Real Analysis

Richard F. Bass

February 2, 2010

1 σ-algebras

Let X be a set. We will use the notation: Ac = x ∈ X : x /∈ A andA−B = A ∩Bc. (The notation A \B is also commonly used.)

Definition 1.1 An algebra is a collection A of subsets of X such that

(a) ∅, X ∈ A;

(b) if A ∈ A, then Ac ∈ A;

(c) if A1, . . . , An ∈ A, then ∪ni=1Ai and ∩n

i=1Ai are in A.

A is a σ-algebra (or σ-field) if in addition

(d) if A1, A2, . . . are in A, then ∪∞i=1Ai and ∩∞i=1Ai are in A.

In (d) we allow countable unions and intersections only; we do not allowuncountable unions and intersections.

Example 1.2 Let X = R and A be the collection of all subsets of R.

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Example 1.3 Let X = R and let

A = A ⊂ R : A is countable or Ac is countable.

Parts (a) and (b) of the definition are easy. Suppose A1, A2, . . . are all in A.If each of the Ai are countable, then ∪iAi is countable, and so in A. If Ac

i0

is countable for some i0, then

(∪Ai)c = ∩iA

ci ⊂ Ac

i0

is countable, and again ∪iAi is inA. Since ∩Ai = (∪iAci)

c, then the countableintersection of sets in A is again in A.

Example 1.4 Let X = [0, 1] and A = ∅, X, [0, 12], (1

2, 1].

Example 1.5 X = 1, 2, 3 and A = X, ∅, 1, 2, 3.

Example 1.6 Let X = [0, 1], and B1, . . . , B8 subsets of X which are pairwisedisjoint and whose union is all of X. LetA be the collection of all finite unionsof the Bi’s as well as the empty set. (So A consists of 28 elements.)

Note that if we take an intersection of σ-algebras, we get a σ-algebra; this isjust a matter of checking the definition.If we have a collection C of subsets ofX, there is at least one σ-algebra containing C, namely, the one consisting ofall subsets of X. We can take the intersection of all σ-algebras that containC; we denote this intersection by σ(C). If A is any σ-algebra containing C,then A ⊃ σ(C).

If X has some additional structure, say, it is a metric space, then we cantalk about open sets. If G is the collection of open subsets of X, then wecall σ(G) the Borel σ-algebra on X, and this is often denoted B. We will seelater that when X is the real line, that B is not equal to the collection of allsubsets of X.

We end this section with the following proposition.

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Proposition 1.7 If X = R, then the Borel σ-algebra is generated by eachof the following collection of sets:

(1) C1 = (a, b) : a, b ∈ R.

(2) C2 = [a, b] : a, b ∈ R;

(3) C3 = (a, b] : a, b ∈ R;

(4) C4 = (a,∞) : a ∈ R;

Proof. (1) Let G be the collection of open sets. Then C1 ⊂ G ⊂ σ(G). σ(G)is the Borel σ-algebra and contains C1. Since σ(C1) is the intersection of allσ-algebras containing C1, then σ(C1) ⊂ σ(G).

To get the reverse inclusion, if G is open, it is the countable union of openintervals. So G ∈ σ(C1), and hence G ⊂ σ(C1). σ(G) is the intersection of allσ-algebras containing G; σ(C1) is one such, so σ(G) ⊂ σ(C1).

(2) If [a, b] ∈ C2, then [a, b] = ∩∞n=1(a − 1n, b + 1

n) ∈ σ(G). So C2 ⊂ σ(G),

and by an argument similar to that in (1), we conclude σ(C2) ⊂ σ(G).

If (a, b) ∈ C1, choose n0 ≥ 2/(b−a) and note (a, b) = ∪∞n=n0[a+ 1

n, b− 1

n] ∈

σ(C2). So the Borel σ-algebra, which is equal to σ(C1) by part (1), is containedin σ(C2).

(3) The proof here is similar to (2), using (a, b] = ∩∞n=1(a, b + 1n) and

(a, b) = ∪∞n=n0(a, b− 1

n], provided n0 is taken large enough.

(4) The proof of this comes from (3), using that (a, b] = (a,∞) − (b,∞)and (a,∞) = ∪∞n=1(a, a + n].

2 Measures

Definition 2.1 A measure on (X,A) is a function µ : A → [0,∞] such that

(a) µ(A) ≥ 0 for all A ∈ A;

(b) µ(∅) = 0;

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(c) if Ai ∈ A are disjoint, then

µ(∪∞i=1Ai) =∞∑i=1

µ(Ai).

Example 2.2 X is any set, A is the collection of all subsets, and µ(A) isthe number of elements in A. This is called counting measure.

Example 2.3 X = R, A the collection of all subsets, x1, x2, . . . ∈ R, anda1, a2, . . . > 0. Set µ(A) =

∑i:xi∈A ai. A particular case of this is if xi = i

and all the ai = 1. We will see later that this allows us to view infinite seriesas functions on this space.

Example 2.4 δx(A) = 1 if x ∈ A and 0 otherwise. This measure is calledpoint mass at x.

We will construct Lebesgue measure on R; this is an extension of thenotion of length. However, the construction is a bit lengthy. We will alsoconstruct Lebesgue measure on Rn; when n = 2, this is an extension of thenotion of area, when n = 3, of volume.

Proposition 2.5 The following hold:

(a) If A, B ∈ A with A ⊂ B, then µ(A) ≤ µ(B).

(b) If Ai ∈ A and A = ∪∞i=1Ai, then µ(A) ≤∑∞

i=1 µ(Ai).

(c) If Ai ∈ A, A1 ⊂ A2 ⊂ · · · , and A = ∪∞i=1Ai, then µ(A) = limn→∞ µ(An).

(d) If Ai ∈ A, A1 ⊃ A2 ⊃ · · · , µ(A1) < ∞, and A = ∩∞i=1Ai, then we haveµ(A) = limn→∞ µ(An).

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Proof. (a) Let A1 = A, A2 = B −A, and A3 = A4 = · · · = ∅. Now use part(c) of the definition of measure.

(b) Let B1 = A1, B2 = A2 −B1, B3 = A3 − (B1 ∪B2), and so on. The Bi

are disjoint and ∪∞i=1Bi = ∪∞i=1Ai. So µ(A) =∑

µ(Bi) ≤∑

µ(Ai).

(c) Define the Bi as in (b). Since ∪ni=1Bi = ∪n

i=1Ai, then

µ(A) = µ(∪∞i=1Ai) = µ(∪∞i=1Bi) =∞∑i=1

µ(Bi)

= limn→∞

n∑i=1

µ(Bi) = limn→∞

µ(∪ni=1Bi) = lim

n→∞µ(∪n

i=1Ai).

(d) Apply (c) to the sets A1 − Ai, i = 1, 2, . . ..

Example 2.6 To see that µ(A1) < ∞ is necessary, let X be the positiveintegers, µ counting measure, and Ai = i, i + 1, . . .. Then the Ai decrease,µ(Ai) = ∞ for all i, but µ(∩iAi) = µ(∅) = 0.

Definition 2.7 A probability or probability measure is a measure such thatµ(X) = 1. In this case we usually write (Ω,F , P) instead of (X,A, µ).

3 Construction of Lebesgue measure

Define m((a, b)) = b−a. If G is an open set and G ⊂ R, then G = ∪∞i=1(ai, bi)with the intervals disjoint. Define m(G) =

∑∞i=1(bi − ai). If A ⊂ R, define

m∗(A) = infm(G) : G open, A ⊂ G. (3.1)

We will show the following.

(A) m∗ is not a measure on the collection of all subsets of R.

(B) m∗ is a measure on a strictly smaller σ-algebra that strictly contains theBorel σ-algebra.

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We will prove these two facts (and a bit more) in a moment, but let’s firstmake some remarks.

A set N is a null set with respect to m∗ if m∗(N) = 0. Let L be thesmallest σ-algebra containing B and all the null sets. More precisely, let Nbe the collection of all sets that are null sets with respect to m∗ and letL = σ(B ∪ N ). L is called the Lebesgue σ-algebra, and sets in L are calledLebesgue measurable.

As part of our proof of (B) we will show that m∗ is a measure on L.Lebesgue measure is the measure m∗ on L. (A) shows that L is strictlysmaller than the collection of all subsets of R.

It is easy to get lost in the construction of Lebesgue measure, so let ussummarize our steps.

First we prove (A), which Proposition 3.1.

We then turn to the construction of Lebesgue measure. It is more conve-nient for technical reasons to define

m∗(A) = inf∞∑i=1

(bi − ai) : A ⊂ ∪∞i=1(ai, bi]. (3.2)

There is no real difference between this and (3.1) since it is clear that thedifference between an open set and a set of the form ∪∞i=1(ai, bi] is countable,and with either definition of m∗, the measure of a point is 0, so the measureof a set consisting of countably many points is 0. However, when we talkabout Lebesgue-Stieltjes measure, then there is a real difference.

We define what it means to be an outer measure (Definition 3.2) and provethat m∗ is an outer measure (Proposition 3.3). We then define what it meansfor a set to be m∗-measurable (Definition 3.4) and prove that the collectionof m∗-measurable sets is a σ-algebra and that m∗ restricted to this σ-algebrais a measure.

This looks promising, but we do not yet know that enough sets are m∗-measurable. That takes one more step. We show in Proposition 3.6 that thecollection of m∗-measurable sets contains the Borel σ-algebra.

Proposition 3.1 m∗ is not a measure on the collection of all subsets of R.

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Proof. Suppose m∗ is a measure. Define x ∼ y if x−y is rational. This is anequivalence relationship on [0, 1]. For each equivalence class, pick an elementout of that class (by the axiom of choice) Call the collection of such pointsA. Given a set B, define B + x = y + x : y ∈ B. Note m∗(A + q) = m∗(A)since this translation invariance holds for intervals, hence for open sets, hencefor all sets. Moreover, the sets A + q are disjoint for different rationals q.

Now[0, 1] ⊂ ∪q∈[−2,2](A + q),

where the sum is only over rational q, so 1 ≤∑

q∈[−2,2] m∗(A + q), and

therefore m∗(A) > 0. But

∪q∈[−2,2](A + q) ⊂ [−6, 6],

where again the sum is only over rational q, so if m∗ is a measure, then12 ≥

∑q∈[−2,2] m

∗(A + q), which implies m∗(A) = 0, a contradiction.

Definition 3.2 A function n on the collection of all subsets satisfying

(a) n(∅) = 0;

(b) if A ⊂ B, then n(A) ≤ n(B);

(c) n(∪∞i=1Ai) ≤∑∞

i=1 n(Ai).

is called an outer measure.

Proposition 3.3 m∗ defined by (3.2) is an outer measure.

Proof. (a) and (b) are obvious. To prove (c), let ε > 0. For each i thereexist intervals Ii1, Ii2, . . ., each of the form (aij, bij], such that Ai ⊂ ∪∞j=1Iij

and∑

j m(Iij) ≤ m∗(Ai) + ε/2i. Then ∪∞i=1Ai ⊂ ∪i,jIij and∑i,j

m(Iij) ≤∑

i

m∗(Ai) +∑

i

ε/2i =∑

i

m∗(Ai) + ε.

Since ε is arbitrary, m∗(∪∞i=1Ai) ≤∑∞

i=1 m∗(Ai).

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Definition 3.4 Let m∗ be an outer measure. A set A ⊂ X is m∗-measurableif

m∗(E) = m∗(E ∩ A) + m∗(E ∩ Ac) (3.3)

for all E ⊂ X.

Theorem 3.5 If m∗ is an outer measure on X, then the collection A of m∗

measurable sets is a σ-algebra and the restriction of m∗ to A is a measure.Moreover, A contains all the null sets.

Proof. By Proposition 3.3,

m∗(E) ≤ m∗(E ∩ A) + m∗(E ∩ Ac)

for all E ⊂ X. So to check (3.3) it is enough to show m∗(E) ≥ m∗(E ∩A) +m∗(E ∩ Ac). This will be trivial in the case m∗(E) = ∞.

If A ∈ A, then Ac ∈ A by symmetry and the definition of A. SupposeA, B ∈ A and E ⊂ X. Then

m∗(E) = m∗(E ∩ A) + m∗(E ∩ Ac)

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

+ m∗(E ∩ Ac ∩Bc).

The first three terms on the right have a sum greater than or equal to m∗(E∩(A ∪B)) because A ∪B ⊂ (A ∩B) ∪ (A ∩Bc) ∪ (Ac ∩B). Therefore

m∗(E) ≥ m∗(E ∩ (A ∪B)) + m∗(E ∩ (A ∪B)c),

which shows A ∪B ∈ A. Therefore A is an algebra.

Let Ai be disjoint sets in A, let Bn = ∪ni=1Ai, and B = ∪∞i=1Ai. If E ⊂ X,

m∗(E ∩Bn) = m∗(E ∩Bn ∩ An) + m∗(E ∩Bn ∩ Acn)

= m∗(E ∩ An) + m∗(E ∩Bn−1).

Repeating for m∗(E ∩Bn−1), we obtain

m∗(E ∩Bn) ≥n∑

i=1

m∗(E ∩ Ai).

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Since Bn ∈ A, then

m∗(E) = m∗(E ∩Bn) + m∗(E ∩Bcn) ≥

n∑i=1

m∗(E ∩ Ai) + m∗(E ∩Bc).

Let n →∞. Recalling that m∗ is an outer measure,

m∗(E) ≥∞∑i=1

m∗(E ∩ Ai) + m∗(E ∩Bc)

≥ m∗(∪∞i=1(E ∩ Ai)) + m∗(E ∩Bc)

= m∗(E ∩B) + m(E ∩Bc)

≥ m∗(E).

This shows B ∈ A.

If we set E = B in this last equation, we obtain

m∗(B) =∞∑i=1

m∗(Ai),

or m∗ is countably additive on A.

If m∗(A) = 0 and E ⊂ X, then

m∗(E ∩ A) + m∗(E ∩ Ac) = m∗(E ∩ Ac) ≤ m∗(E),

which shows A contains all null sets.

Define m(∪ni=1(ai, bi]) =

∑ni=1(bi− ai) if the (ai, bi] are disjoint. Note m is

well-defined (a set might be expressible as a union of such intervals in morethan one way).

The last step in the construction is the following.

Proposition 3.6 Every set in the Borel σ-algebra is m∗- measurable.

Proof. Since the collection of m∗-measurable sets is a σ-algebra, it sufficesto show that every interval J of the form (a, b] is m∗-measurable. Let E beany set with m∗(E) < ∞; we need to show

m∗(E) ≥ m∗(E ∩ J) + m∗(E ∩ J c). (3.4)

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Choose I1, I2, . . . of the form (ai, bi] such that E ⊂ ∪iIi and

m∗(E) ≥∑

i

(bi − ai)− ε.

Since E ⊂ ∪Ii, we have m∗(E ∩ J) ≤∑

m∗(Ii ∩ J) and m∗(E ∩ J c) ≤∑m∗(Ii ∩ J c). Hence we have

m∗(E ∩ J) + m∗(E ∩ J c) ≤∑

i

[m∗(Ii ∩ J) + m∗(Ii ∩ J c)].

Now m∗(Ii ∩ J) is the length of an interval and m∗(Ii ∩ J c) is the length oftwo intervals, so

m∗(Ii ∩ J) + m∗(Ii ∩ J) = m∗(Ii).

Thusm∗(E ∩ J) + m∗(E ∩ J c) ≤

∑m∗(Ii) ≤ m∗(E) + ε.

Since ε is arbitrary, this proves (3.4).

We now drop the asterisks from m∗ and call m Lebesgue measure.

4 Examples and related results

Example 4.1 Recall the Cantor set is constructed by taking the interval[0, 1], removing the middle third, removing the middle thirds of each of thetwo remaining subintervals, and continuing. The Cantor set is what remains;it is closed, uncountable, and every point is a limit point. Moreover, itcontains no intervals.

After one stage, the measure of the two intervals is 2(13), after two stages

4(1/9), and after n stages, (2/3)n. Since the Cantor set C is the intersectionof all these sets, the Lebesgue measure of C is 0.

Suppose we define f0 to be 1/2 on the interval (1/3, 2/3), 1/4 on the in-terval (1/9, 2/9), 3/4 on the interval (7/9, 8/9), and so on. Define f(x) =inff0(y) : y ≥ x for x < 1. Define f(1) = 1. Notice f = f0 on the comple-ment of the Cantor set. f is monotone, so it has only jump discontinuities.But if it has a jump continuity, there is a rational of the form k/2n with

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k ≤ 2n that is not in the range of f . On the other hand, by the construction,each of these values is taken by f0 for some point in the complement of C,and so is taken by f . The only way this can happen is if f is continuous.This function f is called the Cantor-Lebesgue function. We will use it inexamples later on. For now, we can see that it a function that increases onlyon the Cantor set, which is of Lebesgue measure 0, yet is continuous.

Example 4.2 Let q1, q2, . . . be an enumeration of the rationals, let ε > 0,and let Ii be the interval (qi − ε/2i, qi + ε/2i). Then the measure of Ii isε/2i−1, so the measure of ∪iIi is at most 2ε. (It is not equal to that becausethere is a lot of overlap.) So the measure of A = [0, 1] − ∪iIi is larger than1− 2ε. But A contains no rational numbers.

Example 4.3 Let us follow the construction of the Cantor set, with thisdifference. Instead of removing the middle third at the first stage, removethe middle fourth, i.e., remove (3/8, 5/8). On each of the two intervals thatremain, remove the middle sixteenths. On each of the four intervals thatremain, remove the middle interval of length 1/64, and so on. The total thatwe removed is

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+ 2( 116

) + 4( 164

) + · · · = 12.

The set that remains contains no intervals, is closed, every point is a limitpoint, is uncountable, and has measure 1/2. Such a set is called a fat Cantorset or generalized Cantor set. Of course, other choices that 1/4, 1/16, etc.are possible.

Let A ⊂ [0, 1] be a Borel measurable set. We will show that A is al-most equal to the countable intersection of open sets and almost equal tothe countable union of closed sets.. (A similar argument to what follows ispossible for sets that have infinite measure.)

Proposition 4.4 Suppose A ⊂ [0, 1] is a Borel measurable set.

(a) There exists a set H that is the countable intersection of open setswhich contains A and m(H − A) = 0.

(b) There exists a set F that is the countable union of closed sets which iscontained in A and m(A− F ) = 0.

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Proof. (a) For each i, there is an open set Gi that contains A and such thatm(Gi − A) < 2−i. This follows from the fact that m(A) = m∗(A) and thedefinition of m∗. Then Hi = ∩j≤iGj will contain A, is open, and since it iscontained in Gi, then m(Hi − A) < 2−i. Let H = ∩∞i=1Hi. H need not beopen, but it is the intersection of countably many open sets. The set H is aBorel set, contains A, and m(H − A) ≤ m(Hi − A) < 2−i for each i, hencem(H − A) = 0.

(b) If A ⊂ [0, 1], let Fi = [0, 1]−Hi, where Hi is a decreasing sequence ofopen sets containing Ac such that m(Hi−Ac) < 2−i. (The Hi are constructedas in the proof of (a), but in terms of Ac instead of A.) Then Fi is anincreasing sequence of closed sets, Fi ⊂ A for each i, and m(A − Fi) < 2−i

for each i. Our result follows from letting F = ∪iFi since m(A − F ) ≤m(A− Fi) < 2−i for each i, hence m(A− F ) = 0..

The countable intersections of open sets are sometimes called Gδ sets; theG is for geoffnet, the German word for “open” and the δ for Durchschnitt,the German word for “intersection.” The countable unions of closed sets arecalled Fσ sets, the F coming from ferme, the French word for “closed,” andthe σ coming from Summe, the German word for “union.”

Therefore, when trying to understand Lebesgue measure, we can look atGδ or Fσ sets, which are not so bad, and at null sets, which can be quite badbut don’t have positive measure.

Next we prove the Caratheodory extension theorem. We say that a mea-sure µ is σ-finite if there exist E1, E2, . . . , such that µ(Ei) < ∞ for all i andX ⊂ ∪∞i=1Ei.

Theorem 4.5 Suppose A0 is an algebra and m restricted to A0 is a measure.Define

m∗(E) = inf ∞∑

i=1

m(Ai) : Ai ∈ A0, E ⊂ ∪∞i=1Ai

.

Then

(a) m∗(A) = m(A) if A ∈ A0;

(b) every set in A0 is m∗-measurable;

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(c) if m is σ-finite, then there is a unique extension to the smallest σ-algebracontaining A0.

Proof. We start with (a). Suppose E ∈ A0. We know m∗(E) ≤ m(E)since we can take A1 = E and A2, A3, . . . empty in the definition of m∗. IfE ⊂ ∪∞i=1Ai with Ai ∈ A0, let Bn = E ∩ (An − ∪n−1

i=1 Ai). Then the Bn aredisjoint, they are each in A0, and their union is E. Therefore

m(E) =∞∑i=1

m(Bi) ≤∞∑i=1

m(Ai).

Thus m(E) ≤ m∗(E).

Next we look at (b). Suppose A ∈ A0. Let ε > 0 and let E ⊂ X. PickBi ∈ A0 such that E ⊂ ∪∞i=1Bi and

∑i m(Bi) ≤ m∗(E) + ε. Then

m∗(E) + ε ≥∞∑i=1

m(Bi) =∞∑i=1

m(Bi ∩ A) +∞∑i=1

m(Bi ∩ Ac)

≥ m∗(E ∩ A) + m∗(E ∩ Ac).

Since ε is arbitrary, m∗(E) ≥ m∗(E∩A)+m∗(E∩Ac). So A is m∗-measurable.

Finally, suppose we have two extensions to the smallest σ-algebra contain-ing A0; let the other extension be called n. We will show that if E is in thissmallest σ-algebra, then m∗(E) = n(E).

Since m is σ-finite, we can reduce to the case where m is a finite measure:if X = ∪iKi with m(Ki) < ∞ and we prove uniqueness for the measure mi

defined by mi(A) = m(A∩Ki), then uniqueness for m follows. So we supposem(X) < ∞.

Since E must be m∗-measurable,

m∗(E) = inf∞∑i=1

m(Ai) : E ⊂ ∪∞i=1Ai, Ai ∈ A0.

But m = n on A0, so∑

i m(Ai) =∑

i n(Ai). Therefore n(E) ≤∑

i n(Ai),which implies n(E) ≤ m∗(E).

Since we do not know that n is constructed via an outer measure, we mustuse a different argument to get the reverse inequality. Let ε > 0 and choose

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Ai ∈ A0 such that m∗(E) + ε ≥∑

i m(Ai) and E ⊂ ∪iAi. Let A = ∪iAi andBk = ∪k

i=1Ai. Observe m∗(E) + ε ≥ m∗(A), hence m∗(A−E) < ε. We have

m∗(A) = limk→∞

m∗(Bk) = limk→∞

n(Bk) = n(A).

Then

m∗(E) ≤ m∗(A) = n(A) = n(E)+n(A−E) ≤ n(E)+m(A−E) ≤ n(E)+ε.

Since ε is arbitrary, this completes the proof.

Remarks: (1) Uniqueness implies there is only one possible Lebesgue mea-sure.

(2) We will use the Caratheodory extension theorem in the study of prod-uct measures. It is also used in the Riesz representation theorem and in theDaniell-Kolmogorov extension theorem.

We now define Lebesgue-Stieltjes measures. Let α : R → R be nonde-creasing and right continuous (i.e., α(x+) = α(x) for all x, where α(x+) =limy→x,y>x α(y)). Suppose we define mα((a, b]) = α(b)− α(a), define

mα(∪∞i=1(ai, bi]) =∑

i

(α(bi)− α(ai))

when the intervals (ai, bi] are disjoint, and define

m∗α(A) = inf

∑i

(α(bi)− α(ai)) : A ⊂ ∪i(ai, bi].

Very much as in the previous section we can show that m∗α is a measure on

the Borel σ-algebra.

The mα measure of a point x is α(x)− α(x−), where

α(x−) = limy→x,y<x

α(x).

So mα(x) is equal to the size of the jump (if any) of α at x.

Lebesgue measure is the special case of mα when α(x) = x.

Given a measure µ on R such that µ(K) < ∞ whenever K is compact,define α(x) = µ((0, x]) if x ≥ 0 and α(x) = −µ((x, 0]) if x < 0. Then α isnondecreasing, right continuous, and it is not hard to see that µ = mα.

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5 Measurable functions

Suppose we have a set X together with a σ-algebra A.

Definition 5.1 f : X → R is measurable or A-measurable if x : f(x) >a ∈ A for all a ∈ R.

Example 5.2 Suppose f is identically constant. Then x : f(x) > a iseither empty or the whole space, so f is measurable.

Example 5.3 Suppose f(x) = 1 if x ∈ A and 0 otherwise. Then x : f(x) >a is either ∅, A, or X. So f is measurable if and only if A is in the σ-algebra.

Example 5.4 Suppose X is the real line with the Borel σ-algebra andf(x) = x. Then x : f(x) > a = (a,∞), and so f is measurable.

Proposition 5.5 The following are equivalent.

(a) x : f(x) > a ∈ A for all a;

(b) x : f(x) ≤ a ∈ A for all a;

(c) x : f(x) < a ∈ A for all a;

(d) x : f(x) ≥ a ∈ A for all a.

Proof. The equivalence of (a) and (b) and of (c) and (d) follow from takingcomplements. The remaining equivalences follow from the equations

x : f(x) ≥ a = ∩∞n=1x : f(x) > a− 1/n,x : f(x) > a = ∪∞n=1x : f(x) ≥ a + 1/n.

Proposition 5.6 If X is a metric space, A contains all the open sets, andf is continuous, then f is measurable.

15

Proof. x : f(x) > a = f−1(a,∞) is open.

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

Proof. If f(x)+ g(x) < α, then f(x) < α− g(x), and there exists a rationalr such that f(x) < r < α− g(x). So

x : f(x) + g(x) < α =⋃

r rational

(x : f(x) < r ∩ x : g(x) < α− r).

f 2 is measurable since x : f(x)2 > a) = x : f(x) >√

a ∪ x : f(x) <−√

a. The measurability of fg follows since fg = 12[(f + g)2 − f 2 − g2].

x : max(f(x), g(x)) > a = x : f(x) > a ∪ x : g(x) > a, and theargument for min(f, g) is similar.

Proposition 5.8 If fi is measurable for each i, then so are supi fi, infi fi,lim supi→∞ fi, and lim infi→∞ fi.

Proof. The result will follow for lim sup and lim inf once we have the resultfor the sup and inf by using the definitions. We have x : supi fi > a =∩∞i=1x : fi(x) > a, and the proof for inf fi is similar.

Definition 5.9 We say f = g almost everywhere, written f = g a.e., ifx : f(x) 6= g(x) has measure zero. Similarly, we say fi → f a.e., if the setof x where this fails has measure zero.

We saw in Proposition 5.6 that all continuous functions are Borel measur-able. The same is true for monotone functions on the real line.

Proposition 5.10 If f : R → R is nondecreasing or nonincreasing, then fis Borel measurable.

16

Proof. Let us suppose f is nondecreasing. The set Aa = x : f(x) > ais then either a semi-infinite open interval or semi-infinite closed interval.This can be seen by a picture. To be more careful, given a ∈ R, let x0 =supy : f(y) ≤ a. If f(x0) = a, then Aa = (x0,∞), while if f(x0) 6= a, thenAa = [x0,∞). In each case Aa is a Borel set.

Proposition 5.11 Let X be a space, A a σ-algebra on X, and f : X → R aA-measurable function. If A is in the Borel σ-algebra on R, then f−1(A) ∈ A.

Proof. Let B be the Borel σ-algebra on R and C = A ∈ B : f−1(A) ∈ A.If A1, A2, . . . ∈ C, then since f−1(∪iAi) = ∪if

−1(Ai) ∈ A, we have thatC is closed under countable unions. Similarly C is closed under countableintersections and complements, so C is a σ-algebra. Since f is measurable,C contains (a,∞) for every real a, hence C contains the σ-algebra generatedby these intervals, that is, C contains B.

Example 5.12 We want to construct a set that is Lebesgue measurable, butnot Borel measurable. Let F be the Cantor-Lebesgue function of Example4.1 and define

f(x) = infy : F (y) ≥ x.

Although f is not continuous, observe that f is strictly increasing (henceone-to-one) and maps [0, 1] into C, the Cantor set. Since f is nondecreasing,f−1 maps Borel measurable sets to Borel measurable sets.

Let A be the non-measurable set we constructed in Proposition 3.1. LetB = f(A). Since f(A) ⊂ C and m(C) = 0, then f(A) is a null set, henceis Lebesgue measurable. On the other hand, f(A) is not Borel measurable,because if it were, then A = f−1(f(A)) would be Borel measurable, a con-tradiction.

6 Integration

In this section we introduce the Lebesgue integral.

17

Definition 6.1 If E ⊂ X, define the characteristic function of E by

χE(x) =

1 x ∈ E;

0 x /∈ E.

A simple function s is one of the form

s(x) =n∑

i=1

aiχEi(x)

for reals ai and measurable sets Ei.

Proposition 6.2 Suppose f ≥ 0 is measurable. Then there exists a sequenceof nonnegative measurable simple functions increasing to f .

Proof. Let Eni = x : (i − 1)/2n ≤ f(x) < i/2n and Fn = x : f(x) ≥ nfor n = 1, 2, . . . , and i = 1, 2, . . . , n2n. Then define

sn =n2n∑i=1

i− 1

2nχEni

+ nχFn .

It is easy to see that sn has the desired properties.

Definition 6.3 If s =∑n

i=1 aiχEiis a nonnegative measurable simple func-

tion, define the Lebesgue integral of s to be∫s dµ =

n∑i=1

aiµ(Ei). (6.1)

If f ≥ 0 is a measurable function, define∫f dµ = sup

∫s dµ : 0 ≤ s ≤ f, s simple

. (6.2)

18

If f is measurable and at least one of the integrals∫

f+ dµ,∫

f− dµ is finite,where f+ = max(f, 0) and f− = −min(f, 0), define∫

f dµ =

∫f+ dµ−

∫f− dµ. (6.3)

Finally, if f = u + iv and∫

(|u|+ |v|) dµ is finite, define∫f dµ =

∫u dµ + i

∫v dµ. (6.4)

A few remarks are in order. A function s might be written as a simplefunction in more than one way. For example χA∪B = χA + χB is A and Bare disjoint. It is clear that the definition of

∫s dµ is unaffected by how s

is written. Secondly, if s is a simple function, one has to think a moment toverify that the definition of

∫s dµ by means of (6.1) agrees with its definition

by means of (6.2).

Definition 6.4 If∫|f | dµ < ∞, we say f is integrable.

The proof of the next proposition follows from the definitions.

Proposition 6.5 (a) If f is measurable, a ≤ f(x) ≤ b for all x, and µ(X) <∞, then aµ(X) ≤

∫f dµ ≤ bµ(X);

(b) If f(x) ≤ g(x) for all x and f and g are measurable and integrable,then

∫f dµ ≤

∫g dµ.

(c) If f is integrable, then∫

cf dµ = c∫

f dµ for all real c.

(d) If µ(A) = 0 and f is measurable, then∫

fχA dµ = 0.

The integral∫

fχA dµ is often written∫

Af dµ. Other notation for the

integral is to omit the µ if it is clear which measure is being used, to write∫f(x) µ(dx), or to write

∫f(x) dµ(x).

Proposition 6.6 If f is integrable,∣∣∣∫ f∣∣∣ ≤ ∫

|f |.

19

Proof. For the real case, this is easy. f ≤ |f |, so∫

f ≤∫|f |. Also −f ≤ |f |,

so −∫

f ≤∫|f |. Now combine these two facts.

For the complex case,∫

f is a complex number. If it is 0, the inequalityis trivial. If it is not, then

∫f = reiθ for some r and θ. Then∣∣∣ ∫

f∣∣∣ = r = e−iθ

∫f =

∫e−iθf.

From the definition of∫

f when f is complex, we have Re (∫

f) =∫

Re (f).Since |

∫f | is real, we have∣∣∣ ∫

f∣∣∣ = Re

( ∫e−iθf

)=

∫Re (e−iθf) ≤

∫|f |.

We do not yet have that∫

(f + g) =∫

f +∫

g.

7 Limit theorems

One of the most important results concerning Lebesgue integration is themonotone convergence theorem.

Theorem 7.1 Suppose fn is a sequence of nonnegative measurable functionswith f1(x) ≤ f2(x) ≤ · · · for all x and with limn→∞ fn(x) = f(x) for all x.Then

∫fn dµ →

∫f dµ.

Proof. By Proposition 6.5(b),∫

fn is an increasing sequence of real numbers.Let L be the limit. Since fn ≤ f for all n, then L ≤

∫f . We must show

L ≥∫

f .

Let s =∑m

i=1 aiχEibe any nonnegative simple function less than f and

let c ∈ (0, 1). Let An = x : fn(x) ≥ cs(x). Since the fn(x) increases tof(x) for each x and c < 1, then A1 ⊂ A2 ⊂ · · · , and the union of the An is

20

all of X. For each n, ∫fn ≥

∫An

fn ≥ c

∫An

sn

= c

∫An

m∑i=1

aiχEi

= cm∑

i=1

aiµ(Ei ∩ An).

If we let n →∞, by Proposition 2.5(c), the right hand side converges to

cm∑

i=1

aiµ(Ei) = c

∫s.

Therefore L ≥ c∫

s. Since c is arbitrary in the interval (0, 1), then L ≥∫

s.Taking the supremum over all simple s ≤ f , we obtain L ≥

∫f .

Example 7.2 Let X = [0,∞) and fn(x) = −1/n for all x. Then∫

fn =−∞, but fn ↑ f where f = 0 and

∫f = 0. The problem here is that the fn

are not nonnegative.

Example 7.3 Suppose fn = nχ(0,1/n). Then fn ≥ 0, fn → 0 for each x, but∫fn = 1 does not converge to

∫0 = 0. The trouble here is that the fn do

not increase for each x.

Once we have the monotone convergence theorem, we can prove that theLebesgue integral is linear.

Theorem 7.4 If f1 and f2 are integrable, then∫(f1 + f2) =

∫f1 +

∫f2.

Proof. First suppose f1 and f2 are nonnegative and simple. Then it is clearfrom the definition that the theorem holds in this case. Next suppose f1

21

and f2 are nonnegative. Take sn simple and increasing to f1 and tn simpleand increasing to f2. Then sn + tn increases to f1 + f2, so the result followsfrom the monotone convergence theorem and the result for simple functions.Finally in the general case, write f1 = f+

1 − f−1 and similarly for f2, and usethe definitions and the result for nonnegative functions.

Proposition 7.5 Suppose fn are nonnegative measurable functions. Then∫ ∞∑n=1

fn =∞∑

n=1

∫fn.

Proof. Let FN =∑N

n=1 fn and write∫ ∞∑n=1

fn =

∫lim

N→∞

N∑n=1

fn

=

∫lim

N→∞FN = lim

N→∞

∫FN (7.1)

= limN→∞

N∑n=1

∫fn =

∞∑n=1

∫fn,

using the monotone convergence theorem and the linearity of the integral.

The next theorem is known as Fatou’s lemma.

Theorem 7.6 Suppose the fn are nonnegative and measurable. Then∫lim infn→∞

fn ≤ lim infn→∞

∫fn.

Proof. Let gn = infi≥n fi. Then gn are nonnegative and gn increases tolim inf fn. Clearly gn ≤ fi for each i ≥ n, so

∫gn ≤

∫fi. Therefore∫

gn ≤ infi≥n

∫fi.

22

If we take the limit as n →∞, on the left hand side we obtain∫

lim inf fn bythe monotone convergence theorem, while on the right hand side we obtainlim infn

∫fn.

A typical use of Fatou’s lemma is the following. Suppose we have fn → fand supn

∫|fn| ≤ K < ∞. Then |fn| → |f |, and by Fatou’s lemma,

∫|f | ≤

K.

Another very important theorem is the dominated convergence theorem.

Theorem 7.7 Suppose fn are measurable functions and fn(x) → f(x). Sup-pose there exists an integrable function g such that |fn(x)| ≤ g(x) for all x.Then

∫fn dµ →

∫f dµ.

Proof. Since fn + g ≥ 0, by Fatou’s lemma,∫(f + g) ≤ lim inf

∫(fn + g).

Since g is integrable, ∫f ≤ lim inf

∫fn.

Similarly, g − fn ≥ 0, so∫(g − f) ≤ lim inf

∫(g − fn),

and hence

−∫

f ≤ lim inf

∫(−fn) = − lim sup

∫fn.

Therefore ∫f ≥ lim sup

∫fn,

which with the above proves the theorem.

Example 7.3 is an example where the limit of the integrals is not theintegral of the limit because there is no dominating function g.

23

If in the monotone convergence theorem or dominated convergence the-orem we have only fn(x) → f(x) almost everywhere, the conclusion stillholds. For if the fn and f are measurable and A = x : fn(x) → f(x), thenfχA → fχA for each x. And since Ac has measure 0, we see from Proposition6.5(d) that

∫fχA =

∫f , and similarly with f replaced by fn.

8 Properties of Lebesgue integrals

Later on we will need the following two propositions.

Proposition 8.1 Suppose f is measurable and for every measurable set Awe have

∫A

f dµ = 0. Then f = 0 almost everywhere.

Proof. Let A = x : f(x) > ε. Then

0 =

∫A

f ≥∫

A

ε = εµ(A)

since fχA ≥ εχA. Hence µ(A) = 0. We use this argument for ε = 1/n andn = 1, 2, . . . , so µx : f(x) > 0 = 0. Similarly µx : f(x) < 0 = 0.

Proposition 8.2 Suppose f is measurable and nonnegative and∫

f dµ = 0.Then f = 0 almost everywhere.

Proof. If f is not almost everywhere equal to 0, there exists an n such thatµ(An) > 0 where An = x : f(x) > 1/n. But then since f is nonnegative,

0 =

∫f ≥

∫An

f ≥ 1

nµ(An),

a contradiction.

We give a result on approximating a function on R by continuous functions.

24

Proposition 8.3 Suppose f is a measurable function from R to R that isintegrable. Let ε > 0. Then there exists a continuous function that is 0outside some bounded interval such that∫

|f − g| < ε.

Proof. If we have continuous functions g1, g2 such that∫|f+ − g1| < ε/2

and∫|f− − g2| < ε/2, where f+ = max(f, 0) and f− = max(−f, 0), then

taking g = g1 − g2 will prove our result. So without loss of generality, wemay assume f ≥ 0.

By monotone convergence∫

fχ[−n,n] increases to∫

f , so by taking n largeenough, the difference of the integrals will be less than ε/2. If we find g suchthat

∫|fχ[−n,n] − g| < ε/2, then

∫|f − g| < ε. Therefore we may assume

that f is 0 outside some bounded interval.

We can find simple functions increasing to f whose integrals increase to∫f . Let sm be a simple function such that sm ≤ f and

∫sm ≥

∫f − ε/2.

If we find g such that∫|sm − g| < ε/2, then

∫|f − g| < ε. So it suffices to

consider the case where f is a simple function.

If f =∑p

i=1 aiχAiand we find gi continuous such that

∫|aiχAi

−gi| < ε/p,then

∑pi=1 gi will be the desired function. So we may assume f is a constant

times a characteristic function, and by linearity, we may assume f is equalto χA for some A contained in a bounded interval [−n, n].

We can choose G open and F closed such that F ⊂ A ⊂ G and m(G−F ) <ε. We can replace G by G ∩ (−n − 1, n + 1). Gc ∩ [−n − 1, n + 1] and Fare compact sets, so there is a minimum distance between them, say, δ. Letg(x) = max(0, 1− dist (x, F )/δ). Then g is continuous, 0 ≤ g ≤ 1, g is 1 onF , g is 0 on Gc, and g is 0 outside of [−n− 1, n + 1]. Therefore

|g − χA| ≤ χG − χF ,

so ∫|g − χA| ≤

∫(χG − χF ) = m(G− F ) < ε.

The method of proof, where one proves a result for characteristic func-tions, then simple functions, then non-negative functions, and then finallyintegrable functions is very common.

25

We finish this section with a comparison of the Lebesgue integral andthe Riemann integral. Here we are only looking at bounded functions from[a, b] into R. If we are looking at the Lebesgue integral, we write

∫f , while,

temporarily, if we are looking at the Riemann integral, we write R(f). Recallthat the Riemann integral on [a, b] is defined as follows: if P is a partition of[a, b], then

U(P, f) =n∑

i=1

( supxi=1≤x≤x

f(x)) (xi − xi−1)

and

L(P, f) =n∑

i=1

( infxi=1≤x≤x

f(x)) (xi − xi−1).

Set R(f) = infU(P, f) : P is a partition similarly R(f). Then the Rie-mann integral exists if R(f) = R(f), and the common value is the Riemannintegral, which we denote R(f).

Theorem 8.4 A bounded measurable function f on [a, b] is Riemann inte-grable if and only if the set of points at which f is discontinuous has Lebesguemeasure 0, and in that case, the Riemann integral is equal in value to theLebesgue integral.

Proof. If P is a partition, define

TP (x) =n∑

i=1

( supxi−1≤y≤xi

f(y))χ[xi−1,xi)(x),

and

SP (x) =n∑

i=1

( infxi−1≤y≤xi

f(y))χ[xi−1,xi)(x).

We see that∫

TP = U(P, f) and∫

SP = L(P, f).

If f is Riemann integrable, there exists a sequence of partitions Qi suchthat U(Qi, f) ↓ R(f) and a sequence Q′

i such that L(Q′i, f) ↑ R(f). It is not

hard to check that adding points to a partition increases L and decreases U ,so if we let Pi = ∪j≤i(Qj∪Q′

j), then Pi is an increasing sequence of partitions,U(Pi, f) ↓ R(f), L(Pi, f) ↑ R(f). We see also that TPi

(x) decreases at eachpoint, say, to T (x), and SPi

(x) increases at each point, say, to S(x). Also

26

T (x) ≥ f(x) ≥ S(x). Then by dominated convergence (recall that f isbounded)∫

(T − S) = limi→∞

∫(TPi

− SPi) = lim

i→∞(U(Pi, f)− L(Pi, f)) = 0.

We conclude T = S = f a.e. If x is not in the null set where T (x) 6= S(x) norin ∪iPi, which is countable and hence of Lebesgue measure 0, then TPi

(x) ↓f(x) and SPi

(x) ↑ f(x). This implies that f is continuous at such f . Since

R(f) = limi→∞

U(Pi, f) = limi→∞

∫TPi

=

∫f,

we see the Riemann integral and Lebesgue integral agree.

Now suppose that f is continuous a.e. Let ε > 0. Let Pi be the partitionwhere we divide [a, b] into 2i equal parts. If x is not in the null set wheref is discontinuous, nor in ∪∞i=1Pi, then TPi

(x) ↓ f(x) and SPi(x) ↑ f(x). By

dominated convergence,

U(Pi, f) =

∫TPi

→∫

f

and

L(Pi, f) =

∫SPi

→∫

f.

This does it.

9 Modes of convergence

Definition 9.1 If µ is a measure, we say a sequence of measurable functionsfn converges to f almost everywhere (written fn → f a.e.) if there is a setof measure 0 and for x not in this set we have fn(x) → f(x).

We say fn converges to f in measure if for each ε > 0

µ(x : |fn(x)− f(x)| > ε) → 0

as n →∞.

27

Proposition 9.2 Suppose µ is a finite measure.

(a) If fn → f , a.e., then fn converges to f in measure.

(b) If fn → f in measure, there is a subsequence nj such that fnj→ f ,

a.e.

Proof. Let ε > 0. If An = x : |fn(x)− f(x)| > ε, then χAn → 0 a.e., andby dominated convergence,

µ(An) =

∫χAn(x) µ(dx) → 0.

This proves (a).

To prove (b), let n1 = 1 and choose nj > nj−1 inductively so that

µ(x : |fnj(x)− f(x)| > 1/j) ≤ 2−j.

Let Aj = x : |fnj(x)− f(x)| > 1/j. Then µ(Aj) ≤ 2−j, and

A = ∩∞k=1 ∪∞j=k Aj

has measure less than ∪j=kAj for every k, hence less than 2−k+1 for everyk. Therefore A has measure 0. If x /∈ A, then x /∈ ∪∞j=kAj for some k, so|fnj

(x)− f(x)| ≤ 1/j for j ≥ k, which means fnj→ f a.e. on Ac.

Example 9.3 Part (a) of the above proposition is not true if µ(X) = ∞.Let X = R and let fn = χ(n,n+1).

Example 9.4 For an example where fn → f in measure but not almosteverywhere, let X = [0, 1], let µ be Lebesgue measure, and let fn(x) =χFn(x), where Fn = y : (

∑nj=1 1/j)( mod 1) ≤ y ≤ (

∑n+1j=1 1/j)( mod 1).

z( mod 1) is defined as the fractional part of z (where the largest integer lessthan z is subtracted from z). Let f(x) = 0 for all x.

Then µ(Fn) ≤ 1/n → 0, so fn → f in measure. But any x will be ininfinitely many Fn’s, so fn does not converge to f(x) at any point.

The following is known as Egoroff’s theorem.

28

Theorem 9.5 If µ is a finite measure, ε > 0, and fn → f a.e., then thereexists a measurable set A such that µ(A) < ε and fn → f uniformly on Ac.

This type of convergence is sometimes known as almost uniform conver-gence.

Proof. LetEnk = ∪∞m=nx : |fm(x)− f(x)| > 1/k.

for fixed k, Enk decreases as n increases, and the intersection ∩nEnk hasmeasure 0. So µ(Enk) → 0. Then there exists an integer nk such thatµ(Enkk) < ε2−k. Let E = ∪∞k=1Enkk. Then µ(E) < ε, and if x /∈ E andn > nk, then |fn(x)− f(x)| ≤ 1/k. Thus fn → f uniformly on Ec.

10 Product measures

If A1 ⊂ A2 ⊂ · · · and A = ∪∞i=1Ai, we write Ai ↑ A. If A1 ⊃ A2 ⊃ · · · andA = ∩∞i=1Ai, we write Ai ↓ A.

Definition 10.1 M is a monotone class is M is a collection of subsets ofX such that

(a) if Ai ↑ A and each Ai ∈M, then A ∈M;

(b) if Ai ↓ A and each Ai ∈M, then A ∈M.

The intersection of monotone classes is a monotone class, and the inter-section of all monotone classes containing a given collection of sets is thesmallest monotone class containing that collection.

The next theorem, the monotone class lemma, is rather technical, but veryuseful.

Theorem 10.2 Suppose A0 is a algebra, A is the smallest σ-algebra con-taining A0, and M is the smallest monotone class containing A0. ThenM = A.

29

Proof. A σ-algebra is clearly a monotone class, so M⊂ A. We must showA ⊂M.

Let N1 = A ∈ M : Ac ∈ M. Note N1 is contained in M, contains A0,and is a monotone class. So N1 = M, and therefore M is closed under theoperation of taking complements.

Let N2 = A ∈ M : A ∩ B ∈ M for all B ∈ A0. N2 is contained in M;N2 contains A0 because A0 is an algebra; N2 is a monotone class because(∪∞i=1Ai) ∩ B = ∪∞i=1(Ai ∩ B), and similarly for intersections. ThereforeN2 = M; in other words, if B ∈ A0 and A ∈M, then A ∩B ∈M.

Let N3 = A ∈ M : A ∩ B ∈ M for all B ∈ M. As in the precedingparagraph, N3 is a monotone class contained in M. By the last sentence ofthe preceding paragraph, N3 contains A0. Hence N3 = M.

We thus have that M is a monotone class closed under the operations oftaking complements and taking intersections. This shows M is a σ-algebra,and so A ⊂M.

Suppose (X,A, µ) and (Y,B, ν) are two measure spaces, i.e., A and B areσ-algebras on X and Y , resp., and µ and ν are measures on A and B, resp.A rectangle is a set of the form A × B, where A ∈ A and B ∈ B. Define aset function µ× ν on rectangles by

µ× ν(A×B) = µ(A)ν(B).

Lemma 10.3 Suppose A×B = ∪∞i=1Ai×Bi, where A, Ai ∈ A and B, Bi ∈ Band the Ai ×Bi are disjoint. Then

µ× ν(A×B) =∞∑i=1

µ× ν(Ai ×Bi).

Proof. We have

χA×B(x, y) =∞∑i=1

χAi×Bi(x, y),

and so

χA(x)χB(y) =∞∑i=1

χAi(x)χBi

(y).

30

Holding x fixed and integrating over y with respect to ν, we have, using (7.1),

χA(x)ν(B) =∞∑i=1

χAi(x)ν(Bi).

Now use (7.1) again and integrate over x with respect to µ to obtain theresult.

Let C0 = finite unions of rectangles. It is clear that C0 is an algebra. ByLemma 10.3 and linearity, we see that µ× ν is a measure on C0. Let A× Bbe the smallest σ-algebra containing C0; this is called the product σ-algebra.By the Caratheodory extension theorem, µ×ν can be extended to a measureon A× B.

We will need the following observation. Suppose a measure µ is σ-finite.So there exist Ei which have finite µ measure and whose union is X. If welet Fn = ∪n

i=1Ei, then Fi ↑ X and µ(Fn) is finite for each n.

If µ and ν are both σ-finite, say with Fi ↑ X and Gi ↑ Y , then µ× ν willbe σ-finite, using the sets Fi ×Gi.

The main result of this section is Fubini’s theorem, which allows one tointerchange the order of integration.

Theorem 10.4 Suppose f : X×Y → R is measurable with respect to A×B.If f is nonnegative or

∫|f(x, y)| d(µ× ν)(x, y) < ∞, then

(a) the function g(x) =∫

f(x, y)ν(dy) is measurable with respect to A;

(b) the function h(y) =∫

f(x, y)µ(dx) is measurable with respect to B;

(c) we have ∫f(x, y) d(µ× ν)(x, y) =

∫ ( ∫f(x, y) dµ(x)

)dν(y)

=

∫ ( ∫f(x, y) dν(y)

)µ(dx).

31

Proof. First suppose µ and ν are finite measures. If f is the characteristicfunction of a rectangle, then (a)–(c) are obvious. By linearity, (a)–(c) holdif f is the characteristic function of a set in C0, the set of finite unions ofrectangles.

Let M be the collection of sets C such that (a)–(c) hold for χC . If Ci ↑ Cand Ci ∈ M, then (c) holds for χC by monotone convergence. If Ci ↓ C,then (c) holds for χC by dominated convergence. (a) and (b) are easy. SoM is a monotone class containing A0, so M = A× B.

If µ and ν are σ-finite, applying monotone convergence to C ∩ (Fn ×Gn)for suitable Fn and Gn and monotone convergence, we see that (a)–(c) holdsfor the characteristic functions of sets in A× B in this case as well.

By linearity, (a)–(c) hold for nonnegative simple functions. By monotoneconvergence, (a)–(c) hold for nonnegative functions. In the case

∫|f | < ∞,

writing f = f+ − f− and using linearity proves (a)–(c) for this case, too.

11 The Radon-Nikodym theorem

Suppose f is nonnegative, measurable, and integrable with respect to µ. Ifwe define ν by

ν(A) =

∫A

f dµ, (11.1)

then ν is a measure. The only part that needs thought is the countable ad-ditivity, and this follows from (7.1) applied to the functions fχAi

. Moreover,ν(A) is zero whenever µ(A) is. We sometimes write f = dν/dµ for (11.1).

Definition 11.1 A measure ν is called absolutely continuous with respectto a measure µ if ν(A) = 0 whenever µ(A) = 0. This is frequently writtenν µ.

Proposition 11.2 A finite measure ν is absolutely continuous with respectto µ if and only if for all ε there exists δ such that µ(A) < δ implies ν(A) < ε.

32

Proof. If the condition given in the statement of the proposition holds, it isclear that ν µ. Suppose now that ν µ. If the condition does not hold,there exists Ek such that µ(Ek) < 2−k but ν(Ek) ≥ ε. Let F = ∩∞n=1∪∞k=n Ek.Then

µ(F ) = limn→∞

µ(∪∞k=nEk) ≤ limn→∞

∞∑k=n

2−k = 0,

butν(F ) = lim

n→∞ν(∪∞k=nEk) ≥ ε;

This contradicts the absolute continuity.

Definition 11.3 A function µ : A → (−∞,∞] is called a signed measure ifµ(∅) = 0 and µ(∪∞i=1Ai) =

∑∞i=1 µ(Ai) whenever the Ai are disjoint and all

the Ai are in A.

Definition 11.4 Let µ be a signed measure. A set A ∈ A is called a positiveset for µ if µ(B) ≥ 0 whenever B ⊂ A and B ∈ A. We define a negativeset similarly. A null set A is one where µ(B) = 0 whenever B ⊂ A ismeasurable.

Example 11.5 Suppose m is Lebesgue measure and µ(A) =∫

Af dm for

some integrable f . If we let P = x : f(x) ≥ 0, then P is easily seen to be apositive set, and if N = x : f(x) < 0, then N is a negative one. The Hahndecomposition which we give below is a decomposition of our space (in thiscase R) into positive and negative sets. This decomposition is unique, exceptthat C = x : f(x) = 0 could be included in N instead of P , or apportionedpartially to P and partially to N . Note, however, that C is a null set. TheJordan decomposition below is a decomposition of µ into µ+ and µ−, whereµ+(A) =

∫A

f+ dm, and similarly µ−(A) =∫

Af− dµ.

Note that if µ is a signed measure, then µ(∪∞i=1Ai) = limn→∞ µ(∪ni=1Ai).

The proof is the same as in the case of positive measures.

Proposition 11.6 Let µ be a signed measure taking values in (−∞,∞]. LetE be measurable with µ(E) < 0. Then there exists a subset F of E that is anegative set with µ(F ) < 0.

33

Proof. If E is a negative set, we are done. If not, there exists a subset withpositive measure. Let n1 be the smallest positive integer such that thereexists E1 ⊂ E with µ(E1) ≥ 1/n. Let k ≥ 2. If Fk = E − (E1 ∪ · · · ∪ Ek−1)is negative, we are done. If not, let nk be the smallest positive integer suchthat there exists Ek ⊂ Fk with µ(Ek) ≥ 1/nk. We continue.

If the construction stops after a finite number of sets, we are done. If not,let F = ∩kFk = E − (∪kEk). Since 0 > µ(E) > −∞ and µ(Ek) ≥ 0, then

µ(E) = µ(F ) +∞∑

k=1

µ(Ek).

Then µ(F ) ≤ µ(E) < 0, so the sum converges. If G ⊂ F is measurable withµ(G) > 0, then µ(G) ≥ 1/N for some N , which contradicts the construction.Therefore F must be a negative set.

We write A∆B for (A − B) ∪ (B − A). The following is known as theHahn decomposition theorem.

Theorem 11.7 Let µ be a signed measure taking values in (−∞,∞]. Thereexist sets E and F in A that are disjoint whose union is X and such that Eis a negative set and F is a positive set. If E ′ and F ′ are another such pair,then E∆E ′ = F∆F ′ is a null set with respect to µ..

Proof. Let L = infµ(A) : A is a negative set. Choose negative sets An

such that µ(An) → L. Let E = ∪∞n=1An. Let Bn = An− (B1∪· · ·∪Bn−1) foreach n. Since An is a negative set, so is each Bn. Also, the Bn are disjoint.If C ⊂ E, then

µ(C) = limn→∞

µ(C ∩ (∪ni=1Bi)) = lim

n→∞

n∑i=1

µ(C ∩Bi) ≤ 0.

So E is a negative set.

Since E is negative,

µ(E) = µ(An) + µ(E − An) ≤ µ(An).

Letting n →∞, we obtain µ(E) = L.

34

Let F = Ec. If F were not a positive set, there would exist B ⊂ F withµ(B) < 0. By Proposition 11.6 there exists a negative set C contained in Bwith µ(C) < 0. But then E ∪ C would be a negative set with µ(E ∪ C) <µ(E) = L, a contradiction.

To prove uniqueness, if E ′, F ′ are another such pair of sets and A ⊂E − E ′ ⊂ E, then µ(A) ≤ 0. But A ⊂ E − E ′ = F ′ − F ⊂ F ′, so µ(A) ≤ 0.Therefore µ(A) = 0. The same argument works if A ⊂ E ′ − E, and anysubset of E∆E ′ can be written as the union of A1 and A2, where A⊂E −E ′

and A2 ⊂ E ′ − E.

Let us say two measures µ and ν are mutually singular if there exist twodisjoint sets E and F in A whose union is X with µ(E) = ν(F ) = 0. This isoften written µ ⊥ ν.

Example 11.8 If µ is Lebesgue measure restricted to [0, 1/2], that is, µ(A)= m(A ∩ [0, 1/2]), and ν is Lebesgue measure restricted to [1/2, 1], then µand ν are mutually singular. We let E = [0, 1/2] and F = (1/2, 1]. Thisexample works because the Lebesgue measure of 1/2 is 0.

Example 11.9 A more interesting example is the following. Let f be theCantor-Lebesgue function and let ν be the Lebesgue-Stieltjes measure asso-ciated with f . Let µ be Lebesgue measure restricted to [0, 1]. Then µ ⊥ ν.To see this, we let E = C, where C is the Cantor set, and F = [0, 1]−C. Wealready know that m(E) = 0 and we need to show ν(F ) = 0. To do that,we need to show ν(I) = 0 for every open interval contained in F . This willfollow if we show ν(J) = 0 for every interval of the form J = (a, b] containedin F . But f is constant on every such interval, so f(b) = f(a), and thereforeν(J) = f(b)− f(a) = 0.

The following is known as the Jordan decomposition theorem.

Theorem 11.10 If µ is a signed measure, there exist measures µ+ and µ−

such that µ = µ+ − µ− and µ+ and µ− are mutually singular. This decom-position is unique.

Proof. Let E and F be positive and negative sets for µ and let µ+(A) =µ(E ∩ A), µ−(A) = −µ(A ∩ F ). This gives the desired decomposition.

35

If µ = ν+ − ν− is another such decomposition with ν+, ν− mutually sin-gular, let E ′ and F ′ be the sets in the definition of mutually singular. ThenX = E ′ ∪ F ′ gives another Hahn decomposition, hence E∆E ′ is a null setwith respect to µ. Then for any A ∈ A,

ν+(A) = µ(A ∩ E ′) = µ(A ∩ E) = µ+(A),

and similarly for ν−, µ−.

The measure µ+ + µ− is called the total variation measure and is written|µ|.

We now are ready for the Radon-Nikodym theorem.

Theorem 11.11 Suppose µ is a σ-finite measure and ν is a finite measuresuch that ν is absolutely continuous with respect to µ. There exists a µ-integrable nonnegative function f such that ν(A) =

∫A

f dµ for all A ∈ A.Moreover, if g is another such function, then f = g almost everywhere withrespect to µ.

Proof. Let us first prove the uniqueness assertion. For every set A we have∫A

(f − g) dµ = ν(A)− ν(A) = 0.

By Proposition 8.1 we have f − g = 0 a.e. with respect to µ.

Since µ is σ-finite, there exist Fi ↑ X such that µ(Fi) < ∞ for each i. Letµi be the restriction of µ to Fi, that is, µi(A) = µ(A ∩ Fi). Define νi, therestriction of ν to Fi, similarly. If fi is a function such that νi(A) =

∫A

fi dµi

for all A, the argument of the first paragraph shows that fi = fj on Fi ifi ≤ j. If we define f by f(x) = fi(x) if x ∈ Fi, we see that f will be thedesired function. So it suffices to restrict attention to the case where µ isfinite.

Let

F =

g : 0 ≤ g,

∫A

g dµ ≤ ν(A) for all A ∈ A

.

F is not empty because 0 ∈ F . Let L = sup∫

g dµ : g ∈ F, and let gn bea sequence in F such that

∫gn dµ → L. Let hn = max(g1, . . . , gn).

36

If g1 and g2 are in F , then h2 = max(g1, g2) is also in F . To see this, letB = x : g1(x) ≥ g2(x), and write∫

A

h2 dµ =

∫A∩B

h2 dµ +

∫A∩Bc

h2 dµ

=

∫A∩B

g1 dµ +

∫A∩Bc

g2 dµ

≤ ν(A ∩B) + ν(A ∩Bc)

= ν(A).

By an induction argument, hn is in F .

The hn increase, say to f . By monotone convergence∫

f dµ = L and∫A

f dµ ≤ ν(A) (11.2)

for all A.

Let A be a set where there is strict inequality in (11.2); let ε be chosensufficiently small so that if π is defined by

π(B) = ν(B)−∫

B

f dµ− εµ(B),

then π(A) > 0. π is a signed measure; let F be the positive set as constructedin Theorem 11.7. In particular, π(F ) > 0. So for every B∫

B∩F

f dµ + εµ(B ∩ F ) ≤ ν(B ∩ F ).

We then have, using (11.2), that∫B

(f + εχF ) dµ =

∫B

f dµ + εµ(B ∩ F )

=

∫B∩F c

f dµ +

∫B∩F

f dµ + εµ(B ∩ F )

≤ ν(B ∩ F c) + ν(B ∩ F ) = ν(B).

This says that f + εχF ∈ F . However,

L ≥∫

(f + εχF ) dµ =

∫f dµ + εµ(F ) = L + εµ(F ),

37

which implies µ(F ) = 0. But then ν(F ) = 0, and hence π(F ) = 0, contra-dicting the fact that F is a positive set for F with π(F ) > 0.

The proof of the Lebesgue decomposition theorem is almost the same.

Theorem 11.12 Suppose µ and ν are two finite measures. There exist mea-sures λ, ρ such that ν = λ + ρ, ρ is absolutely continuous with respect to µ,and λ and µ are mutually singular.

Proof. Define F and L and construct f as in the proof of the Radon-Nikodym theorem. Let ρ(A) =

∫A

f dµ and let λ = ν−ρ. We have∫

Af dµ ≤

ν(A), so λ(A) ≥ 0 for all A. To keep things straight, we record that we havef = dρ/dµ and ρ + λ = ν. We need to show µ and λ are mutually singular.

Suppose not. Then there exists F ∈ A with µ(F ) > 0 and λ(F ) > 0, andso ν(F ) ≥ λ(F ) > 0. Note that for ε small enough, (λ − εµ)(F ) > 0. Weclaim that there exist ε > 0 and E ⊂ F such that E is a positive set withrespect to λ− εµ and µ(E) > 0. Given the claim, if A ∈ A,

ε

∫A

χE dµ = εµ(A ∩ E) ≤ ν(A ∩ E)

≤ λ(A ∩ E) ≤ λ(A)

= ν(A)−∫

A

f dµ.

This says that ∫A

(f + εχE) dµ ≤ ν(A)

for all A ∈ A, or f + εχE ∈ F . But∫(f + εχE) dµ =

∫f dµ + εµ(E) > L,

a contradiction to the definition of L.

It remains to prove the claim. Let F = Pn∪Nn be a Hahn decompositionfor the measure λ− 1

nµ restricted to F , let P = ∪Pn, and N = ∩Nn = F −P .

Then N is a negative set for λ− 1nµ for each n, or 0 ≤ λ(N) ≤ 1

nµ(N); this

38

implies λ(N) = 0. If µ(P ) = 0, then µ ⊥ λ, and we are supposing that isnot the case. Therefore µ(P ) > 0, hence µ(Pn) > 0 for some n, and Pn is apositive set for λ− 1

nµ. Now take ε = 1/n and E = Pn.

12 Differentiation of real-valued functions

In this section we want to look at when f : R → R is differentiable and whenthe fundamental theorem of calculus holds. Briefly,

(1) Functions of bounded variation are differentiable;

(2) The derivative of∫ x

af(y) dy is equal to f a.e. if f is integrable;

(3)∫ b

af ′(y) dy = f(b)− f(a) if f ′ is absolutely continuous.

Let E ⊂ R be a measurable set and let O be a collection of intervals. Wesay O is a Vitali cover of E if for each x ∈ E and each ε > 0 there existsan interval G ∈ O containing x whose length is less than ε. m will denoteLebesgue measure.

Lemma 12.1 Let E have finite measure and let O be a Vitali cover of E.Given ε > 0 there exists a finite subcollection of disjoint intervals I1, . . . , In

such that m(E − ∪ni=1In) < ε.

Proof. We may replace each interval in O by a closed one, since the set ofendpoints of a finite subcollection will have measure 0.

Let O be an open set of finite measure containing E. Since O is a Vi-tali cover, we may suppose without loss of generality that each set of O iscontained in O. Let a0 = supm(I) : I ∈ O. Let I1 be any element ofO with m(I1) ≥ a0/2. Let a1 = supm(I) : I ∈ O, I disjoint from I1,andchoose I2 ∈ O disjoint from I1 such that m(I2) ≥ a1/2. Continue in this way,choosing In+1 disjoint from I1, . . . , In and in O with length at least one halfas large as any other such interval in O that is disjoint from I1, . . . , In.

If the process stops at some finite stage, we are done. If not, we gener-ate a sequence of disjoint intervals I1, I2, . . . Since they are disjoint and all

39

contained in O, then∑∞

i=1 m(Ii) ≤ m(O) < ∞. So there exists N such that∑∞i=N+1 m(Ii) < ε/5.

Let R = E − ∪Ni=1Ii; we will show m(R) < ε. Let Jn be the interval

with the same center as In but five times the length. Let x ∈ R. Thereexists an interval I ∈ O containing x with I disjoint from I1, . . . , IN . Since∑

m(In) < ∞, then∑

an ≤ 2∑

m(In) < ∞, and an → 0. So I must eitherbe one of the In for some n > N or at least intersect it, for otherwise wewould have chosen I at some stage. Let n be the smallest integer such thatI intersects In; note n > N . We have m(I) ≤ an−1 ≤ 2m(In). Since x is inI and I intersects In, the distance from x to the midpoint of In is at mostm(I) + m(In)/2 ≤ (5/2)m(In). Therefore x ∈ Jn.

Then R ⊂ ∪∞i=N+1Jn, so m(R) ≤∑∞

i=N+1 m(Jn) = 5∑∞

i=N+1 m(In) < ε.

Given a function f , we define the derivates of f at x by

D+f(x) = lim suph→0+

f(x + h)− f(x)

h, D−f(x) = lim sup

h→0−

f(x + h)− f(x)

h

D+f(x) = lim infh→0+

f(x + h)− f(x)

h, D−f(x) = lim inf

h→0−

f(x + h)− f(x)

h.

If all the derivates are equal, we say that f is differentiable at x and definef ′(x) to be the common value.

Theorem 12.2 Suppose f is nondecreasing on [a, b]. Then f is differentiable

almost everywhere, f ′ is integrable, and∫ b

af ′(x) dx ≤ f(b)− f(a).

Proof. We will show that the set where any two derivates are unequal hasmeasure zero. We consider the set E where D+f(x) > D−f(X), the othersets being similar. Let Eu,v = x : D+f(x) > u > v > D−f(x). If we showm(Eu,v) = 0, then taking the union of all pairs of rationals with u > v showsm(E) = 0.

Let s = m(Eu,v), let ε > 0, and choose an open set O such that Eu,v ⊂ Oand m(O) < s+ε. For each x ∈ Eu,v there exists an arbitrarily small interval[x− h, x] contained in O such that f(x)− f(x− h) < vh. Use Lemma 12.1to choose I1, . . . , In which are disjoint and whose interiors cover a subset A

40

of Eu,v of measure greater than s− ε. Suppose In = [xn − hn, xn]. Summingover these intervals,

N∑n=1

[f(xn)− f(xn − hn)] < v

n∑n=1

hn < vm(O) < v(s + ε).

Each point y ∈ A is the left endpoint of an arbitrarily small interval(y, y + k) that is contained in some In and for which f(y + k)− f(y) > uk.Using Lemma 12.1 again, we pick out a finite collection J1, . . . , JM whoseunion contains a subset of A of measure larger than s − 2ε. Summing overthese intervals yields

M∑i=1

[f(yi + ki)− f(yi)] > u∑

ki > u(s− 2ε).

Each interval Ji is contained in some interval In, and if we sum over those ifor which Ji ⊂ In we find∑

[f(yi + ki)− f(yi)] ≤ f(xn)− f(xn − hn),

since f is increasing. Thus

N∑n=1

[f(xn)− f(xn − hn)] ≥M∑i=1

[f(yi + ki)− f(yi)],

and so v(s + ε) > u(s− 2ε). This is true for each ε, so vs ≥ us. Since u > v,this implies s = 0.

This shows that

g(x) = limh→0

f(x + h)− f(x)

his defined almost everywhere and that f is differentiable wherever g is finite.Define f(x) = f(b) if x ≥ b. Let gn(x) = n[f(x + 1/n) − f(x)]. Thengn(x) → g(x) for almost all x, and so g is measurable. Since f is increasing,gn ≥ 0. By Fatou’s lemma∫ b

a

g ≤ lim inf

∫ b

a

gn = lim inf n

∫ b

a

[f(x + 1/n)− f(x)]dx

= lim inf[n

∫ b+1/n

b

f − n

∫ a+1/n

a

f]

= lim inf[f(b)− n

∫ a+1/n

a

f]

≤ f(b)− f(a).

41

For the last inequality, we use the fact that f is increasing. This shows thatg is integrable and hence finite almost everywhere.

A function is of bounded variation if sup∑k

i=1 |f(xi)− f(xi−1)| is finite,where the supremum is over all partitions a = x0 < x1 < · · · < xk = b of[a, b].

Lemma 12.3 If f is of bounded variation on [a, b], then f can be written asthe difference of two nondecreasing functions on [a, b].

Proof. Define

P (y) = sup k∑

i=1

[f(xi)−f(xi−1)]+

, N(y) = sup k∑

i=1

[f(xi)−f(xi−1)]−

,

where the supremum is over all partitions a = x0 < x1 < · · · < xk = y fory ∈ [a, b]. P and N are measurable since they are both increasing. Since

k∑i=1

[f(xi)− f(xi−1)]+ =

k∑i=1

[f(xi)− f(xi−1)]− + f(y)− f(a),

taking the supremum over all partitions of [a, y] yields

P (y) = N(y) + f(y)− f(a).

Clearly P and N are nondecreasing in y, and the result follows by solvingfor f(y).

From this lemma, we see that functions of bounded variation are dif-ferentiable a.e. But the function sin(1/x) defined on (0, 1] is differentiableeverywhere, but is not of bounded variation.

Next we look at when the derivative of∫ x

af(t) dt is equal to f(x) a.e.

Define the indefinite integral of an integrable function f by

F (x) =

∫ x

a

f(t) dt.

42

Lemma 12.4 If f is integrable, then F is continuous and of bounded vari-ation.

Proof. The continuity follows from the dominated convergence theorem.The bounded variation follows from

k∑i=1

|F (xi)− F (xi−1)| =k∑

i=1

∣∣∣∫ xi

xi−1

f(t) dt∣∣∣ ≤ k∑

i=1

∫ xi

xi−1

|f(t)| dt ≤∫ b

a

|f(t)| dt

for all partitions.

Lemma 12.5 If f is integrable and F (x) = 0 for all x, then f = 0 a.e.

Proof. For any interval,∫ d

cf =

∫ d

af −

∫ c

af = 0. By dominated conver-

gence and the fact that any open set is the countable union of disjoint openintervals,

∫O

f = 0 for any open set O.

If E is any measurable set, take On open that such that χOn decreases toχE a.e. By dominated convergence,∫

E

f =

∫fχE = lim

∫fχOn = lim

∫On

f = 0.

This with Proposition 8.1 implies f is zero a.e.

Proposition 12.6 If f is bounded and measurable, then F ′(x) = f(x) foralmost every x.

Proof. By Lemma 12.4, F is continuous and of bounded variation, and soF ′ exists a.e. Let K be a bound for |f |. If

fn(x) =F (x + 1/n)− F (x)

1/n,

then

fn(x) = n

∫ x+1/n

x

f(t) dt,

43

so |fn| is also bounded by K. Since fn → F ′ a.e., then by dominated con-vergence,∫ c

a

F ′(x) dx = lim

∫ c

a

fn(x) dx = lim n

∫ c

a

[F (x + 1/n)− F (x)] dx

= lim[n

∫ c+1/n

c

F (x) dx− n

∫ a+c

a

F (x) dx]

= F (c)− F (a) =

∫ c

a

f(x) dx,

using the fact that F is continuous. So∫ c

a[F ′(x) − f(x)] dx = 0 for all c,

which implies F ′ = f a.e. by Lemma 12.5.

Theorem 12.7 If f is integrable, then F ′ = f almost everywhere.

Proof. Without loss of generality we may assume f ≥ 0. Let fn(x) = f(x)if f(x) ≤ n and let fn(x) = n if f(x) > n. Then f − fn ≥ 0. If Gn(x) =∫ x

a[f − fn], then Gn is nondecreasing, and hence has a derivative almost

everywhere. By Proposition 12.6, we know the derivative of∫ x

afn is equal to

fn almost everywhere. Therefore

F ′(x) = G′n(x) +

[ ∫ x

a

fn

]′≥ fn(x)

a.e. Since n is arbitrary, F ′ ≥ f a.e. So∫ b

aF ′ ≥

∫ b

af = F (b) − F (a). On

the other hand, by Theorem 12.2,∫ b

aF ′(x) dx ≤ F (b) − F (a) =

∫ b

af . We

conclude that∫ b

a[F ′ − f ] = 0; since F ′ − f ≥ 0, this tells us that F ′ = f a.e.

Finally, we look at when∫ b

aF ′(y) dy = F (b)− F (a).

A function is absolutely continuous on [a, b] if given ε there exists δ suchthat

∑ki=1 |f(x′i) − f(xi)| < ε whenever xi, x

′i) is a finite collection of

nonoverlapping intervals with∑k

i=1 |x′i − xi| < δ.

It is easy to see that absolutely continuous functions are continuous andthat the Cantor-Lebesgue function is not absolutely continuous.

44

Lemma 12.8 If F (x) =∫ x

af(t) dt for f integrable on [a, b], then F is abso-

lutely continuous.

Proof. Let ε > 0. Choose a simple function s such that∫ b

a|f − s| < ε/2.

Let K be a bound for |s| and let δ = ε/2K. If (xi, x′i) is a collection of

nonoverlapping intervals, the sum of whose lengths is less than δ, then setA = ∪k

i=1(xi, x′i) and note

∫A|f − s| < ε/2 and

∫A

s < Kδ = ε/2.

Lemma 12.9 If f is absolutely continuous, then it is of bounded variation.

Proof. Let δ correspond to ε = 1 in the definition of absolute continuity.Given a partition, add points if necessary so that each subinterval has lengthat most δ. We can then group the subintervals into at most K collections,each of total length less than δ, where K is an integer larger than (1+b−a)/δ.So the total variation is then less than K.

Lemma 12.10 If f is absolutely continuous on [a, b] and f ′(x) = 0 a.e.,then f is constant.

The Cantor-Lebesgue function is an example to show that we need theabsolute continuity.

Proof. Let c ∈ [a, b], let E = x ∈ [a, c] : f ′(x) = 0, and let ε > 0. Foreach point x ∈ E there exists arbitrarily small intervals [x, x+h] ⊂ [a, c] suchthat |f(x + h) − f(x)| < εh. By Lemma 12.1 we can find a finite collectionof such intervals that cover all of E except for a set of measure less than δ,where δ is the δ in the definition of absolute continuity. If the intervals are[xi, yi] with xi < yi ≤ xi+1, then

∑|f(xi+1)− f(yi)| < ε by the definition of

absolute continuity, while∑|f(yi) − f(xi)| < ε

∑(yi − xi) ≤ ε(c − a). So

adding these two inequalities together,

|f(c)− f(a)| =∣∣∣∑[f(xi+1)− f(yi)] +

∑[f(yi)− f(xi)]

∣∣∣ ≤ ε + ε(c− a).

Since ε is arbitrary, then f(c) = f(a), which implies that f is constant.

45

Theorem 12.11 If F is absolutely continuous, then

F (b)− F (a) =

∫ b

a

F ′(y) dy.

Proof. Suppose F is absolutely continuous on [a, b]. Then F is of boundedvariation, so F = F1 − F2 where F1 and F2 are nondecreasing, and F ′ existsa.e. Since |F ′(x)| ≤ F ′

1(x) + F ′2(x), then

∫|F ′(x)| dx ≤ F1(b) + F2(b) −

F1(a) − F2(a), and hence F ′ is integrable. If G(x) =∫ x

aF ′(t) dt, then G is

absolutely continuous by Lemma 12.8, so F − G is absolutely continuous.Then (F − G)′ = 0 a.e., and therefore F − G is constant. Thus F (x) =∫ x

aF ′(t) dt + F (a). If we set x = b, we get our result.

13 Lp spaces

We assume throughout this section that the measure is σ-finite. For 1 ≤ p <∞, define the Lp norm of f by

‖f‖p =( ∫

|f(x)|pdµ)1/p

.

For p = ∞, define the L∞ norm of f by

‖f‖∞ = infM : µ(x : |f(x)| ≥ M) = 0.

For 1 ≤ p ≤ ∞ the space Lp is the set f : ‖f‖p < ∞.The L∞ norm of a function f is the supremum of f provided we disregard

sets of measure 0.

It is clear that ‖f‖p = 0 if and only if f = 0 a.e.

Proposition 13.1 (Holder’s inequality) If 1 < p, q < ∞ and p−1 + q−1 = 1,then ∫

|f(x)g(x)| dµ ≤ ‖f‖p‖g‖q.

This also holds if p = ∞ and g = 1.

46

Proof. If M = ‖f‖∞, then∫

fg ≤ M∫|g| and the case p = ∞ and q = 1

follows. So let us assume 1 < p, q < ∞. If ‖f‖p = 0, then f = 0 a.e and∫fg = 0, so the result is clear if ‖f‖p = 0 and similarly if ‖g‖q = 0. Let

F (x) = |f(x)|/‖f‖p and G(x) = |g(x)|/‖g‖q. Note ‖F‖p = 1 and ‖G‖q = 1,and it suffices to show that

∫FG ≤ 1.

The second derivative of the function ex is again ex, which is positive, andso ex is convex. Therefore if 0 ≤ λ ≤ 1, we have

eλa+(1−λ)b ≤ λea + (1− λ)eb.

If F (x), G(x) 6= 0, let a = p log F (x), b = q log G(x), λ = 1/p, and 1 − λ =1/q. We then obtain

F (x)G(x) ≤ F (x)p

p+

G(x)q

q.

Clearly this inequality also holds if F (x) = 0 or G(x) = 0. Integrating,∫FG ≤

‖F‖pp

p+‖G‖q

q

q=

1

p+

1

q= 1.

One application of Holder’s inequality is to prove Minkowski’s inequality,which is simply the triangle inequality for Lp.

We first need the following lemma:

Lemma 13.2 If a, b > 0 and 1 ≤ p < ∞, then

(a + b)p ≤ 2p−1ap + 2p−1bp.

Proof. To prove this, we may without loss of generality assume a ≤ b. Thecase a = 0 is obvious, so we assume a > 0. Dividing both sides by a andletting x = b/a, the inequality we want is equivalent to

(1 + x)p ≤ 2p−1 + 2p−1xp, x ≥ 1. (13.1)

Clearly this inequality is valid for x = 1. So to prove (13.1) it suffices toshow that the derivative of

(1 + x)p − 2p−1 − 2p−1xp

47

is less than equal to 0, or

p(1 + x)p−1 ≤ 2p−1pxp−1, x ≥ 1.

This last inequality hods because 2x ≥ 1 + x when x ≥ 1.

Proposition 13.3 (Minkowski’s inequality) If 1 ≤ p ≤ ∞, then

‖f + g‖p ≤ ‖f‖p + ‖g‖p.

Proof. Since |(f + g)(x)| ≤ |f(x)|+ |g(x)|, integrating gives the case whenp = 1. The case p = ∞ is also easy. So let us suppose 1 < p < ∞. If ‖f‖p

or ‖g‖p is infinite, the result is obvious, so we may assume both are finite.The inequality Lemma 13.2 with a = |f(x)| and b = |g(x)| yields, after anintegration,∫

|(f + g)(x)|pdµ ≤ 2p

∫|f(x)|pdµ + 2p

∫|g(x)|pdµ.

So we have ‖f + g‖p < ∞. Clearly we may assume ‖f + g‖p > 0.

Now write|f + g|p ≤ |f | |f + g|p−1 + |g| |f + g|p−1

and apply Holder’s inequality with q = (1− 1p)−1. We obtain∫

|f + g|p ≤ ‖f‖p

( ∫|f + g|(p−1)q

)1/q

+ ‖g‖p

( ∫|f + g|(p−1)q

)1/q

.

Since p−1 + q−1 = 1, then (p− 1)q = p, so we have

‖f + g‖pp ≤

(‖f‖p + ‖g‖p

)‖f + g‖p/q

p .

Dividing both sides by ‖f + g‖p/qp and using the fact that p− (p/q) = 1 gives

us our result.

Minkowski’s inequality says that Lp is a normed linear space, provided weidentify functions that are equal a.e.

We say fn converges to f in Lp if ‖fn − f‖p → 0 as n → ∞. The nextproposition compares convergence in Lp to convergence in measure. Beforewe prove this, we prove an easy preliminary result known as Chebyshev’sinequality.

48

Lemma 13.4 If 1 ≤ p < ∞,

µ(x : |f(x)| ≥ a) ≤‖f‖p

p

ap.

Proof. If A = x : |f(x)| ≥ a, then

µ(A) ≤∫

A

|f(x)|p

apdµ ≤ 1

ap

∫|f |pdµ.

Proposition 13.5 If fn converges to f in Lp, then it converges in measure.

Proof. If ε > 0, by Chebyshev’s inequality

µ(x : |fn(x)−f(x)| > ε) = µ(x : |fn(x)−f(x)|p > εp) ≤‖fn − f‖p

p

εp→ 0.

Letting fn = n2χ(0,1/n) on [0, 1] with the measure being Lebesgue measuregives an example where fn converges to 0 a.e. and in measure, but does notconverge in Lp.

Example 9.4 is an example where fn converges to 0 in Lp but not a.e.

We next show that Lp is complete. This is often phrased as saying thatLp is a Banach space, i.e., a complete normed linear space.

Proposition 13.6 If 1 ≤ p ≤ ∞, then Lp is complete.

Proof. We do only the case p < ∞; the case p = ∞ is easy. Suppose fn

is a Cauchy sequence in Lp. Given ε = 2−(j+1), there exists nj such that ifn, m ≥ nj, then ‖fn − fm‖p ≤ 2−(j+1). Without loss of generality we mayassume nj ≥ nj−1 for each j.

49

Set n0 = 0 and define f0 ≡ 0. If Aj = x : |fnj(x) − fnj−1

(x)| > 2−j/2,then from Lemma 13.4, µ(Aj) ≤ 2−jp/2. We have

µ(∩∞j=1 ∪∞m=j Am) = limj→∞

µ(∪∞m=jAm) ≤ limj→∞

∞∑m=j

µ(Am) = 0.

So except for a set of measure 0, for each x there is a last j for whichx ∈ ∪∞m=jAm, hence a last j for which x ∈ Aj. So for each x (exceptfor the null set) there is a j0 (depending on x) such that if j ≥ j0, then|fnj

(x)− fnj−1(x)| ≤ 2−j.

Set

gj(x) =

j∑m=1

|fnm(x)− fnm−1(x)|.

gj(x) increases for each x, and the limit is finite for almost every x by thepreceding paragraph. Let us call the limit g(x). We have

‖gj‖p ≤j∑

m=1

2−j + ‖fn1‖p ≤ 2 + ‖fn1‖p

by Minkowski’s inequality, and so by Fatou’s lemma, ‖g‖p ≤ 2+‖fn1‖p < ∞.We have

fnj(x) =

j∑m=1

(fnm(x)− fnm−1(x)).

Suppose x is not in the null set where g(x) is infinite. Since |fnj(x)−fnk

(x)| ≤|gnj

(x)− gnk(x)| → 0 as j, k →∞, then fnj

(x) is a Cauchy series (in R), andhence converges, say to f(x). We have ‖f − fnj

‖p = limm→∞ ‖fnm − fnj‖p;

this follows by dominated convergence with the function g defined above asthe dominating function.

We have thus shown that ‖f − fnj‖p → 0. Given ε = 2−(j+1), if m ≥ nj,

then ‖f − fm‖p ≤ ‖f − fnj‖p + ‖fm − fnj

‖p. This shows that fm convergesto f in Lp norm.

Next we show:

Proposition 13.7 The set of continuous functions with compact support isdense in Lp(Rd).

50

Proof. Suppose f ∈ Lp. By dominated convergence∫|f − fχ−[n,n]|p → 0

as n → ∞, the dominating function being |f |p. So we may suppose f hascompact support. By writing f = f+−f− we may suppose f ≥ 0. By takingsimple functions sm increasing to f , we have

∫|f − sm|p → 0 by dominated

convergence, so it suffices to consider simple functions. By linearity, it sufficesto consider characteristic functions with compact support. Given such a χE

and ε > 0 we showed in Proposition 8.3 that there exists g continuous withcompact support and with values in [0, 1] such that

∫|g − χE| < ε. Since

|g − χE| ≤ 1, then∫|g − χE|p ≤

∫|g − χE| < ε.

The following is very useful.

Proposition 13.8 For 1 < p < ∞ and p−1 + q−1 = 1,

‖f‖p = sup∫

fg : ‖g‖q ≤ 1

. (13.2)

When p = 1 (13.2) holds if we take q = ∞, and if p = ∞ (13.2) holds if wetake q = 1.

Proof. The right hand side of (13.2) is less than the left hand side byHolder’s inequality. So we need only show that the right hand side is greaterthan the left hand side.

First suppose p = 1. Take g(x) = sgn f(x), where sgn a is 1 if a > 0, is0 if a = 0, and is −1 if a < 0. Then g is bounded by 1 and fg = |f |. Thistakes care of the case p = 1.

Next suppose p = ∞. Since µ is σ-finite, there exist sets Fn increasingup to X such that µ(Fn) < ∞ for each n. If M = ‖f‖∞, let a be anyfinite real less than M . By the definition of L∞ norm, the measure of A =x ∈ Fn : |f(x)| > a must be positive if n is sufficiently large. Let g(x) =(sgn f(x))χA(x)/µ(A). Then the L1 norm of g is 1 and

∫fg =

∫A|f |/µ(A) ≥

a. Since a is arbitrary, the supremum on the right hand side must be M .

Now suppose 1 < p < ∞. We may suppose ‖f‖p > 0. Let qn be a sequenceof nonnegative simple functions increasing to f+, rn a sequence of nonneg-ative simple functions increasing to f−, and sn(x) = (qn(x) − rn(x))χFn(x).Then sn(x) → f(x) for each x, |sn(x)| ≤ |f(x)| for each x, sn is a simple func-tion, and ‖sn‖p < ∞ for each n. If f ∈ Lp, then ‖sn‖p → ‖f‖p by dominated

51

convergence. If∫|f |p = ∞, then

∫|sn|p → ∞ by monotone convergence.

For n sufficiently large, ‖sn‖p > 0.

Let

gn(x) = (sgn f(x))|sn(x)|p−1

‖sn‖p/qp

.

Since (p− 1)q = p, then

‖gn‖q =(∫|sn|(p−1)q)1/q

‖sn‖p/qp

=‖sn‖p/q

p

‖sn‖p/qp

= 1.

On the other hand, since |f | ≥ |sn|,∫fgn =

∫|f | |sn|p−1

‖sn‖p/qp

≥∫|sn|p

‖sn‖p/qp

= ‖sn‖p−(p/q)p .

Since p− (p/q) = 1, then∫

fgn ≥ ‖sn‖p, which tends to ‖f‖p.

The above proof also establishes

Corollary 13.9 For 1 < p < ∞ and p−1 + q−1 = 1,

‖f‖p = sup∫

fg : ‖g‖q ≤ 1, g simple

.

The space Lp is a normed linear space. We can thus talk about its dual,namely, the set of bounded linear functionals on Lp. The dual of a space Yis denoted Y ∗. If H is a bounded linear functional on Lp, we define the normof H to be ‖H‖ = supH(f) : ‖f‖p ≤ 1.

Theorem 13.10 If 1 < p < ∞ and p−1 + q−1 = 1, then (Lp)∗ = Lq.

What this means is that if H is a bounded linear functional on Lp, thenthere exists g ∈ Lq such that H(f) =

∫fg and that if g ∈ Lq, then H(f) =∫

fg is a bounded linear functional on Lp.

Proof. If g ∈ Lq, then setting H(f) =∫

fg for f ∈ Lp yields a bounded lin-ear functional; the boundedness follows from Holder’s inequality. Moreover,from Holder’s inequality and Proposition 13.8 we see that ‖H‖ = ‖g‖q.

52

Now suppose we are given a bounded linear functional H on Lp and wemust show there exists g ∈ Lq such that H(f) =

∫fg. First suppose µ(X) <

∞. Define ν(A) = H(χA). If A and B are disjoint, then

ν(A ∪B) = H(χA∪B) = H(χA + χB) = H(χA) + H(χB) = ν(A) + ν(B).

To show ν is countably additive, it suffices to show that if An ↑ A, thenν(An) → ν(A). But if An ↑ A, then χAn → χA in Lp, and so ν(An) =H(χAn) → H(χA) = ν(A); we use here the fact that µ(X) < ∞. Thereforeν is a countably additive signed measure. Moreover, if µ(A) = 0, thenχA = 0 a.e., hence ν(A) = H(χA) = 0. By writing ν = ν+ − ν− and usingthe Radon-Nikodym theorem for both the positive and negative parts, wesee there exists an integrable g such that ν(A) =

∫A

g for all sets A. Ifs =

∑aiχAi

is a simple function, by linearity we have

H(s) =∑

aiH(χAi) =

∑aiν(Ai) =

∑ai

∫gχAi

=

∫gs.

By Corollary 13.9,

‖g‖q = sup∫

gs : ‖s‖p ≤ 1, s simple≤ supH(s) : ‖s‖p ≤ 1 ≤ ‖H‖.

If sn are simple functions tending to f in Lp, then H(sn) → H(f), while byHolder’s inequality

∫sng →

∫fg. We thus have H(f) =

∫fg for all f ∈ Lp,

and ‖g‖p ≤ ‖H‖. By Holder’s inequality, ‖H‖ ≤ ‖g‖p.

In the case where µ is σ-finite, but not finite, let Fn ↑ X be such thatµ(Fn) < ∞ for each n. Define functionals Hn by Hn(f) = H(fχFn). Clearlyeach Hn is a bounded linear functional on Lp. Applying the above argument,we see there exist gn such that Hn(f) =

∫fgn and ‖gn‖q = ‖Hn‖ ≤ ‖H‖.

It is easy to see that gn is 0 if x /∈ Fn. Moreover, by the uniqueness part ofthe Radon-Nikodym theorem, if n > m, then gn = gm on Fm. Define g bysetting g(x) = gn(x) if x ∈ Fn. Then g is well defined. By Fatou’s lemma, gis in Lq with a norm bounded by ‖H‖. Since fχFn → f in Lp by dominatedconvergence, then Hn(f) = H(fχFn) → H(f), since H is a bounded linearfunctional on Lp. On the other hand Hn(f) =

∫Fn

fgn =∫

Fnfg →

∫fg

by dominated convergence. So H(f) =∫

fg. Again by Holder’s inequality‖H‖ ≤ ‖g‖p.

53

14 Fourier transforms

Fourier transforms give a representation of a function in terms of frequencies.We give the basic properties here.

If f ∈ L1(Rn), define the Fourier transform f by

f(u) =

∫Rn

eiu·xf(x)dx, u ∈ Rn. (14.1)

We are using u · x for the standard inner product in Rn. Various books haveslightly different definitions. Some put a negative sign before the iu ·x, somehave a 2π either in front of the integral or in the exponent. The basic theoryis the same in any case.

Some basic properties of the Fourier transform are given by

Proposition 14.1 Suppose f and g are in L1. Then

(a) f is bounded and continuous;

(b) (f + g)(u) = f(u) + g(u); (af)(u) = af(u);

(c) if fa(x) = f(x + a), then fa(u) = e−iu·af(u);

(d) if ga(x) = eia·xg(x), then ga(u) = f(u + a);

(e) if ha(x) = f(ax), then ha(u) = a−nf(u/a).

Proof. (a) f is bounded because f ∈ L1 and |eiu·x| = 1. We have

f(u + h)− f(u) =

∫ (ei(u+h)·x − eiu·x

)f(x)dx.

So

|f(u + h)− f(u)| ≤∫ ∣∣∣eiu·x

∣∣∣ ∣∣∣eih·x − 1∣∣∣ |f(x)|dx.

The integrand is bounded by 2|f(x)|, which is integrable, and eih·x − 1 → 0as h → 0, and thus the continuity follows by dominated convergence.

54

(b) is obvious. (c) follows because

fa(u) =

∫eiu·xf(x + a)dx =

∫eiu·(x−a)f(x)dx = e−iu·af(u)

by a change of variables. For (d),

ga(u) =

∫eiu·xeia·xf(x)dx =

∫ei(u+a)·xf(x)dx = f(u + a).

Finally for (e), by a change of variables,

ha(u) =

∫eiu·xf(ax)dx = a−n

∫eiu·(y/a)f(y)dy

= a−n

∫ei(u/a)·yf(y)dy = a−nf(u/a).

One reason for the usefulness of Fourier transforms is that they relatederivatives and multiplication.

Proposition 14.2 Suppose f ∈ L1 and xjf(x) ∈ L1, where xj is the jth

coordinate of x. Then

∂f

∂uj

(u) = i

∫eiu·xxjf(x)dx.

Proof. Let ej be the unit vector in the jth direction. Then

f(u + hej)− f(u)

h=

1

h

∫ (ei(u+hej)·x − eiu·x

)f(x)dx

=

∫eiu·x

(eihxj − 1

h

)f(x)dx.

Since ∣∣∣1h

(eihxj − 1

)∣∣∣ ≤ |xj|

and xjf(x) ∈ L1, the right hand side converges to∫

eiu·xixjf(x)dx by dom-inated convergence. Therefore the left hand side converges. Of course, thelimit is ∂f/∂uj.

55

The convolution of f and g is defined by

f ∗ g(x) =

∫f(x− y)g(y)dy.

By a change of variables, this is the same as∫

f(y)g(x−y)dy, so f ∗g = g∗f .

Proposition 14.3 (a) If f, g ∈ L1, then f ∗ g is in L1 and ‖f ∗ g‖1 ≤‖f‖1‖g‖1.

(b) The Fourier transform of f ∗ g is f(u)g(u).

Proof. (a) We will show f ∗ g is finite a.e. by showing∫|f ∗ g(x)| dx < ∞.

We have ∫|f ∗ g(x)|dx ≤

∫ ∫|f(x− y)| |g(y)|dy dx.

Since the integrand is nonnegative, we can apply Fubini and the right handside is equal to∫ ∫

|f(x− y)|dx |g(y)|dy =

∫ ∫|f(x)|dx |g(y)|dy = ‖f‖1‖g‖1.

The first equality here follows by a change of variables. To verify that we cando a change of variables, we reduce to simple functions and then characteristicfunctions, and then use the translation invariance of Lebesgue measure.

(b) We have

f ∗ g(u) =

∫eiu·x

∫f(x− y)g(y)dy dx

=

∫ ∫eiu·(x−y)f(x− y)dx eiu·yg(y)dy

=

∫f(u)eiu·yg(y)dy = f(u)g(u).

We applied Fubini in the first equality; this is valid because as we saw in (a),the absolute value of the integrand is integrable.

We want to give a formula for recovering f from f . First we need tocalculate the Fourier transform of a particular function.

56

Proposition 14.4 (a) Suppose f1 : R → R is defined by

f1(x) =1√2π

e−x2/2.

Then f1(u) = e−u2/2.

(b) Suppose fn : Rn → R is given by

fn(x) =1

(2π)n/2e−|x|

2/2.

Then fn(u) = e−|u|2/2.

Proof. (a) may be proved using contour integration, but let’s give a realvariable proof. Let g(u) =

∫eiuxe−x2/2dx. Differentiate with respect to

u. We may differentiate under the integral sign because (ei(u+h)x − eiux)/his bounded in absolute value by |x| and |x|e−x2/2 is integrable; thereforedominated convergence applies. We then obtain

g′(u) = i

∫eiuxxe−x2/2dx.

By integration by parts this is equal to

−u

∫eiuxe−x2/2dx = −ug(u).

Solving the differential equation g′(u) = −ug(u), we have

[log g(u)]′ =g′(u)

g(u)= −u,

so log g(u) = −u2/2 + c1, and so then

g(u) = c2e−u2

. (14.2)

Since g(0) =∫

e−x2/2dx =√

2π, c2 =√

2π. Substituting this value of c2 in

(14.2) and dividing both sides by√

2π proves (a).

57

For (b), since fn(x) = f1(x1) · · · f1(xn) if x = (x1, . . . , xn),

fn(u) =

∫· · ·

∫ei

Pj ujxjf1(x1) · · · f1(xn)dx1 · · · dxn

= f1(u1) · · · f1(un) = e−|u|2/2.

One more preliminary before proving the inversion theorem.

Proposition 14.5 Suppose ϕ is in L1 and∫

ϕ(x)dx = 1. Let ϕA(x) =A−nϕ(x/A).

(a) Then ‖f ∗ ϕA − f‖1 → 0 as A → 0.

(b) If f is continuous with compact support, then f ∗ ϕA converges to fpointwise.

Proof. (a) Let ε > 0. Choose g continuous with compact support so that‖f−g‖1 < ε. Let h = f−g. A change of variables shows that ‖ϕA‖1 = ‖ϕ‖1.Observe

‖f ∗ ϕA − f‖1 ≤ ‖g ∗ ϕA − g‖1 + ‖h ∗ ϕA − h‖1

and

‖h ∗ ϕA − h‖1 ≤ ‖h‖1 + ‖h ∗ ϕA‖1 ≤ ‖h‖1 + ‖h‖1‖ϕA‖1 < ε(1 + ‖ϕ‖1).

So since ε is arbitrary, it suffices to show that g ∗ ϕA → g in L1.

We start by writing

g ∗ ϕA(x)− g(x) =

∫g(x− y)ϕA(y)dy − g(x) =

∫g(x− Ay)ϕ(y)dy − g(x)

=

∫[g(x− Ay)− g(x)]ϕ(y)dy.

We used a change of variables and the fact that∫

ϕ(y)dy = 1. Because gis continuous with compact support, then g is bounded, and the integral onthe right goes to 0 by dominated convergence, the dominating function being‖g‖∞|ϕ(y)|. Therefore g ∗ ϕA(x) converges to g(x) pointwise.

58

To show the convergence in L1, we have∫|g ∗ ϕA(x)− g(x)|dx ≤

∫ ∫|g(x− Ay)− g(x)| |ϕ(y)|dy dx

=

∫ ∫|g(x− Ay)− g(x)| |ϕ(y)|dx dy.

Since g is continuous with compact support and hence bounded, for each y

GA(y) =

∫|g(x− Ay)− g(x)|dx

converges to 0 as A → 0 by dominated convergence. Also

GA(y) ≤∫|g(x− Ay)|dx +

∫|g(x)|dx ≤ 2‖g‖1 < ∞.

Then ∫GA(y)|ϕ(y)|dy

converges to 0 as A → 0 by dominated convergence, the dominating functionbeing 2‖g‖1|ϕ(y)|.

(b) This follows from the argument we used for g above.

Now we are ready to give the inversion formula. The proof seems longerthan it might be, but there is no avoiding the introduction of the functionHa or some similar function.

Theorem 14.6 Suppose f, f ∈ L1. Then

f(y) =1

(2π)n

∫e−iu·yf(u)du, a.e.

Proof. If g(x) = a−nf(x/a), then its Fourier transform is f(au). So theFourier transform of

1

an

1

(2π)n/2e−x2/2a2

is e−a2u2/2. Therefore if we let

Ha(x) =1

(2π)ne−|x|

2/2a2

,

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we haveHa(u) = (2π)−n/2ane−a2|u|2/2.

We have ∫f(u)e−iu·yHa(u)du

=

∫ ∫eiu·xf(x)e−iu·yHa(u)dx du

=

∫ ∫eiu·(x−y)Ha(u)du f(x) dx

=

∫Ha(x− y)f(x)dx. (14.3)

We can interchange the order of integration because∫ ∫|f(x)| |Ha(u)|dx du < ∞.

The left hand side of the first line of (14.3) converges to (2π)−n∫

f(u)e−iu·ydy

as a →∞ by dominated convergence and the fact that f ∈ L1. The last lineof (14.3) is equal to ∫

Ha(y − x)f(x)dx = f ∗ Ha(y) (14.4)

since Ha is symmetric. But by Proposition 14.5, f ∗ Ha converges to f in L1

as a →∞.

The last topic that we consider is the Plancherel theorem.

Theorem 14.7 (a) Suppose f is continuous with compact support. Then

f ∈ L2 and‖f‖2 = (2π)−n/2‖f‖2. (14.5)

(b) We can use the result in (a) to define f when f ∈ L2 and so that(14.5) holds.

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Proof. (a) Let g(x) = f(−x). Note

g(u) =

∫eiu·xf(−x)dx =

∫e−iu·xf(−x)dx =

∫eiu·xf(x)dx = f(u).

By (14.3) and (14.4) with y = 0

f ∗ g ∗ Ha(0) =

∫f ∗ g(u)Ha(u)du. (14.6)

Since f ∗ g(u) = f(u)g(u) = |f(u)|2, the right hand side of (14.6) converges

by monotone convergence to (2π)−n∫|f(u)|2du as a → ∞. Since f and g

are continuous with compact support, then it is easy to see that f ∗ g is also,and so the left hand side of (14.6) converges to f ∗ g(0) =

∫f(y)g(−y)dy =∫

|f(y)|2dy by Proposition 14.5(b).

(b) The set of continuous functions with compact support is dense in L2.Given a function f in L2, choose a sequence of continuous functions withcompact support fm such that fm → f in L2. By the result in (a), fm isa Cauchy sequence in L2, and therefore converges to a function in L2, whichwe call f . If f ′m is another sequence of continuous functions with compactsupport converging to f in L2, then fm − f ′m is a sequence of continuousfunctions with compact support converging to 0 in L2; by the result in (a),

fm− f ′m converges to 0 in L2, and therefore f is defined uniquely up to almosteverywhere equivalence. By passing to the limit in L2 on both sides of (14.5),we see that (14.5) holds for f ∈ L2.

Richard F. BassDepartment of MathematicsUniversity of ConnecticutStorrs, CT 06269-3009, USAbass@math.uconn.edu

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