Notes typeset by Adam Gutter - CMUgautam/teaching/2014-15/720-measure/pdfs/a… · Notes typeset by Adam Gutter Lecture 1 - 08/25/2014 Book Recommendations For grad students, use

21-720 Measure and Integration, Fall 2014Notes typeset by Adam Gutter

Lecture 1 - 08/25/2014

Book Recommendations

• For grad students, use Folland. In-depth, comprehensive, challenging.

• Also good but simpler is Cohn. Cohn is a very nice read.

• Rudin is good, but very dense.

Quotes - ”My favourite book is Folland.” ”These books are all isomorphic.”

Here’s some motivation for the development of the Lebesgue measure/Lebegue integral:Challenge - Let pfnq be a sequence of functions defined on r0, 1s, 0 ď fn ď 1 for all n. Show that if

fn Ñ 0 pointwise, then limnÑ8

fnş1

0fn “ 0.

Using Riemann integration, this is easy if fn Ñ 0 uniformly. Not true if we only have pointwiseconvergence. However, the problem becomes easy again if we use the theory of Lebesgue integration.

Intuition - The Riemann integral is calculated by subdividing the domain of a function, then draw-ing rectangles to approximate the area under a curve. The Lebesgue integral is calculated by

subdividing the range of a function. Approximate f by a function of the formnř

i“1

aiχAi , where

χSpxq “

#

1 x P S

0 x R S, where the sets Ai are disjoint and ”nice” and f « ai on each Ai. We then

approximateż b

a

f «nÿ

i“1

ai ¨ ”lengthpAiq”

For example, we might have a1 ă ¨ ¨ ¨ ă an, Ai “ f´1prai, ai`1qq.To get some intuition for this analogy, consider a banker counting money. One way to do this

is to add the bills together in sequence. This is like Riemann integration. Lebesgue integration isanalogous to sorting the bills into stacks of different denomination, then counting the size of eachstack.

We will develop a notion of ”size” for a set A, denoted by λpAq. In one dimension, this willcorrespond to length. In two dimensions, this will correspond to area. In three dimensions, this willcorrespond to volume, etc.

There are some things we must be careful about, lest we get in trouble.

Banach-Tarski Paradox - There is a positive integer n, disjoint sets A1, . . . , An Ď R3, and mapsR1, . . . , Rn : R3 Ñ R3 that are combinations of rotations, reflections, and translations such thatŤ

1ďiďn

Ai “ Bp0, 1q andď

1ďiďn

RipAiq “ Bp0, 1q YBp2, 1q

Quote - ”With this problem, I’ve solved world hunger. Give me one loaf of bread, I’ll give you two!”

Issue - Although it seems like we’re doubling the volume of the ball, there ends up being no sensiblenotion of volume for the sets Ai, so we can’t identify the volume of the ball with the sum of thevolumes of the pieces. We’ll formalize this later.Remark - This can be done with n “ 5 (and this is optimal).

Definition - Let X be a set. We say Σ Ď PpXq is a σ-algebra on the set X if


1. H P Σ

2. A P Σ ùñ Ac P Σ

3. pAnq Ď Σ ùñŤ

nAn P Σ

Intuition - Think of Σ as all subsets of X to which we can assign a meaningful notion of volume.These will be our ”measurable” sets.Remark - Conditions 1 and 2 imply that X P Σ. Conditions 2 and 3 imply that Σ is closed undercountable intersections.

Example - Σ “ tH, Xu, Σ “ PpXq.

Proposition - If Σ1 and Σ2 are σ-algebras on X, then so is Σ1XΣ2. In fact, if tΣαuαPA is a collectionof σ-algebras on a set X, then so is

Ş

αPA

Σα.

Proof - Easy exercise.

Definition - Let E Ď PpXq be arbitrary. Denote by σpEq the σ-algebra on X generated by E . Thisis defined to be the intersection of all σ-algebras on X that contain E .

By the previous proposition, this definition is justified. Intuitively, σpEq is the smallest σ-algebraon X that contains E .

Example/Definition - Take X “ Rd, E “ topen setsu. We call σpEq the Borel σ-algebra on X and

denote it by BpRdq.Remark - BpRdq “ σpthypercubes in Rduq (exercise).Remark - BpRdq “ σptclosed setsuq.

Definition - Let X be a set, Σ a σ-algebra on X. We say µ : Σ Ñ r0,8s is a positive measure onpX,Σq if

1. µpHq “ 0.

2. Given disjoint sets pAnq Ď Σ, we have µpŤ

ně1Anq “

ř

ně1µpAnq

Remark - In general, (2) doesn’t imply (1). Why? Given A P Σ, we have

µpAq “ µpAYHq “ µpAq ` µpHq

We’d like to conclude that µpHq “ 0, but this is only true if µpAq ă 8. Thus, 2 implies 1 if thereis A P Σ such that µpAq ă 8.

Example (Counting measure) - Let X be any set, Σ “ PpXq, and define µpAq “ |A|, the cardinalityof A. That this is a measure is left as an exercise.Example (δ measure) - Let a0 P X, Σ “ PpXq. Define

δa0pXq “

#

1 x P A

0 x R A

That this is a measure is left as an exercise. We call this the δ measure at a0.


Lecture 2 - 08/27/2014

Goal - Construct the Lebesgue measure, which will be a measure on pRd,BpRdqq. This measure willsatisfy

λppa1, b1q ˆ ¨ ¨ ¨ ˆ pad, bdqq “ź

1ďiďd

pbi ´ aiq

for all ai ă bi. That is, the measure will agree on cells in Rd.

Goal 2 - Define a more robust notion of integration. Abstractly, given a measure space pX,Σ, µq:

1. If s : X Ñ R is of the form s “ř

1ďiďN

aiχAi , where the sets Ai P Σ are disjoint, defineş

Xs dµ “

ř

1ďiďN

aiµpAiq.

2. Given an arbitrary function f : X Ñ R, approximate f by functions from (1) to defineş

Xf dµ.

Construction of Lebesgue Measure

Definition - We say I Ď R is a cell if it is a finite interval, i.e. I is of the form pa, bq, pa, bs, ra, bq, orra, bs, where a, b P R.Definition - We say I Ď Rd is a cell if I “ I1 ˆ ¨ ¨ ¨ ˆ Id, where each Ii is a cell in R. Given anonempty cell I in Rd, define

`pIq “ź

1ďiďd

psuppIiq ´ infpIiqq

and define `pHq “ 0. (Intuitively, this is volume.)

Definition - Given A Ď Rd arbitrary, define

λ˚pAq “ inftÿ

ně1

`pInq | A Ďď

ně1

In, In cellsu

We call λ˚ the Lebesgue outer measure. Although λ˚ is defined on all of PpRdq, it is only countablyadditive on a proper subset of PpRdq (as we shall see).

Properties of λ˚

1. λ˚pHq “ 0

2. A Ď B ùñ λ˚pAq ď λ˚pBq (monotonicity)

3. If Ai Ď Rd, then λ˚pŤ

iě1

Aiq ďř

iě1

λ˚pAiq (countable subadditivity)

Proof - (1) and (2) are immediate. (Quote - ”For the entire semester, there’s only one trick we keepusing, which we’ll use now.”) For (3), let ε ą 0. For each i, let pIi,nq be a cover of Ai by cells suchthat

ÿ

ně1

`pIi,nq ď λ˚pAiq `ε2i

Clearly tIi,nui,nPN is a countable cover ofŤ

iě1

Ai by cells, so by definition, we have

λ˚pď

iě1

Aiq ďÿ

i,n

`pIi,nq ďÿ

i

pλ˚pAiq `ε2i q “

ÿ

iě1

λ˚pAiq ` ε

Since ε ą 0 was arbitrary, let εÑ 0` to get λ˚pŤ

iě1

Aiq ďř

iě1

λ˚pAiq.

Next time, we’ll show two things. First, that λ˚pIq “ `pIq for any cell I. Next, we’ll show thatλ˚ has a property called separated additivity. That is, if the distance between two sets A,B Ď Rd ispositive, then λ˚pAYBq “ λ˚pAq ` λ˚pBq.


Lecture 3 - 08/29/2014

Definition - We say A,B Ď Rd are separated if dpA,Bq ą 0, where

dpA,Bq “ inft|a´ b| | a P A, b P Bu

Proposition 1 - If A,B Ď Rd with dpA,Bq ą 0, then λ˚pAYBq “ λ˚pAq ` λ˚pBq.Proof - We may assume λ˚pAq ` λ˚pBq ă 8 (otherwise, the result is clear by monotonicity). Bysubadditivity, we have λ˚pA Y Bq ď λ˚pAq ` λ˚pBq, so it suffices to show that λ˚pA Y Bq ěλ˚pAq ` λ˚pBq.

Denote δ “ dpA,Bq and let tIku be a cover of AYB by cells. By subdividing each Ik into cells ofdiameter at most δ

2 , we obtain a new collection tJku of cells. Note thatř

k

`pIkq ěř

k

`pJkq since the

subdivision process may result in new cells that are duplicated (and we only count each cell once).Since diampJkq ă

δ2 for each k, then Jk X A ‰ H implies Jk X B “ H and vice-versa. Now

partition tJku “ tJ1ku Y tJ

2ku, where J 1k XA ‰ H and J2k XA “ H for each k. Now

ď

k

J 1k Ě A,ď

k

J2k Ě B ùñÿ

k

`pIkq ěÿ

k

`pJkq “ÿ

k

`pJ 1kq `ÿ

k

`pJ2k q ě λ˚pAq ` λ˚pBq

Taking the infimum over all covers tIku, we have λ ˚ pAYBq ě λ˚pAq ` λ˚pBq, as desired.Remark - For sets A and B that aren’t separated, we don’t have disjoint additivity. Intuitively, theproblem is with the boundary of A and B. If A, B are sufficiently complicated, then there is no wayto divide a cover of A Y B into a cover of A and a cover B (like we did above). Covering A maynecessarily entail covering some of B and/or vice-versa, so we get double-counting.

Proposition 2 - If I is a cell, then λ˚pIq “ `pIq.Lemma - λ˚pAq “ inft

ř

n`pInq | A Ď

Ť

nIn, In open cellsu “: µ˚pAq.

Proof - Clearly λ˚pAq ď µ˚pAq. We need to show that µ˚pAq ď λ˚pAq. We may assume λ˚pAq ă 8,otherwise this is clear.

Let ε ą 0 and let pInq be a cover of A by cells such thatř

n`pInq ď λ˚pAq ` ε. For each n, let

Jn Ě In be an open cell with `pJnq ´ `pInq ăε

2n . Now

µ˚pAq ďÿ

n

`pJnq ďÿ

n

`pInq ` ε ď λ˚pAq ` 2ε

Since ε ą 0 was arbitrary, then µ˚pAq ď λ˚pAq, as desired.

Proof (of Proposition 2) - First assume that I is closed. Clearly λ˚pIq ď `pIq by definition. We needto show λ˚pIq ě `pIq. By Lemma 1, it’s enough to consider covers of I by open cells.

Say pInq is a cover of I by open cells in Rd. We need to show thatř

ně1`pInq ě `pIq. Since I

is closed, it is compact, so there is N such that I ĎŤ

1ďnďN

In. Extend the faces of the cells tInu

to hyperplanes and use these to subdivide I into new cells pJnq. Clearlyř

n`pJnq “ `pIq (by the

distributive law for multiplication in R), andř

n`pJnq ď

ř

n`pInq. Thus,

ř

n`pInq ě `pIq. Taking the

infimum over covers of I by open cells, we have λ˚pIq ě `pIq, as desired.If I is not closed, let J Ď I be a closed cell such that `pJq ě `pIq ´ ε. Then

λ˚pIq ě λ˚pJq “ `pJq ě `pIq ´ ε

Taking εÑ 0`, we have the desired result.


Remark (translation invariance) - For any cell I and point x0 in Rd, `pIq “ `pI ` x0q. Therefore,λ˚pAq “ λ˚pA` x0q for any set A and point x0. That is, the Lebesgue outer measure is translationinvariant.Remark - There’s another way to obtain λ˚. Instead of using the volume of cubes, try and definea notion of volume by requiring it to be translation invariant (among other things). Doing this, werecover the Lebesgue measure. More generally, this can be applied to topological groups to definesomething called Haar measure.

We return to the construction of the Lebesgue measure, introducing Caratheodory’s criterionand abstracting our discussion.

Definition - We say µ˚ is an outer measure on X if µ˚ : PpXq Ñ r0,8s and

1. µ˚pHq “ 0

2. A Ď B ùñ µ˚pAq ď µ˚pBq

3. µ˚pŤ

nAnq ď

ř

nµ˚pAnq

Example - λ˚ is an outer measure.

Theorem (Caratheodory) - Let µ˚ be an outer measure on X. Define

Σ “ tE | for all A Ď X, µ˚pAq “ µ˚pAX Eq ` µ˚pAX Ecqu

Then

1. Σ is a σ-algebra.

2. µ æΣ is a measure.

Proof - Next time.


Lecture 4 - 09/01/2014Theorem (Caratheodory) - Let µ˚ be an outer measure on X. Define

Σ “ tE | for all A Ď X, µ˚pAq “ µ˚pAX Eq ` µ˚pAX Ecqu

Then

1. Σ is a σ-algebra, the (µ˚-)measurable subsets of X.

2. µ æΣ is a measure.

Proof - We first show that H P Σ and Σ is closed under complementation.

• H P Σ: For any A Ď X, we have µ˚pAq “ 0` µ˚pAq “ µ˚pAXHq ` µ˚pAXXq.

• Given E P Σ, we have µ˚pAq “ µ˚pAX Eq ` µ˚pAX Ecq “ µ˚pAX Ecq ` µ˚pAX pEcqcq.

To show that Σ is closed under countable unions, we first prove some facts about finite unionsand µ˚.

Claim 1 E,F P Σ ùñ E Y F P Σ. (Thus, Σ is closed under finite unions.)

Proof Let A Ď X. Then, appealing to measurability of E,F , we have

µ˚pAX pE X F qq ` µ˚pAX pE X F qcq

“ µ˚pAX pE Y F q X Eq ` µ˚pAX pE Y F q X Ecq ` µ˚pAX pE X F qcq

“ µ˚pAX Eq ` µ˚pAX F X Ecq ` µ˚pAX F c X Ecq

“ µ˚pAX Eq ` µ˚pAX Ecq

“ µ˚pAq

Claim 2 If E1, . . . , En P Σ are pairwise-disjoint and A Ď X, then

µ˚pAX pď

1ďiďN

Eiqq “ÿ

1ďiďN

µ˚pAX Eiq

Proof By induction, it’s enough to consider N “ 2. Let E,F P Σ be disjoint,A Ď X. Now

µ˚pAXpEYF qq “ µ˚pAXpEYF qXEq`µ˚pAXpEYF qXEcq “ µ˚pAXEq`µ˚pAXF q

To show that Σ is closed under countable unions (proving (1)), it’s enough by Claim 1 to showthat Σ is closed under finite disjoint unions. Let pEnq Ď Σ pairwise-disjoint, A Ď X. We wantŤ

nEn P Σ.

By subadditivity, it’s enough to show that µ˚pAq ě µ˚pAXŤ

nEnq`µ

˚pAXpŤ

nEnq

cq. By Claim

2 and monotonicity, we have

µ˚pAq “ µ˚pAXď

1ďnďN

Enq ` µ˚pAX p

ď

1ďnďN

Enqcq

ě µ˚pAXď

1ďnďN

Enq ` µ˚pAX p

ď

ně1

Enqcq

“ÿ

1ďnďN

µ˚pAX Enq ` µ˚pAX p

ď

ně1

Enqcq


Taking the limit N Ñ8, we have (by subadditivity)

µ˚pAq ěÿ

ně1

µ˚pAX Enq ` µ˚pAX p

ď

ně1

Enqcq ě µ˚pAX p

ď

ně1

Enqq ` µ˚pAX p

ď

ně1

Enqcq

Thus, we have equality above, which proves the result. Moreover, an immediate consequence of thisproof is that µ˚ is countably additive on Σ (take A “

Ť

ně1En).

Let’s return to the construction of the Lebesgue measure on Rd. Let λ˚ be the Lebesgue outermeasure. By Caratheodory,

L :“ LpRdq “ tE Ď Rd | for all A Ď Rd, λ˚pAq “ λ˚pAX Eq ` λ˚pAX Ecqu

is a σ-algebra and λ :“ λ˚ æΣ is a measure, the Lebesgue measure.This seems promising, but there’s one more problem. We don’t know what L is! It may be

trivial!

Proposition - Let I be a cell. Then I P L.

Corollary - BpRdq Ď LpRq.Remark - BpRdq Ĺ LpRdq Ĺ PpRdq. In fact, |BpRdq| “ |R| and |LpRdq| “ |PpRdq|.

Proof (of Proposition) - We need to show that for all A, λ˚pAq ě λ˚pA X Iq ` λ˚pA X Icq. Letε ą 0. Choose a cell I1 Ď I such that λ˚pIzI1q ă ε and dpI1, I

cq ą 0. By separated additivity andmonotonicity of λ˚, we have

λ˚pAX I1q ` λ˚pAX Icq “ λ˚pAX pI1 Y I

cqq ď λ˚pAq p˚q

Moreover,λ˚pAX Iq ď λ˚pAX I1q ` λ

˚pAX pIzI1qq ď λ˚pAX I1q ` ε

Rearranging, we have λ˚pAX I1q ě λ˚pAX Iq ´ ε. By p˚q, we have

λ˚pAq ě λ˚pAX I1q ` λ˚pAX Icq ě λ˚pAX Iq ` λ˚pAX Icq ´ ε

Let εÑ 0` to get λ˚pAq ě λ˚pAX Iq ` λ˚pAX Icq.

We’ve now shown that the Lebesgue measure exists, is nontrivial, and agrees with our intuitivenotion of volume for cells. Next time, we’ll show that the Lebesgue measure is the unique measurethat satisfies our requirements.


Lecture 5 - 09/05/2014

Theorem (Uniqueness) - If µ is a measure on pRd,Lq (or pRd,Bq) and µpIq “ λpIq for all cells I,then

µpAq “ λpAq

for all A P L (or B).Proof - =We proceed in steps.

1. Let A P L be arbitrary. Let pInq be a cover of A by cells. Then by countable subadditivityand monotonicty,

µpAq ďÿ

n

µpInq “ÿ

n

λpInq “ÿ

n

`pInq

Taking the infimum over all possible covers, we have µpAq ď λpAq.

2. Now suppose that A P L is bounded, say A Ď I, a bounded cell. By the previous step,

µpIzAq ď λpIzAq ùñ µpIq ´ µpAq ď λpIq ´ λpAq ùñ λpAq ď µpAq

By the previous step, we have µpAq “ λpAq.

3. For arbitrary A P L, write A “Ť

ně1pAX pń, nqdq. Now

µpAq “ limnÑ8

µpAX pń, nqdq “ limnÑ8

λpAX pń, nqdq “ λpAq,

which finishes the proof.

Our approach above used a neat trick, but now we’ll consider a more general, abstract approachto uniqueness.

Uniqueness Abstractly

Let X be some set, Σ a σ-algebra on X, C Ď PpXq. Let µ, ν be measures on pX,Σq such thatµpXq “ νpXq ă 8. Suppose that µ “ ν on C. Does this imply that µ “ ν on σpCq.

As it turns out, the answer is no! For example, consider two sets A and B. We might haveµpAzA X Bq “ µpBzA X Bq “ 0 and µpA X Bq “ 1, while νpAzA X Bq “ νpBzA X Bq “ 1, whileνpAXBq “ 0. Then µ “ ν on tA,Bu, but µ and ν disagree on AXB.

To prove that µ “ ν on σpCq, we need C to be closed under intersections! Now denote D “

tA | µpAq “ νpAqu. Does D have the properties of a σ-algebra that we want?

1. H P D is clear.

2. A P D ùñ Ac P D follows from µpXq “ νpXq ă 8.

3. We can’t generally show that D is closed under finite unions (let alone countable) unless weknow that D is closed under finite intersections! Instead, let’s consider a property that weknow holds of D.

31 If pAnq Ď D is increasing, thenŤ

nAn P D.

21 While we’re at it, let’s generalize (2): If A,B P D with A Ď B, then BzA P D. So, not only isD closed under complements, but it’s closed under relative complements.


The discussion above motivates a number of definitions.

Definition - We say Π Ď PpXq is a π-system Π ‰ H and

A,B P Π ùñ AXB P Π

Definition - We say Λ Ď PpXq is a λ-system if

1. X P Λ

2. A,B P Λ, A Ď B ùñ BzA P Λ

3. pAnq Ď Λ, An Ď An`1 ùñŤ

nAn P Λ.

Theorem (Dynkin) - If Π is a π-system and Λ is a λ-system, then

Λ Ě Π ùñ Λ Ě σpΠq

Corollary - If µ, ν are finite measures that agree on Π, X P Π, then µ “ ν on σpΠq.Remark 1 - The intersection of an arbitrary family of λ-systems is a λ-system.Remark 2 - If Λ is both a Λ-system and a π-system, then it is a σ-algebra. (Intuition: ”finiteintersection + complements + increasing countable unions = countable unions”)

Proof (of Theorem) - It’s enough to show that λpΠq Ě σpΠq, since Λ Ě λpπq. By Remark 2, it’senough to show that λpπq is a π-system, so then it’s a σ-algebra.

For any C P λpπq, set ΛC “ tA P λpπq | A X C P λpπqu. To show that λpπq is a π-system, weshow that ΛC “ λpπq for each C P λpπq. In particular, it’s enough to show that for any C P λpπq,ΛC is a λ-system that contains π.

Claim ΛC is a λ-system.

Proof 1. Clearly X P ΛC .

2. If A,B P ΛC with A Ď B, then

pBzAq X C “ pB X CqzpAX Cq P λpπq

since AX C,B X C P λpπq.

3. Let pAnq Ď ΛC be increasing. Then

pď

ně1

Anq X C “ď

ně1

pAn X Cq P λpπq

since An X C Ď An`1 X C.

Claim ΛC Ě Π.

Proof We need to show for all B P Π, B X C P λpΠq. If C P Π, this is immediatesince Π is a π-system. If C P λpΠq, fix B P Π. By the previous case, ΛB Ě λpπq, soB X C P λpΠq, so B P ΛC . Thus, ΛC Ě λpπq, so we are done.


Lecture 6 - 09/08/2014

Regularity

Definition - Let pX, dq be a metric space, BpXq the Borel σ-algebra of X. We say µ is a regularBorel measure on X if

1. µ is a measure on pX,BpXqq

2. For every A P BpXq, µpAq “ inftµpUq | U Ě A is openu

3. For every open U Ď X, µpUq “ suptµpKq | K Ď U is compactu

4. For every compact K P BpXq, µpKq ă 8.

A measure satisfying condition 2 is called an outer regular measure. A measure satisfying condition3 is called an inner regular measure. A measure satisfying condition 4 is called a Radon measure.

If A P BpXq satisfies condition 2, we say A is inner regular with respect to µ or µ-inner regular.If A satisfies condition 3, we say A is outer regular with respect to µ or µ-outer regular.

Goal: Regularity of λ.

Theorem (Regular of Finite Borel Measures) - Let X be a compact space. Let µ be a finite Borelmeasure on X. Then µ is regular.Proof - Note, µ is Radon since it is finite. Define

Λ “ tA P BpXq | A is µ-inner and µ-outer regularu

It suffices to show that Λ contains all open sets and that Λ is a λ-system, since then Λ Ě BpXq(follows because open sets form a π-system).

Let U Ď X be open. Then U is trivially µ-outer regular. For each n, define Kn “ tx PU |; dpx, U cq ě 1

nu. Then clearly Kn is closed. Since X is compact, then Kn is compact. Note,since U is open, then x P U if and only if dpx, Uq ą 0. Therefore, U “

Ť

ně1Kn. Since Kn Ď Kn`1,

then limnÑ8

µpKnq “ µpUq, thus U is µ-inner regular, so Λ contains all open sets.

We now argue that Λ is a λ-system. Clearly X P Λ since X is both open and compact. Now letA1, A2 P Λ with A1 Ď A2. Fix ε ą 0. Since µ is finite, then for each i “ 1, 2, there are open Ui andcompact Ki such that

Ki Ď Ai Ď Ui, µpAizKiq ă ε, µpUizAiq ă ε

Note that K2zU1 Ď A2zA1 Ď U2zK1, where K2zU1 is compact and U2zK1 is open. Moreover,

µppU2zK1qzpA2zA1qq “ µppU2zA2q Y pA1zK1qq “ µpU2zA2q ` µpA1zK1q ă 2ε

and

µppA2zA1qzpK2zU1qq “ µppA2 XAc1q X pK2 X U

c1 qcq

“ µppA2 XAc1q X pK

c2 Y U1qq

“ µppA2 XAc1 XK

c2q Y pA2 XA

c1 X U1qq

ď µppA1 XKc1q Y pA

c1 X U1qq

ď µpA1zK1q ` µpU1zA1q

ă 2ε


Letting εÑ 0`, we have A2zA1 P Λ.Finally, let pAnq Ď Λ be increasing. Fix ε ą 0. For each n, let Kn Ď An Ď Un be compact and

open with µpAnzKnq ăε

2n and µpUnzAnq ăε

2n . Denote A “Ť

ně1An and U “

Ť

ně1Un. Then A Ď U ,

U is open, and

µpUzAq “ µpď

ně1

pUnzAqq ďÿ

ně1

µpUnzAq ďÿ

ně1

µpUn Ánq ă ε

Let En “Ť

1ďiďn

Ki, K “Ť

iě1

Ki. Note, each En is compact (since it is closed) and En Ď En`1 Ď K.

Moreover, limnÑ8

µpEnq “ µpKq ă 8, so there is N such that µpEN q ą µpKq ´ ε, or µpKzEN q ă ε.

Now EN Ď A and

µpAzEN q “ µpAzKq`µpKzEN q ă µpď

ně1

pAnzKqq` ε ďÿ

ně1

µpAnzKq` ε ďÿ

ně1

µpAnzKnq` ε ď 2ε,

Letting εÑ 0`, we have A P Λ, so Λ is a λ-system. This finishes the proof.


Lecture 7 - 09/10/2014

Regularity of Lebesgue Measure

Proposition I - For all A P LpRdq, we have

λpAq “ inftλpUq;U Ě A, U openu “ suptλpKq | K Ď A, K compactu

and also λpKq ă 8 for all compact K.Proof - Clearly λpAq ď inf

UĚAλpUq. Moreover, countable subadditivity implies

λpAq “ λ˚pAq “ inftÿ

`pInq | In open cells, A Ďď

n

Inu

ě inftλpď

n

Inq | In open cells, A Ďď

n

Inu

“ inftλpUq | U Ě A, U openu

Thus, λ is outer regular. To show that λ is inner regular, we reduce to the case that A is bounded.(For the general case, consider the bounded sets AX rń, nsd.)

Since A is bounded, there is a closed cell I with I Ě A. Fix ε ą 0. By outer regularity, there isopen U Ě IzA such that

λpIzAq ď λpUq ď λpIzAq ` ε

Rearranging, we haveλpAq ě λpIq ´ λpUq “ λpIzUq ě λpAq ´ ε

We are done, since IzU is compact (closed and bounded).Finally, note that compact sets clearly have finite measure because they are bounded.

Proposition II - For all A P L, there is ε ą 0, U open, and C closed such that

C Ď A Ď U, λpUzCq ă ε

Proof - Case I: Suppose that λpAq ă 8. By Proposition I, given ε ą 0 there are compact K Ď Aand open U Ě A with

λpAq ´ ε2 ă λpKq ď λpAq ď λpUq ă λpAq ` ε

2

Now λpUzKq ă ε and K is closed, so we are done.Case II: A is arbitrary. Set Dn “ Bp0, n` 1qzBp0, nq and write A “

Ť

npAXDnq. By Case I, for

each n and ε ą 0, there are open Un and closed Cn with

Cn Ď AXDn Ď Un, λpUnzCnq ăε

2n

Set U “Ť

nUn, C “

Ť

nCn. Clearly U is open. Moreover, C is closed since Cn Ď Dn for each n.

Finally,λpUzCq ď

ÿ

n

ε2n “ ε

Corollary - Let A Ď Rd. ThenA P L ðñ A “ Fσ YN,

where Fσ is a countable union of closed sets and λpNq “ 0.


Existence of a Nonmeasurable Set

Proposition - LpRq Ĺ PpRqProof - Define an equivalence relation „ on R by x „ y if and only if x ´ y P Q. Partition R intoequivalence classes, say for α P R we write

Cα “ tx P R | x´ α P Qu “ α „

Let A Ď R be a set of representatives for the equivalence classes, i.e. for each α, AXCα is a singleton.

Claim A R LpRq

Proof Suppose not, i.e. A P L. Then R “Ť

qPQpA` qq (disjoint union), so

λpRq “ÿ

qPQλpA` qq “

ÿ

qPQλpAq ùñ λpAq ą 0

We argue that λpAq “ 0, which is a contradiction. By inner regularity, it suffices toshow that λpKq “ 0 for all compact K Ď A. Given such K, set E “

Ť

qPQ,|q|ď1

pK`qq.

Now that E P L and λpEq ă 8 since E is bounded. Now

λpEq “ÿ

qPQ,|q|ď1

λpKq ùñ λpKq “ λpEq “ 0

Remark - There is A Ď R such that

E Ď A, E P L ùñ λpEq “ 0, E Ď Ac, E P L ùñ λpEq “ 0


Lecture 8 - 09/12/2014

Completion

Proposition - Let A Ď Rd. Then

A P LpRdq ðñ DF,G P BpRdq, F Ď A Ď G, λpGzF q “ 0

Proof p ùñ q Suppose A P L. Then for each n P N, there are closed Cn, open Un with

Cn Ď A Ď Un, λpUnzCnq ă1n

Let F “Ť

nCn, G “

Ť

nUn. Now λpGzF q ď λpUnzCnq ă

1n for all n, so λpGzF q “ 0.

p ðù q Write A “ F Y pAzF q. Since AzF Ď GzF and λpGzF q “ 0, then AzF P L, so A P L.

Let pX,Σ, µq be a measure space.Definition - Define N “ tA Ď X | DB P Σ, B Ě A, µpBq “ 0u, the set of all µ-null subsets of X.Definition - We say Σ is complete with respect to µ if Σ Ě N .Remark - The σ-algebra of a measure constructed via an outer measure using the Caratheodoryconstruction is complete.Proof - Let µ be such a measure and let N be any µ-null set, say M Ě N with µpMq “ 0. For anyA P X, we have

µpAq ě µpMq ` µpAXN cq ě µpAXNq ` µpAXN cq,

so N P Σ.Corollary - LpRdq is complete.

Definition - We define the completion Σµ of Σ with respect to µ by

Σµ “ tAYN | A P Σ, N P N u

Definition - Given A P Σµ, define µpAq “ µpBq if A “ B YN , where B P Σ and N P N .Remark - µ is well-defined. (We prove this below.)Lemma - A P Σµ if and only if there are F,G P Σ such that F Ď A Ď G and µpGzF q “ 0.Proof - Let A P Σµ, say A “ B YN where B P Σ, N P N . Since N is µ-null, there is M P Σ withM Ě N and µpMq “ 0. Let F “ B, G “ B YM . Then F Ď A Ď G, µpGzF q ď µpMq “ 0.

Conversely, let F,G P Σ with F Ď A Ď G, µpGzF q “ 0. Write A “ F Y pAzF q. Clearly AzF isµ-null, so A P Σµ.

Proposition - Let pX,Σ, µq be a measure space, pX,Σµ, µq its completion.

1. Σµ is a σ-algebra.

2. µ is a measure on Σµ and µ æΣ“ µ.

3. Σµ is complete with respect to µ.

Proof (1) Clearly H P Σµ. Given A P Σµ, the previous lemma implies there are F,G P Σ withF Ď A Ď G and µpGzF q “ 0. Now F c, Gc P Σ with Gc Ď Ac Ď F c and µpF czGcq “ µpGzF q “ 0.Finally, let pAnq Ď Σµ, say An “ Bn Y Nn, where Bn P Σ and Nn is µ-null. Set B “

Ť

nBn,

N “Ť

nNn. Now B P Σ and by subadditivity, N is µ-null. Thus

Ť

nAn P Σµ.


(2) To see that µ is well-defined, let A P Σµ, say A “ B1YN1 “ B2YN2, where B1, B2 P Σ andN1, N2 are µ-null. Let Mi Ě Ni for each i, µpMiq “ 0. Now

µpB1q ď µpB2 YM2q ď µpB2q ` µpM2q “ µpB2q

and similarly, µpB2q ď µpB1q, so µpB1q “ µpB2q. Thus, µ is well-defined.Clearly µpHq “ 0. Now let pAnq Ď Σµ be disjoint, say An “ Bn YNn, where Bn P Σ and Nn is

µ-null. WriteŤ

nAn “ B YN , where B “

Ť

nBn P Σ and N “

Ť

nNn is µ-null. Now

µpď

n

Anq “ µpBq “ÿ

n

µpBnq “ÿ

n

µpAnq,

since the sets Bn are disjoint. Thus, µ is a measure, and clearly µ æΣ“ µ.(3) Let N be null with respect to Σµ, say A P Σµ with A Ě N , µpAq “ 0. Let F Ď A Ď G

with µpGzF q “ 0, F,G P Σ. Now µpF q “ µpAq “ µpGq “ 0. In particular, we have H Ď N Ď G,µpGzHq “ µpGq “ 0, so N P Σµ. Thus, Σµ is complete.Remark - Taking the completion does not simply correspond to throwing in all sets of measure zero!Example - µ “ λ, X “ Rd, Σ “ tX,Hu. Then Σµ “ tX,Hu.


Lecture 9 - 09/15/2014

Goal: Measurable Functions

Definition - Let pX,Σq be a measure space, pY, τq a topological space (usually R or r´8,8s). Wesay f : X Ñ Y is measurable if f´1pτq Ď Σ, i.e. the preimage of an open set is measurable.

(Remark - The topology on r´8,8s is generated by intervals r´8, aq, pb,8s.)Example - If X is a metric space, with Σ Ě BpXq, then continuous functions are measurable.

Proposition - Let BpY q be the Borel σ-algebra on Y . Then f : X Ñ Y is measurable if and only if

f´1pBpY qq Ď Σ.Proof p ðù q This is clear, since f´1pτq Ď f´1pBpY qq.p ùñ q We prove a useful lemma.

Lemma Let f : X Ñ Y be arbitrary. Then Σ1 “ tA Ď Y | f´1pAq P Σu is aσ-algebra.

Proof Exercise.

By the lemma, if f : X Ñ Y is measurable, then Σ1 is a σ-algebra containing all open sets, thusit contains BpY q.

Proposition - Let f : X Ñ r´8,8s. Then

f is measurable ðñ @a P R, f´1pr´8, aqq P Σ

ðñ @a P R, f´1ppa,8sq P Σ

ðñ @a P R, f´1pra,8sq P Σ

Proof Clearly measurability implies the other three conditions by the previous lemma. Since thehalf-infinite intervals generate the Borel σ-algebra, the reverse implication is clear, since Σ1 “ tA Ďr´8,8s ; f´1pAq P Σu is a σ-algebra.

Proposition - Say f : X Ñ Rd is measurable, g : Rd Ñ Rd is Borel measurable (i.e. g´1pRdq ĎBpRdq). Then g ˝ f is measurable.Proof We have

pg ˝ fq´1pBpRdq Ď f´1pg´1pBpRdqqq Ď f´1pBpRdq Ď Σ

Proposition - Let f, g : X Ñ r´8,8s be arbitrary. Then

f, g are measurable ðñ pf, gq : X Ñ r´8,8s2 is measurable

Proof - Supppose that f, g are measurable. Then for any open U, V Ď r´8,8s, we have

pf, gq´1pU ˆ V q “ f´1pUq X g´1pV q P Σ

In particular, pf, gq´1pIq P Σ for any open cell I Ď r´8,8s. Since the cells generate the Borelσ-algebra, then pf, gqpBpr´8,8s2q Ď Σ, so pf, gq is measurable.

Conversely, if pf, gq is measurable, then so are f “ π1 ˝ pf, gq and g “ π2 ˝ pf, gq (here, πi denotesprojection onto the ith coordinate, which is continuous).Corollary - Let f, g : X Ñ R be measurable. Then f ` g is measurable.Proof - Denote F px, yq “ x ` y, which is continuous, Gpxq “ pfpxq, gpxqq. Then f ` g “ F ˝ G,which is continuous by the previous propositions.

Proposition - Let pfnq be a sequence of measurable functions, fn : X Ñ r´8,8s. Then the followingfunctions are measurable.


1. sup fn

2. lim sup fn

3. inf fn

4. lim inf fn

Moreover: (5) if fn Ñ f pointwise, then f is measurable.Proof - Denote g “ inf fn. Then tg ě αu “

Ť

ntfn ě αu, so since the fn’s are measurable, so is g.

Now sup fn “ ´ infp´fnq, so sup fn is measurable. Combining (1) and (3), we get (2) and (4). Noweither of (2) or (4) implies (5).


Lecture 10 - 09/17/2014

The Devil’s Staircase

Recall - The Cantor set C Ď r0, 1s is compact, has Hausdorff dimension α “ ln 2ln 3 , and Hα denotes

the Hausdorff measure of dimension α.

Definition - The Cantor function (also known as devil’s staircase) is defined by fpxq “ HαpCXr0,xsqHαpCq .

Intuitively, this is the fraction of the Cantor set that lies to the left of x.Remarks

1. f is continuous, since |fpyq ´ fpxq| ď Ci|x´ y|α for some Ci ě 0 and α P p0, 1q (exercise).

2. f is increasing (since Hα is a measure).

Set gpxq “ inftf “ xu “ infty | fpyq “ xu. Note, since f is continuous, then the infimum isattained, so fpgpxqq “ x. However, it’s not hard to see that g isn’t continuous since gp 1

2 q “13 , while

gpyq ě 23 for any y ą 1

2 . Note that

1. For all x P r0, 1s, gpxq P C (exercise).

2. g is strictly increasing, since x ă y ðñ fpgpxqq ă fpgpyqq ùñ gpxq ă gpyq.

3. g is Borel measurable, since

tgpxq ď αu “ tx | gpxq ď αu

“ tx | x “ fpgpxqq ď fpαqu pf is increasing)

“ tx | x ď fpαqu

Proposition - LpRq Ľ BpRq.Proof - Let A Ď r0, 1s, A R L (we know such A exists). Set B “ gpAq. Since B Ď C andλpCq “ 0, then B P L by completeness. On the other hand, B R B, since g´1pBq “ A R B andg is Borel measurable.

Remark - Let h1, h2 : RÑ R, h1 Lebesgue measurable and h2 Borel measurable. Then h1 ˝h2

need not be Lebesgue measurable.

Proof - Let h1 “ χB (B as above), h2 “ g. Then

h1 ˝ h2 “ χB ˝ g “ χg´1pBq “ χA,

which is not measurable since A is not measurable.

Definition - Let pΣ, X, µq be a measure space. We say property P holds almost everywhere(abbreviated a.e.) on some set A if there is a null set N such that P holds on AzN . We sayP holds a.e. if P holds a.e. on X.

Notation - Some say @x instead of a.e. x. We won’t use this notation.

Examples - Almost every real is irrational. The function fpxq “ |x| is differentiable a.e.

Theorem (Rademacher) - Any Lipschitz function f : RÑ R is differentiable almost everywhere.


Example - Continued Fractions - Given x P r0, 1s, one may write x “ 1

a1`1

a2`1...

, where ai P Z

is uniquely determined by x, say ai “ aipxq. Define pnpxqqnpxq

“ 1

a1`1

...1

an`1

in reduced terms. We

know that pnpxqqnpxq

Ñ x as nÑ8. Typically (when x is irrational), qnpxq Ñ 8.

Theorem - limnÑ8

ln qnpxqn exists and is equal to π2

12 lnp2q almost everywhere.

Proof - Comes from ergodic theory.

Proposition - Say pX,Σ, µq is the completion of pX,Σ, µq. Then

f : X Ñ R is Σµ-measurable ðñ Dg : X Ñ R, g is Σ-measurable, f “ g a.e.

Proof - p ðù q Given g, let U Ď R be open. Then

f´1pUq “ ptf “ gu X g´1pUqq Y ptf ‰ gu X f´1pUqq

Since the first set is in Σ and the second is null, then f´1pUq P Σµ.

p ùñ q If f “ χE , where E P Σµ, this is clear, since we may write E “ E1 YN , where E1 P Σand N is null. Now g “ χE1 does the trick. Since the result holds for indicator functions, itholds for simple functions.

The result for measurable functions will follow from a theorem next time about approximatingmeasurable functions by simple functions.


Lecture 11 - 09/19/2014

Approximation of Measurable Functions

Definition - We say s : X Ñ R is simple if it is Borel measurable and has finite range.

Definition - Given A P Σ, the characteristic function of A is χA, χApxq “

#

1 x P A

0 x R A.

Remark - If s has a finite range, then s has the form

s “ÿ

1ďiďN

ais´1ptsiuq,

where rangepsq “ ts1, . . . , sNu. Hence, s is simple if and only if s´1ptsiuq is measurable for each i.

Proposition 1 - If f : X Ñ r0,8q is measurable, then there exists an increasing sequence psnq ofsimple functions such that sn Õ f pointwise.Proof - Define Ak,n :“ f´1pr k2n ,

k`12n qq and let sn “

ř

1ďkďn2n

k2nχAk,n ` χtfěnu. Check that this

works.

Proposition 2 - If f : X Ñ R is measurable, then there exists a sequence psnq of simple functionssuch that |sn| ď f for all n and sn Ñ f pointwise.Proof - Denote f` “ maxpf, 0q, f´ “ maxp´f, 0q, so f “ f`´f´. Find psnq and ptnq with sn Õ f`

and tn Õ f´, so now sn ´ tn Ñ f . Moreover,

|sn ´ tn| ď |sn| ` |tn| ď f` ` f´ “ |f |

Theorem (Lusin’s Theorem) - Let X be a metric space, µ a finite regular measure. If f : X Ñ R isµ-measurable, then for every ε ą 0, there is g : X Ñ R continuous such that µptf ‰ guq ă ε.Lemma 1 - For each ε ą 0, there is closed C Ď X such that µpXzCq ă ε and f æC : C Ñ R iscontinuous.Lemma 2 (Tietze Extension Theorem: Easy Version) - Let C Ď X be closed, h : C Ñ R continuous.Then there is continuous H : X Ñ R such that H æC“ h.Proof (Lusin’s Theorem) - Follows from Lemmas 1 and 2.

Proof (Lemma 2) - Define Hpxq “

#

hpxq x P C

infcPCthpcq ` dpx,cq

dpx,Cq ´ 1u x R C. Not hard to check that this is

continuous. Note, this only works if h is bounded below on C. If h is unbounded, then considerarctanphq instead.

Proof (Lemma 1) - Case I: Suppose that f is bounded. Without loss of generality, f : X Ñ r0, 1s.

Set Ak,n “ f´1pr k2n ,k`12n qq, where 0 ď k ď 2n. For all k, n, there exist compact sets Kk,n Ď Ak,n

such thatµpAk,nzKk,nq ă

ε2np2n`1q

Now set Cn “Ť

0ďkď2nKk,n. Note that by construction, we have µpCcnq ă

ε2n by construction, since

µpCcnq ďř

k

µpAk,nzKk,nq. Now set C “Ş

nCn.

Claim This set C works.


Proof Clearly C is closed. Moreover, µpCcq ďř

nµpCcnq “ ε. To see that f æC

is continuous, denote sn “ř

k

k2nχAk,n . Note that sn : Cn Ñ R is continuous (by

the pasting lemma for continuous functions) since Cn “Ť

K

Kk,n and sn æKk,n is

constant. Since C Ď Cn, then sn : C Ñ R is continuous.

On C, |f ´ sn| ď1

2n , so sn Ñ f uniformly. Thus, f is continuous on C.

Remark - This theorem is still true if X “ Rd, µ “ λ because we only need the sets Kk,n to be closed(not necessarily compact) such that λpAk,nzKk,nq ă

ε2np2n`1q and the rest of the proof is identical.


Lecture 12 - 09/22/2014

Integration

Definition - Let s : X Ñ R is simple, measurable, and nonnegative, say s “ř

i

aiχAi , where

Ai “ s´1paiq. Defineż

X

s dµ “ÿ

i

aiµpAiq

with the convention 0 ¨ 8 “ 0.Definition - Given f : X Ñ r0,8s measurable, define

ż

X

f dµ :“ sup0ďsďf

s simple meas.

ż

X

s dµ

Definition - Given f : X Ñ r´8,8s measurable, define f` “ maxpf, 0q, f´ “ maxp´f, 0q (note:these are measurable functions since x ÞÑ |x| is continuous and continuous˝measurable is measur-able).

1. Ifş

Xf` dµ,

ş

Xf´ dµ ă 8, we say f is integrable (or L1) and define

ż

X

f dµ “

ż

X

f` dµ´

ż

X

f´ dµ

2. If eitherş

Xf` dµ or

ş

Xf´ dµ is finite, we say f is integrable in the extended sense and defineż

X

f dµ “

ż

X

f` dµ´

ż

X

f´ dµ P r´8,8s

Notation - Depending on whether the measure or measure space are clear from context, we willsometimes write

ż

X

f dµ “

ż

f dµ,

ż

X

f “

ż

f

Example 1 - Let X “ N, µ the counting measure. All sets are measurable, so all functions f : X Ñ Rare measurable. Note a function a : N Ñ R is just a sequence of real numbers. What is

ş

N a dµ?Well, we’d like to say it’s just

ř

nan. However, since the integral is defined as the positive part minus

the negative part, we can only conclude thatş

N a dµ “ř

nan in the case that

ř

|an| ă 8.

Remark - The above example illustrates that we’ve somehow lost cancellation. Even a conditionally

convergent sum (such asř

n

p´1qn

n won’t haveş

N a dµ defined, since we split up a “ a` ´ a´.

Proposition 1 - Let s ě 0 be simple and measurable with spXq “ ta1, . . . , anu, Ai “ s´1paiq. Then

sup0ďtďs

t simple meas.

ż

X

t dµ “ÿ

i

aiµpAiq

Proof - The left-hand side is clearly at least as large as the right-hand side. For the reverse inequality,it’s enough to prove the following lemma.

Lemma Let s, t be simple and measurable with 0 ď s ď t. Thenż

X

s dµ ď

ż

X

t dµ


Proof Exercise.

Proposition 2 - If f, g are integrable, thenş

Xpf ` gq dµ “

ş

Xf dµ`

ş

Xg dµ.

Proof - We can’t do this yet! Let’s see what goes wrong. Start by assuming that f, g ě 0. Let0 ď s ď f , 0 ď t ď g, so that s` t ď f ` g. Now it’s not hard to show that

ż

X

s dµ`

ż

X

t dµ “

ż

X

ps` tq dµ ď

ż

X

pf ` gq dµ

Taking the supremum over 0 ď s ď f , 0 ď t ď g, we have

ż

X

f dµ`

ż

X

g dµ ď

ż

X

pf ` gq dµ

The problem is that we can’t easily get the reverse! What to do? (To be continued)Remark - To finish this off, we need some sort of limit theorem to take care of the sup in the definitionof

ş

Xf dµ.

Theorem (Lebesgue’s Monotone Convergence Theorem) - Say fn : X Ñ r0,8s are measurable withfn ď fn`1. Denote f “ lim fn (this exists because f is monotone). Then

lim

ż

X

fn dµ “

ż

X

f dµ

Proof - For all n, we haveş

fn ďş

fn`1 ďş

f (this follows from monotonicity of the integral. Inparticular, lim

ş

fn exists and limş

fn ďş

f .We need to show that lim

ş

fn ěş

f . To do this, let 0 ď s ď f simple, measurable. We wantlim

ş

fn ěş

s. First, we take a naıve approach (which we’ll fix in a moment). Write spXq “ta1, . . . , aNu, Ai “ s´1paiq. Set En “ tfn ě su. Then pEnq is increasing and now

ż

X

fn dµ ě

ż

X

χEnf dµ ě

ż

X

χEns dµ “Nÿ

i“1

aiµpAi X Enq

We’d like to take the limit n Ñ 8 and sy µpAi X Enq Ñ µpAiq to get the desired result. Theproblem is that this only works if

Ť

nEn “ X. This may not happen if the sequence pfnq doesn’t

cross s everywhere.The solution is to instead let 0 ă ε ă 1 and set En “ tfn ě p1 ´ εqsu. Now

Ť

nEn “ X. We

modify the above inequality and get

ż

X

fn dµ ě

ż

X

χEnf dµ ě

ż

X

χEnp1´ εqs dµ “ p1´ εqNÿ

i“1

aiµpAi X Enq

Taking the limit n Ñ 8, we have limş

fn ě p1 ´ εqş

s. Take ε Ñ 0` to get limş

fn ěş

s, asdesired.

Proposition (Linearity) - If f, g are integrable, then f ` g is integrable andş

pf ` gq “ş

f `ş

g.Proof - We do this in four steps.

(1) Let s, t ě 0 be simple and integrable. Thenş

ps` tq “ş

s`ş

t (exercise).


(2) Now let f, g ě 0 be measurable and integrable. Wantş

pf`gq “ş

f`ş

g. Last time, we showedthat there exist increasing sequences psnq, ptnq of nonnegative measurable simple functions withsn Ñ f and tn Ñ g pointwise. By MCT,

lim

ż

sn “ f, lim

ż

tn “ g ùñ

ż

pf ` gq “ lim

ż

psn ` tnq “ limp

ż

sn `

ż

tnq “

ż

f `

ż

g

(3) Lemma - Let f be integrable, f “ g ´ h, where g, h ě 0 are integrable. Then

ż

f “

ż

g ´

ż

h

Proof - Write f “ f` ´ f´ “ g ´ h. Now

f` ` h “ g ` f´ ùñ

ż

f` `

ż

h “

ż

g `

ż

f´ ùñ

ż

f “

ż

pf` ´ f´q “

ż

g ´

ż

h

(4) Now let f, g be arbitrary and integrable, say f “ f` ´ f´, g “ g` ´ g´. Now

pf ` gq “ pf` ` g`q ´ pf´ ` g´q

Applying the lemma from (3), we have

ż

pf ` gq “

ż

pf` ` g`q ´

ż

pf´ ` g´q “ p

ż

f` ´

ż

f´q ` p

ż

g` ´

ż

g´q “

ż

f `

ż

g


Lecture 13 - 09/24/2014

Last time, we defined the integral, proved the Monotone Convergence Theorem, and used it to provelinearity.

Quote - ”The selling point of the Lebesgue integral is the interchangeability ofş

and lim.”

Remark - Note that the MCT fails if we don’t require fn ě 0. Even if we assume fn ě ´1, forexample, consider the case X “ R, fn “ ´

1n . Now

ş

fn “ ´8 for all n, butş

lim fn “ 0.

Goal - If fn ě 0 and lim fn exists, want to say limş

fn “ş

lim fn.

Two Counterexamples

1. Mass escaping to infinity: Set fn “ χrn,n`1q. Then

lim

ż

fn “ 1 ‰ 0 “

ż

lim fn

2. Mass collects at a point: Set fn “ nχp0,

1n s

. Then

lim

ż

fn “ 1 ‰ 0 “

ż

lim fn

To see what goes wrong here, consider the following theorem.

Theorem (Lebesgue Dominated Convergence Theorem) - Suppose that

1. fn : X Ñ r0,8s is measurable.

2. fn Ñ f pointwise.

3. There is integrable F such that |fn| ď F for all n.

Then limş

fn “ş

f .Remark - This addresses the counterexamples above. In the first example, the ’envelope’ of pfnqis the function fpxq “ 1, which is not integrable. In the second example, the envelope of pfnq isfpxq “ 1

x , also not integrable.

To prove LDCT, we first need a lemma.

Fatou’s Lemma - Let fn ě 0 be measurable, fn Ñ f pointwise. Then

lim inf

ż

fn ě

ż

f

Quote - ”Mass can escape, but cannot be created.”Proof - Set gk “ inftfn | k ě nu. Note that

gk Ñ f and pgkq is increasing, nonnnegative, so by MCT,

lim

ż

gk ě

ż

f


Now fn ě gn for all n, soş

fn ěş

gn by monotonicity. Thus,

lim inf

ż

fn ě lim inf

ż

gn “ lim

ż

gn “

ż

f

Remark - We cannot conclude that limş

fn ěş

f , since limş

fn may not exist.

Proof (LDCT) - Set gn “ f ` fn. Now gn ě 0, so Fatou’s lemma implies

ż

F ` lim inf

ż

fn “ lim inf

ż

gn ě

ż

pF ` fq “

ż

F `

ż

f

Cancellingş

F , we have lim infş

fn ěş

f .Now set hn “ F ´ fn. Since hn ě 0, Fatou’s lemma implies

ż

F ´ lim sup

ż

fn “ lim inf

ż

hn ě

ż

pF ´ fq “

ż

F ´

ż

f

Rearranging, we have lim supş

fn ďş

f . Now

lim sup

ż

fn ď

ż

f ď lim inf

ż

fn ùñ lim

ż

fn “

ż

f

Quote - ”The only work we’ve done to prove this was in the Monotone Convergence Theorem, andeven that was only a few lines. The key is that we’ved coded all these results into the definition ofmeasure.”

Application - The Laplace Transform

Notation - We denoteş

Xfpxq dµpxq “

ş

Xf dµ. For example,

ş1

0x2 dλpxq is the integral of x2 with

respect to λ on r0, 1s. We denote integration with respect to λ by dλpxq “ dx. Why do we do this?IOU - If f : r0, 1s Ñ R is Riemann integrable, then f is L-integrable. Moreover, the Riemannintegral and Lebesgue integral agree with one another.Remark - An improper Riemann integral may disagree with the Lebesgue integral, since the Lebesgueintegral doesn’t allow for cancellation of an infinite positive part and an infinite negative part.

Suppose now that f : r0,8q Ñ R is integrable. Define the Laplace transform of f by

F psq “

ż 8

0

e´stfptq dt

Note: F : p0,8q Ñ R. Here’s an example of where Lebesgue integration is helpful.

Proposition

1. Ifş8

0|f | ă 8, then F is continuous.

2. Ifş8

0t|fptq| ă 8, then F is differentiable (in fact, C1).


Proof - For (1), given s0 P p0,8q, we have limsÑs0

“ limsÑs0

ş8

0e´stfptq dt. We want to apply LDCT. To

get F ps0q on the right-hand side, let psnq be a sequence with sn Ñ s0. Note that |e´sntfptq| ď |fptq|.Since

ş8

0|fptq| dt ă 8, then by LDCT, we have

limnÑ8

F psnq “ F ps0q

To prove (2), write F psq´F ps0qs´s0

“ş8

0e´sté´s0t

s´s0fptq dt. For any s ‰ s0, the Mean Value Theorem

implies that there is s1 such thate´stés0t

s´s0“ ´te´s1t

Thus, | e´stés0t

s´s0| ď t|fptq|, so since

ş8

0t|fptq| dt ă 8, then the LDCT implies that

limsÑs0

F psq´F ps0qs´s0

“

ż 8

0

´te´stfptq dt “ Lptfptqq

By (1), Lptfptqq is continuous, so we are done.


Lecture 14 - 09/26/2014

We begin with an application of MCT.

Example (Beppo Levi’s Theorem) - Let pfnq be a sequence of nonnegative measurable functions.Then

ż

X

ÿ

n

fn dµ “ÿ

n

ż

X

fn dµ

Proof - Set sN “ř

1ďnďN

fn. Note, 0 ď sn ď sn`1. Now apply the MCT.

Example - Let φ : RÑ r0,8s withş

R φ ă 8. Define

fpxq “ÿ

mPN,nPZ`φppx´ m

n q2m`nq

Note that

1. f “ 8 at all rationals if φp0q ą 0.

2. f ă 8 a.e. To see this, note that Beppo Levi impliesż

Rf “

ÿ

m,n

ż

Rφppx´ m

n q2m`nq “

ÿ

m,n

2´pm`nqż

Rφ ă 8

In particular, f ă 8 a.e.

We’ll spend most of the next week thoroughly studying convergence. Before that, let’s mentiona few things.

Pushforward Measures

Definition - Let pX,Σ, µq be a measure space, f : X Ñ Y arbitrary. Define Σ1 Ď PpY q by

Σ1 “ tA | f´1pAq P Σu

Define ν on Σ1 by νpAq “ µpf´1pAqq.Proposition - ν is a measure on pY,Σ1q. Moreover, if g : Y Ñ r´8,8s is integrable, then

ż

Y

g dν “

ż

X

g ˝ g dµ

Proof - We’ve already seen that Σ1 is a σ-algebra. Moreover, it’s easy to check that ν is a measure.To prove the statement about integrability, it’s enough to show it for simple functions by the MCTand linearity. In fact, we may consider s “ χA, where A P Σ1. Now

ż

Y

s dν “ νpAq “ µpf´1pAqq “

ż

X

χf´1pAq dµ “

ż

X

s ˝ f dµ

Nowş

Yg dν “

ş

Xg ˝ f dµ for all g positive and simple. Thus, this is true for any integrable g.

Application - Fix x0 P Rd, f : Rd Ñ Rd integrable. Then

ż

Rdfpxq dx “

ż

Rdfpx` x0q dx


Proof - Let gpxq “ x`x0. Equip the second Rd with the pushforward measure µ induced by pRd, λq.Now µpAq “ λpg´1pAqq “ λpAq, so by translation invariance, µ “ λ. The previous proposition nowyields

ż

Rdfpxq dλpxq “

ż

Rdfpxq dµpxq “

ż

Rdfpx` x0q dλpxq

It’s time for one last remark about integration.

Proposition - Let pX,Σ, µq be complete and let f : X Ñ R be integrable, f “ g a.e. Then g isintegrable and

ş

f “ş

g.Proof - We already know g is measurable. To prove that g is integrable, set h “ f ´ g. Then h “ 0a.e. and h is measurable.

Claimş

h “ 0.

Proof Decompose h “ h` ´ h´. We’ll show thatş

Xh` “

ş

Xh´ “ 0. To do this,

let 0 ď s ď h` be simple, say s “ř

aiχAi , where the sets Ai are disjoint andmeasurable. If ai ‰ 0, then Ai Ď th

` ą 0u, so µpAiq “ 0. Thus,ş

Xs dµ “ 0, so

nowş

Xh` dµ “ 0. Similarly,

ş

Xh´ “ 0. This finishes the proof.

Now h integrable implies f ´ h “ g integrable, andş

g “ş

pf ´ hq “ş

f .Remark - In light of this remark, the MCT and LDCT (for Rd) are still true if we replace fn Ñ fwith fn Ñ f a.e.

Convergence

Theorem (Egorov’s Theorem) - Let pX,Σ, µq be a measure space, µpXq ă 8. Let pfnq be a sequenceof measurable functions with fn Ñ f pointwise a.e. Then for any ε ą 0, there is A Ď X such thatµpXzAq ă ε and fn Ñ f uniformly on A.Remark - In general, we can’t do better, i.e. we can’t get µpAcq “ 0.Proof (of theorem) - For each k, there is n0 such that n ą n0 implies |fn ´ f | ă 1

k a.e. This isbecause

µpď

m

č

něm

t|fn ´ f | ă1k uq “ µpXq

In particular, for each k there is nk such that

µpč

něnk

t|fn ´ f | ă1k uq ě µpXq ´ ε

2k

Now set A “Ş

k

Ak. Then

• µpAq ě µpXq ´ ε.

• Fix k P N. For all a P A, we have a P Ak, so |fnpaq ´ fpaq| ă 1k for all n ě nk. Since k was

arbitrary, then fn Ñ f uniformly on A.


Lecture 15 - 09/29/2014

Definition (a.e. pointwise convergence) We say fn Ñ f µ-a.e. if there exists a µ-null set N P Σ suchthat fn Ñ f pointwise on N c, i.e. if tfn Û fu is null.

Definition (convergence in measure) We say fn Ñ f in µ-measure (and write fnµÑ f) if for every

ε ą 0, we havelimnÑ8

µptx P X | |fn ´ f | ą εuq “ 0

Definition (Lp convergence) We say fn Ñ f in L1 ifş

X|fn ´ f | dµÑ 0 as nÑ 8. More generally,

for p ě 1, we say fn Ñ f in Lp (and write fnLpÑ f) if

limnÑ8

ż

|fn ´ f |p “ 0

Later we will define a similar notion for p “ 8.

Intuition - Think of Lp as convergence in a metric space, where the metric is

dpf, gq “ p

ż

X

|f ´ g|p dµq1p

(We need p ě 1, otherwise the triangle inequality fails.)

Example - To see that fn Ñ f a.e. does not imply fnµÑ f , consider fn “ χrn,n`1s. Then fn Ñ 0

a.e. but µpt|fn ´ 0| ą εuq “ 1 for all n and 0 ă ε ă 1.

Proposition 1 - If µpXq ă 8 and fn Ñ f a.e., then fn Ñ f in measure.Proof - Let ε ą 0. Since fn Ñ f a.e., then χt|fn´f |ąεu Ñ 0 a.e. Moreover, χt|fn´f |ąεu ď 1, which isintegrable since µpXq ă 8. By LDCT, we have

limnÑ8

µpt|fn ´ f | ą εuq “ limnÑ8

ż

χt|fn´f |ąεu “ 0

Remark - This can also be proved using Egorov’s theorem.

Question - Does the converse hold? I.e. does fnµÑ f imply fn Ñ f a.e.?

Answer - No! For example, f1 “ χr0,

12 q

, f2 “ χr12 ,1q

, f3 “ χr0,

14 q

, f4 “ χr14 ,

12 q

, etc. Set f “ 0.

Then clearly µpt|fn ´ f | ą εuq Ñ 0 for all ε ą 0, so fn Ñ f in measure. However, for all x P r0, 1q,fnpxq “ 0 infinitely often and fnpxq “ 1 infinitely often, so fn fails to converge a.e.

Proposition 2 - Suppose fn Ñ f in measure. Then there is a subsequence pfnkq Ď pfnq with fnk Ñ fa.e.Proof - Let n1 be such that µpt|fn ´ f | ą 1uq ă 1

2 for n ě n1. Let n2 ą n1 such that µpt|fn ´ f | ą12uq ă

14 . Repeat this process, choosing nk ą nk´1 such that µpt|fnk ´ f | ą

1k uq ă

12k

.

Claim fnk Ñ f a.e.

Proof Let Ak “ t|fnk ´ f | ą 1k u, A “ tx | x P Ak only finitely many timesu. By

definition, if x P A, then fnkpxq Ñ fpxq. Now it suffices to show that µpAcq “ 0.Write

Ac “ tx | x P Ak infinitely oftenu “č

kě1

ď

něk

An


Now for any k,µpAcq ď µp

ď

něk

Anq ďÿ

něk

12n “

12k´1

Let k Ñ8, then µpAcq “ 0.

Now let’s investigate Lp convergence.

Definition - Let V a vector space. We say ¨ is a norm on V if

1. v ě 0 for all v P V and v “ 0 if and only if v “ 0.

2. For all α P R, αv “ |α|v.

3. u` v ď u ` v for all u, v P V (triangle inequality).

Given a norm ¨ on V , define d : V ˆ V Ñ R by dpu, vq “ u´ v. It’s not hard to check that d isa metric, so now pV, dq is a metric space.

Definition - Define L1pX,Σ, µq “ tf : X Ñ r´8,8s |ş

X|f | dµ ă 8u. Clearly L1 is a vector space.

We attempt to define a norm on L1 by

f1 “

ż

X

|f | dµ

This clearly satsfies properties (2) and (3) of norms, but f1 “ 0 only implies that f “ 0 a.e., notthat f “ 0. This motivates the following definition.Definition - Define an equivalence relation „ on L1 by f „ g if f “ g a.e. Set L1 “ L1 „. Now forF P L1, F represented by some function f , define F 1 “ f1. This is well-defined by definition of„.


Lecture 16 - 10/01/2014

Recall - In Lp spaces, convergence fnLpÑ f is defined as lim

nÑ8

ş

X|fn ´ f |

p dµ “ 0.

Exercise - |f | ď f8 a.e.

Henceforth, assume that pX,Σ, µq is complete.Definition - LppX,µq :“ tf : X Ñ r´8,8s | fp ă 8u.

Note that fp “ 0 only implies f “ 0 a.e., not f “ 0. To make ¨p a norm, define an equivalencerelation „ on Lp by f „ g if and only if f “ g a.e. Now define LppX,µq “ LppX,µq „. For F P Lp,define F p “ fp, where f is any representative of F . This is well-defined by definition of „.

Given F,G P LppX,µq, define F `G “ rf ` gs, where F “ f „, G “ g „. This is well-defined.Remark - We can’t meaningfully talk about F pxq since f „ g doesn’t imply fpxq “ gpxq. We canmeaningfully define

ş

AF , tF ą au, etc. In general, we treat elements of Lp as functions, only doing

operations that are constant on equivalence classes.

Goal 1 - Lp is a Banach space.Definition - Let pV, ¨ q be a vector space, dpu, vq “ u´ v. Now pV, dq is a metric space. We saypV, ¨ q is a Banach space if pV, dq is complete.E.g. Rd is a Banach space (with the standard norm).

Theorem For all p P r1,8s, LppX,µq is a Banach space.Proof - We only need to show

1. f ` gp ď fp ` gp

2. Completeness

The rest is easy. IOU this proof.

Lemma (Holder’s Inequality) - Let p, q P r1,8s such that 1p `

1q “ 1. If f P Lp, g P Lq, then

|ş

fg| ď fpgq. In particular, fg P L1.Remark - p “ q “ 2 is Cauchy-Schwarz.Intuition - The relationship between p and q follows by ”counting dimensions”. Say X “ Rd, f, gare dimensionless quantities. Then

ş

Rd fg has dimension Ld, fp has dimension Ldp, and gq has

dimension Ldq. We want Ld “ LdpLdq, which holds if and only if 1p `

1q “ 1.

To formalize this argument (see homework), replace fpxq by fpλxq. Do a change of variablesand then take λ large and λ small to see that the inequality only holds when 1

p `1q “ 1.

Induction Proof (Holder) - The main idea is to show that for real numbers xi, yi, i “ 1, . . . , N , we

have |ř

i

xiyi| ď př

i

|xi|pq1pp

ř

i

|yi|qq1q. Now let s, t be simple functions and write s “

ř

i

xiχEi ,

t “ř

i

yiχEi , where xi, yi P R and the sets tEiu are disjoint. Writing µpEiq “ µpEiq1pµpEiq

1q, we

havest1 “ |

ÿ

i

µpEiqxiyi| ď |ÿ

i

|xi|pµpEiq|

1p|ÿ

i

|yi|qµpEiq|

1q “ sptq

Now the result holds for simple functions, so use approximation and monotone convergence to finishthis off.

Better Proof - First assume that p, q P p1,8q. The proof starts with Young’s Inequality, which is asfollows.


Claim (Young’s Inequality) Let x, y ě 0 and let p, q P p1,8q, 1p `

1q “ 1. Then

xy ď xp

p `yq

q .

Proof This follows from convexity of the function hpzq “ lnpzq, since

lnpxyq “ lnpxpqp `

lnpxqqq ď lnpx

p

p `yq

q q

To prove Holder, first note that if fp “ 0 or gq “ 0, then f “ 0 or g “ 0, so we are done.

Assume now that fp, gq ‰ 0. Set F “ ffp

, G “ ggq

, so that F p “ Gq “ 1. By monotonicity

and Young’s inequality, we have

|

ż

FG| ď

ż

|F ||G| ďş

|F |p

p `

ş

|G|q

q “ 1

Multiplying by fpgq, we have |ş

fg| ď fpgq.The case p “ 1, q “ 8 is left as an exercise.

Lemma (Duality) - Let p P r1,8q, f P Lp, 1p `

1q “ 1. Then

fp “ supgPLqzt0u

1gq

ż

fg p˚q

Remark - If p “ 8, this is still true provided that X is σ-finite.Proof - By Holder, for any g P Lqzt0u, we have 1

gq

ş

fg ď fp. Thus, LHS ě RHS in p˚q. For

the reverse inequality, set F “ ffp

. It’s enough to show that supGq“1

ş

FG “ 1. To get this, choose

G “ |F |p´1signpF q. Clearlyş

FG “ş

|F |p “ 1. Moreover,

ż

|G|q “

ż

|F |pp´1qq “

ż

|F |pq´q “

ż

|F |p “ 1


Lecture 17 - 10/03/2014

IOU - Lp is a Banach space. We still need the triangle inequality and completeness. We’ll prove thefirst today.

Last time, we saw the following:

1. If p, q P r1,8s, 1p `

1q “ 1, then |

ş

fg| ď fpgq.

2. We also have a duality principle: Given p P r1,8q, f P Lp, we have

fp “ supqPLqzt0u

1gq

ż

fg

3. The previous statement holds for p “ 8 if X is σ-finite.

Proposition - Let f, g P Lp. Then f ` g P Lp and

f ` gp ď fp ` gp

Proof - If p “ 8, this is easy. So assume p P r1,8q. The function x ÞÑ |x|p is convex, so

12p |f ` g|

p “ | 12f `12g|

p ď 12 |f |

p ` 12 |g|

p

Since the right-hand side is integrable, so is the left-hand side, thus f ` g P Lp.For the triangle inequality, let q be such that 1

p `1q “ 1. By duality, we have

f ` gp “ suphPLqzt0u

1hq

ż

pf ` gqh

ď suphPLqzt0u

1hq

ż

fh` suphPLqzt0u

1hq

ż

gh

“ fp ` gp

Quote - ”Duality gets you an amazing amount of mileage.”

Before we prove completeness, let’s discuss another useful inequality. On your homework, you’llshow that this can be used to prove Holder’s inequality. First recall that if φ : R Ñ R is convex,then given a finite set of points ci P r0, 1s with

ř

i

ci “ 1, we have φpř

i

cixiq ďř

i

ciφpxiq. Loosely

speaking, Jensen’s inequality extends this result to integrals.Jensen’s Inequality - Let φ be a convex real-valued function, µpXq “ 1. Then for any f P L1pXq,we have

φp

ż

X

f dµq ď

ż

X

φ ˝ f dµ

Remark - If φ is convex, then φ is continuous. In particular, φ is Borel, so φ˝f is measurable. (Thisjustifies well-definedness of the expression on the right.)

Completeness of Lp

Lemma - Let p P r1,8q. Say pfnq Ď Lp such thatř

nfnp ă 8. Then

1.ř

nfn converges pointwise a.e., say to f .


2. f P Lp andř

nfn converges in Lp to f .

Proof - Define sN “ř

1ďnďN

fn, tN “ř

1ďnďN

|fn|. For (1), we must show that sN converges and is

finite a.e., for which it’s enough to show that tN converges and is finite a.e.Since the functions ptnq are increasing and nonnegative, we can set F “

ř

n|fn| “ lim

NÑ8tN . By

the triangle inequality, we have

tN p ďÿ

1ďnďN

fnp ďÿ

n

fnp ă 8

Nowş

tpN “ tN pp ď p

ř

nfnpq

p ă 8, so by the MCT, we have

limNÑ8

ż

tpN “

ż

F p ă 8

Thus F P Lp. In particular, F is finite a.e., so now (1) holds.For (2), note that |f | ď F implies fp ď F p, so now f P Lp. Moreover, by the MCT and the

triangle inequalities for Lp and R, we have

f ´ sN p “ ÿ

něN`1

fnp ď ÿ

něN`1

|fn|p “ limmÑ8

ÿ

měněN`1

|fn|p

ď limmÑ8

ÿ

měněN`1

fnp

“ÿ

něN`1

fnp Ñ 0 as N Ñ8

Thus, sN Ñ f in Lp.

Proposition - Lp is complete.Proof - Let pfnq Ď Lp be Cauchy. It’s enough to find a subsequence pfnkq of pfnq such that pfnkqconverges in Lp. To do this, choose n1 ă n2 ă ¨ ¨ ¨ such that fnk`1

´ fnkp ă12k

. By the lemma,since

ř

k

fnk`1´ fnkp ă 1, then

ř

k

pfnk`1´ fnkq converges both a.e. and in Lp. Set

f “ÿ

k

pfnk`1´ fnkq ` fn1

Let the sum telescope, so that fnkLpÑ f .

Exercise - Prove that L8 is complete.

Convergence - We have convergence a.e., convergence in measure, convergence in Lp. What’s therelationship between these notions?

• We know a.e. convergence implies convergence in measure if µpXq ă 8 (but not in general).

• We know convergence in measure implies a.e. convergence along a subsequence.

• In general, a.e. convergence does not imply convergence in Lp, even on finite measure spaces.For example, X “ p0, 1q, fn “ 2nχ

p0,1n q

.

• However, convergence in Lp does imply a.e. convergence along a subsequence (similar to seconditem above).


• Convergence in Lp implies convergence in measure (next time, using Chebyshev).

• Convergence in measure does not imply convergence in Lp (next time).

Lemma (Chebyshev’s Inequality) - Let f be meaurable. Then for all λ ą 0,

µp|f | ą λq ď 1λf1

Proof - Clearly λχ|f |ąλ ď |f |, so integrate both sides and we’re done.


Lecture 18 - 10/06/2014

Proposition - pfnqLpÑ f , p P r1,8q implies fn

µÑ f .

Proof - Use Chebyshev. We have

µpt|fn ´ f | ą εuq “ µpt|fn ´ f |p ą εpuq ď 1

εp fn ´ fpp Ñ 0 as nÑ8

IOU - fnµÑ f + ”mystery assumption” is equivalent to fn

LpÑ f .

Proposition 1 - S “ ts | s simple, µpts ‰ 0uq ă 8u is dense in Lp for all p P r1,8s.Proof - Let f P Lp. If p P r1,8q, let psnq be sequence of simple functions with sn Ñ f a.e. and|sn| ď f a.e. Note, f P Lp implies sn P L

p, so then µptsn ‰ 0uq ă 8 for each n. Now by the LDCT,

sn ´ fpp “

ż

|sn ´ f |p Ñ 0

since |sn ´ f |p ď p|f | ` |f |qp “ 2p|f |p, which is integrable.

If p “ 8, then partition r´f8, f8s into finitely many subintervals tAiu of length at most ε,then set s “

ř

i

xiχf´1pAiq, where xi is the midpoint of Ai. Now |f ´ s| ă ε a.e., so f ´ s8 ď ε.

Proposition 2 - Let X be a σ-compact metric space (i.e. X “Ť

nKn, where the sets pKnq are

increasing and compact). If µ is regular on X, then CcpXq is dense in Lp.Remark/Definition - CcpXq “ tcontinuous functions on X with compact supportu.Proof - First assume that X is compact. Let f P Lp and fix ε ą 0. For each n ě 1, define

fn “

$

’

&

’

%

f |f | ď n

n f ą n

ń f ă ń

(Think of this as a truncated version of f .) By LDCT, fnLpÑ f since fn ď |f |. Let N sufficiently

large so that fN ´ fp ăε2 . Choose δ ą 0 sufficiently small so that δp2nqp ă p ε2 q

p. By Lusin’stheorem, there is continuous gn such that µptfn ‰ gnuq ă δ. By truncating if necessary, we mayassume |gn| ď n, so now gn has compact support. Moreover,

fn ´ gnpp “

ż

tfn‰gnu

|fn ´ gn|p ď δp2nqp ă p ε2 q

p

Now gn ´ fp ď gn ´ fnp ` fn ´ fp ăε2 `

ε2 “ ε, so we’re done.

The case that X is σ-compact is left as an exercise.

Proposition 3 - If Σ “ σpCq, where C is countable, then LppXq is separable for p P r1,8q.Proof - Homework.

Theorem (Vitali) - Let f P L1pXq, pfnq Ď L1pXq. Then fnL1

Ñ f if and only if

1. fnµÑ f

2. For any ε ą 0, there is δ ą 0 such that µpAq ă δ impliesş

A|fn| ă ε for all n.

3. For any ε ą 0, there is F P Σ such that µpF q ă 8 andş

F c|fn| ă ε for all n.


Quotes ”Condition (2) prevents mass collecting at a point. Condition (3) prevents mass escaping toinfinity.”Remark - Condition (2) is sometimes referred to as equi-integrability. Condition (3) is sometimesreferred to as tightness.Proof (of Vitali) - Suppose that fn Ñ f in L1. We already have fn

µÑ f .

To show equi-integrability, let n0 be such that n ą n0 impliesş

X|f ´ fn| ă

ε2 . Now for any set

A Ď X, we haveż

A

|fn| ď

ż

A

|f ´ fn| `

ż

A

|f | ă

ż

A

|f | ` ε2

If µpAq ă δ, then for any λ ą 0,

ż

A

|f | “

ż

AXt|f |ďλu

|f | `

ż

AXt|f |ąλu

|f | ď λδ `

ż

t|f |ąλu

|f |

By LDCT, we have limλÑ8

ş

t|f |ąλu|f | “ lim

λÑ8

ş

Xχt|f |ąλu|f | “ 0. Now let λ be sufficiently large such

thatş

t|f |ąλu|f | ă ε

4 and let δ ą 0 be sufficiently small such that λδ ă ε4 . Now

µpAq ă δ, n ą n0 ùñ

ż

A

|fn| ă ε

To get the result for n ď n0, repeat the above argument (the one used to minimizeş

A|f |) for the

finitely many functions fn, n ď n0.Next time, we’ll prove that pfnq is tight and prove the converse of the theorem.


Lecture 19 - 10/08/2014

Proof (of Vitali, continued) - We need to show that fnL1

Ñ f implies pfnq is tight. Fix ε ą 0. For anyF , we have

ż

F

|fn| ď

ż

F

|f | `

ż

F

|fn ´ f | ă

ż

F

|f | ` ε2

for n sufficiently large, say n ě n0 (where n0 is independent of F ).

Set F “ t|f | ą λu (we’ll choose λ later). By Chebyshev’s inequality, we have µpF q ď f1λ ă 8.

Nowş

F c|f | “

ş

|f |χt|f |ďλu. As λÑ 0`,ş

|f |χt|f |ďλu Ñ 0 by LDCT, so let λ be sufficiently small sothat

ş

|f |χt|f |ďλu ăε2 . Repeat the same trick for the finitely many functions f1, . . . , fn0

, then we’redone.

Conversely, suppose that fnµÑ f , pfnq is equi-integrable, and pfnq is tight. Fix ε ą 0. Then for

any set F and λ ą 0, we have

ż

|fn ´ f | “

ż

F

|fn ´ f | `

ż

F c|fn ´ f |

“

ż

FXt|fn´f |ďλu

|fn ´ f | `

ż

FXt|fn´f |ąλu

|fn ´ f | `

ż

F c|fn ´ f | p˚q

Since pfnq is tight and f P L1, then pfnq Y tfu is tight. Choose F such that µpF q ă 8 andş

F cp|f | ` |fn|q ă

ε3 , which is an upper bound for the third term of p˚q.

The first term is bounded above by λµpF q, so let λ ą 0 sufficiently small such that λµpF q ă ε3 .

This bounds the first term.Since pfnq is equi-integrable and f P L1, then pfnq Y tfu is equi-integrable. Let δ ą 0 such that

µpAq ă δ impliesş

Ap|fn| ` |f |q ă

ε3 . Finally, since fn

µÑ f , let n0 be sufficiently large such that

n ą n0 implies µpt|fn ´ f | ą λuq ă δ. This bounds the second term above by ε3 , finishing the

proof.Remark - If µpXq ă 8, then any family of L1 functions is tight.Remark - Let F P L1, |fn| ď F for all n. Then

1. pfnq is equi-integrable.

2. pfnq is tight.

Proposition

1. Say limλÑ8

supn

ş

t|fn|ąλu|fn| “ 0. Then pfnq is equi-integrable.

2. (de la Vallee-Poussin) Suppose there is increasing φ : r0,8q Ñ r0,8q such that limxÑ8

φpxqx “ 8

and supn

ş

Xφp|fn|q ă 8. Then pfnq is equi-integrable.

Proof - (1) Fix ε ą 0. Suppose µpAq ă δ (we’ll choose δ ą 0 later). Then

ż

A

|fn| “

ż

AXt|fn|ďλu

|fn| `

ż

AXt|fn|ąλu

|fn| ď λµpAq `

ż

t|fn|ąλu

|fn|

Let λ be sufficiently large so thatş

t|fn|ąλu|fn| ă

ε2 . Now let δ ą 0 be sufficiently small so that

λδ ă ε2 .


(2) By (1), it’s enough to show that limλÑ0

supn

ş

t|fn|ąλu|fn| “ 0. Let C “ sup

n

ş

Xφp|fn|q ă 8. Let

λ0 be such that λ ą λ0 implies λ ă εφpλq. Now

ż

t|fn|ąλ0u

|fn| ď

ż

t|fn|ąλ0u

εφp|fn|q ď εC

Take the supremum over n and then take the limit λ0 Ñ8, so now

limλ0Ñ8

supn

ż

t|fn|ąλ0u

|fn| “ 0

”Classic” Example - Say µpXq ă 8, fn Ñ f a.e., and supnfnp ă 8 for some p ą 1. Then fn Ñ f

in L1. This follows from (2) above with φpxq “ xp. Why?Since µpXq ă 8, then a.e. convergence implies convergence in measure, and we also have

tightness for free. We get equi-integrability by (2) with φpxq “ xp.


Lecture 20 - 10/13/2014

Plan - Signed measures, Hahn decompositionSetting - pX,Σq a measurable space

Definition - Let µ : Σ Ñ p´8,8s or r´8,8q. We say µ is a signed measure if µ is countablyadditive and µpHq “ 0.E.g. Let µ1, µ2 be positive measures, µ2pXq ă 8. Then µ :“ µ1 ´ µ2 is a signed measure.E.g. 2 - Let ν be a positive measure, f : X Ñ R measurable, and suppose that f´ “ maxp0,´fq P

L1. Then define

µpAq “

ż

A

f` dν ´

ż

A

f´ dν, A P Σ

µ is a signed measure.Proof - Countable additivity follows from monotone convergence.

Motivation - We’d like to form a normed vector space of measures. We can’t do this with positivemeasures alone.Remark - This construction has applications to finding solutions of stochastic differential equations.

Definition - Let µ be a signed measure on pX,Σq. We say A P Σ is µ-positive if for every E Ď Awith E P Σ, we have µpAq ě 0, i.e. when µ is restricted to A, we get a positive measure. We defineµ-negative sets similarly.Notation - Denote Ppµq “ tµ-positive setsu, N pµq “ tµ-negative setsu.

Remarks

1. If A P Σ such that µpAq is finite, then µpEq is finite for any E Ď A in Σ. This follows fromwriting

µpAq “ µpEq ` µpAzEq

2. Ppµq and N pµq are closed under countable unions and intersections, as well as (relative)complements.

3. For any P P Ppµq and N P N pµq, P and N are µ-disjoint. That is, for all E Ď P X N ,µpEq “ 0.

Theorem (Hahn Decomposition) - Let µ be a signed measure on pX,Σq. Then there exist P P Ppµqand N P N pµq such that X “ P YN .Proof - Without loss of generality, assume µ ą ´8. Denote

L “ inftµpAq : A P N pµqu

(Note: There is at least one such A, namely A “ H.)

Claim L is finite. Moreover, the infimum is witnessed.

Proof Let pAnq Ď N pµq such that µpAnq Ñ L. Consider the union N “Ť

nNn. By

a previous remark, N P N pµq. Thus, L ď µpNq by definition. On the other hand,µpNq “ µpAnq ` µpNzAnq ď µpAnq for each n, so µpNq ď L. Thus, µpNq “ L. Inparticular, L is finite (since µ ą ´8).


Set P “ XzN . It remains to show that P P Ppµq. We show this by contradiction.Suppose that P R Ppµq. Let A Ď P in Σ such that µpAq ă 0. If A P N pµq, we have a

contradiction, sinceµpAYNq ď µpAq ` µpNq ă µpNq “ L,

contradicting the definition of L. If not, the following lemma will yield the same contradiction.

Lemma Let A P Σ, ´8 ă µpAq ă 0. Then there is E Ď A, E P N pµq, such thatµpEq ď µpAq.

Proof The idea is to cut out the parts of A that are positive. We do this as follows.Set

δ1 “ suptµpEq : E Ď A, E P Σu

If δ1 “ 0, then A P N pµq, so take E “ A. If not, δ1 P p0,8s, so choose A1 Ď A inΣ such that µpA1q ě minp δ12 , 1q.

Continue recursively. At step n, we have

(a) δ1 ě δ2 ě ¨ ¨ ¨ ě δn´1 ą 0

(b) disjoint subsets pAkqn´1k“1 of A such that µpAkq ě minp δk2 , 1q.

Define δn “ suptµpEq : E Ď AzpŤ

iAiq, E P Σu. If δn “ 0, stop and set E “

AzpŤ

iAiq. Otherwise, δn P p0,8s, so let An Ď pAzŤ

iAiq with µpAnq ě minp δn2 , 1q.

If this process stops at any point, we are done. Otherwise, set E :“ AzŤ

ně1An.

Why does this work? First write

µpEq “ µpAq ´ÿ

n

µpAnq ă µpAq p˚q

It remains to argue that E P N pµq. This boils down to showing that δn Ñ 0 asn Ñ 8. Let D Ď E in Σ with µpDq ě 0. Then for each n, D Ď Az

Ť

1ďkďn

Ak. In

particular, µpDq ď δn. We are done, provided we show that δn Ñ 0. But this followsfrom p˚q above. Since µpEq is finite, then

ř

nµpAnq is finite, so

ř

nminp δn2 , 1q ă 8,

so δn Ñ 0.


Lecture 21 - 10/15/2014

Plan: Jordan decomposition, MpX,ΣqSetting - pX,Σq is a measurable space, µ a signed measure with Hahn decomposition pP,Nq, whereP XN “ H, X “ P YN , P µ-positive and N µ-negative.

Definition - We say signed measures µ1 or µ2 are singular, denoted µ1 K µ2, if there exist disjointX1, X2 P Σ such that for all A P Σ,

µ1pAq “ µ1pAXX1q, µ2pAq “ µ2pAXX2q

In this case, we say µ1 is concentrated on X1 and µ2 is concentrated on X2.Theorem (Jordan Decomposition) - Given µ, there exist unique (positive) measures µ` and µ´ suchthat

(a) µ “ µ` ´ µ´.

(b) At least one of µ`, µ´ is finite.

(c) µ` K µ´.

Remark This uniqueness is in some sense more advantageous than the Hahn decomposition, since wemay able to change the sets P and N in the Hahn decomposition, i.e. P and N are only essentiallyunique. By contrast, µ` and µ´ are unique by (c).Proof - Existence is easy. Let pP,Nq be a Hahn decomposition for µ. Define

µ`pAq “ µpAX P q, µ´pAq “ ´µpAXNq

By definition, µ` and µ´ satisfy (a)-(c).For uniqueness, let µ` and µ´ satisfy (a)-(c). We show that for any A P Σ, we have

µ`pAq “ supEĎA,EPΣ

µpEq p˚q

(ě) For any E Ď A in Σ, we have

µ`pAq ě µ`pEq “ µpEq ` µ´pEq ě µpEq

(ď) Let pX`, X´q be such that µ` is concentrated on X` and µ´ is concentrated on X´. Now

µ`pAq “ µ`pAXX`q “ µpAXX`q ` µ´pAXX`q “ µpAXX`q

Take E “ AXX` and p˚q is proven.From p˚q, we get µ`. To get µ´, simply apply p˚q again, replacing µ by ´µ. This works because

´µ “ µ´ ´ µ`, so p´µq` “ µ´.

Definition - Given the Jordan decomposition µ “ µ` ´ µ´, the total variation of µ is defined by

|µ| :“ µ` ` µ´

If |µ|pXq ă 8, then µ “ |µ|pXq is called the total variation norm of µ.Remark - We’ll se that ¨ is a norm on the space of bounded total variation measures.E.g. - Let ν be a measure, f P L1pνq. Define µpAq “

ş

Af dν “

ş

Af` dν ´

ş

Af´ dν. In this case,

|µ|pAq “

ż

A

|f | dν, µ “

ż

X

|f | dν “ fL1pνq


Remark/Quote - ”The space of signed measures is an extension of L1-space.”

Theorem - The spaceMpX,Σq of finite signed measures, equipped with the total variation norm, isa Banach space.Proof - Everything but Banach is easy. Let pµnq ĎMpX,Σq be Cauchy, i.e.

µn ´ µm Ñ 0 as n,mÑ8 p˚˚q

In this case, supnµn ă 8 (by the triangle inequality, Cauchy sequences are bounded).

Observe that for any ν PMpX,Σq, we have

ν “ ν` ` ν´ “ ν`pXq ´ ν´pXq

“ supdisjointA1,A2PΣ

tνpAq ´ νpA2qu

“ supdisjointA1,A2PΣ

t|νpA1q| ` |νpA2q|u

ď 2 supAPΣ

|νpAq|

In particular, we have shown that supAPΣ

|νpAq| ď ν ď 2 supAPΣ

|νpAq|. Thus, p˚˚q is equivalent to

supAPΣ

|µn ´ µmpAq| Ñ 0 as n,mÑ8 p˚ ˚ ˚q

Thus for fixed A P Σ, pµnpAqq is a Cauchy sequence of real numbers, so define µpAq :“ limnÑ8

µnpAq.

By p˚ ˚ ˚q, we havesupAPΣ

|µpAq ´ µnpAq| Ñ 0 as n,mÑ8

It remains to show that µ is a signed measure.To get finite additivity, it suffices to consider two disjoint sets A,B P Σ. We have

|µpAYBq ´ pµpAq ` µpBqq|

ď |µpAYBq ´ µnpAYBq| ` |µnpAq ´ µpAq| ` |µnpBq ´ µpBq|

ď 3 supEPΣ

|µpEq ´ µnpEq| Ñ 0

as nÑ8. Now for σ-additivity, let pAkq Ď Σ be disjoint. Write

|µpď

k

Akq ´mÿ

k“1

µpAkq|

ď pm` 1q supEPΣ

|µpEq ´ µnpEq| ` |µnpď

kěm`1

Akq|

ď pm` 1q supEPΣ

|µpEq ´ µnpEq| ` supEPΣ

|µnpEq ´ µ`pEq| ` |µ`pď

kěm`1

Akq|

Take lim supn

of the right-hand side and the first term disappears. Then take mÑ 8 and the third

term disappears. Finally, take the limit ` Ñ 8 so that the middle term disappears. Thus, µ isσ-additive.


Lecture 22 - 10/20/2014

Definition - Let µ, ν be positive measures. We say ν is absolutely continuous with respect to µ,denoted ν ! µ, provided that

µpAq “ 0 ùñ νpAq “ 0 for all A P Σ

Example - Let pX,Σ, µq be a measure space, f : X Ñ r0,8s measurable, and define dν “ f dµ, i.e.νpAq “

ş

Af dµ. (This was on a previous homework.)

Theorem (Radon-Nikodym) - Let µ, ν ě 0 be σ-finite, ν ! µ. Then there exists a unique (up toµ-a.e.) measurable function g : X Ñ r0,8q such that dν “ g dµ.Notation - g is called the Radon-Nikodym derivative of ν with respect to µ. We write g “ dν

dµ .Proof - Case I: Suppose µ, ν ă 8. Define

F “ tf : X Ñ r0,8s | f measurable and

ż

A

f dµ ď νpAq for all A P Σu

Note, F ‰ H since 0 P F .

Claim 1 For any f1, f2 P F , maxpf1, f2q P F .

Proof For all A P Σ,ż

A

maxpf1, f2q “

ż

AXtf1ąf2u

f1 `

ż

AXtf1ďf2u

f2

ď νpAX tf1 ą f2uq ` νpAX tf1 ď f2uq

“ νpAq

Claim 2 Let pfnq Ď F be increasing. Then limnÑ8

fn P F .

Proof Follows by Monotone Convergence Theorem.

Now let α “ supfPF

ş

Xf dµ ď νpXq ă 8. Let pfnq Ď F with

ş

Xfn dµ Ñ α as n Ñ 8. By Claim

1, we may replace each fn with max1ďiďn

tfiu and assume that pfnq is increasing. Set g “ limnÑ8

fn, now

g P F by Claim 2. In particular,ş

Xg dµ ď νpXq ă 8 implies that g ă 8 µ-a.e., so we may assume

g : X Ñ r0,8q.To finish Case 1, it remains to show that dν “ g dµ. To do this, define dν0 “ dν ´ g dµ, i.e.

ν0pAq “ νpAq ´ş

Ag dµ for all A P Σ. Clearly ν0 ě 0 since g P F . We want to show that ν0 ď 0, so

then ν0 “ 0. It’s enough to prove the following claim.

Claim For any ε ą 0, ν0 ď εµ.

Proof Fix ε ą 0. Since ν0 ´ εµ is a signed measure, there is a corresponding Hahndecomposition X “ P Y N . To get ν0 ď εµ, we show that µpP q “ 0. Then byν ! µ, we have νpP q “ 0, so ν0 ´ εµ is negative and we’re done.

To show that µpP q “ 0, it’s enough to show that g ` εχP P F (by definition ofg). Write

ż

A

pg ` εχP q dµ “

ż

A

g ` εµpAX P q

“ νpAq ´ ν0pAq ` εµpAX P q

“ νpAq ´ ν0pAXNq ´ pν0 ´ εµqpAX P q

ď νpAq


Thus g ` εχP P F , so µpP q “ 0 and we are done.

We still need to show uniqueness. Suppose that dν “ g1 dµ “ g2 dµ. Then for all A P Σ, wehave

νpAq “

ż

A

g1 dµ “

ż

A

g2 dµ ùñ

ż

A

pg1 ´ g2q dµ

Let A “ tg1 ą g2u. We get µptg1 ą g2uq “ 0. By symmetry, µptg1 ă g2uq “ 0. Thus, g1 “ g2 µ-a.e.

Case II: Suppose µ, ν are σ-finite. Write X “Ť

nFn, where Fn Ď Fn`1, µpFnq, νpFnq ă 8. For

each n, define µnpAq “ µpA X Fnq, νnpAq “ νpA X Fnq. Note that µn, νn ă 8 and νn ! µn. ByCase I, for each n there is a unique (µn-a.e.) fn such that dνn “ fn dµn. By this uniqueness, wemust have fn`1 æFn“ fn for each n (µn-a.e.). Thus, g “ lim

nÑ8fn is well-defined and has the desired

property (by the MCT).


Lecture 23 - 10/22/2014

Lebesgue Decomposition

Definition - Let µ be a positive σ-finite measure, ν a signed finite measure. We write ν ! µ if

µpAq “ 0 ùñ |ν|pAq “ 0 for all A P Σ

Theorem - Let µ be positive and σ-finite, ν signed and finite, ν ! µ. Then there exists a uniqueg P L1pµq such that dν “ g dµ.Proof - By Jordan, we can write ν “ ν` ´ ν´. Now Radon-Nikodym yields g`, g´ such thatdν˘ “ g˘ dµ. Set g “ g` ´ g´.(Boring) Example - For any finite signed measure µ, we have µ ! |µ|. By Radon-Nikodym, there isg such that dµ “ g d|µ|. But we know this already! If X has Hahn decomposition X “ P YN withrespect to µ, then g “ χP ´ χN .

Later, we’ll see a really surprising and nontrivial application of Radon-Nikodym.

Theorem (Lebesgue Decomposition) - Let µ, ν be positive measures, ν σ-finite. Then there existpositive measures νS and νAC such that

1. ν “ νS ` νAC

2. νAC ! µ

3. νS K µ

Recall - νS K µ if and only if there is N P Σ such that µpNq “ 0, νSpNcq “ 0.

Proof - Case I: Suppose ν is finite. Set

N “ tN P Σ | µpNq “ 0u, α “ suptνpNq | N P N u ă 8

Let pNnq Ď N with νpNkq ą α´ 1k for each k. Set N “

Ť

k

Nk. Note that νpNq “ α and µpNq “ 0.

Define νSpAq “ νpAXNq, νACpAq “ νpAXN cq. Clearly ν “ νS ` νAC . Moreover, νS K µ sinceµpNq “ 0 and νSpN

cq “ 0. It remains to show that νAC ! µ.Suppose µpAq “ 0. We need νACpAq “ 0, i.e. νpAzNq “ 0. Since AYN P N , then

α “ νpNq ď νpAYNq ď α

Since ν is finite, then νpAzNq “ 0, so we’re done.Case II: Suppose that ν is σ-finite. Write X “

Ť

nFn, where νpFnq ă 8 and Fn Ď Fn`1. Set

νpnqpAq “ νpA X Fnq, so νpnq ă 8. By Case I, there exists NN Ď Fn and positive measures νpnqS ,

νpnqAC such that νpnq “ ν

pnqS ` ν

pnqAC and µpNnq “ 0, ν

pnqS pN c

nq “ 0.Define N “

Ť

nNn, νSpAq “ νpAXNq, νACpAq “ νpAXN cq, and ”suffer” (i.e. check that it all

works out).

Proposition - The Lebesgue decomposition is unique.

Proof - Suppose that ν “ νp1qS ` ν

p1qAC “ ν

p2qS ` ν

p2qAC . Now

νp1qS ´ ν

p2qS “ ν

p2qAC ´ ν

p1qAC

The left-hand side is singular with respect to µ. The right-hand side is absolutely continuous with

respect to µ. Since these are equal, they must both be equal to zero, so that νp1qS “ ν

p2qS and

νp1qAC “ ν

p2qAC .


Corollary (Radon-Nikodym-Lebesgue Decomposition) - Let µ, ν be positive, σ-finite measures. Thenthere is unique g : X Ñ r0,8q and unique νS such that

dν “ dνS ` g dµ, νS K µ

Application of Radon-Nikodym: The Dual of Lp

Let X be a Banach space. We denote by X˚ the (continuous) dual of X, defined by

X˚ “ tx˚ | x˚ : X Ñ R linear and continuousu

Theorem - Let p P r1,8q, µ a positive σ-finite measure, 1p `

1q “ 1. Then Lq is isometric to pLpq˚.

Remark - Given g P Lq, define Λg : Lp Ñ R by

fΛgÑ

ż

X

fg dµ

By Holder, this is well-defined. We’ll prove the theorem by showing that g ÞÑ Λg maps Lq to pLpq˚

isometrically.


Lecture 24 - 10/24/2014

Theorem - Let p P r1,8q, µ a σ-finite positive measure, 1p `

1q “ 1. Then Lq is isometrically

isomorphic to pLpq˚.Lemma - Let U be a Banach space, T : U Ñ R linear. Then

T is continuous ðñ DC ě 0 s.t. |Tu| ď Cu for all u

Definition - We say T : U Ñ R is bounded if there is C ě 0 such that |Tu| ď Cu for all u P U .Proof (of Lemma) - Say T is bounded. Then |Tu ´ Tv| “ |T pu ´ vq| ď Cu ´ v. In particular, Tis Lipschitz, so T is continuous.

Conversely, suppose T is continuous. Let δ ą 0 such that u ă δ implies |T puq| “ |T puq´T p0q| ă1. Then for all v ‰ 0, we have

|Tv| “ 2vδ |T p

δ2vv q| ă

2δ v

Notation - For u P U , denote by uU the norm of u in U . Similarly, for u˚ P U˚, denote by u˚U˚the norm of u˚ in U˚ (to be defined).Definition - Given u˚ P U˚, define

u˚U˚ :“ supuPU,uU“1

u˚puq

(Note: This is finite since each u˚ is bounded.)Exercise ¨ U˚ is a norm on U˚ and pU˚, ¨ U˚q is a Banach space.

Proof (of Theorem) - Consider the map Λ : Lq Ñ pLpq˚ which takes g P Lq to Λg P pLpq˚ defined

by Λgpfq “ş

fg. The map Λg is clearly linear. Moreover, it is continuous by Holder’s inequalitybecause |Λgpfq| ď gqfp (so it is bounded). Thus, for all g P Lq, Λg P pL

pq˚.It remains to show that Λ is a linear bijective isometry. Linearity is clear. For isometry, we want

ΛgpLpq˚ “ gLq for all g P Lq. Note,

ΛgpLpq˚ “ supfPLp,f‰0

|Λgpfq|fp

“ supfPLp,f‰0

1fp

|

ż

fg| “ gq

(by the duality lemma from long ago). Thus, Λ is an isometry.Since Λ is an isometry, it is automatically injective. We need to show Λ is surjective. Let

φ P pLpq˚. Want g P Lq such that φ “ Λg.Case I: Suppose µ is finite. The main idea is to note that for any A P Σ, χA P L

p, so φpχAq isdefined. Define νpAq “ φpχAq. We show that

1. ν is a signed, finite measure.

2. ν ! µ, so that by Radon-Nikodym, dν “ g dµ

3. φ “ Λg

For (1), note that νpHq “ φp0q “ 0. Given pAnq Ď Σ countable and pairwise disjoint, we havefor each N

νpď

1ďiďN

Aiq “ φpχ Ť

1ďiďN

Aiq “ÿ

1ďiďN

φpχAiq “ÿ

1ďiďN

νpAiq

Thus, we have finite additivity for ν. For countable additivity, it’s enough to show that

ÿ

1ďiďN

χAiNÑ8Ñ χ Ť

iě1Ai in Lp


Writeχ Ť

iě1Ai ´

ÿ

1ďiďN

χAipp “ χ

Ť

iěN`1

Aipp “

ÿ

iěN`1

µpAiq Ñ 0

Thus, ν is countably additive.For (2), suppose µpAq “ 0. Then χA “ 0 in Lq, so νpAq “ φpχAq “ 0. By Radon-Nikodym,

there is g such that dν “ g dµ.For (3), note that for any simple function s, we have

Λgpsq “

ż

sg dµ “

ż

s dν “ÿ

i

aiνpAiq “ φpsq,

To show that g P Lq, let f P Lp be arbitrary. Let psnq Ď Lp with snLpÑ f . Then by continuity of

φ, φpsnq Ñ φpfq in R. But φpsnq “ş

sn dν Ñş

f dν (definition ofş

). Hence for all f P Lp,

φpfq “

ż

f dν “

ż

fg dµ “ Λgpfq

Now φ bounded implies g P Lq, since

supfp“1

ż

fg dµ “ φpLpq˚ ă 8

Case II: If µ is σ-finite, reduce to Case I.


Lecture 25 - 10/27/2014

Last Time - If p P r1,8q, µpXq ă 8, and 1p`

1q “ 1, there is a bijective isometry pLppXqq˚ » LqpXq.

Remark - Why is this useful? Many important normed vector spaces of functions, e.g. pLppXq, ¨pq,are infinite-dimensional. As a result, we don’t get compactness of the closed unit ball (contrast withRn). The Banach-Alaoglu theorem (from functional analysis) says that the closed unit ball of thedual space X˚ is weak-˚ compact (compact with respect to something called the weak-˚ topology).In particular, we have some sort of compactness for LqpXq, since it is the dual space of LppXq.Remark - For p “ 8, pL8q˚ » tν | ν a finitely additive measure, ν ! µu, a ”boring space”. Insearch of a better result, let’s consider a subspace of L8 (which will have a bigger, hopefully moreinteresting space as its dual).

Suppose that X is a compact metric space, so now CpXq Ď L8pXq. (Note: pCpXq, ¨ 8q is aBanach space since a uniform limit of continuous functions is continuous.) Let µ be a finite, signed,regular Borel measure on X. Define

Λµ P CpXq˚, Λµpfq “

ż

X

f dµ for all f P CpXq

Theorem (Riesz) - DenoteM “ tµ | µ a finite, signed, regular Borel measure on Xu. Then the mapΛ :MÑ CpXq˚ defined by

pΛpµqqpfq “ Λµpfq “

ż

X

f dµ

is a bijective linear isometry.Proof - Linearity is immediate. To show that µ is an isometry, first write

µ “ |µ|pXq “ µpP q ´ µpNq, Λµ “ supf8“1

|Λµpfq|,

where pP,Nq is a Hahn decomposition of X with respect to µ, P XN “ H. Clearly µ ě Λµ. Toget the converse, we’d like to choose f “ χP ´ χN above, but this may not be continuous, so useregularity/Lusin’s theorem to approximate f .

It remains to show that µ ÞÑ Λµ is bijective. Since this map is an isometry, it is injective, so onlysurjectivity remains.

Suppose I P pCpXq˚q (”I for integral”). We need to find µ PM such that Ipfq “ş

Xf dµ for all

f P CpXq.Case I: Suppose I is positive. That is, given f P CpXq with f ě 0, we have Ipfq ě 0 (intuitively,

I looks like integration with respect to a positive measure). For any open U Ď X, define

µ˚pUq “ suptIpfq | f P CpXq, 0 ď f ď 1, supppfq Ď Uu

(We’re trying to recover the measure of any Borel set, but our only means of approximation iscontinuous functions.) Now for any A Ď X, define

µ˚pAq “ inftµ˚pUq | U Ě A is openu

It’s not hard to check that µ˚ is indeed an outer measure, so define, as in Caratheodory’s theorem,Σ “ tE | µ˚pAq “ µ˚pAX Eq ` µ˚pAX Ecq for all A Ď Xu. Now µ˚ æΣ is a measure.

Lemma 1 Σ Ě BpXq.

Lemma 2 For all f P CpXq, Ipfq “ş

Xf dµ.

Proof (Lemma 1) It’s enough to show U P Σ for all U Ď X open. Let A Ď X.We need to show that µ˚pAq ě µ˚pAX Uq ` µ˚pAX U cq.


• If A is open, write A “ V . Fix ε ą 0. Let f, g P CpXq such that 0 ď f, g ď 1,and

supppfq Ď V X U, Ipfq ě µ˚pV X Uq ´ ε2

supppgq Ď V zsupppfq, Ipgq ě µ˚pV zsupppfqq ´ ε2 ě µ˚pV X U cq ´ ε

2

Now f ` g P CpXq, 0 ď f ` g ď 1, and supppf ` gq Ď V , so

µ˚pV q ě Ipf ` gq “ Ipfq ` Ipgq ě µ˚pV X Uq ` µ˚pV X U cq ´ ε

Taking the limit εÑ 0`, we are done.

• Let A be arbitrary. Fix ε ą 0 and let A Ď V Ď X open with µ˚pV q ď µ˚pAq`ε.Now

µ˚pAq ` ε ě µ˚pV q “ µ˚pV X Uq ` µ˚pV X U cq ě µ˚pAX Uq ` µ˚pAX U cq

Taking εÑ 0`, we are done.

Proof (Lemma 2) Need to show f P CpXq implies Ipfq “ş

Xf dµ.

• Step 1: Suppose f ě χA, where A P BpXq. We show that Ipfq ě µpAq. Givenε ą 0, define U “ tf ą 1 ´ εu. Note that U is open and U Ě A. If g P CpXqwith 0 ď g ď 1 and supppgq Ď U , then

g ď f1´ε ùñ Ipgq ď 1

1´εIpfq,

since we are assuming Ip¨q is a positive operator. Taking the supremum over all

such g, we have µpUq ď Ipfq1´ε . Taking the limit εÑ 0`, we have µpAq ď Ipfq.

(Note: Although U “ Upεq, we clearly have A Ď Upεq for any ε ą 0, soµpAq ď µpUpεqq.)

To be continued...


Lecture 26 - 10/29/2014

We continue with the proof of Riesz’s theorem (refer to previous lecture’s notes).

Lemma 2 Ipfq “ş

Xf dµ for all f P CpXq.

Proof • Step 1: We showed that χA ď f implies µpAq ď Ipfq for A P BpXq.• Step 2: Given s “

ř

i

aiχAi ě 0, where tAiu are disjoint, we show that

s ď f ùñ

ż

s

dµ ď Ipfq

To do this, fix ε ą 0. For each i select Ki Ď Ai compact, Ui Ě Ki open suchthat the sets tUiu are disjoint and f

1´ε ě ai on Ui. (Note: Disjointness ispossible since disjoint compact subsets of a metric space can be separated.)Now let φi : X Ñ r0, 1s be continuous such that φi “ 1 on Ki and supppφiq ĎUi. For each i, we have

aiχKi ďφif1´ε ùñ aiµpKiq ď

Ipφifq1´ε

Summing over i, we have

ÿ

i

aiµpKiq ďÿ

i

Ipφifq1´ε “

Ippř

iφiqfq

1´ε ďIpfq1´ε ,

since the supports of the functions tφiu are disjoint. Taking the supremumover compact Ki Ď Ai, we have

ż

s dµ ď Ipfq1´ε

Taking the limit εÑ 0`, we haveş

s dµ ď Ipfq, as desired.

• Step 3: Exercise: If 0 ď f ď t and t is simple, then Ipfq ďş

Xt dµ. (Appeal

to Step 2 and finiteness of µ.)

• Step 4: By the Exercise, given f P CpXq, we have

Ipf`q ě sup0ďsďf`

s simple

ż

X

s dµ “

ż

X

f` dµ “ inf0ďf`ďtt simple

ż

X

t dµ ě Ipf`q

which impliesş

Xf` dµ “ Ipf`q, so

ş

Xf dµ “ Ipfq.

Case II: Let I P CpXq˚ be arbitrary. Given f ě 0, define I`pfq “ suptIpgq | 0 ď g ď f, g PCpXqu. Note I`pfq ě 0 (take g “ 0). Moreover, this is finite since I is bounded. Check that this islinear. Define

I´pfq “ inftIpgq | 0 ď g ď f, g P CpXqu

or alternately define I´ “ I` ´ I (we want I “ I` ´ I´). Now for f P CpXq arbitrary, define

I`pfq “ I`pf`q ´ I`pf´q, I´pfq “ I`pfq ´ Ipfq

Check that everything works out! We get I` P pCpXqq˚ by applying Case I, same for I´.This finishes the proof.

Remark - The decomposition I “ I` ´ I´ corresponds to the Jordan decomposition. Explicitly, ifI “ µ, then I` “ µ` and I´ “ µ´.

Two Extensions of the Theorem


1. If X is locally compact and I : CcpXq Ñ R is positive, then there is a unique regular Borelmeasure µ such that Ipfq “

ş

Xf dµ for all f P CcpXq.

2. If X is locally compact, let M be the set of finite, signed, regular Borel measures on X. Wehave CcpXq Ď L8pXq, so we can define C0pXq “ CcpXq in L8pXq. We call C0pXq the set ofcontinuous functions that vanish at infinity. It can be shown that pC0pXqq

˚ “M, where themap

Λ :MÑ pC0pXqq˚, Λµpfq “

ż

X

f dµ

is a bijective linear isometry.


Lecture 27 - 10/31/2014

Exercise: Continuity of Regular Measures - Let X be a compact metric space, µ a finite regularpositive Borel measure on X such that µptxuq “ 0 for all x P X. Then for any α P p0, µpXqq, thereis A Ď X with µpAq “ α. (Hint: Use Riesz.)

Product Measures and Fubini’s Theorem

Setting - pX,Σ, µq and pY, τ, νq are measure spaces, µ, ν σ-finite.Definition - We denote Σ ˆ τ “ tA ˆ B | A P Σ, B P τu, the set of all ”rectangles” in A ˆ B. Theproduct σ-algebra on AˆB is defined by

Σb τ :“ σpΣˆ τq

Theorem - There is a unique measure π on pX ˆ Y,Σb τq such that

πpAˆBq “ µpAqνpBq

for all A P Σ, B P τ . We call π the product measure on pX ˆ Y,Σb τq.Remark - This fails if µ or ν is not σ-finite.

We begin by stating a series of theorems which we will then proceed to prove. We take thisapproach because the final result (Fubini’s theorem) plays a decisive role in our method of proof forthe first result (existence of the product measure).

Theorem - Let f : X ˆ Y Ñ r0,8s be Σb τ -measurable.

(a) For all x P X, the map y ÞÑ fpx, yq is τ -measurable. Similarly, for all y P Y , the mapx ÞÑ fpx, yq is Σ-measurable.

(b) The functions x ÞÑş

Yfpx, yq dνpyq and y ÞÑ

ş

Xfpx, yq dµpxq are Σ- and τ -measurable respec-

tively.

(c) (Tonelli)

ż

XˆY

f dπ “

ż

X

´

ż

Y

fpx, yq dνpyq¯

dµpxq “

ż

Y

´

ż

X

fpx, yq dµpxq¯

dνpyq p˚q

Theorem (Fubini) - Suppose f P L1pX ˆ Y, πq.

(a) For a.e. x P X,ş

Yfpx, yq dνpyq is finite and µ-integrable.

(b) For a.e. y P Y ,ş

Xfpx, yq dµpxq is finite and ν-integrable.

(c) The equality p˚q holds and the integrals are finite.

Proof (existence of product measure)Main Idea - Given E Ď X ˆ Y , for each x P X define SxpEq “ ty P Y | px, yq P Eu. For each y P Y ,define TypEq “ tx P X | px, yq P Eu. Now define

πpEq “

ż

X

νpSxpEqq dµpxq

We need to show three things:

1. SxpEq P τ for all x P X.


2. The function x ÞÑ νpSxpEqq is Σ-measurable.

3. π is a measure.

First, we prove uniqueness. Say π1, π2 are two such measures. If µ, ν are finite, then the settπ1 “ π2u is a λ-system containing Σˆ τ , which is a π-system. By Dynkin’s theorem, tπ1 “ π2u “

Σb τ . If µ, ν are σ-finite, then take limits and use the previous case.For existence, we prove a sequence of lemmas.

Lemma 1 For all E P Σb τ , x P X, y P Y , SxpEq P τ and TypEq P Σ.

Proof Define F “ tE P Σbτ | @x P X, SxpEq P τu. This is a σ-algebra containingthe rectangles Σˆ τ , so F “ Σb τ . The argument is the same for TypEq.

Lemma 2 For all E P Σbτ , the function x ÞÑ νpSxpEqq is Σ-measurable. Similarly,the function y ÞÑ µpTypEqq is τ -measurable.

Proof Denote Λ “ tE P Σ b τ | x ÞÑ νpSxpEqq is Σ-measurableu. This is notobviously a σ-algebra (because of the problem of non-disjoint unions). Instead, weshow that Λ is a λ-system! Since Λ contains the rectangles Σ ˆ τ , we then haveΛ “ Σb τ .

First suppose that µ, ν are finite.

(1) Clearly X ˆ Y P Λ, since νpSxpX ˆ Y qq “ νpY q, a constant function.

(2) Suppose A,B P Λ, A Ď B. We have

νpSxpBzAqq “ νpSxpBqq ´ νpSxpAqq,

a difference of measurable functions, which is measurable. Thus BzA P Λ.

(3) Let pAnq Ď Λ with An Ď An`1. We have

νpSxpď

n

Anqq “ νpď

n

SxpAnqq “ limnÑ8

νpSxpAnqq,

a limit of measurable functions, which is measurable. ThusŤ

nAn P Λ. Hence,

Λ is a λ-system.

If µ, ν are σ-finite, take limits and use the previous case.

To be continued...


Lecture 28 - 11/03/2014

Quotes (A bowl of candy is sitting at the front of the classroom) Student: ”Are we going over theReese’s Representation Theorem with candy?” Professor: ”What’s the connection between Rieszand candy?” (slaps forehead) ”I’m slow.”Proof (of existence of product measure, continued) - We still need to show that πpEq “

ş

XνpSxpEqq dµpxq

is a measure. Let pEnq Ď Σb τ be pairwise-disjoint. Now

πpď

n

Enq “

ż

X

νpSxpď

n

Enqq dµpxq

“

ż

X

ÿ

n

νpSxpEnqq dµpxq

“ÿ

n

ż

X

νpSxpEnqq dµpxq (Beppo-Levi/MCT)

“ÿ

n

πpEnq

Hence, π is a measure. Moreover, for any AˆB P Σˆ τ , we have

πpAˆBq “

ż

X

νpSxpAˆBqq dµpxq “

ż

X

νpBqχApxq dµpxq “ νpBq

ż

A

dµpxq “ µpAqνpBq

Remark - We could have defined the product measure more constructively, starting with some outermeasure that uses coverings of E P Σb τ by countable collections of rectangles pAn ˆBnq Ď Σˆ τ .Then do some sort of Caratheodory argument. There are two drawbacks to this: First of all, thiswould be very tedious. Second of all, this construction makes it difficult to recover the ”nicer”definition of π in terms of integration. We will see that with our definition of π, it doesn’t take toomuch work to prove Tonelli’s and Fubini’s theorems.

Proof (Tonelli) - Let f : X ˆ Y Ñ r0,8s be Σb τ -measurable. We need to show that

(a) For all x P X, y P Y , the maps fpx, ¨q and fp¨, yq are τ -measurable, Σ-measurable respectively.

(b) The mapsş

Yfp¨, yq dνpyq and

ş

Xfpx, ¨q dµpxq are Σ- and τ -measurable respectively.

(c) (Tonelli)

ż

XˆY

f dπ “

ż

X

´

ż

Y

fpx, yq dνpyq¯

dµpxq “

ż

Y

´

ż

X

fpx, yq dµpxq¯

dνpyq p˚q

We proceed in several steps.

1. Say f “ χE , where E P Σb τ . By definition, we haveż

XˆY

f dπ “ πpEq “

ż

X

νpSxpEqq dµpxq “

ż

X

´

ż

Y

χEpx, yq dνpyq¯

dµpxq

and we have already shown that (a) and (b) hold. By symmetry, the function πpEq “ş

YµpTypEqq dνpyq is a measure that satisfies πpA ˆ Bq “ µpAqνpBq. By uniqueness of the

product measure, we have π “ π, so p˚q holds.

2. By linearity and the previous step, p˚q holds if f is a positive simple function. Moreover, (a)and (b) are clearly preserved under positive linear combinations.


3. Now let f : X ˆ Y Ñ r0,8s be arbitrary Σ b τ -measurable. Let sn Õ f be a sequence ofpositive, simple, Σbπ-measurable functions. By MCT, (a) and (b) clearly hold. For p˚q, write

ż

XˆY

f dπ(MCT)“ lim

nÑ8

ż

XˆY

sn dπ “ limnÑ8

ż

X

ż

Y

snpx, yq dνpyq dµpxq

“

ż

X

´

limnÑ8

ż

Y

snpx, yq dνpyq¯

dµpxq (MCT)

“

ż

X

ż

Y

limnÑ8

snpx, yq dνpyq dµpxq (MCT)

“

ż

X

ż

Y

f dνpyq dµpxq

Proof (Fubini) - Let f P L1pX ˆ Y, πq. Write f “ f` ´ f´. Nowż

XˆY

f dπ “

ż

XˆY

pf` ´ f´q dπ

“

ż

XˆY

f` dπ ´

ż

XˆY

f´ dπ

“

ż

X

ż

Y

f`px, yq dνpyq dµpxq ´

ż

X

ż

Y

f´px, yq dνpyq dµpxq (Tonelli)

“

ż

X

´

ż

Y

f` dνpyq ´

ż

Y

f´ dνpyq¯

dµpxq p˚˚q

“

ż

X

ż

Y

fpx, yq dνpyq dµpxq

Why does step p˚˚q hold? This is because

µptx |

ż

Y

f`px, yq dνpyq “ 8uq “ 0, µptx |

ż

Y

f´px, yq dνpyq “ 8uq “ 0

In particular,ş

Yf`px, yq dνpyq ´

ş

Yf´px, yq dνpyq is well-defined except on µ-null set.

Application of Fubini/Tonelli - Let f : X Ñ r0,8q be measurable, µ σ-finite on X. Then

1.ş

Xf dµ “

ş8

0µptf ą xuq dλpxq

2. If φ : r0,8q Ñ r0,8q is C1, increasing, and satisfies φp0q “ 0, thenż

X

φpfq dµ “

ż 8

0

µptf ą xuqφ1pxq dλpxq

Note that (1) implies (2), since (1) impliesż

X

φpfq “

ż 8

0

µptφpfq ą xuq dλpxq “

ż 8

0

µptf ą xuqφ1pxq dλpxq

by change of variables. To prove (1), define E “ tpx, λq | λ P r0, fpxqsu. Then by Tonelli,ż

XˆRχE dπ “

ż

X

ż 8

0

χE dλ dµpxq “

ż

X

fpxq dµpxq

ż

XˆRχE dπ “

ż 8

0

ż

X

χE dλ dµpxq “

ż 8

0

µptf ě yuq dλpyq

Remark - We also haveş

f “ş8

0µptf ą yuq dλpxq, since we’re only cutting out the graph of the

function (which has measure zero).


Lecture 29 - 11/05/2014

Setting - All functions are functions on Rd.Motivation - Consider a continuous but erratic function fptq, e.g. the price of a stock over the courseof a year, which typically moves up and down quite a lot. Instead of studying fptq, we can study a

”moving average” of fptq, defined by F ptq “şt`ε

t´εfptq ¨ 1

2ε dt (note: F ptq is the average value of thefunction fptq over the interval pt´ ε, t` εq). This function generally looks smoother and is easier tointerpret.

Definition - Let f, g be functions on Rd. We define a new function f ˚ g (”f star g” or ”f convolvedwith g”) by

pf ˚ gqpxq “

ż

Rdfpx´ yqgpyq dy

Remark - By making a change of variables, we see that f ˚ g “ g ˚ f .Remark - In the Motivation above, F ptq “ f ˚ p 1

2εχr´ε,εsq.

When is f ˚ g defined? To answer this, we must make assumptions on f and g. For simplicity,let’s suppose first that f, g P L1. Then by Tonelli’s theorem, we have

ż

Rd|f | ˚ |g| dx “

ż

Rd

ż

Rd|fpx´ yq||gpyq| dydx

“

ż

Rd|gpyq|

ż

Rd|fpx´ yq| dxdy

“

ż

Rd|gpyq|f1 dy

“ f1g1

In particular,ş

Rd |f | ˚ |g| dx is finite, so not only is f ˚ g defined a.e., but f ˚ g P L1.The previous result is nice. How does it generalize? That is, given f P Lp and g P Lq, is f ˚ g

defined? Can we find r such that f ˚ g P Lr?We seek an inequality of the form f ˚ gr ď fpgq. As we’ve done previously, let’s try

”counting dimensions”. View f, g as dimensionless functions, ` denotes length, so that Rd hasdimension `d. We note that:

• f ˚ g has dimension `d, so then f ˚ gr has dimension `d`dr.

• fpgq has dimension `dp`dq.

If f ˚ g P Lr, then we should expect d ` dr “

dp `

dq , or 1 ` 1

r “1p `

1q . This agrees with the result

we just derived above; take p “ q “ r “ 1.

Theorem (Young’s Inequality) - Let p, q P r1,8s. If f P Lp, g P Lq, then f ˚ g P Lr, where

1` 1r “

1p `

1q . In fact,

f ˚ gr ď fpgq

Proof - Let r1 be the conjugate of r. By the duality lemma, it’s enough to show that for all h P Lr1

,we have

ż

pf ˚ gqh ď fpgqhr1 p˚q

For simplicity, assume f, g, h ě 0. Otherwise, replace f, g, h by |f |, |g|, |h| and use monotonicity ofthe integral. Let p1, q1 be the conjugates of p, q respectively. We have

1p `

1q “ 1` 1

r ,1p “ 1´ 1

p1 ,1q “ 1´ 1

q1 ùñ1p1 `

1q1 `

1r “ 1 p˚˚q


In light of p˚˚q, we would like to apply Holder’s inequality (the general version for n functions,n ě 2), followed by Tonelli’s theorem. To start, write

ż

Rdpf ˚ gqh “

ż

Rd

ż

Rdfpx´ yqgpyqhpxq dydx

“

ż

Rd

ż

Rdpfpx´ yqαgpyqβq ¨ pfpx´ yq1´αhpyqγq ¨ pgpyq1´βhpxq1´γq dydx

(Note: Applying Holder’s inequality will separate the three functions f, g, h into three pairs of twofunctions, which will then permit us to use Tonelli’s theorem.) We’d like one of each of the threefactors above to be in Lp

1

, Lq1

, and Lr. We know f P Lp, g P Lq, h P Lr1

, so it’s enough to findα, β, γ such that

αr “ p, βr “ q, p1´ αqq1 “ p, γq1 “ r1, p1´ βqp1 “ q, p1´ γqp1 “ r1

From the first, second, and fourth equation, we require that α “ pr , β “ q

r , γ “ r1

q1 . It is straight-forward to verify that these values satisfy the third, fifth, and sixth equations. By Holder, wehave

ż

Rdpf ˚ gqh ď p

ĳ

fpx´ yqpgpyqqq1rp

ĳ

fpx´ yqphpxqr1

q1q1

p

ĳ

gpyqqhpxqr1

q1p1

“ fprp gqrq fpq1

p gr1q1

r1 gqp1

q hr1p1

r1

“ fα`p1´αqp gβ`p1´βqq ` hγ`p1´γqr1

“ fpgqhr1

This finishes the proof.

Mollification

Definition - Let tφnu be a sequence of functions on Rd. We say tφnu is an approximate identity if

1. φn ě 0 for all n.

2.ş

Rd φn “ 1 for all n¿

3. For each ε ą 0, we have limnÑ8

ş

|x|ąεφnpxq dx “ 0.

Example - Let φ P L1 with φ ě 0,ş

Rd φ “ 1. For any ε ą 0, define φεpxq “1εdφpxε q. Then tφεuεÑ0

is an approximate identity.

Approximate identities are useful because they provide a means to approximate arbitrary func-tions by a sequence of much nicer functions. For example, given an approximate identity consisting ofsmooth functions, we can use these to approximate many non-smooth functions by smooth functions.In particular...

Proposition - Let tφnu be an approimate identity.

1. If f P Lp for some p P r1,8q, then f ˚ φnLpÑ f , i.e. f ˚ φn ´ fp Ñ 0.

2. If f P L8 and f is continuous at some x P Rd, then f ˚ φnpxq Ñ fpxq.


Proof - We prove (2), leaving (1) for next time. Fix ε ą 0. Write

|φn ˚ fpxq ´ fpxq| “ |

ż

φnpyqfpx´ yq dy ´ fpxq|

“ |

ż

φnpyqpfpx´ yq ´ fpxqq dy|

ď

ż

φnpyq|fpx´ yq ´ fpxq| dy

“

ż

|y|ăδ

p¨ ¨ ¨ q dy `

ż

|y|ěδ

p¨ ¨ ¨ q dy

Pick δ ą 0 sufficiently small so that |fpx´ yq ´ fpxq| ă ε2 for all |y| ă δ. This bounds the first term

by ε2 . Now let N be sufficiently large so that n ě N implies

ş

|y|ěδφnpyq dy ď

ε4f8

. This finishes

the proof of (2).


Lecture 30 - 11/07/2014

Remark - Given a measure µ on Rd, define f ˚ µpxq “ş

Rd fpx ´ yq dµpyq. For example, we havef ˚ δ0pxq “ fpxq (i.e. δ0 is the identity for this operation ˚). This justifies the nomenclature”approximate identity”, since f ˚ φn Ñ f “ f ˚ δ0.

Recall - From last time, we still need to prove that f P Lp, p P r1,8q implies f ˚ φnLpÑ f .

Proof - We make use of two facts from previous homeworks. First, that if p P r1,8q and f P Lp, then

τyfLpÑ f as |y| Ñ 0 (here, τyfpxq “ fpx ´ yq). The second is Minkowski’s inequality for integrals.

Fix ε ą 0 and write

f ˚ φnpxq ´ fpxq “

ż

fpx´ yqφnpyq dy ´ fpxq “

ż

pτyfpxq ´ fpxqqφnpyq dy

Now

f ˚ φn ´ fp “ p

ż

|

ż

pτyfpxq ´ fpxqqφnpyq dy|p dxq1p

ď

ż

τyf ´ fpφnpyq dy

“

ż

|y|ăδ

τyf ´ fpφnpyq dy `

ż

|y|ěδ

τyf ´ fpφnpyq dy

Take δ ą 0 small enough so that τy ´ fp ăε2 for all |y| ă δ. Then take N sufficiently large so that

ş

|y|ěδφnpyq dy ă

ε4fp

. This finishes the proof.

Application: Fourier SeriesQuote - ”Whatever I say for the first lecture and a half probably is OK with just the Riemannintegral, but after that you’ll see how much more you can get.”Setup - We consider periodic functions on r0, 1s. Denote enpxq “ e2πinx, n P Z. Define

L2perr0, 1s “ tf : RÑ C | fpx` 1q “ fpxq,

ż 1

0

|f |2 ă 8u

If f, g P L2per, define xf, gy “

ş1

0fg. (Note: To integrate a complex-valued function, simply integrate

the real and imaginary parts separately, then add them back together. ”This is the only sensibleway to integrate such functions.”) This defines an inner product on L2

per that induces the norm

f2 “ş1

0|f |2.

Definition - Given f P L2perr0, 1s, define fpnq “ xf, eny “

ş1

0fpxqe´2πinx dx (Fourier coefficients).

Definition - Given N ě 0, define SNf “ř

|n|ďN

fpnqen (partial sums of Fourier series).

Remark - The set tenunPZ is orthonormal, i.e. xen, emy “ δnm.

Goal 1 - Show that SNfL2

Ñ f for all f P L2per.

One useful observation is that xSNf, f ´ SNfy “ 0 (easy to check). More generally, we have thefollowing:

Lemma - Let pN P spanteŃ , . . . , eNu. Then xf ´ SNf, pN y “ 0.Proof - By linearity, it’s enough to check this for pN “ em, where |m| ď N . Write

xf, emy “ fpmq, xSNf, emy “ÿ

|n|ďN

fpnqxen, emy “ fpmq


Corollary 1 (Bessel’s inqeuality) - f`2 ď fL2

Proof - Set EN “ f ´ SNf . Then by the Pythagoras theorem for inner products,

SNf ` EN “ f, ; xSNf,EN y “ 0 ùñ f22 “ SNf22 ` EN

22 ě SNf

22 “

ÿ

|n|ďN

|fpnq|2

Taking the limit N Ñ8, we have the result.

Corollary 2 - Suppose there is a sequence ppN q such that for each N , pN P spanteŃ , . . . , eNu and

pNL2

Ñ f . Then SNfL2

Ñ f .Proof - It’s enough to show that f´SNf2 ď f´pN . This follows by basic facts about projectionsfor inner product spaces. Explicitly, write

f ´ pN “ pf ´ SNfq ` pSNf ´ pN q ùñ f ´ pN 22 “ f ´ SNf

22 ` SNf ´ pN

22 ě f ´ SNf

22,

where we have used the fact that f ´ SNf and SNf ´ pN are orthogonal (by the lemma).

Let’s relate SNf back to convolutions and approximate identities. Write

SNfpxq “ÿ

|n|ďN

fpnqe2πinx “ÿ

|n|ďN

ż 1

0

fpyqe´2πiny dy e2πinx “

ż 1

0

ÿ

|n|ďN

e2πinpxýqfpyq dy

This motivates the definition DN pxq “ř

|n|ďN

e2πinx, so that SNf “ DN ˚ f . We call DN the (Nth

order) Dirichlet kernel.By interpreting DN as a geometric series, we may express it as a ratio of sines. What we’d like

is to say that tDNu is an approximate identity. Unfortunately, the functions tDNu are not evenpositive! Next try: Maybe t|DN |u? This also fails, since

ş

|DN | « lnN . What can we do?

Instead of considering the partial sums, let’s try the Cesaro sums! That is, instead of consideringSNf , let’s consider

σNf “1N

Nÿ

n“0

Snf,

the average of the first N partial sums. We may write

σNf “1N

Nÿ

n“0

Dn ˚ f “ p1N

Nÿ

n“0

Dnq ˚ f

Denote KN “1N

Nř

n“0Dn, the Fejer kernel.

Fact - tKNu is an approximate identity!

Corollary - KN ˚ fL2


Corollary - SNfL2


Quote - ”Even though the functions KN ˚f are a strictly worse approximation to f than the functionsSNf , it’s much easier to show that they converge to f .”


Lecture 31 - 11/10/2014

Motivation/Intuition - The functions tenu are an orthonormal ”basis” of L2per. We expect/hope that

f “ř

nfpnqen.

Remark 1 - Fourier coefficients can be defined much more generally. Certainly the definition extends

to f P L1. In fact, given a finite measure µ on r0, 1s, define µpnq “ş1

0en dµ.

Consequently, if f P Lpr0, 1s for any p P r1,8s, we can define fpnq “ş1

0fpxqenpxq dx. With that

in mind, we ask:

Question - For which p P r1,8s do we have SNfLpÑ f?

Answer - For p “ 8, this fails. E.g. let f P L8 be discontinuous on a set of positive measure. Infact, it is even possible to find continuous f P L8 such that Snf Ñ f in L8. However, on yourhomework you will show that if f P Cαr0, 1s, where α ą 0, then SNf Ñ f in L8.

Proposition If p P p1,8q, f P Lp, then SNfLpÑ f .

Proof - Hard harmonic analysis.Fact - This result is false for p “ 1.

Question 2 - We know f P L2per implies SNf

L2

Ñ f . When do we have pointwise convergence?Answer - Theorem (Carleson-Hunt) - For p ą 1, f P Lp, we have SNf Ñ f a.e.

˚ ˚ ˚

Theorem - The map L2perr0, 1s Ñ `2, f ÞÑ f “ pfpnqq is a bijective linear isometry.

Proof - We showed last time that f P L2 implies f P `2, so the map is well-defined. Moreover, since

SNfL2

Ñ f , thenf`2 “ lim

NÑ8

ÿ

|n|ďN

|fpnq|2 “ limNÑ8

SNfL2 “ fL2

Thus, the map is an isometry, so it is also injective. Linearity is also clear. It remains to provesurjectivity.

Let panq P `2. We’d like to define f “

ř

nanen, but is this in L2

per? Yes, because

ÿ

Mď|n|ďN

anenL2 “ pÿ

Mď|n|ďN

|an|2q12,

so the sequence of partial sums is Cauchy, thus f P L2per.

Next Goal - We seek to establish the following correspondence:

decay of f Ø ”regularity”/”smoothness” of f

Intuition - For n large, en is ”squiggly”. Too much squiggliness limits the smoothness of f .

Prop (Riemann-Lebesgue Lemma) - If f P L1, then f P L8. Moreover, f vanishes at infinity. That

is, lim|n|Ñ8

fpnq “ 0.

Remark - This does not extend to finite measures. That is, µ a finite measure on r0, 1s does notimply µpnq Ñ 0 at infinity. For example, let µ “ δ0, the delta measure centered at 0. Then for any

n, we have µpnq “ş1

0en dδ0 “ enp0q “ 1.


However, the Riemann-Lebesgue Lemma says that if µ ! λ, then µpnq Ñ 0 at infinity.Proof - The idea is to approximate f by trigonometric polynomials. Let f P L1. Fix ε ą 0. Thenthere is N such that f ´ σNf1 ă ε. For n ą N , we have

fpnq “ pf ´ σNfqpnq ` pσNfqpnq “ pf ´ σNfqpnq

Thus, by Holder’s inequality, we have

|fpnq| “ |

ż 1

0

pf ´ σNfqen dx| ď f ´ σNf1en8 ă ε

Remark - Previously we showed that f P L2 implies f P `2. This can be seen as another regularityresult.

Differentiability

Definition - We say f P L2per is weakly differentiable with weak derivative g if, for all φ P C8per, we

haveż 1

0

fφ1 “ ´

ż 1

0

gφ

(Intuition: We require that integration by parts works.)Example - If f P C1

per, then f 1 is certainly a weak derivative of f (as we would expect).

Proposition - If f P L2per and has weak derivative f 1 P L2

per, then

f 1pnq “ 2πinfpnq

Remark - This is what we would hope for. If f “ř

nfpnqe2πinx, then differentiation term-by-term

gives us the result above.


Lecture 32 - 11/12/2014

Last Time - Goal: ”Regularity” of f Ø Decay of f .Proposition - If f P L2

perr0, 1s, and has a weak derivative f 1 P L2perr0, 1s, then f 1pnq “ 2πinfpnq.

Proof - By definition of a weak derivative,

f 1pnq “ xf 1, eny “ ´xf, e1ny “ 2πinxf, eny “ 2πinfpnq

Corollary - If f P L2per has a weak derivative in L2

per, then

ÿ

n

p1` |n|2q|fpnq|2 ă 8

Proof - Since f 1 P L2per, then pf 1pnqq P `2. Now use f 1pnq “ 2πinfpnq.

Definition - Let s ě 0. Define the (periodic) Sobolev space of order s by

Hs :“ tf P L2perr0, 1s |

ÿ

n

p1` |n|2qs|fpnq|2 ă 8u

“ tf P L2perr0, 1s | pp1` |n|

2qs2fpnqq P `2u

For f P Hs, denote fHs “ p1` |n|2qs2fpnq`2 .

Remark - For s ě 0, s P N, we have

Hs “ tf | f P L2 and has sth order weak derivatives in L2u

(If s “ 1, note that f P H1 implies that p2πinfpnqq P `2, so this must correspond to some g P L2per.

By definition, this g must be a weak derivative for f . For s ą 1, this follows by induction.)

Theorem (Sobolev Embedding Theorem) - If f P Hs and s ą 12 , then f is continuous. That is, f

agrees a.e. with a continuous function.Corollary - If f P Hs, s ą n` 1

2 , then f P Cn.Proof (of Theorem) - It’s enough to prove the following result:

DC ą 0, f8 ď CfHs @f P Hs X C8per p˚q

That is, the inclusion map Hs X C8per ãÑ L8 is continuous.

Assuming p˚q, let f P Hs and let pφnq Ď Hs X C8per such that φnHsÑ f (e.g. φn “ Snf). Now

pφnq is Cauchy in Hs, so by p˚q, pφnq is Cauchy in L8, thus it converges uniformly. In particular,φn Ñ f uniformly, so f is continuous.

It remains to prove p˚q. For any x P r0, 1s, we have fpxq “ř

nfpnqenpxq. Therefore,

f8 ďÿ

n

|fpnq| “ÿ

n

p1` |n|2qs2|fpnq| ¨ 1p1`|n|2qs2

ď fHspÿ

n

1p1`|n|2qs q

12,

where the final step follows from the Cauchy-Schwarz inequality and the fact that C :“ př

n

1p1`|n|2qs q

12 ă

8 for s ą 12 .

Goal - Lebesgue differentiation


Suppose µ is a finite measure on Rd and µ ! λ. By Radon-Nikodym, there is f P L1 such thatdµ “ f dλ. We expect that

fpxq “ limrÑ0`

µpBpx,rqqλpBpx,rqq

Lebesgue proved that this is true a.e.Step 1: Considering the maximal function, defined by

Mµpxq :“ suprą0

|µ|pBpx,rqqλpBpx,rqq

If f P L1, then define Mfpxq :“ suprą0

1λpBpx,rqq

ş

Bpx,rq|f | dλ. What can we prove about these objects?

Theorem - If µ is a finite measure, then λptMµ ą αuq ď 3d

α µ. If f P L1, then λptMf ą αuq ď3d

α f1.

Want - To prove this theorem, we’d like f P L1 ùñ Mf P L1. Unfortunately this is false, but wewill get Mf P L1,8, which is enough to prove the theorem. Incidentally, we get something nicer forLp, p ą 1.Fact (Homework) - If f P Lp, p P p1,8s, then Mf P Lp and Mfp ď Cpfp, where Cp dependsonly on d and p, not on f .Lecture 33 - 11/14/2014

Homework Discussion - Given f P C1perr0, 1s, it’s easy to see that f´τhf8 « Ophq, since f´τhf ď

f 18h. On your homework, you’ll show something more general: If f P L2 and has weak derivativeDf P L2, then f ´ τhf2 « Ophq.

Recall - Lebesgue Differentiation Theorem: Given µ a finite signed measure on Rd, µ ! λ, we

have dµdλ pxq “ lim

rÑ0`

µpBpx,rqqλpBpx,rqq holds λ-a.e. To prove this, we first introduce new quantities that will

(eventually) provide upper bounds in our proof.Definition - Let µ be a finite measure on Rd, f P L1pRq. Define

Mµpxq “ suprą0

|µ|pBpx,rqq|λpBpx,rqq

Mfpxq “ suprą0

1λpBpx,rqq

ż

Bpx,rq

|f |

Theorem (Hardy-Littlewood Maximal Inequality) - There is a constant C ą 0 (can choose C “ 3d)

such that if µ is a finite measure on Rd, then for all α ą 0 we have

λptMµ ą αuq ď Cα µ

Proof - IOU.Remark - Given f P L1, we’d like to conclude that Mf P L1, perhaps Mf1 ď Cf1 for someconstant C. This is false! For example, take f “ χr0,1s. For x large, we have Mfpxq ě 1

2x R L1!

Proof (of Lebegue Differentiation) - By Radon-Nikodym, it’s enough to show that for any positive

function f P L1, we have

limrÑ0`

1λpBpx,rqq

ż

Bpx,rq

|f | dλ “ fpxq λ-a.e.


The main idea is approximate f by a continuous function, since the result clearly holds for continuousfunctions.

Fix f P L1. Define

Ωfpxq “ lim suprą0

1λpBpx,rqq

ż

Bpx,rq

|fpyq ´ fpxq| dy

To prove the result, it’s enough to show that Ωf “ 0 a.e. (sinceş

|f | ďş

|f ´ fpxq| ` fpxq). Notethat if f is continuous, we are done. If not, fix ε ą 0 and write f “ g`h, where g P L1 is continuousand h1 ă ε. Note that

Ωf ď Ωg ` Ωh “ Ωh

To estimate Ωh, it’s enough to consider the following (surprisingly loose) upper bound:

Ωhpxq “ lim suprą0

1λpBpx,rqq

ż

Bpx,rq

|hpyq ´ hpxq| dy

ď

´

suprą0

1λpBpx,rqq

ż

Bpx,rq

|h|¯

` |hpxq|

“ pMh` |h|qpxq

By the Hardy-Littlewood Maximal Inequality and Chebyshev’s Inequality, we have, for α ą 0,

λptΩf ą αuq ď λptΩh ą αuq

ď λptMh ą α2 uq ` λpt|h| ą

α2 uq

ď 3d

α2h1 `1α2h1

ă 2¨3d`2α ¨ ε

The quantity λptΩf ą αuq is independent of ε, so take the limit ε Ñ 0` to get λptΩf ą αuq “ 0.Thus, the set tΩf ą 0u is λ-null, so Ωf “ 0 a.e.

It remains to prove the Hardy-Littlewood Maximal Inequality. To do this, we state a lemma(which we’ll prove next time).

Lemma (Vitali Covering Lemma) - Let E Ď Rd, tB1, . . . , BNu a set of balls such that E ĎŤ

1ďiďN

Bi.

Then there is a subcollection tB1iu Ď tBiu of disjoint balls such that E ĎŤ

i

3B1i.

Notation - If B “ Bpx, rq, then 3B :“ Bpx, 3rq.Proof (Hardy-Littlewood) - It’s enough to prove the result for µ positive (exercise). Let α ą 0and define E “ tMµ ą αu. Fix K Ď E compact. For each x P K, there is rx ą 0 such thatµpBpx, rxqq ą αλpBpx, rqq. These balls cover K, so let tB1, . . . , BNu be a finite subcover of K. ByVitali, there is a disjoint subcollection tB11, . . . , B

1Mu such that K Ď 3B1i. Now

λpKq ďMÿ

i“1

λp3Biq “ 3dMÿ

i“1

λpB1iq ă3d

α

Mÿ

i“1

µpB1iq “3d

α µpMď

i“1

B1iq ď3d

α µ

Taking the supremum over compact K Ď E, we have λpEq ď 3d

α µ, as desired.


Lecture 34 - 11/17/2014

Lemma (Vitali) - Let E Ď Rd, tB1, . . . , BNu a collection of balls whose union covers E. Then there

is a disjoint subcollection tB1iu such that the balls t3B1iu cover E. (Here, B “ Bpx, rq ùñ 3B :“Bpx, 3rq.)Proof - By reordering, we may assume that radiuspB1q ě radiuspB2q ě ¨ ¨ ¨ . Choose B11 “ B1.Consider all balls disjoint from B11. Choose B12 to be the one of largest radius. Note that if BkXB

11 ‰

H for some k, then 3B11 Ě Bk.Given tB11, . . . , B

1ku, let B1k`1 be the ball of largest radius that is disjoint from B11, . . . , B

1k (if

there is such a ball). Eventually, this process terminates. Now

• These balls are clearly disjoint.

• These balls cover E. Why? Given Bi, either Bi “ B1j for some j, or Bi meets some B1j oflarger radius, in which case Bi Ď 3B1j .

Corollary (of Lebesgue Differentiation) - Let A P LpRdq. Then

limrÑ0`

λpAXBpx,rqqλpBpx,rqq “ χApxq λ-a.e.

Proof - Apply Lebesgue differentiation with f “ χA.Remark - Given µ a finite measure on Rd, we know

µ ! λ ùñ limrÑ0`

µpBpx,rqqλpBpx,rqq “

dµdλ pxq λ-a.e.

What if µ! λ? Decompose µ “ µS ` µAC , where µAC ! λ and µS K λ. We know

limrÑ0`

µACpBpx,rqqλpBpx,rqq “

dµACdλ pxq λ-a.e.

Claim (Homework) - limrÑ0`

|µS |pBpx,rqqλpBpx,rqq “

#

0 λ-a.e.

8 µS-a.e.

˚ ˚ ˚

The Fundamental Theorem of Calculus

Recall - For Riemann integration, if we define F pxq “şx

0f dx, where f is continuous, then F is

differentiable and F 1 “ f .Theorem - Let f P L1pRq and define F pxq “

şx

0f . Then F is differentiable and F 1 “ f a.e.

Proof - For h ą 0, we have F px`hq´F px´hq2h “ 1

2h

şx`h

x´hf “ µpBpx,hqq

λpBpx,hqq , where µpAq “ş

Af . By the

Lebesgue Differentiation Theorem, we have

fpxq “ dµdλ pxq “ lim

hÑ0`

µpBpx,hqqλpBpx,hqq “ F 1pxq λ-a.e.

Recall - For Riemann integration, if f P C1, thenşb

af 1 “ fpbq´ fpaq. Another way to express this is

f P C1 ùñ @x

ż x

0

f 1 “ fpxq ´ fp0q

We would like to a version of this result for differentiable functions, dropping the C1 assumption.


• Attempt 1: Is it enough to assume that f is differentiable a.e.? The answer is no, since (forexample), we may have f 1 R L1r0, 1s. For example, let fpxq “ lnx for x P p0, 1s, fp0q “ 0. Thenf is differentiable a.e. with f 1 “ 1

x , butşx

0f 1 “ 8 for all x ą 0. In particular, f 1 R L1r0, 1s.

• Attempt 2: What if we assume f is differentiable a.e. and f 1 P L1? This is still notenough. For example, let fpxq be the Cantor function, which satisfies f 1 “ 0 a.e. In particular,ş1

0f 1 “

ş1

00 “ 0 ‰ 1 “ fp1q ´ fp0q.

To obtain the desired result, we introduce a new notion:

Definition - We say f : ra, bs Ñ R is absolutely continuous if, for any ε ą 0, there is δ ą 0 such thatfor all finite sets tpxi, yiquiďN of disjoint subintervals of ra, bs, we have

Nÿ

i“1

|xi ´ yi| ă δ ùñNÿ

i“1

|fpxiq ´ fpyiq| ă ε

Note - f absolutely continuous implies f continuous.

Proposition - Let g P L1ra, bs such that fpxq ´ fpaq “şx

ag. Then f is absolutely continuous.

Proof - Fix ε ą 0. We know g is equi-integrable since λpra, bsq is finite. Then there is δ ą 0 such

that λpAq ă δ impliesş

A|g| ă ε. Now suppose that tpxi, yiquiďN are disjoint with

Nř

i“1

|xi ´ yi| ă δ.

Then λpNŤ

i“1

pxi, yiqq ă δ, soş

NŤ

i“1pxi,yiq

|g| ă ε. Now

Nÿ

i“1

|fpxiq ´ fpyiq| “Nÿ

i“1

|

ż yi

xi

g| ď

ż

NŤ

i“1pxi,yiq

|g| ă ε

Theorem - A function f : ra, bs Ñ R is absolutely continuous if and only if

1. f is differentiable a.e.

2. f 1 P L1

3. For all x P ra, bs, fpxq “ fpaq `şx

af 1

Proof - We just proved the reverse direction. We will prove the converse next time.


Lecture 35 - 11/19/2014

Theorem - A function f : ra, bs Ñ R is absolutely continuous if and only if

1. f is differentiable a.e.

2. f 1 P L1

3. For all x P ra, bs, fpxq “ fpaq `şx

af 1

Proof - We already proved the reverse direction last time. For the converse, suppose that f isabsolutely continuous.

Lemma 1 Suppose f is absolutely continuous and strictly increasing. Then (1),(2), and (3) hold.

Remark We need absolute continuity. Continuity + strictly increasing isn’t enough,e.g. let fpxq “ gpxq ` x, where gpxq is the Cantor function.

Proof Define µpAq “ λpfpAqq. Note, f´1 is increasing, thus measurable, so A P Bimplies fpAq P B.

Claim µ ! λ

Proof Suppose λpAq “ 0. We need to show µpAq “ λpfpAqq “ 0.

Fix ε ą 0 and let K Ď fpAq be compact. We know f´1pKq iscompact and f´1pKq Ď A, so λpf´1pKqq “ 0. Choose δ ą 0 suchthat, given disjoint tpxi, yiquiďN with

ř

1ďiďN

|xi ´ yi| ă δ¡ we haveř

1ďiďN

|fpxiq ´ fpyiq| ă ε. Since λpf´1pKqq, we may choose such

tpxi, yiquiďN which cover f´1pKq. Now

λpKq ďÿ

1ďiďN

λpfpxi, yiqq “ÿ

1ďiďN

|fpxiq ´ fpyiq| ă ε

Since ε ą 0 was arbitrary, then λpKq “ 0. Since K Ď fpAq wasarbitrary compact, then λpfpAqq “ 0.

By Radon-Nikodym, there is g P L1pra, bsq such that dµ “g dλ, so (1) and (2) hold with f 1 “ g. For (3), write

fpxq ´ fpaq “ λppfpaq, fpxqqq “ µppa, xqq “

ż x

a

g dλ “

ż x

a

f 1 dλ

Lemma 2 Suppose f is absolutely continuous and increasing. Then (1), (2), and(3) hold.

Proof Apply Lemma 1 to the function gpxq “ x ` fpxq. Now g is differentiablea.e., g1 P L1, so the same holds for f . For (3), write

fpxq “ gpxq ´ x “ gpaq `

ż x

a

g1 ´ x

“ fpaq ` a`

ż x

a

pf 1 ` 1q ´ x

“ fpaq ` a`

ż x

a

f 1 ` px´ aq ´ x

“ fpaq `

ż x

a

f 1


Definition - We say f : ra, bs Ñ R has bounded variation if the function

F pxq :“ sup∆

ÿ

xiP∆

|fpxiq ´ fpxi´1q|, x P ra, bs

is finite, where ∆ ranges over all partitions of ra, xs. We call F pxq the variation off .

Remark - If f : ra, bs Ñ R has bounded variation, then f can be written as adifference g´h of increasing functions (in fact, the converse holds). This is because

f “ pF`fq´pF´fq2 , where F˘f

2 are both increasing.

Lemma 3 Suppose f is absolutely continuous. Then (1), (2), and (3) hold.

Proof Claim If f is absolutely continuous, then f has boundedvariation. Moreover, the variation of f is absolutely continuous.

Proof Let F denote the variation of f . Set ε “ 1 and let δ ą 0such that if tpxi, yiquiďN are disjoint and

ř

1ďiďN

|xi ´ yi| ă δ, thenř

1ďiďN

|fpxiq ´ fpyiq| ă 1. Let ∆ be any partition of ra, bs. Refine

it by setting∆1 “ ∆Y ptnδ | n P Zu X ra, bsq

Nowř

xiP∆

|fpxiq ´ fpxi´1q| ă r báδ s. In particular, F is finite. The

proof that F is absolutely continuous is left as an exercise.

By the Claim, we may write f “ pF`fq´pF´fq2 , a difference of increasing, ab-

solutely continuous functions. By linearity of the integral and of differentation, wehave the desired result.


Lecture 36 - 11/21/2014

Change of Variables

Theorem - Let U, V Ď Rd be open, φ : U Ñ V a C1 bijection. Then for all f P L1pV q, we haveż

V

f dλ “

ż

U

f ˝ φ|det∇φ| dλ,

where ∇φ “ pBjφiq is the Jacobian of φ.Proof - For A Ď U , A P BpRdq, define µpAq “ λpφpAqq. (Here, µ is defined so that λ on V is thepushforward of µ with respect to φ.) To prove the theorem, we need only prove a series of lemmas.

Lemma 1 For all A P BpUq, φpAq P BpV q. Moreover, µ is a Borel measure on U .

Lemma 2 µ ! λ

Lemma 3 dµdλ “ |det∇φ| λ-a.e.

Once these lemmas are proven, we have the result. This is becauseż

V

f dλ “

ż

U

f ˝ φ dµ (pushforward)

“

ż

U

f ˝ φdµdλ dλ pµ ! λq

“

ż

U

f ˝ φ| det∇φ| dλ

Proof (Lemma 1) Suppose A P BpUq. Define

Λ “ tA P U | φpAq P BpV qu

This is a Λ-system because

• φpUq “ V P BpV q• If A Ď B in Λ, then φpBzAq “ φpBqzφpAq P BpV q (φ is bijective).

• pAnq Ď Λ increasing implies φpŤ

nAnq “

Ť

nφpAnq P BpV q.

Moreover, Λ Ě K :“ tK Ď U | K is compactu, which is a π-system (Λ Ě Kbecause if K Ď U is compact, then so is φpKq Ď V , so φpKq P BpV q). Thus,Λ Ě σpKq “ BpUq. (Compact sets generate the Borel σ-algebra since for anynonempty closed C and x P C, we have C “

Ť

nC X Bpx, nq, a union of compact

sets.)

Proof (Lemma 3) Assuming Lemma 2, we have dµdλ pxq “ Dµpxq :“ lim

rÑ0`

µpBpx,rqqλpBpx,rqq

holds λ-a.e., so it’s enough to prove that Dµpxq “ |det∇φx| for all x P U . Below,we consider simple cases for φ and proceed to generalize.

• Step 0: Suppose φ is linear, i.e. φpxq “ Mx for some invertible M P MdpRq.Then µ “ λ˝φ is translation invariant on Rd and finite on bounded sets, so bya previous result there is C ą 0 such that µpAq “ CλpAq, so now Dµpxq “ C.Let A be the unit cube. Then by a previous homework, we have

C “ µpAq “ |detM | “ |det∇φ| ùñ Dµpxq “ |det∇φx|


• Step 1: We may assume without loss that x “ 0 and φpxq “ 0 (we can alwaysreduce to this case by composing φ with translations).

– Case I: Suppose det∇φ0 “ 0, i.e. ∇φ0 is not invertible. We want to showDµp0q “ 0. To do this, denote T “ ∇φp0q, T : Rd Ñ Rd. Since detT “ 0,then dimprangepT qq ă d, so λprangepT qq “ 0.Now let ε ą 0. By definition of differentiation, we have

0 “ lim|x|Ñ0

|φpxq´φp0q´∇φ0px´0q||x| “ lim

|x|Ñ0

|φpxq´Tx||x|

Then there is r0 ą 0 such that |x| ă r0 implies |φpxq´Tx| ă ε|x|. We knowthat for any r, T pBp0, rqq is contained in a pd ´ 1q-dimensional subspaceof Rd and diampT pBp0, rqqq ď T ¨ 2r. Thus,

µpBp0, rqq “ λpφpBp0, rqqq ď pT 2rqd´1 ¨ εr

Now0 ď Dµp0q ď lim

rÑ0`

p2T qd´1εrd

rdλpBp0,1qq“ p2T qd´1ε

Taking the limit εÑ 0`, we have Dµp0q “ 0 “ det∇φ0

To be continued...


Lecture 37 - 11/24/2014

Theorem - Let U, V Ď Rd be open, φ : U Ñ V a C1 bijection. Then for all f P L1pV q, we have

ż

V

f dλ “

ż

U

f ˝ φ|det∇φ| dλ,

where ∇φ “ pBjφiq is the Jacobian of φ.Proof (continued from last time) - We defined µpAq “ λpφpAqq, a measure on BpUq. We still need

to show that µ ! λ and Dµ :“ limrÑ0`

µpBp¨,rqqλpBp¨,rqq “ |det∇φ|. We consider Dµpxq for x “ 0, φpxq “ 0.

• We’ve already proven that Dµ “ |det∇φ| if φ is linear or det∇φ0 “ 0.

• Case II(a): Suppose ∇φ0 “ I, the identity matrix. Then lim|x|Ñ0

|φpxq´x||x| Ñ 0, so there is r0 ą 0

such that |φpxq ´ x| ă ε|x| for |x| ă r0. Now φpBp0, rqq Ď Bp0, p1` εqrq implies

µpBp0, rqq “ λpφpBp0, rqqq ď λpBp0, p1` εqrqq “ p1` εqdλpBp0, rqq

In particular, Dµp0q “ limrÑ0`

µpBp0,rqqλpBp0,rqq ď p1` εq

d. Taking the limit εÑ 0`, we have Dµp0q ď 1.

Remark - It turns out to be difficult to show that Dµ ě 1. What we’d like is to show thatφpBp0, rqq Ě Bp0, p1 ´ εqrq (easy if φ´1 P C1). This involves Brouwer’s fixed-point theoremfrom topology, so we omit the proof here. (For more details, see pp. 150-151 of Rudin’s Realand Complex Analysis, 3rd edition.) If you’re unsatisfied, simply assume that φ´1 P C1 andthen the result follows (see Remark after this proof).

• Case II(b): det∇φ0 ‰ 0. Set T “ ∇φ0 and consider ψ “ T´1φ. Certainly ∇ψ0 “ I, so byCase II(a),

1 “ limrÑ0`

λpψpBp0,rqqqλpBp0,rqq

“ limrÑ0`

λpT´1φpBp0,rqqqλpBp0,rqq

“ |detT |´1 limrÑ0`

λpφpBp0,rqqqλpBp0,rqq

ùñ |detT | “ limrÑ0`

µpBp0,rqqλpBp0,rqq ,

so we are done. This finishes the proof of Lemma 3 from last time.

It remains to show that µ ! λ. By writing U as an increasing union of compact sets, we havethat µ is regular (Homework 3, Question 2). Thus, it suffices to show that if K Ď U is compact andλpKq “ 0, then µpKq “ 0.

Given such K, let ε ą 0 and let W Ě K be open with λpW q ă ε and C :“ supxPW

|∇φx| ă 8 (this

is possible since we may choose W to have compact closure). By the Mean Value Theorem, for anyx, y P Bpx0, r0q ĎW , we have, for i “ 1, . . . , d,

φipxq ´ φipyq “ ∇φipziq ¨ px´ yqùñ |φipxq ´ φipyq| ď C|x´ y|

ùñ |φpxq ´ φpyq| ď dC|x´ y|

ùñ λpφpBpx0, r0qqq ď pdCqdλpBpx0, r0qq


Now cover K with balls tBpx, rxq | x PW,Bpx, 3rxq ĎW u. By compactness, we may pass to a finitesubcover. By the Vitali covering lemma, we may pass to a finite subset of disjoint balls tB1iu suchthat K Ď

Ť

i

3B1i. Now

µpKq “ λpφpKqq ď λpď

i

φp3B1iqq

ďÿ

i

λpφp3B1iqq

ďÿ

i

pdCqd3dλpB1iq

“ pdCqd3dλpď

i

B1iq

ď pdCqd3dλpW q

ă pdCqd3dε

Taking the limit ε Ñ 0`, we have µpKq “ 0, so we are done. This finishes the proof of thetheorem.Remark - Another way to show that the theorme holds for φ´1 P C1 is through an appeal tosymmetry. If we only prove that Dµ ď 1 in Case II(a) above, we end up showing that

ż

V

f ď

ż

U

f ˝ φ|det∇φ| dλ for all f ě 0

Applying this result to φ´1, we have

ż

U

f ˝ φ|det∇φ| dλ ďż

V

f ˝ φ ˝ φ´1|det∇φφ´1pxq|| det∇φ´1| dλ “

ż

V

f dλ,

which implies equality.

Notes typeset by Adam Gutter - CMUgautam/teaching/2014-15/720-measure/pdfs/a… · Notes typeset by Adam Gutter Lecture 1 - 08/25/2014 Book Recommendations For grad students, use

Documents