ, for all B ∈ Dµ. Recall from the def-inition of the completion Dµ of the σ-algebra D that any set B ∈ Dµ can bewritten as B = B ∩M for some subset M of a µ-measurable null set N in D.Therefore

f−1(B) = f−1(B ∪M) = f−1(B) ∪ f−1(M)

Because N ⊃M we also have that f−1(N) ⊂ f−1(M) and µ(f−1(N)) = µ(N) = 0by the definition of f . This means that f−1(M) is also a subset of a µ-measurablenull set in D and since f−1(B) ∈ D by definition of f we have

f−1(B) = f−1(B)´¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¶∈D

∪ f−1(M)´¹¹¹¹¹¹¹¹¹¹¹¹¹¸¹¹¹¹¹¹¹¹¹¹¹¹¶

⊂f−1(N), µ(f−1(N))=0

∈ Dµ

Now we need to verify that µ(f−1(B)) = µ(B) for all B ∈ Dµ. Recall thatthe extension µ is well-defined in Dµ with µ(B) ∶= µ(B) for any B = B∪M ∈ Dµ.Hence,

µ(f−1(B)) = µ(f−1(B ∪M))= µ(f−1(B) ∪ f−1(M))= µ(f−1(B))= µ(B) =∶ µ(B)

Exercise 13.1.3. Let (X,A) be a measurable space and µ ∶ A → [0,∞) afunction satisfying

1. µ is finitely additive

2. µ is σ-subadditive

Show that µ is σ-additive. (4 pts)

Proof. The plan for the proof is to sandwich µ(⋃n∈NAn) between two sums thatare the same when taking the limit.

First of all, because of σ-subadditivity we have that, for any countable col-lection (An)n∈N ⊂ A,

µ(⋃n∈N

An) ≤ ∑n∈N

µ(An), (13.1)

13.2. EXERCISE SET 2 83

which in particular holds for pairwise disjoint unions, which we will assume fromhere on. We can rewrite the union as

⊍n∈N

An =N

⊍n=1⊍

∞⊍

n=N+1An.

Because of finite additivity we can introduce the measure as

µ(⊍n∈N

An) = µ(N

⊍n=1) + µ(

∞⊍

n=N+1An) .

And applying finite additivity again we get

µ(⊍n∈N

An) =N

∑n=1

µ(An) + µ(∞⊍

n=N+1An) .

Since µ ≥ 0 we can rearrange the previous expression to obtain

N

∑n=1

µ(An) ≤N

∑n=1

µ(An) + µ(∞⊍

n=N+1An) = µ(⊍

n∈NAn) (13.2)

By combining ?? and ?? we get

N

∑i=1

µ(An) ≤ µ(⊍n∈N

An) ≤ ∑n∈N

µ(An)

Taking the limit as N →∞, which we can do since N is not inside the argumentsto µ (in that case it would require that µ was already a measure, which it isn’t,yet), we have

∑n∈N

µ(An) ≤ µ(⊍n∈N

An) ≤ ∑n∈N

µ(An) Ô⇒ µ(⊍n∈N

An) = ∑n∈N

µ(An),

or that µ is σ-additive.

13.2 Exercise set 2Due September 27th, 2019.

Exercise 13.2.1. Let Q be the set of all real rational numbers and let IQ =[a, b)Q ∣ a, b ∈ Q where [a, b)Q = q ∈ Q ∣ a ≤ q < b.

1. Prove that σ(IQ) = ℘(Q) where ℘(Q) is the collection of all subsets of Q.(1.5 pts.)

2. Let µ be the counting measure on ℘(Q) and let ν = 2µ. Show that ν(A) =µ(A) for all A ∈ IQ, but ν ≠ µ on σ(IQ) = ℘(Q). Why doesn’t thiscontradict Theorem 5.7 in your book? (1.5 pts.)

Proof.


1. We shall prove the double containment. First, recall that ℘(Q) is a σ-algebra on Q. Also, by Remark 3.5 we have that IQ ⊆ ℘(Q) Ô⇒ σ(IQ) ⊆σ(℘(Q)) ⊆ ℘(Q). The last inclusion comes from the fact that ℘(Q) isalso a σ-algebra on Q so it must contain the smallest σ-algebra on Q thatcontains information about ℘(Q). For the reverse containment, we shallprove that any subset A ∈ ℘(Q) is also in σ(IQ). For any A ⊂ Q define(qn)n∈N to be an enumeration of the rationals in A. This is posible since#Q =#N. Therefore, we can write

A = ⋃n∈Nqn, where qn ∈ σ(IQ)

Therefore, A ∈ σ(IQ) because σ-algebras are closed under countable union.

2. It is clear that any interval A ∈ IQ contains infinitely many rationals,except if the interval is empty, i.e. a = b Ô⇒ [a, b) = ∅. Therefore,

µ(A) = ν(A) =⎧⎪⎪⎨⎪⎪⎩

0 if A = ∅∞ otherwise

But, if we consider A ∈ σ(IQ) = ℘(Q) then we have some finite sets whereµ(A) =#A, and clearly ν(A) = 2#A. The equality between µ and ν onlyholds when the set is either empty or infinite, but not for finite sets suchas A = 1,2 where µ(A) = 2 but ν(A) = 4.Why doesn’t this contradict Theorem 5.7? Even though there is an ex-hausting sequence in the generator, namely (An)n∈N where An = [−n,n)Q,the measure is not finite for any An. Moreover, there cannot be any ex-hausting sequence in IQ with a finite measure because we already saw thatµ(A) =∞, ∀A ∈ IQ, A ≠ ∅.

Exercise 13.2.2. Let X be a set and µ, ν ∶ ℘(X) → [0,∞) two outer measureson X. Define ρ ∶ ℘(X) → [0,∞) by ρ(A) = max(µ(A), ν(A)). Show that ρ isanother outer measure on X.

Proof. Firstly, from the definition of ρ we can see that the domain and codomainare compatible with the definition of an outer measure. Next, we prove each ofthe properties of an outer measure.

1. ρ(∅) =max(µ(∅), ν(∅)) =max(0,0) = 0

2. For any A,B ∈ ℘(X), A ⊆ B we have

ρ(A) =max(µ(A), ν(A)) ≤max(µ(B), ν(B)) = ρ(B)

3. For any sequence (An)n∈N ⊂ ℘(X) we must verify that ρ(⋃n∈NAn) ≤∑n∈N ρ(An).Let us prove the following first. For any function f ∶X × Y → R we have

maxx∈X

∑y∈Y

f(x, y) ≤ ∑y∈Y

maxx∈X

f(x, y)


Choose any x0 ∈X and any y0 ∈ Y and we have that f(x0, y0) ≤max f(x0, y0).Hence ∑y∈Y f(x0, y) ≤ ∑y∈Y f(x, y). Because this is true for all x0 ∈X wehave maxx∈X ∑y∈Y f(x, y) ≤ ∑y∈Y maxx∈X f(x, y).Using this, i.e. choosing Y = N, X = µ, ν and defining f(µ,n) = µ(An)and f(ν, n) = ν(An) we have

ρ(⋃n∈N

An) =maxµ(⋃n∈N

An) , ν (⋃n∈N

An) =max∑n∈N

µ(An),∑n∈N

ν(An)

≤ ∑n∈N

maxµ(An), ν(An) = ∑n∈N

ρ(An)

Exercise 13.2.3. Let (X,A, µ) be a measure space. For A ∈ A let

S(A) ∶= B ∈ A ∣ B ⊂ A,µ(B) <∞.

Define ν ∶ A→ [0,∞] by ν(A) ∶= supµ(B) ∣ B ∈ S(A).

1. Show that ν is monotone, i.e. if A1,A2 ∈ A such that A1 ⊆ A2, thenν(A1) ≤ ν(A2). (0.5 pts).

2. Show that if A ∈ A with µ(A) <∞, then ν(A) = µ(A). (1 pt.)

3. Show that ν is a measure on A. (2.5 pts.)

4. Show that if µ is σ-finite, the µ = ν. (1 pt.)

Proof.

1. We first show that if A1 ⊆ A2, then S(A1) ⊆ A2. Let C ∈ S(A1), thenC ∈ A ∧C ⊂ A1 ∧ µ(C) <∞. Clearly C ⊆ A2 since A1 ⊆ A2 so C ∈ S(A2).Finally,

ν(A1) = supµ(B) ∣ B ∈ S(A1) ≤ supµ(B) ∣ B ∈ S(A2) = ν(A2).

2. Clearly the domain and range of ν are compatible with the definition ofa measure, i.e. ν ∶ A → [0,∞]. We need to show that the two propertiesfrom ?? hold.

(a) ν(∅) = supν(B) ∣ B ∈ S(∅) = µ(∅) = 0 since S(∅) = ∅.(b) Let (An)n∈N ⊂ A be a pairwise disjoint sequence of sets in A. Then,

ν (⊍n∈N

An) =


13.3 Exercise set 3Due October 4th, 2019.

Exercise 13.3.1. Let (X,A, µ) be a measure space and G = A1,A2, . . . acountable partition of X with Ak ∈ A, ∀k ∈ N. Define a function u ∶ X → R byu(x) = ∑∞k=1 k ⋅ 1Ak

.

1. Show that u is A/B(R)-measurable. (1.5 pts)

2. Show that σ(µ) = σ(G), where σ(µ) is the smallest σ-algebra making umeasurable. (2.5 pts)

3. Suppose that 0 < µ(An) <∞ for all n ∈ N. Define ν on A by

ν(B) =∞∑n=1

3−n ⋅ µ(B ∩An)µ(An)

.

Show that ν is a finite measure on (X,A). (1.5 pts)

4. Under the assumptions of part 3, prove that if B ∈ A, then µ(B) = 0 ifand only if ν(B) = 0. (1 pt)

Proof.

1. Observe that the function u(x) always evaluates to a natural number. Inparticular, u(x) = k for the k ∈ N that satisfies x ∈ Ak. u is well-definedsince G is a countable partition of X.We must check whether for any B ∈ B(R), u−1(B) ∈ A. Given a B ∈ B(R)we rewrite it as B = B′ ∪Bnat where Bnat contains all the naturals in Band B′ = B ∖Bnat. Then,

u−1(B) = u−1(B′ ∪Bnat) = u−1(B′) ∪ u−1(Bnat) = ∅ ∪ ⋃n∈Bnat

An ∈ A.

2. By definition of σ(u) we have

σ(u) ∶= σ(u−1(B(R))) = u−1(B(R)),

since the preimage of any σ-algebra is a σ-algebra .Also, recall from exercise set 1 that

σ(G) = ⋃i∈I

Ai ∣ Ai ∈ G, I ⊂ N .

Now we prove the double containment. Take any A ∈ σ(u). Then there isa B ∈ B(R) such that A = u−1(B). By part one we already now that

A = ∅ ∪ ⊍n∈Bnat

An ∈ σ(G).

For the reverse containment, let G ∈ σ(G), therefore there is a set I ⊂ Nsuch that G = ⋃i∈I Ai. Furthermore, I ∈ B(R) and thus, by part one,

⋃i∈I

Ai = u−1(I) ∈ σ(u).


3. First let us check that ν is well defined. ν(B) is well defined for anyB ∈ A since B ∩An ∈ A as A is ∩-stable. Also, ν is non-negative since itis computed as a sum of products of non-negative numbers.Now we check the two properties in the definition of measure:

(a)

ν(∅) =∞∑n=1

3−n ⋅ µ(∅ ∩An)µ(An)

=∞∑n=1

3−n ⋅ µ(∅)µ(An)

=∞∑n=1

3−n ⋅ 0

µ(An)= 0

(b) For any pairwise disjoint collection (Bn)n∈N ⊂ A we have

ν⎛⎝⊍j∈N

Bj

⎞⎠=∞∑n=1

3−n ⋅µ ((⊍j∈NBj) ∩An)

µ(An)

=∞∑n=1

3−n ⋅µ (⊍j∈N(Bj ∩An))

µ(An)

=∞∑n=1

3−n ⋅ ∑j∈N µ (Bj ∩An)µ(An)

= ∑j∈N

∞∑n=1

3−n ⋅ µ (Bj ∩An)µ(An)

= ∑j∈N

ν(Bj)

Additionally, we must check that ν is finite. Since µ is a measure, it ismonotone so µ(B ∩An) ≤ µ(An) and thus

ν(B) =∞∑n=1

3−n ⋅ µ(B ∩An)µ(An)

≤∞∑n=1

3−n <∞

4. It is clear that if µ(B) = 0 then, by monotonicity, since B ∩ An ⊂ B, wehave that 0 ≤ µ(B ∩An) ≤ µ(B) = 0 and therefore ν(B) = ∑∞n=1 3−n ⋅ 0 = 0.For the reverse, we have that if ν(B) = 0, then it must be because µ(B ∩An) = 0, ∀n ∈ N since 3−n > 0, ∀n ∈ N. We can rewrite µ(B) as

µ(B) = µ(B ∩X) = µ⎛⎝B ∩ ⊍

j∈NAj

⎞⎠= µ⎛⎝⊍j∈N

B ∩Aj

⎞⎠

= ∑j∈N

µ(B ∩Aj) = ∑j∈N

0 = 0.

Therefore µ(B) = 0 ⇐⇒ ν(B) = 0, ∀B ∈ A.

Exercise 13.3.2. Consider the measure space ([0,1],B, λ) where B = B(R) ∩[0,1), i.e. the restriction of the Borel σ-algebra to the interval [0,1), and λdenotes the Lebesgue measure restricted to B. Define a map T ∶ [0,1) → [0,1)by

T (x) =⎧⎪⎪⎨⎪⎪⎩

4x if 0 ≤ x ≤ 14,

43(x − 1

4) if 1

4≤ x < 1.


1. Show that T is B/B-measurable (in short, Borel measurable). (1.5 pts)

2. Consider the image measure T (λ) defined by T (λ)(B) = λ(T −1(B)), forall B ∈ B. Show that T (λ) = λ. (2 pts)

Proof.

1. It is enough to check if T −1([a, b)) ∈ B, since J ∩ [0,1) = [a, b) ∣ 0 ≤ a ≤b ≤ 1 is a generator of B.

T −1([a, b)) = x ∈ [0,1) ∣ T (x) ∈ [a, b)

= x ∈ [0, 14) ∣ T (x) ∈ [a, b) ∪ x ∈ [1

4,1) ∣ T (x) ∈ [a, b)

= x ∣ 0 ≤ a ≤ x < b ≤ 1

4 ∪ x ∣ 1

4≤ a ≤ 4

3(x − 1

4) < b ≤ 1

= [a4,b

4) ∪ [3a

4+ 1

4,3b

4+ 1

4) ∈ B,

since each of the intervals is itself in J ∩ [0,1). (This is because 0 ≤a4, b4, 3a

4+ 1

4, 3b

4+ 1

4< 1 since a, b ∈ [0,1).)

2. First we will prove that T (λ)([a, b)) = λ([a, b)),∀[a, b) ∈ J ∩ [0,1). Usingthe same as in part one we have

T (λ)([a, b)) = λ(T −1[a, b))

= λ([a4,b

4) ⊍ [3a

4+ 1

4,3b

4+ 1

4))

= b

4− a

4+ 3b

4+ 1

4− 3a

4− 1

4= b − a = λ([a, b)).

Recall that B = σ(J ∩ [0,1)) and that there exists an exhausting se-quence Bn ↑ [0,1) where Bn = [0,1), ∀n ∈ N and that λ([0,1)) < ∞ andT (λ)([0,1)) = λ(T −1([0,1))) = λ([0,1)) < ∞. Then, by uniqueness, wehave that T (λ) = λ on all B.

13.4 Exercise set 4Due October 25th, 2019.

Exercise 13.4.1. Let (X,A, µ) be a measure space and u ∈ L1R(µ). Define

Bn = x ∈X ∣ 2−n ≤ ∣u(x)∣ < 2n, for n ≥ 1. Set B = ⋃∞n=1Bn.

1. Show that ∫X udµ = ∫B udµ. (1.5 pts)

2. Prove that limn→∞ ∫Bnudµ = ∫X udµ. (1.5 pts)


3. Show that for every ε > 0, there exists a positive integer N , such thatµ(BN) <∞ and ∣∫Bc

Nudµ∣ < ε. (1.5 pts)

Proof.

1. We have that u ∶ X → R so for every n ∈ N, Bn ⊂ X. Therefore, B =⋃∞n=1Bn ⊂X and we may write

∫Xudµ = ∫

Budµ + ∫

Bcudµ.

We will show that ∫Bc udµ = 0. We have that

Bc = (∞⋃n=1

Bn)c

=∞⋂n=1

Bcn.

For each n ∈ N, Bcn = ∣u∣ < 2−n ⊍ ∣u∣ ≥ 2n and Bc

1 ⊃ Bc2 ⊃ . . . , so (Bc

n) isa decreasing sequence with Bc

n ↓ Bc. We can further split ∫Bc udµ as

∫Bc

udµ = limn→∞

(∫∣u∣<2−n udµ + ∫∣u∣≥2n udµ)

= limn→∞∫∣u∣<2−n udµ + lim

n→∞∫∣u∣≥2n udµ.

For the first integral we have that

limn→∞∫∣u∣<2−n udµ ≤ lim

n→∞∫∣u∣<2−n 2−ndµ ≤ lim

n→∞∫ 2−ndµ.

Since 2−n ≥ 2−(n+1) ≥ ⋅ ⋅ ⋅ ≥ 0 is a decreasing sequence of µ-integrable func-tions, the Monotone Convergence Theorem applies and

limn→∞∫∣u∣<2−n udµ ≤ lim

n→∞∫ 2−ndµ = ∫ infn∈N

2−ndµ = ∫ 0dµ = 0.

For the second integral, we use Markov’s inequality to get

µ(∣u∣ ≥ 2n) ≤ 1

2n∫ udµ.

Since u ∈ L1R(µ), we have that ∫ udµ is a finite number so we can take the

limit to get

limn→∞

µ(∣u∣ ≥ 2n) ≤ limn→∞

1

2n∫ udµ = lim

n→∞1

2n= 0,

so ∣u∣ ≥ 2n is a µ-null set and ∫∣u∣≥2n udµ = 0. We conclude that∫Bc udµ = 0 and therefore ∫X udµ = ∫B udµ.


2. Define (un)n∈N by un = u1Bn . Clearly, un ↑ u1B and un ≤ un+1. So theMonotone Convergence Theorem applies and

limn→∞∫Bn

udµ = limn→∞∫ u1Bndµ

= limn→∞∫ undµ

= ∫ limn→∞

undµ

= ∫ u1Bdµ

= ∫Budµ = ∫

Xudµ.

3. As before, we write,

∫Bc

N

udµ = ∫Xudµ − ∫

BN

udµ.

By the definition of limit and part 2 we know that for every ε > 0, thereexists an N ∈ N such that

∣∫BN

udµ − ∫Xudµ∣ = ∣∫

Xudµ − ∫

BN

udµ∣ < ε.

Therefore,

∣∫Bc

N

udµ∣ = ∣∫Xudµ − ∫

BN

udµ∣ < ε.

We just need to check that µ(BN) < ∞. For that we use the fact thatBN ⊂ ∣u∣ ≤ 2N and a version of Markov’s inequality:

µ(BN) ≤ µ(∣u∣ ≤ 2N) ≤1

2N∫ ∣u∣dµ <∞,

since u ∈ L1R(µ) Ô⇒ ∣u∣ ∈ L

1R(µ).

Exercise 13.4.2. Consider the measure space ([0,1],B([0,1]), λ) where B([0,1])is the restriction of the Borel σ-algebra to [0,1], and λ is the restriction of theone-dimensional Lebesgue measure to [0,1].

1. Show that limn→∞ ∫[0,1]xn

(1+x)2 dλ(x) = 0. (1.5 pts)

2. Show that limn→∞ ∫[0,1]nxn−1

1+x dλ(x) = 12. (1 pt)

Proof.

1. Let un = xn

(1+x)2 . We have that, for every n ∈ N and every x ∈ [0,1],

xn ≤ 1 ≤ (1 + x)2 Ô⇒ ∣un∣ ≤ 1,∀x ∈ [0,1], n ∈ N.


We can rewrite un as

un(x) =xn

(1 + x)21[0,1) +xn

(1 + x)211 =xn

(1 + x)21[0,1) +1

411,

and therefore

limn→∞

un(x) = limn→∞

( xn

(1 + x)21[0,1)(x) +1

411(x))

= 0 ⋅ 1[0,1)(x) +1

4⋅ 11(x)

= 1

411(x), ∀x ∈ [0,1].

Therefore (un)n∈N is a sequence of µ-integrable functions bounded by 1and we can apply the Lebesgue Dominated Convergence theorem to obtain

limn→∞∫[0,1] undλ(x) = ∫[0,1] limn→∞

undλ(x)

= ∫1

411(x)dλ(x)

= ∫11

4dλ(x) = 0,

since λ(1) = 0.

2.

Exercise 13.4.3. Let (X,A, µ) be a measure space, and u ∈ Lp(µ) for somep ∈ [1,∞). For n ≥ 1, define un =minmax(u,−n), n.

1. Prove that limn→∞ ∥un∥p = ∥u∥p. (2 pts)

2. Prove that for any ε, there exists an integer n ≥ 1 such that ∫ ∣u − un∣p ≤ ε.(1 pt)

Proof.

1. Observe that we may rewrite un as

un(x) =⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩

−n if u(x) < nu(x) if − n ≤ u(x) ≤ nn if n < u(x)

and therefore

∣un(x)∣ =⎧⎪⎪⎨⎪⎪⎩

u(x) if 0 ≤ ∣u(x)∣ ≤ nn if n < ∣u(x)∣

.


From this we can see that ∣un∣ ≤ ∣un+1∣ ≤ ∣u∣ and thus (∣un∣)n∈N is an increas-ing sequence of non-negative functions which converges to ∣u∣. Thereforewe can use Beppo-Lévi to get

limn→∞

∥un∥p = limn→∞

(∫ ∣un∣p dµ)1p

= ( limn→∞∫ ∣un∣p dµ)

1p

??= ∫ ( limn→∞

∣un∣p dµ)1p

= ∫ (∣u∣pdµ)

1p = ∥u∥p

2. From part one we know that limn→∞ ∥un∥p = ∥u∥p. Therefore, by Riesz’stheorem, we have that limn→∞ ∥un − u∥p = 0. From the definition of limitwe get that for any ε′ > 0, there exists an N ∈ N such that

(∫ ∣uN − u∣p dµ)1p

< ε′.

Raising everything to the power of p we get

∫ ∣uN − u∣p dµ = ∫ ∣u − uN ∣p dµ < (ε′)p.

Choose ε′ = ε 1p and the proof follows.

13.5 Practice mid-term 2019-2020Exercise 13.5.1. Let (X,A) be a measure space such that A = σ(G) where Gis a collection of subsets of X such that ∅ ∈ G. Show that for any A ∈ A thereexists a countable collection GA ⊆ G such that A ∈ σ(GA).

Proof. This exercise presents a good opportunity to apply the good set principle(cf. ??).

Let

B ∶= A ∈ A ∣ ∃GA ⊂ G countable, A ∈ σ(GA).

First, we will show that B is a σ-algebra .

1. Clearly X ∈ A. Let GX = ∅ which is clearly finite and GX ⊂ G byhypothesis. Then σ(GX) = ∅,∅c = ∅,X Ô⇒ X ∈ B.

2. Let A ∈ B. Then A ∈ A by definition of B and there exists GA ⊂ Gcountable with A ∈ σ(GA). Since σ(GA) is a σ-algebra it contains Ac andclearly Ac ∈ A. Let GAc = GA and therefore Ac ∈ B.

13.5. PRACTICE MID-TERM 2019-2020 93

3. Finally, let (An)n∈N ⊂ B. Clearly, (An)n∈N ⊂ A and, for each n ∈ N, thereexists GAn ⊂ G countable such that An ∈ σ(GAn). Also, since A is aσ-algebra , we have that ⋃n∈NAn ∈ A. Let

G⋃n∈N An = ⋃n∈NGAn .

G⋃n∈N An ⊂ G is countable since every GAn ⊂ G is countable and G⋃n∈N An isa countable union of countable sets. Therefore ⋃n∈NAn ∈ B.

Now we show that B = A. Clearly, from the definition of B we see thatB ⊆ A. For the reverse containment, we will show that G ⊂ B and thereforeA = σ(G) ⊆ σ(B) = B. Let A ∈ G ⊂ A. Define GA = A ⊂ G and clearly finiteand A ∈ σ(GA). Therefore A ∈ B, ∀A ∈ G Ô⇒ G ⊂ B.

Exercise 13.5.2. Let X be a set and F a collection of real-valued functions onX satisfying the following properties:

1. F contains the constant functions,

2. if f, g ∈ F and c ∈ R, then f + g, fg, cf ∈ F .

3. if fn ∈ F , and f = limn→∞ fn, then f ∈ F .

For A ⊆X, denote by 1A the indicator function of A, i.e.

1A(x) =⎧⎪⎪⎨⎪⎪⎩

1 if x ∈ A,

0 if x /∈ A.

Show that the collection A = A ⊆X ∣ 1A ∈ F is a σ-algebra .

Proof. We shall prove that the three properties of a σ-algebra hold in A:

1. X ∈ A since 1X(x) = 1 constantly, and therefore 1X ∈ F .

2. if A ∈ A, then 1A ∈ F . We can write 1Ac = 1−1A. Since 1,1A ∈ F we have1Ac ∈ F Ô⇒ Ac ∈ A.

3. Let (An)n∈N ⊂ A. Then (1An)n∈N ∈ F . We can write

1⋃n∈N An =∞∑n=1

1Ai

To prove 1⋃n∈N Ai ∈ F we construct the following sequence of functions(fn)n∈N ∈ F given by

fn =n

∑i=1

1Ai = 1An + fn−1

Clearly, fn ∈ F for all n because of property 2. Then,

f = limn→∞

fn =∞∑i=1

1Ai = 1⋃n∈N An ∈ F ,

and thus, ⋃n∈NAi ∈ A.


Exercise 13.5.3. Consider the measure space ([0,1],B([0,1]), λ) where B([0,1])is the restriction of the Borel σ-algebra to to [0,1], and λ is the restriction of theLebesgue measure to [0,1]. Let E1, . . . ,Em be a collection of Borel measurablesubsets of [0,1] such that every element x ∈ [0,1] belongs to at least n sets inthe collection Ejmj=1, where n ≤ m. Show that there exists a j ∈ 1, . . . ,msuch that λ(Ej) ≥ n

m.

Proof. Observe that if every x ∈ [0,1] belongs to at least n sets in (Ej) then wehave that

m

∑j=1

1Ej(x) ≥ n, ∀x ∈ [0,1].

We proceed by contradiction. Suppose λ(Ej) < nm, ∀j ∈ 1, . . . ,m. Then

n = ∫[0,1] ndλ??≤ ∫[0,1]

m

∑j=1

1Ej(x)dλ =m

∑j=1

λ(Ej) <m ⋅n

m= n.

This is a contradiction, so there must exists at least one j ∈ 1, . . . ,m such thatλ(Ej) ≥ n

m.

Exercise 13.5.4. Consider the measure space (R,B(R), λ) where B(R) is theBorel σ-algebra over R, and λ is the one-dimensional Lebesgue measure. Letfn ∶ R→ R be defined by

fn(x) =2n−1∑k=0

3k + 2n2n

1[k/2n,(k+1)/2n](x), n ≥ 1.

1. Show that fn is measurable, and fn(x) ≤ fn+1(x), for all x ∈ R.

2. Show that ∫ supn≥1 fndλ = 52.

Proof.

1. For a given n and k we define

ank =3k + 2n

2nand An

k = [k

2n,k + 12n) .

Clearly, Ank ∈ B(R) for any n and k and ank > 0. Therefore, fn is a simple

function and thus fn is A/B(R)-measurable.Notice that

Ank = [

k

2n,k + 12n) = [ 2k

2 ⋅ 2n ,2k + 22 ⋅ 2n )

= [ 2k

2n+1,2k + 12n+1

) ⊍ [2k + 12n+1

,2k + 22n+1

) = An+12k ⊍An+1

2k+1,

and

ank =3k + 2n

2n= 6k + 2n+1

2n+1= an+12k ≤ an+12k + an+12k+1.

13.5. PRACTICE MID-TERM 2019-2020 95

Thus,

fn =2n−1∑k=0

ank1Ank≤

2n−1∑k=0

an+12k 1An+12k+ an+12k+11An+1

2k+1= fn+1.

2. Since (fn) is an increasing sequence of simple functions we now that

∫ supn≥1

fndµ = supn≥1∫ fndµ = sup

n≥1Iµ(fn) = lim

n→∞Iµ(fn),

and

Iµ(fn) =2n−1∑k=0

ankµ(Ank) =

2n−1∑k=0

3k + 2n2n

⋅ (k + 12n− k

2n)

=2n−1∑k=0

3k

2n ⋅ 2n +1

2n= 1 + 3

2

2n − 12n

. Thus,

∫ supn≥1

fndµ = limn→∞

Iµ(fn) = 1 +3

2= 5

2.

Exercise 13.5.5. Let µ and ν be two measures on the measure space (E,B)sucht that µ(A) ≤ ν(A) for all A ∈ B. Show that if f is any non-negative functionon (E,B), then ∫E fdµ ≤ ∫E fdν.

Proof. 1By ?? we now that there exists a sequence of non-negative simple func-tions (fn)n∈N ⊂ E+(B) such that fn ↑ f . Moreover, each of this simple functionshas a representation of the form

fn =Mn

∑j=0

aj1Aj

Thus,

∫Efdµ = lim

n→∞∫E fndµ = limn→∞

Iµ(fn)

= limn→∞

n

∑i=1

Mn

∑j=0

ajµ(Aj) ≤ limn→∞

n

∑i=1

Mn

∑j=0

ajν(Aj)

= limn→∞

Iν(fn) = limn→∞∫E fndν = ∫ fdν

1Now that the solutions to this practice midterm have been published, I noticed there isa much more cleaner (and modular!) way of doing this. First prove it for f = 1A for someA ∈ A. Then, prove it for f ∈ E+(A), for some simple function f . Finally, prove it for thegeneral case. Each of these builds on the previous and you’re left with something much morereadable and useful.


13.6 Practice final 2019-2020Exercise 13.6.1. Consider the measure space ([1,∞),B([1,∞)), λ) where B([1,∞])is the Borel σ-algebra and λ is the Lebesgue measure restricted to [1,∞). Showthat

limn→∞∫

n sin(x/n)x3

dλ(x) = 1.

(Hint: limn→∞ sin(x)/x = 1).

Proof. Since it is not clear how we would go about directly evaluating theLebesgue integral, let us check if we can use the Riemann integral to calculatethe value. Let fn(x) = n sin(x/n)

x3 . Recall from ?? that if (R) ∫N1 ∣fn(x)∣d(x) <∞

for all N ∈ N then (R) ∫∞1 fn(x)dx = ∫[1,∞) fn(x)dλ(x). So first, we show that

fn is Riemann integrable from 1 to N for any N . We have that sin y ≤ y for anyy ≥ 0 so

∣fn(x)∣ = ∣n sin(x/n)

x3∣ ≤ n ⋅ (x/n)

x3= 1

x2, ∀x ∈ [1,∞)

and thus

(R)∫N

1∣fn(x)∣dx ≤ (R)∫

∞

1

1

x2= 1 <∞, ∀N ∈ N.

So far, we have proven that for each n ∈ N, fn ∈ L1. Moreover, by using the hintwe get that

limn→∞

fn(x) = limn→∞

sin(x/n)x2 ⋅ (x/n) =

1

x2limn→∞

sin(x/n)x/n = 1

x2.

We reuse the argument from before to show that ∀n ∈ N, fn(x) ≤ 1x2 . We

do this to get a function w ∈ L1 such that ∣fn∣ ≤ w, ∀n ∈ N. We are now in aposition to apply Lebesgue’s Dominated Convergence theorem (cf. ??). Withno further ado,

limn→∞∫[1,∞)

n sin(x/n)x3

dλ(x) ??= ∫[1,∞) limn→∞n sin(x/n)

x3dλ(x)

= ∫[1,∞)1

x2dλ(x)

??= (R)∫∞

1

1

x2dx = 1.

Exercise 13.6.2. Consider (R,B(R), λ) where B(R) is the Borel σ-algebra andλ is the one-dimensional Lebesgue measure.

1. Prove that for f ∈ L1(λ) and n ∈ Z one has

∫[0,1] f(x + n)dλ(x) = ∫[n,n+1] f(x)dλ(x).

13.6. PRACTICE FINAL 2019-2020 97

2. Let f ∈ L1(λ) and define g(x) = 1[0,1]∑n∈Z f(x+n). Prove that g ∈ L1(λ)and that

∫Rg(x)dλ(x) = ∫

Rf(x)dλ(x).

This exercise is not too difficult but it is very technical and we have to becareful that we choose the correct theorem for each step and that its hypothesesare met.

Proof.

1. At first, one might be tempted to split f = f+−f−, approximate by increas-ing sequences simple functions and go all the way back to the definitionof the integral for simple functions. Even though this is possible, it mightas well make one go nuts with the indices. Therefore, we use a standardargument in this course: prove it first for indicator functions, then forf ∈ E+, then for f ∈M+, and finally for f ∈ L1.

(a) Let f = 1A where A ∈ B(R). Then,

∫[0,1] f(x + n)dλ(x) = ∫ 1[0,1](x) ⋅ 1A(x + n)dλ(x)

= ∫ 1[0,1](x) ⋅ 1A−n(x)dλ(x)

= ∫ 1[0,1]∩(A−n)(x)dλ(x)

= λ([0,1] ∩ (A − n)).

We can now use that the Lebesgue measure is translation invariantto get

λ([0,1] ∩ (A − n)) = λ(([0,1] + n) ∩A) = λ([n,n + 1],∩A),

and therefore

∫[0,1] f(x + n) = λ([n,n + 1] ∩A)

= ∫[n,n+1] 1A(x)dλ(x)

= ∫[n,n+1] f(x)dλ(x).


(b) Now let f(x) = ∑Nj=0 aj1Aj(x) ∈ E+(B(R)). We have

∫[0,1] f(x + n)dλ(x) = ∫[0,1]N

∑j=0

aj1Aj(x + n)dλ(x)

=N

∑j=0

aj ∫[0,1] 1Aj(x + n)dλ(x)

=N

∑j=0

aj ∫[n,n+1] 1Aj(x)dλ(x)

= ∫[n,n+1]N

∑j=0

aj1Aj(x)dλ(x)

= ∫[n,n+1] f(x)dλ(x).

(c) Next let f ∈M+(B(R)). By the Sombrero lemma (??) we have thatthere exists an increasing sequence of non-negative simple functions(fj)j∈N ⊂ E+ such that fj ↑ f . Thus, using Beppo-Lévi we get

∫[0,1] f(x + n)dλ(x) = ∫[0,1] supj∈Nfj(x + n)dλ(x)

??= supj∈N∫[0,1] fj(x + n)dλ(x)

= supj∈N∫[n,n+1] fj(x)dλ(x)

??= ∫[n,n+1] supj∈Nfj(x)dλ(x) = ∫[n,n+1] f(x)dλ(x).

(d) Finally let f ∈ L1(λ). Then f = f+ − f− with f+, f− ∈M+(B(R))and therefore

∫[0,1] f(x + n)dλ(x) = ∫[0,1] f+(x + n)dλ(x) − ∫[0,1] f

−(x + n)dλ(x)

= ∫[n,n+1] f+(x)dλ(x) − ∫[n,n+1] f

−(x)dλ(x)

= ∫[n,n+1] f(x)dλ(x).

2. First we prove that g ∈ L1(λ). To be extra precise, we can rewrite

g(x) = ∑n∈N

gn(x), gn(x) = 1[0,1] ⋅ f(x + n) + 1[0,1]f(x − n).

Notice how we have changed the summation index to make it easier forus to apply ?? and we will use it just for that. Keep in mind that thistheorem only works for non-negative functions, so we will use ∣g∣ to check

13.7. MID-TERM 2019-2020 99

integrability:

∫R∣g(x)∣dλ(x) = ∫

R∣∑n∈N

gn(x)∣dλ(x)

??≤ ∫

R∑n∈N∣gn(x)∣dλ(x)

??= ∑n∈N∫R∣gn(x)∣dλ(x)

??≤ ∑

n∈Z∫R1[0,1] ⋅ f(x + n)dλ(x)

= ∑n∈Z∫[0,1] f(x + n)dλ(x)

= ∑n∈Z∫[n,n+1] f(x)dλ(x)

= ∫Rf(x)dλ(x) <∞,

where we have used the generalised Minkowski inequality twice, first in itsinfinite form and second in the standard one, when changing the summa-tion index back to Z. Thus, ∣g∣ ∈ L1(λ) Ô⇒ u ∈ L1(λ).Finally we check the equality. As we have seen that g ∈ L1 we now havean integrable bound for [the absolute value of] our sequence ∑n∈N gn(x)so we can apply Lebesgue Dominated Convergence to get

∫Rg(x)dλ(x) = ∫

R∑n∈N

gn(x)dλ(x)

??= ∑n∈N∫Rgn(x)dλ(x)

= ∑n∈Z∫[0,1] f(x + n)dλ(x)

= ∑n∈Z∫[n,n+1] f(x)dλ(x)

= ∫Rf(x)dλ(x),

where we have used the previous part.

13.7 Mid-term 2019-2020The exam took place on Friday, October 11 and we had two hours to completeit.

13.8 Final 2019-2020The exam took place on Wednesday, November 6 and we had three hours tocomplete it.

List of Theorems

101

Notes on Measure and Integration

Documents

Transcript of Notes on Measure and Integration