Real Analysis Oral Exam study notes

Notes transcribed by Mihai Nica

3

Abstract. These are some study notes that I made while studying for my oralexams on the topic of Real Analysis, mostly covering the theory of integration.I took these notes from parts of the textbooks by Richard Bass [1] and a fewother sources which are indicated throughout. Please be extremely cautionwith these notes: they are rough notes and were originally only for me to helpme study and are not complete or guaranteed to be free of errors. I have madethem available to help other students on their oral exams.

Contents

Families of sets 41.1. Algebras and σ − algebras 41.2. Monotone Class Theorem 41.3. Pi-Lambda Theorem 5

Measures 7

Construction of Measures 93.4. Outer Measures 93.5. Lebesgue-Stieltjes Measure 113.6. Examples and related results 133.7. Nonmeasurable Sets 143.8. The Caratheodory Extension Theorem 15

Measurable Functions 174.9. Measurability 174.10. Approximation of functions 184.11. Lusin's Theorem 18

The Lebesgue Integral 20

Limit Theorems 216.12. Monotone Convergence Theorem 216.13. Linearity of the Integral 226.14. Fatou's Lemma 226.15. Dominated Convergence Theorem 22

Product Measures 247.16. Product σ−algebras 247.17. The Fubini Theorem 25

Signed Measures 268.18. Positive and Negative sets 268.19. Hahn Decomposition Theorem 268.20. Jordan Decomposition Theorem 27

The Radon-Nikodym Theorem 289.21. Absolute Continuity 289.22. The main theorem 289.23. Lebesgue Decomposition Theorem 29

Di�erentiation 3010.24. Di�erentiation of Monotone Functions 3210.25. Functions of Bounded Variation 3410.26. Di�erentiation of an Integral 3510.27. Absolute Continuity 37

Lp spaces 3911.28. Norms 3911.29. Completeness 4011.30. Convolutions 4111.31. Bounded Linear Functionals 41

Fourier Transforms 4312.32. Basic Properties 4312.33. The Inversion theorem 44

Fourier Series 4513.34. More Fourier Facts 48

Bibliography 50

Families of sets

These are notes from Chapter 2 of [1].

1.1. Algebras and σ − algebras

Fix a universal set X to work in.

Definition. (2.1) An algebra is a collection A of subsets of X that:i) ∅, X ∈ Aii) A ∈ A =⇒ Ac ∈ Aiii) closed under �nite unions and intersectionsA σ−algebra has:iv) closed under countable unions and intersections.

Lemma. (2.7.) Any arbitrary intersection of σ−algebras is a sigma algebra

Definition. We de�ne the σ−algebra generated by a subset C to be thesmallest σ− algebra cotaining C: i.e. the intersection of all σ− algebras containingC. This is denoted by σ(C).

Definition. The Borel σ − algebra is the σ − algebra that is generated byany of the following: 1) open intervals 2) closed intervals 3) half open intervals 4)semi-in�nite open intervals (a,∞]

Proof. Its easy to check by doing some intersections unions that these allgenerate the same thing. �

1.2. Monotone Class Theorem

Definition. (2.9.) A monotone class is a collection of subsets of X thatis closed under increasing, and decreasing sets. I.e. An ↑ A =⇒ A ∈ M andAn ↓ A =⇒ A ∈M.

Theorem. (2.10) Suppose A0is an algebra and, A = σ(A0) is the smallestσ − algebra containing A0 andM is the smallest monotone class containing A0.

ThenM = A.(Mihai's version) In other words: for an algebra A0, the smallest monotone

class containing A0 is in fact, a σ − algebra.

Proof. We must show A ⊂ M as the other inclusion is clear. To do this wehave to show that the monotone class is closed under intersections and unions, forthenM will be an algebra containgn A0 and hence A = σ(A0) ⊂M.

The idea of the proof is the usual trick with sigma algebras and so on: De�nethe set you are interested in and show it contains what you want.

Claim 1: Closed under complements

4

1.3. PI-LAMBDA THEOREM 5

Let N = {A ∈M : Ac ∈M}. Since A0 is an algebra and closed under com-plements, A0 ⊂ N . We claim now that N is a monotone class, for if An ↑ A withAn ∈ N then by def'n Acn ∈ M and we have Acn ↓ Ac shows Ac ∈ M since M isa monotone class. Similar for An ↓ A. Hence N is a monotone calss containg A0

and consequentlyM⊂ N i.e. M = N .Claim 2: Closed under intersections with A0

Let N = {A ∈M : A ∩B ∈M ∀B ∈ A0}. Again N contains A0 since A0 isan algebra, and N is a monotne class since if An ↑ A then An∩B ↑ A∩B for everyB ∈ A0, so the monotone-class-ness of M gives us A ∩ B ∈ M and this holds forevery B ∈ A0. (same idea for An ↓ A). As before, we get N =M.

Claim 3: Closed under intersectionsLet N = {A ∈M : A ∩B ∈M ∀B ∈M}. This contain A0 by the previous

claim! It is a monotone class by the same argument as in claim 2. Hence, we haveM = N .

This shows that M is in fact a σ − algebra (�nite unions/intersections +monotone limits yields countable unions/intersections by splitting up any countableunion/intersection up into a monotne limit of �nite unions/intersections) �

Remark. The sneaky thing in the proof is that you can't jump to closed underintersections right away, because you don't know that A0 is in there a priori. Thelittle extra step is to �rst prove that A0 is closed under intersections fromM andthen move on to the whole space.

1.3. Pi-Lambda Theorem

I'm going to include a little thinger on the Pi-Lambda theorem, as this comesup occasionally in Probability theory.

Definition. A π−system is a collection of subsets which is closed under �niteintersections.

Definition. A λ−system C is a collection of subsets which has:i) X ∈ Cii) A,B ∈ C and A ⊂ B =⇒ B −A ∈ Ciii) An ∈ C , and An ↑ A =⇒ A ∈ C

Proposition. (Alternative de�nition of a λ−system)a) X ∈ Cb) A ∈ C =⇒ Ac ∈ Cc) If An ∈ C are disjoint then

⋃∞n=1An ∈ C

Proof. Its easy to see that i), ii) and iii) imply a,b,c. We prove the conversestatements one at a time assuming a), b) and c):

ii): For any A,B with A ∩B = ∅, we have that:A ∪B ∈ C

=⇒ Ac ∩Bc ∈ CBut Ac∩Bc = Ac−B so this shows that whenver A∩B = ∅, we have Ac−B ∈ C.

Noticing that A ∩B = ∅ ⇐⇒ A ⊂ Bc and relabeling Bc = C gives the result ii)iii) Take set di�erences Bn = An −An−1which are disjoint. �

Theorem. If C is both a π−system and a λ−system, then C is a σ−algebra

1.3. PI-LAMBDA THEOREM 6

Proof. For any arbitarty sequence An ∈ C, we can create Bn ∈ C which aredisjoint with ∪nk=1Bk = ∪nk=1Ak by doing intersections (ok since C is a π−system)and complements (ok since C is a λ-system). Then, since C is a λ−system, we havethat ∪∞k=1Bk ∈ C and so ∪∞k=1Ak = ∪∞k=1Bk ∈ C too! �

Theorem. (7.4.) (Dynkin's π−λ theorem) Suppose Pis a π-system and, A =σ(P) is the smallest σ−algebra containing P and L is the smallest monotone classcontaining P.

Then L = A.(Mihai's version) In other words: for a π−system P0, the smallest λ−system

containing P0 is in fact, a σ − algebra. )

Remark. This is very similar in feel to the monotone class theorem, it has thesame �two step� trick to it.

Proof. We will show that L is a σ-algebra. By the above theorem, it su�cesto show that is is a π-system, so we will do this easier task instead.

Claim 1: For a �xed C ∈ L, de�ne LC := {D ∈ L : C ∩D ∈ L}. Then LC is aλ-system.

Pf: Use the �rst def'n of λ−system (because this one is easier to check) andthe result follows wihtout too much di�cutly because L itself is a λ−system.

Now, choose any C ∈ P. Since P is a π−system, it is closed under intersectionsand we know at least that P ⊂ LC since L contains P. Now, since LC is a λ−systemcontaining P, we have that L = LC since L is the smallest such λ-system.

Since LC = L for every C ∈ P, this is saying that C ∩D ∈ L for every C ∈ Pand every D ∈ L.

Now for any E ∈ L, we claim that LE contain P. Indeed, to check this wewould need E ∩ C ∈ L for every C ∈ P, but this is exactly the above property.Hence LE = L.

Since this holds for all E ∈ L, we have that L is in fact a π−system. �

Remark. There should be a nicer way to make the proofs of the monotoneclass theorem and the π − λ theorem look more similar, but I'm not going to dothat now.

Measures

These are notes from Chapter 3 of [1].

Definition. (3.1) A measure on a set X and an σ− algebraA is a a functionµ : A → [0,∞] such that:

i) µ(∅) = 0ii) Countably additive for disjoint sets:

µ(⋃∞

i=1Ai) =

∞∑i=1

µ(Ai) for disjoint A′is

Proposition. (3.5.) (Basic Properties of Measures)i) A ⊂ B =⇒ µ(A) ≤ µ(B)ii) A = ∪∞i=1Ai =⇒ µ(A) ≤

∑∞i=1 µ(Ai)

iii) Ai ↑ A =⇒ limn→∞ µ(Ai) = µ(A)iv) If µ(A1) <∞, then ,Ai ↓ A =⇒ limn→∞ µ(Ai) = µ(A)

Proof. These all follow by doing constructions involving complements and setdi�erences and so on in such a way as to reduce the sets in question as unions ofdisjoint sets, for which we know that measures play nice.

Most of them use the �Treat Disjointly Trick� (See remark below)i) Holds since µ(A−B) ≥ 0ii) Make Bi ⊂ Ai by taking set di�erences so that ∪ni=1Bi = ∪ni=1Ai for each n

and the Bi's are disjoint. Then µ(A) = µ(∪Bi) =∑µ(Bi) ≤

∑µ(Ai)

iii) µ(An) is a montone increasing function and is bounded above by µ(A), soit is convergent (to possibly ∞if µ(A) =∞). Hence, by the monotone convergencetheorem for real numbers, this sequence has a limit. To see that the limit is actuallyµ(A), take Bi = Ai+1 −Ai so the Bi's are disjoint and have ∪ni=1Bn = ∪ni=1Ai andapply the countable additivity to get the result. Explicitly:

µ(A) = µ(∪∞n=1An)

= µ(∪∞n=1Bn)

=

∞∑n=1

µ(Bn)

= limn→∞

n∑i=1

µ(Bi)

= limn→∞

µ (∪ni=1Bi)

= limn→∞

µ (∪ni=1Ai)

iv) Take complements of the result in iii) (or, more generally if you arent in a�ntis measures spaceBk = A1 −Ak). The hypothesis that µ(A1) <∞ is needed to

7

MEASURES 8

avoid a �∞−∞” problem. Counterexample to keep in mind: An = (n,∞) whichis decreasing to the empty set but always has in�nte Lebesgue measure. �

Remark. A common theme that comes up for this type of thing: The proofis easy for disjoint sets. The arbitary case can be reduced to the disjoint case bymaking sets Bi which are disjoint and have ∪ni=1Bi = ∪ni=1Ai for every n. I willlabel this trick as [Treat Disjointly]

Definition. (3.7.) De�nition of �nite and σ-�nite measure spaces. A null

set is a set which is a subset of a 0 measure measurable set. (i.e. null sets neednot be measurable) A complete measurable space is one where all null sets aremeasurable. The completion of A is the smallest sigma algebra containing all thenull sets.

Construction of Measures

These are notes from Chapter 4 of [1].Recall the di�uculty with measures is that if you try to de�ne them on all

subsets of a space X, you get problems. The solution is to carefully choose theright σ − algebra to de�ne the measure on. As a consequence, �measurable sets�will be nice enough to work with! This is outlined here.

3.4. Outer Measures

Definition. (4.1.) An outer measure is a function µ∗ de�ned on a collectionof subsets of X satisfying:

i) µ∗(∅) = 0ii) A ⊂ B =⇒ µ∗(A) ⊂ µ∗(B)iii) µ∗(∪∞i=1Ai) ≤

∑∞i=1 µ

∗(Ai)

Remark. There is no notion of �measurable� or σ-algebra or whatever foran outer measure. It has to be de�ned for every subset. It is called an �outermeasure� because of the following common way to construct an outer measure byapproximation with sets �from the outside�. (Sometimes you'll see the constructionhere as the de�nition of an outer measure, and the properties we demand in thede�nition here would be something to be proven)

Proposition. (4.2.) Suppoce C is a collection of subsets of X containg both ∅and X. Suppse that ` : C → [0,∞] with `(∅) = 0. De�ne:

µ∗(E) = inf

{ ∞∑i=1

`(Ci) : Ci ∈ C∀i and E ⊂ ∪∞i=1Ci

}Then µ∗is an outer measure.

Proof. The points i) and ii) from the de�ntion are very easy. To prove iii)consider as follows. Given any collection of subsets A1, A2, . . . and any ε > 0, �nd acollection of subsets Cij ∈ C so that for each �xed i theC ′ijs cover Ai andµ

∗(Ai) ≥∑`(Cij)+ ε/2

i for each i. (This is by de�nition of �inf�). Notice that the collectionCij now covers all of ∪Ai and so we have that:

µ∗(∪Ai) ≤∑ij

`(Cij)

≤ . . . ≤∞∑i=1

µ∗(Ai) + ε

Since this holds for any ε, the desired inequality must be true. �

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3.4. OUTER MEASURES 10

Example. Lebesgue outer measure is what you get when you put in C =halfopen intervals and `(a, b] = b − a . The Lebesgue-Stieltjes outer measure is whatyou get if you again choose C =half open intervals and you have a non-decreasingand right continuous function α.

Definition. (4.5) A set A ⊂ X is measurable with respect to an outer

measure µ∗ (or more simply µ∗−measurable....I like the �rst phrasing better be-cause it reminds me that outer measures are de�ned on all subsets of a space)if:

µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac) ∀E ⊂ X

Remark. By the de�nition of an outer measure, it is ALWAYS TRUE for anyset A that:

µ∗(E) ≤ µ∗(E ∩A) + µ∗(E ∩Ac) ∀E ⊂ XSo the property of being measurable is really about the other inequality. This

is most often how one proves certain sets are measurable.One way to rememember the direction of the sign is to think that �E ∩A and

E ∩ Ac together form a cover of E. Hence, since µ∗(E) is the inf over the measurof all covers, µ∗(E) ≤the sum�. Of coruse, this is a mnemonic device only sincestricly speaking the measure µ∗ is de�ned only as the inf over covers formed by thespecial sets C.

Theorem. (4.6.) Let A = {measurble w.r.t. µ∗ sets} .Then A is a σ−algebra.If we de�ne µ := µ∗|A, then µ is a measure on A. Finally, A contains all the setsof outer measure 0.

Remark. One of the main themes to keep in mind for this proof is that weare trying hard to get the inequalities of the form µ∗(E) ≥ . . . , because this isthe non-trivial inequality for being measurable. Where will these inequalities comefrom? The only way is to use the trivial side of the inequality in a clever way:you must manipulate µ∗(E) = . . . = µ∗(U ∩ V ) + µ∗(U ∩ V c) and then remarkµ∗(U ∩ V ) + µ∗(U ∩ V c) ≥ µ∗(U) since this always holds. (i.e. use property iii)from the de�nition of outer measure.)

Proof. Claim 1: A is closed under complements.Pf: Follows by the symmetry of the de�nition of measurable w.r.t. µ∗.Claim 2: A,B ∈ A =⇒ A ∪B ∈ APf: For any E, use the measurablity property of A at the set E and use the

measurability property of the set B once at the set E∩A and once at the set E∩Ac:µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac)

= (µ∗(E ∩A ∩B) + µ∗(E ∩A ∩Bc) +(µ∗(E ∩Ac ∩B) + µ∗(E ∩Ac ∩Bc)))

Now notice that the �rst three terms are E ∩A ∩B, E ∩Ac ∩B, E ∩A ∩Bc.Since A ∪ B = (A ∩ B) ∪ (A ∩ Bc) ∪ (B ∩ Ac), so consequently the �rst three setshere cover E ∩ (A∪B). Hence the sum of their measures is ≥ µ∗(E ∩ (A∪B)) andwe get (alspo use Ac ∩Bc = (A ∪B)cby DeMorgan):

µ∗(E) ≥ µ∗(E ∩ (A ∪B)) + µ∗(E ∩ (A ∪B)c)

So indeed, A∪B satisfy the non-trivial direction of the measurable inequality,and are hence measurable.

3.5. LEBESGUE-STIELTJES MEASURE 11

Claim 3: An ∈ A =⇒ ∪∞n=1An ∈ APf: We �rstly remark that is su�ces to check this for disjoint sets An, because

by using the property of Claim 2, we can rewrite any countable union as a countableunion of disjoint sets modulo some �nite union operations.

For disjoint sets An let Bn = ∪ni=1Ai and B = limn→∞Bn. For E ⊂ X have:

µ∗(E ∩Bn) = µ∗(E ∩Bn ∩An) + µ∗(E ∩Bn ∩Acn)= µ∗(E ∩An) + µ∗(E ∩Bn−1)

Moving the µ∗(E∩Bn−1) to the other side, and then using this as a telescopingsum, we get that:

µ∗(E ∩Bn)− µ∗(E ∩B1) =

n∑i=1

µ∗(E ∩Ai)

=⇒ µ(E ∩Bn) ≥n∑i=1

µ∗(E ∩Ai)

Hence:

µ∗(E) = µ∗(E ∩Bn) + µ∗(E ∩Bcn) ≥n∑i=1

µ∗(E ∩Ai) + µ∗(E ∩Bc)

Taking n→∞, (ok since the sum is monotone increasing) we get:

µ∗(E) ≥∞∑i=1

µ∗(E ∩Ai) + µ∗(E ∩Bc)

≥ µ∗ (∪∞i=1E ∩Ai) + µ∗(E ∩Bc)= µ∗(E ∩B) + µ∗(E ∩Bc)≥ µ∗(E)

So indeed B is µ∗-measurable!These claims together show that A is a sigma algebra.Claim 4: µ := µ∗|A is a measure.Pf: We have only to show that it is countable additive. Following the discussion

above, we had that µ∗(E) =∑∞i=1 µ

∗(E ∩Ai) + µ∗(E ∩Bc) for any set E. TakingE = B gives exactly the result for countable additivity.

Claim 5: If µ∗(A) = 0 then A is µ∗measurable.Pf: This follows from the monotone property of the outer measure, for if

µ∗(A) = 0 then:

µ∗(E ∩A) + µ∗(E ∩Ac) ≤ µ∗(A) + µ∗(E ∩Ac)≤ µ∗(A) + µ∗(E)

= µ∗(E)

Shows the non-trivial direction of the measurability inequality. �

3.5. Lebesgue-Stieltjes Measure

Let α(x) be an increasing right continuous function, let C = {(a, b] : a,< b} andde�ne `((a, b]) = α(b)− α(a). Then de�ne the outer measure:

m∗(E) = inf{∑

`(Ai) : Ai cover E}

3.5. LEBESGUE-STIELTJES MEASURE 12

By Prop 4.2 this is an outer measure, and by theorem 4.6 this de�nes a measureon the collection of m∗−measurable sets.

The convenience of using semi open intervals is due to the following fact:

Lemma. If K and L are disjoint semi open intervals, (of the form (a, b] ) andK ∪ L is also a semi open interval then:

`(K) + `(L) = `(K ∪ L)

Proof. Both are equal to α(c) − α(a) = (α(c) − α(b)) + (α(b) − α(a)) wherea, b, c denote the endpoints of the intevrals in question. �

What kinds of sets are m∗−measurable? Here is a useful fact!

Proposition. (4.7.) Every set in the Borel σ−algebra on R ism∗−measurable.

Proof. Since the collection of m∗−measurable sets is a σ−algebra (proven inProp 4.6) it su�ces to show that every interval J of the form (c, d] ism∗−measurable.Let E be any set with m∗(E) <∞. We need to show that:

m∗(E) ≥ m∗(E ∩ J) +m∗(E ∩ Jc)

Choose I1, I2, . . . of the form (ai, bi] that cover J and so thatm∗(E) ≥∑i[α(bi)−

α(ai)]− ε. Now, since E ⊂ ∪Ii we have that:

m∗(E ∩ J) ≤∑i

m∗(Ii ∩ J)

m∗(E ∩ Jc) ≤∑i

m∗(Ii ∩ Jc)

So we get in sum that:

m∗(E ∩ J) +m∗(E ∩ Jc) ≤∑i

m∗(Ii ∩ J) +m∗(Ii ∩ Jc)

Now we get to the convenience of using semi-open intervals! Since each Ii andJ are semi open intervals, Ii ∩ J is an interval too! Ii ∩ Jc is the union of zero, oneor two such intervals. This is exactly the setup for us to apply the Lemma before.Applying this zero, one or two times we have:

m∗(Ii ∩ J) +m∗(Ii ∩ Jc) = m∗(Ii)

Thus our inequality above is:

m∗(E ∩ J) +m∗(E ∩ Jc) ≤∑i

m∗(Ii) ≤ m∗(E) + ε

Since this holds for any ε > 0, we have the desired inequality. �

The next thing to verify is that m∗ and ` agree on semi-open sets. This is notat all surprising, but it needs to be veri�ed.

Lemma. (4.8.) Let Jk = (ak, bk) be a collection of �nte open intervals whichcovera �nite closed interval [C,D] then:

n∑k=1

[α(bk)− α(ak)] ≥ α(D)− α(C)

3.6. EXAMPLES AND RELATED RESULTS 13

Proof. By looking at a subset of the intervals, we may suppose WOLOG thatthe intervals are �in order� so to speak so that:

a1 ≤ C ≤ b1 and , an ≤ D ≤ bn and ak < bk−1 < bk

Then just write out both sums and see by comparison to a telescoping sumthat the inequality holds. �

Proposition. (4.9) If e and f are �nite and I = (e, f ] then m∗(I) = `(I)

Proof. (This is the one that uses compactness of closed intervals. The ideais to convert any cover by semi-open intervals to a cover using open intervals (thisgives an ε of error by the continuity of α) and then use compactness to get downto �nitely many intervals of interest. The apply the previous lemma.)

Clearly m∗(I) ≤ `(I) by the inf de�nition of m∗ since I is a cover for itself.For the other inequalty, suppose that I ⊂ ∪Ai where each Ai = (ci, di] is a aninterval. By the right continuity of the function α, choose C ∈ (e, f) so that

α(C) < α(e) + ε/2. Let D = f . Choose d′

i > d so that α(d′

i) < α(di) + ε/2i+1 andlet Bi = (ci, d

′i)

(Each (ci, d′i) aproximates the semi open interval (ci, di] and the whole thing

is done so that the total error from the point of view of the α function is no morethan ε.)

Now use the compactness of [C,D] to �nd a �nite set of the Bi's that cover allof [C,D] still. By the previous lemma we get that:

`(I) ≤ α(D)− α(C) + ε/2 ≤∑

`(Ai) + ε/2i+1 + ε/2

and since ε is arbitary the result follows. �

Now that all the major points have been hit, we will drop the ∗ and refer to mas the Lebesgue-Stieltjes measure corresponding to α.

3.6. Examples and related results

3.6.1. Approximating Sets.

Example. (4.10) For the Lebesgue measure its not hard to show that singletonsare measure 0 and so the open, half-open, closed, half-closed intervals all have thesame measure. All countable sets are Lebesgue measure 0 too.

Example. (4.11) The middle thirds Cantor set is uncontable but still hasLebesgye measure 0. The Cantor ternary function is also brei�y described in thisexample.

Example. (4.13) By changing the fraction of removed set as you go, you canend up with a positive measure generalized Cantor set. This set is closed andcontains no intervals, and every point is a limit point.

Proposition. (4.14) Let A ⊂ [0, 1] be a Borel measurable set and let m be theLebesgue measure. Then:

i) Given any ε > 0 there is an open set G so that A ⊂ G and m(G−A) < εii) Given any ε > 0 there is a closed set F so that F ⊂ A and m(A− F ) < εiii) There is a Gδ set H so that A ⊂ H and m(H −A) = 0iv) There is an Fσ set F so that F ⊂ A and m(F −A) = 0

3.7. NONMEASURABLE SETS 14

Remark. We used half open sets in our construction of the Riemann-Stietijesmeasure, so we need to adapt from our half open intervals to the open intervals thatmake up open sets. If one had used open intervals to construct the measure (whichis totally legit btw) then this proposition would be straight from the inf de�nitionof m∗; but some early results (mostly the proof that m∗ = ` for intervals would beslightly harder)

Proof. i) Approximate A be a union of semi-open intervals so that the erroris no more than ε/2 (this holds by the inf de�nition of m∗) Then approximate eachsemi-open interval by an open interval so that th error is no more than ε/2n+1.

ii) Take complements and use the result in i)iii) Choose ε = 1

n and apply the result in i) to get a sequence of open set Gn.WOLOG they are decreasing. Take their intersection to get the desired Gδset.

iv) Same procedure as in iii) applied to closed sets Fn generated from ε = 1n in

part ii) �

3.7. Nonmeasurable Sets

Theorem. (4.15) [Existence of a non-measurable set] Let m∗be the outer mea-sure de�ned in the usual way with the collection C = {(a, b]} and `((a, b]) = b − a.The m∗ is not a measure on the collection of all subsets of R.

Remark. Since we showed thatm∗IS a measure when restricted to them∗measurablesets, this is saying that there are some sets that are not m∗−measurable, and thatthese sets are badly behaived enough that they prohibit m∗from being a measurehere.

Proof. (Here we have the construction using shifting by rationals. There isalso the �x ∼ y if x − y = kα mod 1 proof where α is irrational� that works in asomeqhat similar way.)

Suppose m∗ is a measure. De�ne x ∼ y if x − y is a rational. This is anequivalence relation. Use the axiom of choice to get a representative from eachequivalence class from the points in [0, 1]. Let A be the set of these representatives.Clearly:

[0, 1] ⊂ ∪q∈[−1,1]∩QA+ q

because every point in [0, 1] belongs to some equivalence class, and is hence arational shift away from some point in A by de�nition. Moreover, each set A+ q1is disjoint from A + q2 for q1 6= q2 for if there was a point w in there intersectionthen w − q1 ∈ A and w − q2 ∈ A are both in the same equivalence class and thiscontradicts the choice of A. Hence:

1 ≤∑q

m∗(A+ q)

=∑q

m∗(A)

(m∗(A) = m∗(A+q) comes from the fact that `(I) = `(I+q) and the de�nitionof m∗). On the other hand, ∪q∈[−1,1]∩QA+ q ⊂ [−1, 2] since A ⊂ [0, 1]. So we get:

1 ≤∑q

m∗(A) ≤ 3

3.8. THE CARATHEODORY EXTENSION THEOREM 15

There is no value for m∗(A) that satis�es these inequalites! Either m∗(A) = 0and the lower bound is violated, or m∗(A) > 0 and the upper bound is viloated. �

3.8. The Caratheodory Extension Theorem

Definition. Recall the de�nition of an algebra. (It was like a σ−algebra but�ntie unions instead of countable ones.) A premeasure on an algebra A0 is afunction µ : A0 → [0,∞] satisfying:

i) µ(∅) = 0ii) Countably additive for disjoint sets that happen to stay in the algebra:If A1, . . . are disjoint and it happens that ∪∞i=1Ai ∈ A0, then

µ(⋃∞

i=1Ai) =

∞∑i=1

µ(Ai)

The next theorem repackages the constructuion of the Lebesgue-Stieltjes mea-sure in a slightly more general framework:

Theorem. (4.16) [CARATHEODORY EXTENSION THEOREM] Supposethat A0 is an algebra and that ` : A0 → [0,∞] is a premeasure on A0. De�ne:

µ∗(E) = inf

{ ∞∑i=1

`(Ai) : each Ai ∈ A0 cover E

}Then:1) µ∗ is an outer measure2) µ∗(A) = `(A) if A ∈ A0

3) Every set in A0 is µ∗−measurable4) If `is σ−�nite then there is a unique extension to σ(A0)

Proof. (1) follows by proposition 4.2(2) µ∗(E) ≤ `(E) for E ∈ A0 by the inf def'n of µ∗ since E covers itself. The

other inequality holes because if E ⊂ ∪Ai, then we can let Bn = E∩(An − ∪n−1i=1 Ai

)so that the ∪nBk = ∪nAk and since the Bi's are pairwise disjoing and each in A0

and their union is E which also happens to be in A0, so by the premeasure property:

`(E) =

∞∑i=1

`(Bi) ≤∞∑i=1

`(Ai)

So clearly then by taking inf's we have that`(E) ≤ µ∗(E)(3) To see that every set A ∈ A0 is µ∗measurable, consider as follows. For any

E ⊂ X, take B1, B2, . . . ∈ A0 that cover E and∑`(Bi) ≤ µ∗(E) + ε. Then, since

`(Bi) = `(Bi ∩ A) + `(Bi ∩ Ac) because ` is a premeasure and A,B ∈ A0. Havethen:

µ∗(E) + ε ≥∞∑i=1

`(Bi) =

∞∑i=1

`(Bi ∩A) +∞∑i=1

`(Bi ∩Ac)

≥ µ∗(E ∩A) + µ∗(E ∩Ac)

Since ε arbitrary, we get the desired non-trivial half of the measurability in-equality. �

3.8. THE CARATHEODORY EXTENSION THEOREM 16

(4) The fact that the measure exists follows from our previous work: We knowthat A = {µ∗ −measurable sets}is a σ−algebra, it contains A0 by (3), and weshowed earlier that µ = µ∗|A is a measure. Uniqueness is most easily done withthe π − λ theorem:

If µ, ν are two measures so that µ(A) = ν(A) = `(A) for all A ∈ A0, letP = {A : µ(A) = ν(A)}. This contains the algebra A0 by construction, and weverify that it is a λ−system. Hence, by the π−λ theorem, it is in fact a σ-algebra!Since it contains A0, it must in fact be all of σ(A0).

Measurable Functions

These are notes from Chapter 5 of [1].

4.9. Measurability

Fix a measurable space (X,A)

Definition. (5.1.) A function f : X → R is measurable or A-measurable

if f−1 ((a,∞)) ∈ A for all a ∈ R

Proposition. (5.5.)The following are equivalent (i) f−1 ((a,∞)) ∈ A for alla ∈ R ii) f−1 ([a,∞)) ∈ A for all a ∈ R iii) f−1 ((∞, a)) ∈ A for all a ∈ R iv)f−1 ((∞, a]) ∈ A for all a ∈ R

Proof. Use compleemnts and intersections/unions with a fudge factor of 1n�

Proposition. (5.6.) If X is a metric space and A contains all the open sets,then continuous functions are always measurable

Proof. The preimage through any continuous function of any open set is open,so f−1 ((a,∞)) ∈ Open Sets ⊂ A �

Proposition. (5.7.) If f, g are measurable then so are f+g,−f, cf, fg,max(f, g)and min(f + g)

Proof. i) {f + g < a} = ∪r∈Q {f < r} ∪ {g < a− r}ii) {−f > a} = {f < −a}iii) Assume WOLOG by ii that c > 0 and then:{cf > a} = {f > a/c}iv) fg = 1

2

((f + g)2 − f2 − g2

)v) {max(f, g) < a} = {f < a} ∩ {g < a}vi) {min(f, g) > a} = {f > a} ∩ {g > a} �

Proposition. (5.8.) If fn are all measurable then so are sup fn, inf fn, lim sup fn, lim inf fn

Proof. {sup fn > a} = ∩n {fn > a} and {lim sup fn > a} = ∪n∩k>n{fk > a} =lim sup {fn > a}. Similar for inf �

Proposition. (5.10) If f is monotone, then f is Borel measurable.

Proof. SupposeWOLOG that f is increasing. Look at x0 = sup {y : f(y) ≤ a}since f is increasing, for x > x0 we have f(x) ≥ a by uisng the de�niton of inf andthe fact that f is increasing. Hence (x0,∞) ⊂ {f > a}. Similarly, (−∞, x0) ⊂{f ≤ a}. So then {f > a} = [x0,∞) or (x0,∞) and in either case it is measur-able. �

17

4.11. LUSIN'S THEOREM 18

Proposition. (5.11.) f is measurable if and only if f−1(A) ∈ A for all Borelmeasurable A

Proof. Check that the set{B : f−1(B) ∈ A

}is a sigma algebra. It is equal

to the Borel sigma algebra if and only if it contains the open intervals. �

4.9.1. Non-Borel set of Lebesgue measure 0.

Example. (5.12) Let f be the Cantor ternary function and let F be its �inverse�

F (x) := inf {y : f(y) ≥ x}This is an increasing function, so it is measurable. Morevoer, F [0, 1] = C the

Cantor set. Since F is Borel measurable, F−1 maps Borel sets to Borel sets.Now, take any non-measurable set A. Let B = F (A). Since B = F (A) ⊂ C, B

is a Lebesgue-measure-zero set. On the other hand, B cannot be Borel measureable,for it it were then we would have A = F−1(B) and that would mean that A ismeasurable, a contradiction.

4.10. Approximation of functions

Definition. (5.13) De�ne the characteristic function χE to be the 1 on Eand 0 o� E. A simple function is a a �nite linear combination of char functions.

Proposition. (5.14) Suppose f is a non-negative and measurable function.Then there exists a sequence of non-negative measurable simple functions sn in-creasing to f

Proof. Perform the �diadic decompostion�, En,i ={i−12n ≤ f <

i2n

}, and then

�round down� to the diadic sets. (You also have to increase the �ceiling� with n asyou go) �

4.11. Lusin's Theorem

Theorem. (5.15) Suppose f : [0, 1] → R is Borel measurable and m is theLebesgue measure. For any ε > 0 there exists a continuous function g : [0, 1] → Rwhich is continuous and which has {f = g} is a closed set with m {f 6= g} < ε.

Proof. [Standard ladder proof]First prove that it is true for characteristic functions. Find a closed set E

and an open set G so that E ⊂ A ⊂ G with measure m(G − E) < ε and put

δ = inf {|x− y| : x ∈ E, y ∈ Gc} and de�ne g(x) = (1− d(x,E)/δ)+. Then g = f

on F = E ∩Gc (both have the value 1 on E and 0 on Gc) and m([0, 1]−F ) < ε bythe choice of E,G .

For simple functions, f =∑Mi=1 aiχAi approximate each aiχAi by a continuous

function gi on a closed set Fi so that m ([0, 1]− Fi) < ε/M and then put F = ∩iFiwill be the set we want.

For arbitrary non-negative bounded functions, do the diadic decompositionfn =

∑i2nχAi,n

which is increasing to f . Notice that hn = fn+1 − fn is a simplenon-negative function which is bounded by 2−n and we have that (by telescoping)fn =

∑nk=1 hk. Now approximate each hk by a continous function gk so that they

agree on a set Fk so that m([0, 1] − F ) ≤ ε/2k. Then we will have that fn and∑nk=1 gk will agree on a set F = ∩kFk so thatm([0, 1]−F ) < ε . Finally, notice that

since each gk ≤ 12kthe in�nte sum

∑∞k=1 gk is uniformly converging (by say M-test)

4.11. LUSIN'S THEOREM 19

and since each gk is continuous, it converges to a continous function g. This functionis equal to

∑∞k=1 hk on all of F . But by de�nition,

∑∞k=1 hk = limn→∞ fn = f so

this is exactly what we want.Finally, for arbitarty functions, split up f into f+ and f− and do the dance for

both pieces. �

Example. (5.16) A good exmaple to see what this theorem is and is not sayingis to look at f = χ[0,1]−Q. Since the irrationals are dense in [0, 1], the function ftakes values of both 1 and 0 in any interval, so you might be tempted to thingthat its impossible for a continuous function to be equal to this on a set of positivemeasure.

The resolution is that, since m(Q) = 0, we can �nd an closed set F thatcontains no rationals withm(F ) > 1−ε (this can be done explicitly in this case: putG = ∪q∈Q∩[0,1]

(q − ε

2n , q +ε2n

)and F = [0, 1]−G contains no rational numbers.)

Put g ≡ 1 everywhere on [0, 1]. Now since F contains no rationals, g and f agreeon F (they are identically 1 there) and m([0, 1]− F ) < ε.

So in this case, Lusin's theorem gives us a very useless approximation otχ[0,1]−Q, namely the constant function 1. This example nicely illustrates that thetype of approximation �there exists a continuous function that agree except for asmall measure set� is actually not so useful.

The Lebesgue Integral

These are notes from Chapter 6 of [1].

Definition. (6.1.) Let X,A, µ be a measure space. If s =∑ni=1 aiχEi

is asimple function, we de�ne the integral of a simple function to be:ˆ

sdµ =

n∑i=1

aiµ(Ei)

We de�ne the integral of a non-negative measurable function to be:ˆfdµ = sup

{ˆsdµ : s ≤ f and s is simple

}For arbitary functions we de�ne

´fdµ =

´f+dµ−

´f−dµ

Definition. (6.2.) We say f is integrable when´|f | dµ <∞

Proposition. (6.3.) Some basic conseqeunces of the de�nitionsi) If a ≤ f(x) ≤ b for all x ∈ X then aµ(X) ≤

´fdµ ≤ bµ(X)

ii) If f ≤ g then´fdµ ≤

´gdµ

iii) If f is integrable, then´cfdµ = c

´fdµ for all c

iv) If µ(A) = 0 and f is integrable then´fχAdµ = 0

Proof. These are pretty straightforward, most of them are proven by �rstproving it for simple functions, and then using the de�nition of the integral as thesup over simple functions to get the result (sometimes the trick you have to do is todo sup

{´sdµ : s ≤ f and s is simple

}= sup

{´sdµ : s ≤ f and s is simple and has a certain property

}where the certain property helps you get the result. (For example in part i) theproperty is s ≥ a) �

Proposition. (6.4.) If f is integrable, then:∣∣∣∣ˆ f

∣∣∣∣ ≤ ˆ |f |Proof. For the real case its easy using f ≤ |f | and −f ≤ |f | �

20

Limit Theorems

These are notes from Chapter 7 of [1].

6.12. Monotone Convergence Theorem

Theorem. (7.1.) Suppose fn is a sequence of non-negative measurable func-tions with f1 ≤ f2 ≤ . . . for all x and with:

limn→∞

fn = f

Then: ˆfndµ→

ˆfdµ

\

Proof. Since fn is an increasing sequence of functions, by the result in thelast chapter that says integration is monotone (i.e. f ≤ g =⇒

´f ≤´g), we know

that´fn is an increasing sequence of numbers. Moreover, it is bounded above by´

f (because fn ≤ f and again because integration is monotone).By the monotone convergence theorem for real number sthen, there is a limit

limn→∞´fndµ = L . L ≤

´fdµ since each term in the sequence has this property,

so we desire only to show that L ≥´fdµ. By the de�nition of the integral, it is

su�cient to show that L ≥´sdµ for any simple function s with s ≤ f .

To do this we give ourselves a multiplicative factor of 1− ε of room. Fix ε > 0.For any simple function s with f ≤ s look at the set An = {fn > (1 − ε)s}. Sincefn ↑ f and s < f we know that An ↑ X. Now notice that (from the basic propertiesof integration we proved) that:

ˆfn ≥

ˆfnχAn

≥ (1− ε)ˆsχAn

= (1− ε)∑

aiµ (Ei ∩An)

→ (1− ε)∑

aiµ(Ei)

= (1− ε)ˆsdµ

Since this holds for any ε > 0 we have´f ≥´s and the result then follows by

the de�ntion of the integral! �

21

6.15. DOMINATED CONVERGENCE THEOREM 22

6.13. Linearity of the Integral

Theorem. (7.4.) If f, g are non-negative and measurable, then:ˆ(f + g) dµ =

ˆfdµ+

ˆgdµ

Proof. (Ladder Proof)We �rst prove this for non-negative simple functions, which is easy from the

de�ntion.Then we use the diadic decomposition to get sn ↑ f and rn ↑ g and then use

the monotone convergence theorem to get the result for non-negative f,g.For arbitarty f, g, split it into f+ and f− �

6.14. Fatou's Lemma

Theorem. (7.6.) Suppose that fn are non-negative and measurable. Then:ˆlim inf fn ≤ lim inf

ˆfn

Proof. Let gn = infk≥n fk so that gn ↑ lim inf fn. Clearly, gn ≤ fk for eachk ≥ n so we have: ˆ

gn ≤ infk≥n

ˆfk

Now take the limit of this inequality as n→∞. By the Monotone covnergencetheorem, the LHS goes to

´lim inf fn (since gn ↑ lim inf fn) while the RHS goes to

lim inf´fn just by the de�nition of the it �

Remark. Let's compare this to Fatou's lemma for sets:

P (lim inf An) ≤ lim inf P (An)

Proof. Let Bn = ∩k≥nAk so that Bn ↑ lim inf An by the de�nition of theliminf of sets. Also clearly, Bn ⊂ Ak for any k ≥ n so in particular:

P (Bn) ≤ infk≥n

P(Ak)

Take limit as n→∞now. By the continuity of probability, Bk ↑ lim inf An =⇒P(Bk) ↑ P (lim inf An) and on the other side tends to lim inf P(An) �

Remark. So the proof is exactly the same, with the continuity of probabilitydoing the work in the set case, and the monotone convergence theorem doing thework in the function case. (notice actually that An ↑ X =⇒ µ(Ei ∩An)→ µ(Ei)is also used explicitly in the MCT proof)

Once the set one is proven btw, I think you can do a standard ladder to get tothe Monotone convergence theorem for arbitary functions.

6.15. Dominated Convergence Theorem

Theorem. (7.7.) Suppose fn are measurable real-valued functions and fn → fa.e.. Suppose there exists an integrable function g such that |fn| ≤ g for all x. Then:

limn→∞

ˆfn →

ˆf

6.15. DOMINATED CONVERGENCE THEOREM 23

Proof. Use Fatous's lemma once on fn+g which is non-negative and once ong − fn which is also non-negative. You get a limsup/liminf sandwhich and get thedesired equality. �

Theorem. (Exercise 7.2.) If fn, gn, f, g are integrable with fn → f a.e., gn →g a.e. and |fn| ≤ gn for each n and

´gn →

´g, then

´fn →

´f

Proof. Since fn − gn and gn − fn are non-negative we can apply Fatou toknow that:

lim inf

ˆ(fn − gn) ≥

ˆ(lim inf fn − gn)

=

ˆf − g

=

ˆf −ˆg

On the other hand, lim inf(´fn − gn

)= lim inf(

´fn) −

´g by the fact that´

gn →´g

(Use the lemma: If bn → b then lim inf (an + bn)→ lim inf(an) + b Pf: (Subse-quences) Take a subsequence nk so that ank

is convergent and . Then bnk→ b since

bn is convergent. Have then lim (ank+ bnk

) = lim(ank) + b ≥ lim inf(an) + b. Since

every limit point of an+ bn is ≥ lim inf(an)+ b the liminf obeys this inequailty too.By taking a subsequence so that lim(ank

) = lim inf(an), we see the other inequality.)

From here the proof procedes the same as the ordinary Fatou lemma. �

Product Measures

These are notes from Chapter 5 of [1].

7.16. Product σ−algebras

Suppose (X,A, µ) and (Y,B, ν) are two measure σ-�nte measure spacdes.

Definition. A measurable rectangle is a set of the form A × B with A ∈A and B ∈ B. Let C0 be the collection of �nite unions of disjoint measurablerectangles, i.e. C0 = {∪ni=1Ai ×Bi}. One can easily verify that C0 is an algebra.(Some people say the measurable rectangles are a semi algebra and then have atheory based on that, but its exactly the same as this)

We de�ne the product sigma algebra by:

A× B = σ (C0)

For E ∈ A× B we de�ne its x-section and y-section by:

sx(E) := {y ∈ Y : (x, y) ∈ E}ty(E) := {x ∈ X : (x, y) ∈ E}

Given a function f : X × Y → R that is A× B measurable, we de�ne for eachx and y Sxf : Y → R and Tyf : X → R by:

Sxf(y) := f(x, y)

Tyf(x) := f(x, y)

Lemma. (11.1) i) If E ∈ A× B then sx(E) ∈ B for each x and ty(E) ∈ A foreach y

ii) If f is A × B measurable, then Sxf and Tyf are B and A measurable re-spectively.

Proof. (Sigma Albgebra proof)Let C be the collection of sets for which sx(E) ∈ B. Check that its a sigma

algebra and that it contains the rectangles.For ii), do a ladder proof. The result holds for simple functions by i) and so

on. �

Proposition. (11.2) For E ∈ A × B, de�ne h(x) = ν (sx(E)) and k(y) =µ (ty(E)) then:

i) h, k are measurableii) We have: ˆ

h(x)µ(dx) =

ˆk(y)ν(dy)

24

7.17. THE FUBINI THEOREM 25

In other words:ˆ [ˆSxχE(y)ν(dy)

]µ(dx) =

ˆ [ˆTyχE(x)µ(dx)

]ν(dy)

This is sometimes written more succinctly as:ˆ ˆχE(x, y)ν(dy)µ(dx) =

ˆ ˆχE(x, y)µ(dx)ν(dy)

Proof. (This is where the monotone class theorem is �nally used)Let C be the collection of sets where this holds. We check that it contains the

rectangles in A× B and also similarly for a �nite union of rectangles (can assumeWOLOG they are disjoint)

Now suppose En ↑ E and each En ∈ C. The result will follow by the monotoneconvergence theorem.

If En ↓ E and each En ∈ C, then the result will follow by using the dominatedconvergence theorem. (To make this one work you must assume that µ and ν are�nite measures....for σ−�nite it will still work by proving it for each �nite piece�rst)

Hence C is a monotone class containing the algebra of �nite unions of rectangles,therefore it actually contains the whole sigma algebra generated by these, which isall of A× B. �

Definition. The product measure µ× ν is de�ned as:

µ× ν(E) =

ˆh(x)µ(dx) =

ˆk(y)ν(dy)

This is indeed a measure. One can check it is �nitely additive, and then use amonotone convergence theroem to get that it is countable additive.

7.17. The Fubini Theorem

Theorem. (11.3.) If f : X × Y → R is measurable with respect to A× B. Ifeither:

a) f is non-negativeb)´|f(x, y)| d (µ× ν) (x, y) <∞

Then:

ˆ ˆf(x, y)d (µ× ν) =

ˆ (ˆf(x, y)dµ(x)

)dν(y)

=

ˆ (ˆf(x, y)dν(y)

)dµ(y)

Remark. Even if (X,A, µ) and (Y,B, ν) are complete, the product µ × ν onA×B will not be complete. (Example a singleton cross with a non-measurable setis not measurable) For this reason

Signed Measures

These are notes from Chapter 12 of [1].

8.18. Positive and Negative sets

Definition. (12.1) Let A be a sigma algebra. A signed measure is a functionµ : A → (−∞,∞] such that µ(∅) = 0 and µ

(∪∞i=1Ai

)=∑∞i=1 µ(Ai) whenever the

Ai's are pairwise disjoint.

Remark. The only di�erence between this and an ordinary measure is that anordinary measure is required to be non-negative as well.

Definition. (12.2.) Let µ be a signed measure, a set A ∈ A is called apositive set for µ if µ(B) ≥ 0 for all B ⊂ A and B ∈ A. Similarly a negative setA is one where µ(B) ≤ 0 for all B ⊂ A, B ∈ A.

Example. (12.3) If m is the Lebesgue measure then :

µ(A) =

ˆA

fdm

for some integrable f is a signed measure. The sets P = {f > 0} and N ={f < 0} are positive and negative sets for µ.

Proposition. (12.4.) Let µ be a signed measure which takes values in (−∞,∞]let E be measureable with µ(E) < 0. Then there exists a measurable subset F of Ewhich cis measurable and which is a negative set with µ(F ) < 0.\

Proof. (The idea is to start with E and cut out all the positive measure setsof size 1

nk. Any subset of the remaining will be negative or else it would have been

cut out)I'm going to skip the actual details here. �

8.19. Hahn Decomposition Theorem

Definition. A4B := (A−B) ∪ (B −A).

Theorem. (12.5.) (1) Let µ be a signed measure taking values in (−∞,∞].There exists disjoint measurable sets E and F in A whose union is X and suchthat E is a negative set and F is a positive set.

2) The decomposition is essentially unique in the sense that: If E′ and F ′ areanother pair of positive and negative sets whose union is X, then E4E′ = F4F ′is a null set w.r.t. µ

3) If µ is not a positive measure then µ(E) 6= 0. If −µ is not a positive measurethen µ(F ) 6= 0.

26

8.20. JORDAN DECOMPOSITION THEOREM 27

Proof. (1) Let L = inf {µ(A) : A is a negative set } Choose sets An so thatµ(An) → L. Let E = ∪∞n=1An. Let Bn = An − (B1 ∪ . . . ∪Bn) so that ∪nk=1Bk =∪nk=1Ak and the Bk's are pairwise disjoint. Since the An's are all negative sets, soare the Bn's. We claim that E = ∪nBn = ∪nAn is a negative set. This follows bythe continuity of measure, sicne for any C ⊂ E we have::

µ(C) = limn→∞

µ (C ∩ (∪ni=1Bi)) = limn→∞

n∑i=1

µ(C ∩Bi) ≤ 0

We claim that µ(E) = L since L ≤ µ(E) = µ (An) + µ(E − An) ≤ µ(An)→ Lmakes the desired sammich.

Now we claim that F = Ec is a positive set. Otherwise, there is a subset C ⊂ Fof negative measure. But then C contains a negative set (prop 12.4.) and so E ∪Cmakes a negative set with measure stricly less than L, contradicting L as the sizeof the �largest� negative set.

(2) Rewritting E4E′ as (E∩F ′)∪(E′∩F ) makes it clear that E4E′ = F4F ′.Any subset of this can be written as a subset of E ∩ F ′ union with a subset ofF ∩ E′. Any subset of either of these must simultaneuous have non-negative andnon-positive measure, so it is indeed a null set.

(3) Skip this for now. �

Definition. We say two (non-negative) measures µ and ν are mutually sin-

gular if there exist two disjoint subsets E and F in A whose union is all of X andwith µ(E) = 0 and µ(F ) = 0. This is often written µ ⊥ ν.

Example. (12.6) The Lebesgue measure restricted to (0, 12 ] and the Lebesgue

measure restrived to [ 12 , 1) are mutually singular. This works because the Lebsegue

measure of { 12} is 0.

Example. (12.7.) Look at the Cantor Ternary function f on [0, 1] and let ν bethe assoiciated Lebesgue-Stieltjes measure. Let µ be the Lebesgue measure. Thenµ ⊥ ν by the partitioning of the [0, 1] = C ∪Cc where C is the Cantor set. On Cc,ν is measure zero because f is constant on every interval there.

8.20. Jordan Decomposition Theorem

Theorem. (12.8.) If µ is a signed measure on a measurable space (X,A),then there exist positive measure µ+ and µ− such that µ = µ+ − µ− and µ+ ⊥ µ−.Moreover this decomposition is unique.

Proof. Use the Hanh decomposition to �nd the negative set E and positiveset F that decompose X/ Let µ+(A) = µ(A ∩ F ) and µ−(A) = −µ(A ∩ E). Thisgives the desired decomposition.

The uniqueness follows with some work from the �uniqueness up to null sets�result for the Hanh decomposition. �

Definition. The measure:

|µ| = µ+ + µ−

Is called the total variation measure of µ and |µ| (X) is called the totalvariation of µ.

The Radon-Nikodym Theorem

These are notes from Chapter 13 of [1].

9.21. Absolute Continuity

Definition. (13.1) A measure ν is said to be absolutly continuous withrespect to a measure µ if µ(A) = 0 =⇒ ν(A) = 0. This is written ν � µ

Proposition. (13.2) Let ν be a �nite measure. Then ν is absolutely continuouswith respect to µif and only if for all ε > 0, there exists δ > 0 so that µ(A) < δ =⇒ν(A) < ε

Proof. (⇒) Suppose by contradiction this is not the case. Then there existsε0 so that for every δ there is a bad set Aδ with µ(Aδ) < δ and ν(Aδ) > ε0.Take δ = 2−n to get a sequence of these sets and let A = lim supAn. Then since∑µ(An) < ∞ we know that µ(A) = 0 by Borel-Ceantelli but on the other hand

ν(A) = ν (lim supAn) ≥ lim sup ν(An) > ε0 which contradicts absolute continuity.(⇐) If µ(A) = 0 then µ(A) < δ for all δ and so using the hypothesis we can

conclude that ν(A) < ε for any ε, meaning that ν(A) = 0 too. �

Remark. This proposition gives the connection with the notion of absolutecontinuity for a function. Recall that a function is said to be absolutely continuousif for all ε > 0, there exists δ > 0 so that any pairwise disjoint sequence of intervals(xk, yk) has

∑k |yk − xk| < δ =⇒

∑k |f(yk)− f(xk)| < ε.

9.22. The main theorem

Lemma. (13.3.) Let µ and ν be �nite positive measure on a measurable space(X,A). Either µ ⊥ ν or else there exists ε > 0 and G ∈ A such that µ(G) > 0 andG is a positive set for ν − εµ. (i.e. ν(A) ≤ εµ(A) for all A ⊂ G)

Proof. Conisder the Hahn decomposition for ν − 1nµ. Let F = ∪nFn be the

union over all the positive sets and E = ∩nEn be the intersection of all the negativesets. Notice that Ec = F by DeMorgan's laws since Ecn = Fn for each n.

Claim: ν(E) = 0Pf: ν(E) ≤ ν(En) ≤ 1

nµ(En) ≤1nµ(X)→ 0

If µ(Ec) = µ(F ) = 0 then the sets E and F divide up X to show that µ ⊥ ν.\Otherwise, µ(Ec) = µ(F ) > 0. In this case, µ(Fn) > 0 for some Fn. Choosing

G = Fn and ε = 1n gives the result. �

Theorem. (13.4) Suppose that µis a σ−�nite positve measure on a measurablespace (X,A) and ν is a �nite positive measure on (X,A) such that ν � µ. Then

28

9.23. LEBESGUE DECOMPOSITION THEOREM 29

there exists a µ−integrable non-negative f which is measurable with respect to Asuch that:

ν(A) =

ˆA

fdµ

Moreover the f is unique up toµ−a.e.

Proof. (The idea is to look at the set of functions f so that´Afdµ ≤ ν(A)

for all A ∈ A, and then to take the �largest� of these) The uniqueness part is easysince if

´Afdµ =

´Agdµ for all A ∈ A then

´A(f − g)dµ = 0 for all A ∈ A and

then f = g a.e. (for example look at the set A = {|f − g| > ε} must be measure 0for all ε)

Let F ={g measurable : g ≥ 0,

´Agdµ ≤ ν(A) for all A ∈ A

}Let L = sup

{´gdµ : g ∈ F

}and take a sequence gn ∈ F so that

´gn → L.

Let hn = max (g1, . . . , gn) . Check that this is still in F . Let f = suphn = limhnsince the hn's are increasing. By the monotone convergence theorem,

´fdµ =

limn

´hndµ ≤ ν(A) since hn ∈ F which shows that f ∈ F and also

´fdµ ≥´

hndµ ≥´gndµ→ Lshows that

´fdµ = L.

Now there is some technical work (in partticular using the previous lemmas)to show that f is the desired function and also the fact that ν � µ to make itwork. �

9.23. Lebesgue Decomposition Theorem

Theorem. (13.5.) If µ is a sigma �nite measure and ν is a �nite measure,then there exists positive measure λand ρ so that:

i) ν = λ+ ρii) ρ� µiii) λ ⊥ µ

Proof. Take f as in the proof of the R-N theorem and set ρ(A) =´Afdµ and

λ = ν − ρ and it will work out (technical details are omitted) �

Di�erentiation

These are notes from Chapter 5 of [3]. Here is a mind map of the proofs of thissection:

30

DIFFERENTIATION 31

Vita

li C

over

A c

olle

ctio

n of

inte

rval

s is

cal

led

a V

itali

cove

r if e

very

poi

nt is

cov

ered

by

an in

terv

al o

f arb

itrar

ily s

mal

l siz

e

Vita

li C

over

Lem

ma

Giv

en ε

> 0

and

a V

itali

cove

r of a

se

t E, t

here

is a

fini

te s

ubse

t of

inte

rval

s fro

m th

e co

ver t

hat c

over

E

excp

et fo

r a s

et o

f siz

e <

ε

Incr

easi

ng F

unct

ions

are

Diff

Pf H

int:

Che

ck th

at th

e "fo

ur

deriv

ativ

es" a

re e

qual

pai

r by

pair

exce

pt fo

r nul

l set

s. L

ook

at th

e se

t w

here

D-<

u<v<

D+

with

u,v

, rat

iona

l. M

ake

a V

itali

cove

r her

e w

here

the

endp

oint

s ar

e co

ntro

lled.

Use

in

crea

sing

to g

et a

n in

equa

lity.

Fato

u B

ound

for

Incr

easi

ng f

⌠f '

< f(b

) - f(

a)

Func

tions

of B

ound

ed V

aria

tion

are

the

Diff

of T

wo

Mon

oton

e Fu

nctio

nsP

f Hin

t: Th

e tw

o fu

nctio

ns a

re th

e cu

mul

ativ

e po

sitiv

e an

d ne

gativ

e va

riatio

n

Func

tions

of

Bou

nded

Var

iatio

n A

re D

iffer

entia

ble

Inde

finite

Inte

gral

s ar

e of

bou

nded

va

riatio

n

diff

( int

f(t)

dt) =

f(x)

Pf h

int:

Pro

ve it

for b

ound

ed fu

nctio

ns

first

usi

ng th

e co

verg

ence

of f

n =

seca

nt c

urve

of s

ize

1/n

to F

' . T

hen

boot

stra

p to

inte

grab

le F

.

Inde

finite

Inte

gral

s ar

e ab

s. c

ont.

Def

n of

Abs

olut

ely

Con

tinuo

us

Abs

. Con

t. is

bo

unde

d va

riatio

n

f abs

con

t and

f' =

0 a

.e. t

hen

f =

cons

tP

f hin

t: Fi

x ε

and

δ . U

se th

e fa

ct th

at |f

' | <

δ to

m

ake

a V

italli

cov

er s

o th

at f

can'

t mov

e m

uch

on th

e in

terv

als.

The

n us

e co

ver l

emm

a to

get

fin

itely

man

y th

at c

over

it e

xcpe

t for

a s

mal

l set

. U

se a

bs c

ont t

o co

ntro

l the

mov

emen

t of f

on

the

smal

l set

.

f abs

con

t the

n f =

int (

diff

f ) d

tP

f hin

t: C

heck

that

g(x

) = f(

x) -

int f

'(t)

dt

has

zero

der

ivat

ive

ever

ywhe

re.

10.24. DIFFERENTIATION OF MONOTONE FUNCTIONS 32

10.24. Di�erentiation of Monotone Functions

Definition. Let I be a collection of intervals we say that I is a Vitali cov-ering of a set E if for all ε > 0 and any x ∈ E, ∃I ∈ I with x ∈ I and `(I) < ε.The intervals may be open, closed or half-opne but no degenerate intervals (i.e.singletons) are allowed.

Remark. Another way to say this heurstically: �I is a Vitali covering if itcovers every point in E with an arbitarily small interval�

Lemma. (Vitali) Let E be a set of �nte outer measure and I a Vitali coveringof E. Then given ε > 0 there is a �nite disjoint collection {I1, . . . , IN} of intervalsin I such that:

m∗

[E −

N⋃n=1

In

]< ε

Proof. (Pf Idea only)Assume WOLOG all the intervals are closed and O ⊃ E is an open set containg

E and every interval is a subset of O.The idea is to recusrvly construct a sequence {I1, . . .} by adding in intervals

one at a time, chosen so that they are disjoint and every new added inteval is closeto being as large (in the sense of length) as it can be. (i.e. its length is at least 1

2

the sup over the lengths of all candidate intervals) Have then∑∞k=1 `(Ik) ≤ m(O).

For any ε > 0 take N so large so that∑k≥N `(Ik) < ε then and then show that

∪Nn=1In is the set we are looking for. The fact that its a Vitalli covering, and thatwe chose the intervals to be close to as large as possible means we could not havemissed a piece of E of size more than ε. �

Definition. The derivatives (there are four of them) of a function f arede�ned as: (Can come from the left or right (denoted by+ or −) and can be thelimsup or the liminf (denoted by either a superscript or a subscript)

D+f(x) = lim suph→0+

f(x+ h)− f(x)h

D−f(x) = lim suph→0−

f(x+ h)− f(x)h

D+f(x) = lim infh→0+

f(x+ h)− f(x)h

D−f(x) = lim infh→0−

f(x+ h)− f(x)h

Clearly D+ ≥ D+ and D− ≥ D−. If all four of these are equal, then we say thefunction is di�erentiable and call its derivative f ′. If D+f = D+f we say that fhas �right-hand� derivatevs and de�ne f ′(x+) to be its value. Similary for f ′(x−)

Remark. The connection between the Vitali covering lemma and these deriva-tives is that at each point x, we can �nd (by de�niton of limsup/liminf) we can�nd a points hn → 0 so that f(x+ hn) is controlled. The intervals [x, x+ hn] willbe a nice vitalli covering for us.

Proposition. (2) If f is continuous on [a, b] and one of its derivatives (forconcreteness say D+) is everywhere non-negative on (a, b), then f is non-decreasingon [a, b] (i.e. f(x) ≤ f(y) for all x ≤ y in the interval)

10.24. DIFFERENTIATION OF MONOTONE FUNCTIONS 33

Proof. ....I tried for a bit to come up with a proof by contradiction but coudlntmake it work. �

Theorem. Let f be an increasing real-valued function on the interval [a, b].Then f is di�erentiable almost everywhere. The derivative f ′ is measurable and:ˆ b

a

f ′(x)dx ≤ f(b)− f(a)

Remark. The Cantor ternary function is an example where f ′ = 0 a.e. andso´f ′ = 0 < f(b)− f(a)

Proof. The idea is to show that any two of the the four derivateves agreeexcept on a set of measure 0. For concreteness we will handle D+f = D−f . Toshow that thesea re not equal except on a set of measure zero, it su�ces to show bycountability that Euv = {D+f > u > v > D−f} is a measure zero set for all u, vrational.

Let O be an open set so that m(O) < m(Eu,v) + ε. For each x ∈ Eu,v there isan arbitrarily small intervale [x− h, x] contained in O such that (since D− < v):

f(x)− f(x− h) < vh

This is a Vitalli covering! By the Vitali covering lemma, we choose a �nitecollection {I1, . . . , IN} of them whose interiors cover Eu,v with measure at leastm(Eu,v)− ε. Denote A = ∪Ik. Then summing over these intervals, we have:

N∑n=1

[f(xn)− f(xn − h)] ≤ v

N∑n=1

hn

≤ v (m(O) + ε)

≤ v (m(Eu,v) + ε)

Now, each point y ∈ A is the left endpoint of some interval (y, y + k)so thatf(y+ k)− f(y) > uk. Supopose WOLOG that each of these is a subset of some Ik.Using this as a Vitalli covering lemma for THIS new vitalli covering J1, . . . witheach interval being a subset of some I ′is. Have (similar to before):

M∑i=1

f(yi + ki)− f(yi) > u∑

ki

> u(m(Eu,v) + ε)

Now, SINCE f is increasing, and every interval Ji is contained in some intervalIi we have for each Jithat:∑

Ji⊂In

f(yi + ki)− f(yi) ≤ f(xn)− f(xn − hn)

so summing these up we get:

N∑n=1

[f(xn)− f(xn − h)] ≥M∑i=1

f(yi + ki)− f(yi)

But the LHS ≤ v (m(Eu,v) + ε) and the RHS > u(m(Eu,v) + ε). Since thisholds for every ε > 0 we conclude that vm(Eu,v) ≥ um(Eu,v). But since u < v thiscan only hold if m(Eu,v) = 0, as desired.

10.25. FUNCTIONS OF BOUNDED VARIATION 34

This shows that:

g(x) = limh→0

f(x+ h)− f(x)h

Is de�ned almost everywhere and that f is di�erentiable when g is �nite.To get the desired inequality on the integral of the derivative, we use Fatous

lemma and the sequence gn(x) =f(x+h)−f(x)

h

∣∣∣h= 1n

(Notice that g is non-negative

since f is increasing, so we can indeed use Fatou's lemma!) which converges point-wise to g. Have: ˆ b

a

g(x) =

ˆ b

a

lim inf gn

≤ lim inf

ˆ b

a

gn

≤ . . .

= f(b)− f(a)

�

10.25. Functions of Bounded Variation

Let f be a real-valued function de�ned on [a, b] and de�ne, for any subdivisionx0 < x1 < . . . < xn:

p =

k∑i=1

[f(xi)− f(xi−1)]+

n =

k∑i=1

[f(xi)− f(xi−1)]−

t = n+ p

=

k∑i=1

|f(xi)− f(xi−1)|

Here x+ = |x| 1x>0 and x− = |x| 1x<0. Clearly, f(b)− f(a) = p− n. Set:

P = sup p

N = supn

T = sup t

These are called the positive, negative and total variations of f . We sometimeswrite Pa,b to remind ourselves of the interval we are looking over.

Definition. We say f is bounded varaition if T <∞.

Theorem. If f is of bounded variation, then:

Ta,b = Pa,b +Na,b

and:

f(b)− f(a) = Pa,b −Na,bCorollary. If f is of bounded variation, then we can write it as di�erence of

two monotone functions

10.26. DIFFERENTIATION OF AN INTEGRAL 35

Proof. By the theorem, we can write:

f(x) = f(a) + Pa,x −Na,xPa,x and Na,x are seen to be monotone functions of x. �

Remark. The converse is also true since any monotone function is of boundedvaraition, and sums of bounded variation functions are bounded variation.

Proof. For any subdivision we have:

p = n+ f(b)− f(a)and taking suprema over all possible subdivisions we get:

P = N + f(b)− f(a)Now we can write:

t = p+ n = 2p− (f(b)− f(a))And taking superma gives:

T = 2P − (f(b)− f(a)) = P +N

�

Corollary. Every function of bounded variation is di�erentiable.

10.26. Di�erentiation of an Integral

We will show that the derivative of F (x) =´ xaf(t)dt is di�erentiable whenever

f is integrable.

Lemma. If f is integrable on [a, b] then the function F de�ned by:

F (x) =

ˆ x

a

f(t)dt

is a continuous function of bounded variation on [a, b]

Proof. The fact that its continuous follows from the �uniform integrability�property of integrals; for any ε > 0 there is a δ > 0 so that the integral over a setof size < δ is less than ε.

To show that f is bounded variation, notice that:∑|F (xi)− F (xi−1)| =

∑∣∣∣∣ˆ xi−1

xi

f(t)dt

∣∣∣∣≤

∑ˆ xi+1

xi

|f(t)| dt

=

ˆ|f(t)| dt

So taking sup, we have an upper bounder for the total varaition. �

Lemma. If f is integrable on [a, b] and:ˆ x

a

f(t)dt = 0

for all x ∈ [a, b] then f(t) = 0 a.e. in [a, b]

10.26. DIFFERENTIATION OF AN INTEGRAL 36

Proof. I think I have a creative proof using the monotone class theorem. Let:

M =

{A :

ˆf1A = 0

}We see (with a bit of work) from the hypothesis thatM contains the algebra

of �nite collections of intervals.Now M is a monotone class, since if An ↑ A or ↓ A then f1An

→ f1A a.e.and is dominated by the integrable function |f |, so by LDCT 0 =

´f1An →

´f1A

shows A ∈M. (I guess I could have done this with σ-algebra instead too)Hence M is the Borel sets. This means f is 0, becuase for example the set{

f > 1n

}must be measure 0 for each n. �

Lemma. (9) If f is bounded, |f | ≤ K and measureable on [a, b] and:

F (x) =

ˆ x

a

f(t)dt+ F (a)

Then F ′(x) = f(x) for almost all x ∈ [a, b]

Proof. Since F is of bounded variation, we know it is di�erentiable. Set fn =F (x+h)−F (x)

h

∣∣∣h= 1n

so that fn → F ′(x) a.e.. Notice also that fn = 1h

´ x+hx

f(t)dt ≤K is bounded, so we are ok to use a bounded convergence theorem. We now checkthat: ˆ

F ′dx = limn→∞

ˆfn

= limn→∞

n

ˆF (x+

1

n)− F (x)dx

= F (b)− F (a)

=

ˆ b

a

f(x)dx

So then F ′ − f integrates to zero on every interval, and consequently mustvanish everywhere. �

Theorem. The above works if f is integrable instead of boudned:

d

dx

ˆ x

a

f(t)dt = f(x)

Proof. Take fn ↑ f with each fn = f ∧ n bounded by say n. We know thatddx

´ xafn = fn(x) a.e. since these are bounded. and so we have that:

F′(x) =

d

dx

ˆf

=d

dx

ˆfn +

d

dx

ˆ(f − fn)

≥ fn(x)

Since f − fn ≥ 0. But then´ (

F′ − fn

)≥ 0 for any n and consequently´

F ′ − f ≥ 0 too (LDCT justi�es us) For any interval. On the other hand, we

showed (it was a Fatou's lemma in the end) that´F ′n ≤ F (b)−F (a) =

´ baf . So we

have cobining these inequalites that´(F ′− f) = 0 over any interval and so F ′ = f

a.e. �

10.27. ABSOLUTE CONTINUITY 37

10.27. Absolute Continuity

Definition. A real valued function f is said to be absolutly continuous on[a, b] if and only if, for all ε > 0 there is a δ > 0 such that:

n∑i=1

|x′i − xi| < δ =⇒n∑i=1

|f(x′i)− f(xi)| < ε

Remark. Compare this with the �de�nition� of absolute continuity for a mea-sure, ν � µ if for all ε > 0 there exists δ > 0 so that:

ν(A) < δ =⇒ µ(A) < ε

(The more usual way to think about this is ν(A) = 0 =⇒ µ(A) = 0, whichis equivalent.) This is how you could think of absolutly continus functions tooheuristically: the change in the function over a tiny set is tiny.

Theorem. For f non-negative integrable, the function g(x) =´ xaf(x)dx is

absolutly continuous. More precisly, for all ε > 0 there exists δ > 0 so that µ(A) <δ =⇒

´Afdx < ε

Proof. Can do this using the equivalence with absoltule continuity for measursabove, thinking of

´Af as a measure. Its not hard by bare hands either though:

take fn = f ∧ n so that fn ↑ f . By LDCT we have:ˆfn →

ˆf

Hence, for any ε > 0, there exits n large enough so that´f1{f>n} < ε/2. Now

take δ = ε/2n and we �nd that for any set A with µ(A) < δ that:ˆ

A

f =

ˆA∩{f≤n}

f +

ˆA∩{f>n}

f

≤ nµ(A) + ε/2

< ε/2 + ε/2

�

Lemma. If f is absolutly continuous on [a, b] then it is of bounded variation on[a, b]

Proof. Take ε = 1 to get a δ in the de�nition of absolute continuity. LetK = 1 + b−a

δ be the largest number of intervals of size δ one could cram into a, b.The varaition on each interval of size δ is no more than 1 (by the de�nition ofabsolutle continiuity) and so the total variation on the whole [a, b] can be no morethan K, the total number of such intervals. �

Corollary. If f is absolutly continuos then f has a derivative almost every-where.

Lemma. If f is abs. continuous on [a, b] and f ′(x) = 0 a.e. then f is constant.

Proof. (Uses vitalli covering lemma) Fix ε and η arbitrarily small. Get theδ from absolute continuity of f . The idea is as follows, for any interval (a, c) weuse the fact that |f ′(x)| < η everywhere to make a Vitalli covering of [a, c] with|f(x+ h)− f(x)| ≤ ηh on each interval [x, x + h]. Then by the Vitalli coveringlemma, extract a �nite set of intervals that cover all of [a, c] except a set of size δ.

10.27. ABSOLUTE CONTINUITY 38

Beacuse |f(x+ h)− f(x)| ≤ ηh on the intervals, the total contribution to|f(c)− f(a)| on the intervals is no more than η(c− a).

Because the δ was chosen by the abs. cont. of f at ε, the total the totalcontribution to |f(c)− f(a)| on the complement of the intervals (which is size nomore than δ) is < ε.

In sum |f(c)− f(a)| < ε + η(c − a) which can be made arbitarily small as weplease. �

Corollary. Every absolutly continuous function can be written as:

f(x) = f(a) +

ˆ x

a

f ′(x)dx

Proof. Let g(x) = f(x) −´ xaf ′(x)dx. Since inde�nite integrals are abs con-

tinuous, g is the di�ernce of absolutely continuous functions and is hence abs.continuous. Moreover, g′ = f ′ − f ′ = 0 a.e. By the previous lemma, then g is aconstant! g(a) = f(a) then gives the result. �

Lp spaces

These are notes from Chapter 15 of [1].Let (X,A, µ) be a σ−�nite meausre sapce. For 1 ≤ p <∞ de�ne the Lpnorm

of f by:

‖f‖p =(ˆ|f(x)|p dx

)1/p

For p = ∞ ‖f‖∞ is the essential supremum of f . The space Lp is the set offunctions whose Lpnorm is �nite.

11.28. Norms

Proposition. (15.1.) (Holder's Inequality) If 1 < p, q <∞ and p−1+q−1 = 1then: ˆ

|fg| dµ ≤ ‖f‖p ‖g‖q

Proof. Assume WOLOG that ‖f‖p = ‖g‖q = 1 and then integratF (x)G(x) ≤F (x)p

p + G(x)q

q to get the result. �

Proposition. (Minkowski's Inequality) If 1 ≤ p ≤ ∞ then:

‖f + g‖p ≤ ‖f‖p + ‖g‖p

Proof. Write:

|f + g|p = |f + g| |f + g|p−1

≤ |f | |f + g|p−1 + |g| |f + g|p−1

Now apply Holder's inequality with p and q =(1− 1

p

)−1to get:

ˆ|f + g|p ≤

ˆ|f | |f + g|p−1 +

ˆ|g| |f + g|p−1

≤ ‖f‖p

(ˆ|f + g|(p−1)q

)1/q

+ ‖g‖p

(ˆ|f + g|(p−1)q

)1/q

This gets us to:

‖f + g‖pp ≤(‖f‖p + ‖g‖p

)‖f + g‖p/qp

And dividing out gives the resutl. �

39

11.29. COMPLETENESS 40

11.29. Completeness

Theorem. (15.4.) If 1 ≤ p ≤ ∞ then Lp is complete.

Proof. Say fn is a Cauchy sequence. Look at the subsequence nk so that∥∥fnk− fnk+1

∥∥ ≤ 2−k. We will show that∑m

(fnm − fnm−1

)converges absolutely

and this will be our limit point for the Cauchy sequence. Let gk =∑ki=1

∣∣fnm− fnm−1

∣∣ByMinkowski we have:

‖gj‖p ≤k∑i=1

∥∥fnm− fnm−1

∥∥p

≤j∑

m=1

2−m < 1

Let g be the pointwise limit g = limn→∞ gn. By Fatou, we have´|g(x)|p ≤

limm→∞ ‖gj‖pp ≤ 1. Hence g must be �nite a.e.

Now have f(x) = limK→∞∑Km=1

(fnm− fnm−1

)(x) = limK→∞ fnK

(x) and byFatou we have:

‖f − fnk‖pp =

ˆ ∣∣f − fnj

∣∣p ≤ lim infK→∞

ˆ ∣∣fnK− fnj

∣∣p ≤∑∥∥fnk+1− fnk

∥∥pp≤ 2(−j+1)p → 0

So fnkl→ f in Lp as desired. �

Proof. To prove a space is complete, it su�ces to prove that every absolutlyconvergent sequence is convergent. Suppose hn is a sequence with

∑‖hn‖Lp <∞.

We want to de�ne f(x) =∑∞n=1 hn(x) and show that this is �nite a.e. and is the

Lp limit of∑∞n=1 hn(x).

Claim 1: The pointwise limit f(x) = limN→∞∑Nn=1 hn(x) exists and is �nite

a.e.Pf: Let gN (x) =

∑Nn=1 |hn(x)| and let g(x) = supN gN (x) = limN→∞ gN (x).

By Fatou's lemm:

ˆ(g(x))

pdx =

ˆlimN→∞

(gN (x))pdx

≤ lim infN→∞

ˆ(gN (x))

pdx

≤ lim infN→∞

N∑n=1

‖hn‖pp by Minkowski inequality

< ∞ since hn is absolutly summable

Hence g(x) is �nite a.e.. Hence for a.e. x the sum f(x) =∑∞n=1 hn(x) converges

absolutly and is �nite.

Claim 2: f(x) = limN→∞∑Nn=1 hn(x) in L

p

11.31. BOUNDED LINEAR FUNCTIONALS 41

Pf: De�ne fN (x) =∑Nn=1 hn(x) so fN → f pointwise for a.e. x. Now again by

Fatou: ˆ|f − fN |p dx =

ˆlim |fM − fN |p dx

≤ lim infM→∞

ˆ|fM − fN |p dx

= lim infM→∞

ˆ ∣∣∣∣∣M∑

k=N+1

hk(x)

∣∣∣∣∣p

dx

≤ lim infM→∞

M∑n=N+1

‖hn‖pp

≤M∑

n=N+1

‖hn‖pp

→ 0 as N →∞

�

Proposition. The set of continuous functions with compact support is densein Lp(R)

Proof. Compact sets is clear since´ ∣∣f − f1[−n,n]∣∣ → 0 by LDCT. To do

continuous function, approximate f by simple functions. Approximate each mea-surable set by a closed set. Then approximate the characteristic function of a closedset by a continuous function. �

11.30. Convolutions

The convolution of two measurable functions is de�ned by:

f ∗ g =

ˆf(x− y)g(y)dy

Proposition. (15.7.) If f, g ∈ L1 then f ∗ g is in L1 and:

‖f ∗ g‖1 ≤ ‖f‖1 ‖g‖1Proof. Write:

‖f ∗ g‖1 ≤ˆ ˆ

|f(x− y)| dx |g(y)| dy

=

ˆ ˆ|f(x)| dx |g(y)| dy

= ‖f‖1 ‖g‖1�

11.31. Bounded Linear Functionals

Theorem. (15.8) For 1 < p <∞ and p−1 + q−1 = 1 we have:

‖f‖p = sup

{ˆfgdµ : ‖g‖q ≤ 1

}

11.31. BOUNDED LINEAR FUNCTIONALS 42

Proof. ≥ follows by the Holder inequality. To see the other way choose some-

thing like g = (sgn(f(x)) |f(x)|p−1

‖f‖p/qp

. �

Proposition. Actually it su�ces to consider only simple functions:

‖f‖p = sup

{ˆfgdµ : ‖g‖q ≤ 1, g is a simple function

}Proof. Approximate f by simple functions and put again g = (sgn(f(x)) |sn(x)|

p−1

‖sn‖p/qp

�

Proposition. (15.10) If 1 < p < ∞ with p−1 + q−1 = 1 and g ∈ Lq thenintegration against g de�nes a bounded linear functional on Lp with ‖H‖ = ‖g‖q

Proof. By Holder H is bounded and by the above prop ‖H‖ = ‖g‖q. �

Theorem. (15.11) If 1 < p < ∞ and H is a real valued bounded linear func-tional on Lp then there exists g ∈ Lq such that H(f) =

´fg and ‖g‖q = ‖H‖

Proof. The idea is to de�ne the measure ν(A) = H(1A) then show that νis a measure with ν � µ. The function we want is g = dν

dµ . Once we have estab-

lished thatH(1A) =´gdµ it is easy to see that for simple functions we haveH(s) =´

sgdµ. Bu then ‖g‖q = sup(´gs : ‖s‖ ≤ 1, s is simple

)= sup (H(s) : ‖s‖ ≤ 1, s is simple) ≤

‖H‖. Hence g ∈ Lq. Finally then, by approximating any function f by simple func-tions one can check H(sn)→ H(f) which shows that H(f) =

´fg. �

Fourier Transforms

These are notes from Chapter 16 of [1].

12.32. Basic Properties

For f ∈ L1(Rn) de�ne:

f(u) =

ˆ

Rn

eiu·xf(x)dx

Proposition. If f, g in L1 then:

i) f is bounded and continuousii)· is linear

Proof. Write:

f(u+ h)− f(u) =

ˆ (ei(u+h)·x − eiu·x

)f(x)dx∣∣∣f(u+ h)− f(u)

∣∣∣ ≤ ˆ ∣∣eiu·x∣∣ ∣∣eih·x − 1∣∣ |f(x)| dx

which is bounded by 2 |f(x)|. Since eix·h− 1→ 0 as h→ 0 the integral → 0 bythe LDCT. �

12.32.1. Derivatives.

Proposition. (16.2.) If f ∈ L1 and xf ∈ L1 then f is di�erentiable with:

f ′ = i

ˆeiu·xxf(x)dx

Remark. In Rn the condition is that xjf ∈ L1 =⇒ f has a partial ujderiavive.

Proof. Write:

f(u+ h)− f(u)h

=

ˆeiu·x

(eihx − 1

h

)f(x)dx

And since xf(x) is integrable, we use∣∣∣ eihx−1

h

∣∣∣ ≤ |x| then apply LDCT and get

the result. �

Proposition. If f ∈ L1, f is absolutly continuous and f ′ ∈ L1 then the

Fourier transform of f ′ is −iuf(u).\

43

12.33. THE INVERSION THEOREM 44

Proof. The trick is to do integration by parts:

f ′ =

∞

−∞

eiuxf ′(x)dx =

ˆiueiuxf(x)dx+ boundaryterms

The fact that f ′is in L1 can be used to help show that f → 0 as x→ ±∞ thatthe boundary terms die. �

Proposition. If f, g ∈ L1 then f ∗ g = f g

Proof. Write it out and use ‖f ∗ g‖L1 ≤ ‖f‖L1 ‖g‖L1 �

12.33. The Inversion theorem

Proposition. Let Ha : R → R be the pdf of a Gaussian with mean 0 andvariance a2 then:

Ha(x) =1

a√2πe−x

2/2a2

Then, Ha is the pdf of a Gaussian with mean 0 and variance 1/a2 and NOPRECONSTANTS:

Ha(u) = e−a2u2/2

Moreover, Ha is an approximate identity as a→ 0, i.e. we have:

sup ‖Ha‖L1 < ∞ˆHa = 1

ˆ

|x|>δ

Ha → 0 as a→ 0 for any �xed δ > 0

Consequenctly we have:

Ha ∗ f‖·‖∞−−−→ f for all f ∈ C∞(R)

Ha ∗ f‖·‖p−−−→ f for all f ∈ Lp(R)

Also notice that Ha(u)→ 1 as a→ 0.

Theorem. If f and f are in L1 then:

f(y) =1

2π

ˆe−iu·y f(u)du converges a.e.

Proof. The trick is to hit things with Ha, then unload the integration ontoH, which we know is its own Fourier transform essentially.

Write:1

2π

ˆe−iu·y f(u)du = lim

a→0

ˆe−iu·y f(u)Ha(u)du =

ˆ ˆeiuxe−iu·y f(u)Ha(u)dxdu

= C

ˆHa(x− y)f(x)dx

= CHa ∗ f→L1

f

�

Fourier Series

These are notes from Chapter 1 of [2]. These are rather informal and do notfollow very closely with the textbook.

Definition. For a function f ∈ L1(T) we de�ne its n-th Fourier coe�cient by:

f(n) =1

2π

ˆf(t)e−intdt

And tis Fourier series is the formal trigonmentric series:

S[f ] =

∞∑n=−∞

f(n)eint

Proposition. Have: ∣∣∣f(n)∣∣∣ ≤ ‖f‖L1

Proof. Holds since∣∣eint∣∣ = 1 �

Definition. A summability kernal is a seqeunce {kn} of 2π periodic functionssuch that:

1

2π

ˆkn(t)dt = 1

1

2π

ˆ|kn(t)| dt = ‖kn‖L1 ≤ C for some �xed C and for all n

limn→∞

2π−δˆ

δ

|kn(t)| dt = 0 for every �xed δ

Lemma. If k is a summability kernal then for continuous ϕ ∈ C (T) we have::

limn→∞

1

2π

ˆkn(s)ϕ(s)ds = ϕ(0)

Proof. Given any ε > 0, �nd δ so small by the continuity of ϕ so that|ϕ(x)− ϕ(0)| < ε whenever |x| < δ. Then:

45

FOURIER SERIES 46

∣∣∣∣ 12πˆkn(s)ϕ(s)ds− ϕ(0)

∣∣∣∣ =

∣∣∣∣ 12πˆkn(s) (ϕ(s)− ϕ(0)) ds

∣∣∣∣≤ 1

2π

2π−δˆ

δ

|kn(s)| |ϕ(s)− ϕ(0)| ds+1

2π

δˆ

−δ

|kn(s)| |ϕ(s)− ϕ(0)| ds

≤ 2 ‖ϕ‖∞

1

2π

2π−δˆ

δ

|kn(s)|

+ ε1

2π

δˆ

−δ

|kn(s)| ds

→ 0 + εC

�

Proposition. If f ∈ L1(T), and kn a summability kernal, then we have:

kn ∗ fL1

−−→ f

Proof. We �rst prove it for f continuous. If f is continuous, then for any ε�nd δ so small so that |f(x)− f(y)| < ε for |x− y| < δ. Consider:∥∥∥∥ 1

2π

ˆkn(s)f(· − s)ds− f(·)

∥∥∥∥L1

=1

(2π)2

ˆ ∣∣∣∣ˆ kn(s) (f(t− s)− f(t)) ds∣∣∣∣ dt

=1

(2π)2

ˆ ∣∣∣∣∣∣ δˆ

−δ

kn(s) (f(t− s)− f(t)) ds+2π−δˆ

δ

kn(s) (f(t− s)− f(t)) ds

∣∣∣∣∣∣ dt≤ 1

(2π)2

ˆ ∣∣∣∣∣∣‖kn‖L1 ε+ 2 ‖f‖∞

2π−δˆ

δ

kn(s)ds

∣∣∣∣∣∣ dt→ 0 + ε sup

n‖kn‖L1

This proves it for continuous f . For f ∈ L1, �nd g continuous with ‖f − g‖L1 <ε and consider:

‖kn ∗ f − f‖L1 ≤ ‖kn ∗ f − kn ∗ g‖L1 + ‖kn ∗ g − g‖L1 + ‖g − f‖L1

≤(supn‖kn‖L1

)‖f − g‖L1 + ‖g − f‖L1 + ‖kn ∗ g − g‖L1

�

Definition. The Fejer Kernal is the summability kernal:

Kn(t) =1

n+ 1

n∑j=0

j∑k=−j

eikt

=

n∑j=−n

(1− |j|

n+ 1

)eijt

Proposition. Have the identity:

Kn(t) =1

n+ 1

(sin n+1

2

sin 12 t

)2

FOURIER SERIES 47

Proof. Use the identity:

cos(t) = cos

(2

(t

2

))= cos

(1

2t

)2

− sin

(1

2t

)2

= 1− 2 sin

(1

2t

)2

So:

sin

(1

2t

)2

=1

2(1− cos(t)) = −1

4e−it +

1

2− 1

4eit

Then check that:(−1

4e−it +

1

2− 1

4eit) n∑j=−n

(1− |j|

n+ 1

)eijt

turns it into a telescoping sum. �

Definition. We de�ne:

σn(f)(t) = (Kn ∗ f) (t) =n∑

k=−n

(1− |j|

n+ 1

)f(j)eijt

Remark. Since Kn is a summability kernal, we know that:

σn(f)ptwise−−−−→ f for f ∈ C (T)

σn(f)L1

−−→ f for f ∈ L1(T)The pointwise convergence is actually uniform (go back to our argument we

used) and the convergence in L1 can be moved up to Lp convergence with thesame method of proof as before (just approximate the Lp function by a contunousfunction)

σn(f)‖·‖∞−−−→ f for f ∈ C(T)

σn(f)Lp

−−→ f for f ∈ Lp(T)

Remark. (This is kind of o� topic...) Using the same kind of naive-ish esti-mates as above, one can show that the Fourier series converge in Cα, the space ofHolder continuous functions:

Sn(f)‖·‖∞−−−→ f for f ∈ Cα(T)

13.33.1. Corollaries to Fejer Convergence.

Theorem. (Fejer Convergence Thm)

σn(f)‖·‖∞−−−→ f for f ∈ C(T)

σn(f)Lp

−−→ f for f ∈ Lp(T)

Corollary. Trig. polynomials are dense in C(T), Lp(T) for 1 ≤ p <∞

Proof. Each σn(f) is a trig. polynomial. �

13.34. MORE FOURIER FACTS 48

Corollary. The functions{eint

}∞n=−∞ form a maximal orthonormal set (i.e.

an orthonormal set) for L2(T).

Proof. Its easy to check they are orthonormal. To see they are maximal,suppose by contradiction that f ⊥ eint for all n. Find g a trig poly. so that‖f − g‖L2 < ε. Then we have that:

‖f‖2 = 〈f, f〉 = 〈f, f − g〉+ 〈f, g〉 ≤ ‖f‖ ‖f − g‖+ 0 ≤ ε ‖f‖

So we get ‖f‖ = 0. �

Corollary. We have:

‖f‖22 =∑n∈Z

∣∣∣f(n)∣∣∣2f = lim

n→∞

n∑i=−n

f(n)eint = limn→∞

n∑i=−n

⟨f, eint

⟩eint

ˆf(x)g(x)dx = 〈f, g〉L2 =

∑n∈Z

⟨f, eint

⟩ ⟨eint, g

⟩=∑n∈Z

f(n)g(n)

Proof. These follow by basic Hilbert space theory and are equivalent to thefact that eint are a basis. The �rst one is Plancharel's identity, the second one isRiesz-Fisher, and the last one is Parseval's identity. �

Corollary. (Uniquenss Theorem) If f(n) = 0 for all n and f ∈ L1(T) thenf ≡ 0

Proof. σn(f) = 0 for every f and σn(f)→ f �

Corollary. (Riemann-Lebesgue) If f ∈ L1(T) then f(n)→ 0 as n→ ±∞

Proof. Given any ε > 0, choose N so large so that ‖σNf − f‖L1 < ε. Then

for any n > N , we have that σn(f) = 0 and so:b vb∣∣∣f(n)∣∣∣ = ∣∣∣ σNf(n)− f(n)∣∣∣ ≤ ‖σNf − f‖L1 < ε

�

Remark. As we have seen in Hang's harmonic class, given any sequence an → 0there is an L1 function whose fourier seires go to 0 slower than an. (You canconstruct it as long as its convex)

13.34. More Fourier Facts

Proposition. Fejer's Theorem: For continuous function the Fejer kernal Fn(x) =1n

∑n−1k=0 Dk(x) where Dk is the Dirichlet kernal Dk(x) =

∑ks=−k e

isx.(This can be

written as Fn(x) = 1n

1−cos(nx)1−cos(x) ) has Fn ∗ f → f uniformly in [−π, π] if f is

continuous.

13.34. MORE FOURIER FACTS 49

Proof. Write

(Fn ∗ f − f) (y) =1

2π

π

−π

Fn(x)f(y − x)dx− f(y)

=1

2π

π

−π

Fn(x) [f(y − x)− f(y)] dx since Fejer kernal integrates to 1

=1

2π

π

−π

Fn(x) [fy−·(x)− fy−·(0)] dx

Where fy−·(z) := f(y−z). Now divide the region of integration into two zones,one where x is small and one where x is large. When x is small, the continuity off can be used to control it. When x large, the fact that the the Fejer kernal has´|x|>δ |Fn| → 0. �

Proposition. If f ∈ L2 then the Fourier series approx for f converges in theL2 sense Dn ∗ f → f in L2

Proof. This amounts to showing that the trigonometric polynomials einx forma BASIS for the Hilbert space L2. Indeed, we �rst remark that these are dense inL2 (by Stone-Weirestrass or by the Fejer kernal we know that these are dense inthe continuous functions....and the continuous functions are dense in L2). Next,suppose by contradiction that einx is not a basis for L2. Then there exists anf ∈ L2 perp to all einx. But then �nd a sequence of trigonometric polynomialsfn → f in L2. We have then:

‖f‖2 = 〈f, f〉= 〈f − fn, f〉 since 〈fn, f〉 = 0

≤ ‖f − fn‖ ‖f‖

→ 0 since fnL2

−−→ f

�

Proposition. The fourier series of a continuous function does not nessisarilyconverge pointwise.

Proof. (Using the Banach Steinhouse principle). Notice that the Dirichletkernal Dn has ‖Dn‖1 ∼ log n → ∞. If Dn ∗ f → f pointwise, then we wouldhave that {x : supn |Dn ∗ f(x)| < ∞} = [−π, π]. By the uniform boundenessprinciple, we would have that supn ‖Dn ∗ f‖∞ is uniformly bounded! However,‖Dn ∗ f‖∞ = ‖f‖∞ ‖Dn‖1 →∞ so this is a contradiction. �

Bibliography

[1] R.F. Bass. Real Analysis for Graduate Students. CreateSpace Independent Publishing Plat-form, 2013.

[2] Y. Katznelson. An Introduction to Harmonic Analysis. Cambridge Mathematical Library.Cambridge University Press, 2004.

[3] H.L. Royden. Real Analysis. Mathematics and statistics. Macmillan, 1988.

50