Measure Theory & Integrationrvhassel/Onderwijs/Old-Onderwijs/2DE08-1011/... · A measure on a...

76
Measure Theory & Integration Lecture Notes, Math 320/520 Fall, 2004 D.H. Sattinger Department of Mathematics Yale University

Transcript of Measure Theory & Integrationrvhassel/Onderwijs/Old-Onderwijs/2DE08-1011/... · A measure on a...

Page 1: Measure Theory & Integrationrvhassel/Onderwijs/Old-Onderwijs/2DE08-1011/... · A measure on a topological space for which the measurable sets is the Borel algebra BX is called a Borel

Measure Theory

& Integration

Lecture Notes, Math 320/520

Fall, 2004

D.H. Sattinger

Department of Mathematics

Yale University

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Contents

1 Preliminaries 1

2 Measures 32.1 Area and Measure . . . . . . . . . . . . . . . . . . . . . . . . 32.2 Caratheodory’s Theorem . . . . . . . . . . . . . . . . . . . . . 72.3 Borel measures . . . . . . . . . . . . . . . . . . . . . . . . . . 10

3 Integration 133.1 Measurable functions . . . . . . . . . . . . . . . . . . . . . . . 133.2 Integration of non-negative functions . . . . . . . . . . . . . . 153.3 The Dominated Convergence Theorem . . . . . . . . . . . . . 203.4 The Riemann Integral . . . . . . . . . . . . . . . . . . . . . . 233.5 Some Convergence Theorems . . . . . . . . . . . . . . . . . . 25

4 Product Spaces 294.1 Fubini’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 294.2 Lebesgue measure in Rn . . . . . . . . . . . . . . . . . . . . . 32

5 Differentiation Theory 375.1 Differentiation Theory of Functions . . . . . . . . . . . . . . . 375.2 Signed Measures . . . . . . . . . . . . . . . . . . . . . . . . . 415.3 The Lebesgue-Radon-Nikodym theorem . . . . . . . . . . . . 445.4 Differentiation on Rn . . . . . . . . . . . . . . . . . . . . . . . 49

6 Lp spaces 556.1 Banach spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . 556.2 Duality . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 586.3 Distribution functions . . . . . . . . . . . . . . . . . . . . . . 606.4 Linear Transformations . . . . . . . . . . . . . . . . . . . . . 62

i

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ii CONTENTS

7 Exercises 677.1 Set Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677.2 Measures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 677.3 Integration . . . . . . . . . . . . . . . . . . . . . . . . . . . . 687.4 Differentiation theory . . . . . . . . . . . . . . . . . . . . . . 697.5 Lp Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 70

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List of Figures

3.1 Fundamental solution of the heat equation . . . . . . . . . . . 21

iii

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iv LIST OF FIGURES

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Chapter 1

Preliminaries

Let an be a sequence of real numbers. The limit superior and limit inferiorof the sequence are defined as

lim supn

an = infk≥1

supn≥k

an, lim infn

an = supk≥1

infn≥k

an.

The following is left as an exercise:

lim infn

an ≤ lim supn

an. (1.1)

Let En be a countable collection of sets. We define

E = lim supn

En =∞⋂

k=1

∞⋃

n=k

En = x : x ∈ Enfor infinitely many n.

E = lim infn

En =∞⋃

k=1

∞⋂

n=k

En = x : x ∈ Enfor all but finitely many n.

It is a simple exercise to show that E = lim infn En ⊂ lim supn En = E. IfE = E we say the limit of the sequence of sets exists and is equal to thiscommon set.

A partial ordering of a set X is a relation ”≤” which satisfies the followingaxioms: Given x, y, z ∈ X,

• x ≤ y, y ≤ z implies x ≤ z.

• x ≤ y, y ≤ x implies x = y.

• x ≤ x.

1

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2 CHAPTER 1. PRELIMINARIES

If, in addition,

• for all x, y ∈ X either x ≤ y, or y ≤ x

then the relation ≤ is called a linear or total ordering. For example, the realline is linearly ordered; the space of continuous real-valued functions on atopological space X, denoted by C(X), is partially ordered by the relation:f ≤ g if g(x)− f(x) ≥ 0 for all x ∈ X; the collection of all subsets of a setX, called the power set of X and denoted by P(X) is partially ordered bythe inclusion relation.

A maximal element of a partially ordered set X is an element M suchthat if x ∈ X and x ≥ M , then x = M . A minimal element is definedsimilarly. A maximal or minimal element may or may not exist, and ingeneral is not unique.

The Hausdorff Maximal Principle states that very partially ordered sethas a maximal linearly ordered subset. Zorn’s Lemma states that if Xis a partially ordered set for which every linearly ordered subset has anupper bound, then X has a maximal element. The Cartesian product of acollection of sets Xαα∈A, denoted by

∏α Xα is defined to be the set of all

maps f : A 7→ ∪αXα such that f(α) ∈ Xα. The Axiom of Choice statesthat if Xα is a non-empty collection of non-empty sets, then

∏α Xα is

non-empty. That is, it is possible to construct a set with one element fromeach of the Xα.

The Hausdorff Maximal principle, Zorn’s Lemma and the Axiom ofChoice are all equivalent.

A well-known introduction to set theory is Paul Halmos’ Naive Set The-ory [3]

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Chapter 2

Measures

2.1 Area and Measure

The modern notion of measure, developed in the late Nineteenth and earlyTwentieth centuries, is an extension of the notion of area, developed by theGreeks. The area of a planar rectangle is defined to be the product of itslength and width. By elementary geometric arguments, the area of a trian-gle, and hence the area of any polygon, can be determined. To determinethe area of a circle, the Greeks constructed sequences of inscribed and cir-cumscribed regular polygons, with the number of sides tending to infinity, toobtain inner and outer approximations to the circle. This gives a sequenceof lower and upper estimates of the area of the circle, and the area of thecircle is defined to be the common limit as the number of sides tends toinfinity. This procedure, known as the Method of Exhaustion (You couldGoogle it), was formulated by Euxodus of Cnidos (408-355 B.C.E.), andwas developed systematically by Archimedes (287-212 B.C.E.). Archimedesdeveloped systematic algorithms for calculating not only the areas, but thevolumes of a wide variety of geometric figures, and laid the foundations ofintegration theory that are still relevant today. In fact, the Riemann integralis really an application of the method of exhaustion and the principles ofArchimedes.

What properties do we want of a notion of ”Area”? Suppose A denotesthe area of a set E. The most obvious properties to require of the setfunction A is that it be non-negative and ”additive”. That is, A(E ∪ F ) =A(E)+A(F ), provided that E∩F = ∅. In addition, in the case of Euclideangeometry, we require A to be invariant under the group of Euclidean motions,that is, translations, rotations, and reflections; but this is not an essential

3

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4 CHAPTER 2. MEASURES

property of a measure.In the theory of measures, the assumption of additivity is replaced by a

slightly stronger assumption, that of countable additivity:

µ(∪∞j=1Ej

)=

∞∑

j=1

µ(Ej), if Ej ∩ Ek = ∅, j 6= k.

The sequence of sets Ej is said to disjoint.

Definition 2.1.1 Let X be a set and let M ⊂ P(X) be a collection ofsubsets of X. A set function µ defined on M is said to be a measure if µ isnon-negative and countably additive.

We point out immediately that the assumption of countable additivity,together with the axiom of choice, leads in general to the existence of non-measurable sets – that is, that µ is not defined everywhere on P(X). Wedemonstrate the existence of non-measurable sets for any measure µ on theunit interval [0, 1] which is invariant under translations (this happens to beLebesgue measure, as we shall see).

Define an equivalence relation on [0, 1] by x ∼ y if x−y is rational. Thispartitions the unit interval into equivalence classes. By the Axiom of Choice,there is a set N consisting of one element from each equivalence class. Foreach rational number r, let Nr = [N+r] mod1, that is, the translate of N by rwith wrapped around around 1. (Or, do the construction on the unit circle).Now clearly [0, 1] = ∪rNr and the sets Nr are disjoint. Hence by countableadditivity, and translation invariance, µ([0, 1]) =

∑r µ(Nr) =

∑r µ(N).

This sum must be infinity or zero, depending on whether µ(N) is positiveor zero. Thus µ cannot be defined on N .

We now turn our attention to the collection M, which are called mea-surable sets. In view of our requirement of countable additivity of µ it isnatural to require that M be closed under countable disjoint unions. Inaddition, it is natural to require that the complement of any measurable setalso be measurable. Then if A and B are measurable, A ∩ B = (Ac ∪ Bc)c

is measurable.

Definition 2.1.2 A σ-algebra M is a collection of subsets of which is closedunder countable unions and complementations.

The verification of the following proposition is left to the reader.

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2.1. AREA AND MEASURE 5

Proposition 2.1.3 Any countable union of sets may be replaced by a count-able disjoint union of sets. The intersection of any family of σ-algebras isagain a σ-algebra. If E ⊂ P(X) there is a smallest σ-algebra containing E.

The smallest σ-algebra containing a collection of sets E is called theσ-algebra generated by E . For example, if X is a topological space, theopen sets on X generate a σ-algebra, called the Borel σ-algebra, which isdenoted by BX . The Borel algebra of sets on R is generated by the openintervals, closed intervals, half-open intervals, or by rays (a,∞), (−∞, a), orby half-open rays, such as (−∞, a].

A measure space is denoted by X,M, µ), where X is the space of points,M is the σ algebra of measurable sets, and µ is the measure, defined on M.A measure on a topological space for which the measurable sets is the Borelalgebra BX is called a Borel measure. Borel measures play a pre-eminentrole in measure theory on Rn.

Borel measures on the line are constructed by specifying the measure ofeach open or half-open interval on the line. If F (x) is a monotone increasing,right continuous function on the line, then the measure of the half-openinterval (a, b] is defined by µ0((a, b]) = F (b)− F (a). The extension of µ0 toBR will be developed below.

A number of important examples arise in this manner. For example,

• Lebesgue measure on the real line, which extends the notion of length,is obtained by taking F (x) = x;

• In probability theory, the distribution function of a random variableX, defined by F (x) = Pr(X ≤ x), generates a probability measure,and Pr(a < X ≤ b) = F (b)− F (a) = µF ((a, b]).

• The Dirac measure is obtained by taking F equal to the Heavisidestep function: H(x) = 0 for x < 0 and H(x) = 1 for x ≥ 1. The Diracmeasure assigns the measure 1 to any set containing the origin, and 0to all other sets.

• The Cantor function F generates a measure µF which assigns themeasure 1 to the Cantor set, which has Lebesgue measure 0; and 0 thethe union of open middle thirds, which has total length 1.

Theorem 2.1.4 Let (X,M, µ) be a measure space. Then µ is

• monotone: If E ⊂ F and both measurable, then µ(E) ≤ µ(F ).

• countably subadditive: For Ej ∈M, µ(∪jEj) ≤∑

j µ(Ej);

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6 CHAPTER 2. MEASURES

• continuous from below: If Ej ⊂ Ej+1 ⊂ is an increasing sequence ofmeasurable sets, then

µ(∪∞j=1Ej

)= lim

j→∞µ(Ej);

• µ is continuous from above: If · · · ⊃ Ej ⊃ Ej+1 ⊃ and µ(E1) < ∞,then

µ(∩∞j=1Ej

)= lim

j→∞µ(Ej).

Proof: If E ⊂ F then F = E ∪ (F\E) so µ(F ) = µ(E) + µ(F\E) ≥ µ(E).If Ej is a sequence of sets with non-empty intersections, then for all

n,

µ(∪n

j=1Ej

)= µ

(∪nj=1Fj

)=

n∑

j=1

µ(Fj) ≤n∑

j=1

µ(Ej),

whereF1 = E1, Fk = Ek\

(∪k−1

j=1Ek

).

Then the Fj are disjoint and Fj ⊂ Ej ; letting n →∞ we obtain the secondstatement in the theorem.

To prove continuity from below, let E0 = ∅ and note that

µ(∪∞j=1Ej

)=

∞∑

j=1

µ(Ej\Ej−1) = limn→∞

n∑

j=1

µ(Ej\Ej−1) = limn→∞µ(En).

Finally, to prove continuity from above, put Fj = E1\Ej . Then thesequence Fj is increasing, and

E1 =( ∪∞j=1 Fj

) ∪ ( ∩∞j=1 Ej

).

Therefore by the previous result,

µ(E1) =µ(∪∞j=1Fj

)+ µ

(∩∞j=1Ej

)

= limj→∞

µ(Fj) + µ(∩∞j=1Ej

)

= limj→∞

µ(E1)− µ(Ej) + µ(∩∞j=1Ej

).

Since µ(E1) < ∞ it follows that

µ(∩∞j=1Ej

)= lim

j→∞µ(Ej).

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2.2. CARATHEODORY’S THEOREM 7

A null set of a measure space (X,M, µ) is a set N for which µ(N) = 0. Bysubadditivity, countable unions of null sets are null sets. If P is a propositionwhich is true except on a set of measure zero, we say that P holds almosteverywhere (a.e.) If the σ-algebra M includes all null sets, we say that themeasure µ is complete. Any measure can be extended to be complete.

2.2 Caratheodory’s Theorem

In this section we describe a general constructive procedure for obtainingcomplete measures. Recall that the area enclosed by a circle is obtained asthe limit of the areas of a sequence of circumscribed polygons. In fact, thearea of a disk is by definition the infimum of the areas of all circumscribedpolygons. This leads to the general notion of outer measure, as follows.

Definition 2.2.1 An outer measure on a non-empty set X is a set func-tion µ∗ defined on P(X) which is non-negative, monotone, and countablysubadditive.

In practice, an outer measure is obtained by starting with an elementaryconcept of measure, called a pre-measure, which is defined on a restrictedclass of sets E ⊂ P(X) and approximating general subsets X from theoutside. We assume that E is an algebra, that is, that it is closed undercomplementation and finite unions. For example, Lebesgue measure in theplane is constructed beginning with the notion of area of a rectangle, ex-tending it to polygons, and defining the ”outer area” of a plane figure bytaking the infimum of the areas of all circumscribed polygons.

Proposition 2.2.2 Let E ⊂ P(X) be an algebra of sets and ρ a non-negative function defined on E, such that ρ(∅) = 0. For any A ⊂ X define

µ∗(A) = inf

∞∑

j=1

ρ(Ej) : A ⊂ ∪∞j=1Ej , Ej ∈ E .

Then µ∗ is an outer measure.

Proof: By its definition, µ∗ vanishes on the empty set. If A ⊂ B then anycovering of B is also a covering of A; hence the infimum for A is taken over alarger collection of coverings, and therefore smaller. Hence µ∗ is monotone.

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8 CHAPTER 2. MEASURES

It remains to demonstrate countable additivity. Let A = ∪jAj ⊂ X, andlet ε > 0. For each j there is a covering of Aj by sets Ejk ∈ E such that

∞∑

k=1

ρ(Ejk) < µ∗(Aj) + 2−jε;

hence

µ∗(A) <

∞∑

j,k=1

ρ(Ejk) <

∞∑

j=1

µ∗(Aj) + ε.

Since ε is arbitrary, µ∗(A) ≤ ∑j µ∗(Aj), and the countable subadditivity of

µ∗ is established. k. .

A measure is obtained by restricting the outer measure to a subclass ofsets, called measurable sets, on which it is countably additive.

Definition 2.2.3 Let µ∗ be an outer measure on a set X. A subset A ⊂ Xis said to be µ∗-measurable if

µ∗(E) = µ∗(E ∩A) + µ∗(E ∩Ac), E ⊂ X.

By subadditivity µ∗(E) ≤ µ∗(E ∩A) + µ∗(E ∩Ac) for all A and E; hence Ais measurable iff

µ∗(E) ≥ µ∗(E ∩A) + µ∗(E ∩Ac), E ⊂ X.

This definition of measurability which is due to the Twentieth centuryGreek mathematician Caratheodory [1].

Theorem 2.2.4 [Caratheodory] If µ∗ is an outer measure on X, then thecollection M of measurable sets is a σ-algebra, and the restriction of µ∗ toM is a complete measure.

Proof: First note that M is closed under complementation, since the defi-nition of measurability is symmetric in A and Ac.

If A,B ∈M and E ⊂ X, then

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

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

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2.2. CARATHEODORY’S THEOREM 9

Since A ∪B = (A ∩B) ∪ (A ∩Bc) ∪ (Ac ∩B) we have, by subadditivity,

µ∗(E ∩A ∩B) + µ∗(E ∩A ∩Bc) + µ∗(E ∩Ac ∩B) ≥ µ∗(E ∩ (A ∪B)).

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

and A ∪B ∈M. This shows that M is an algebra. If A and B are disjointsets in M then

µ∗(A ∪B) = µ∗((A ∪B) ∩A) + µ∗((A ∪B) ∩Ac) = µ∗(A) + µ∗(B),

so µ∗ is finitely additive on M.Now suppose that Aj is a sequence of disjoint measurable sets, and let

Bn = ∪n1Aj , with B = ∪∞1 Aj . Then for any E ⊂ X

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

=µ∗(E ∩An) + µ∗(E ∩Bn−1) = . . .

=n∑

j=1

µ∗(E ∩Aj).

Therefore

µ∗(E) = µ∗(E ∩Bn) + µ∗(E ∩Bcn) ≥

n∑

j=1

µ∗(E ∩Aj) + µ∗(E ∩Bc).

Letting n →∞ and using the countable subadditivity of µ∗, we obtain

µ∗(E) ≥∞∑

j=1

µ∗(E ∩Aj) + µ∗(E ∩Bc) ≥ µ∗(E ∩B) + µ∗(E ∩Bc ≥ µ∗(E).

Thus B ∈ M; and, taking E = B we obtain also that µ∗ is countablyadditive on M.

Finally, if µ∗(A) = 0, then

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

hence A is measurable, and µ∗ restricted to M is a complete measure. Thiscompletes the proof of Theorem 2.2.4.

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10 CHAPTER 2. MEASURES

The Caratheodory extension theorem is a primary method of construct-ing measures. Typically, the set function ρ of Proposition 2.2.2 is taken tobe a pre-measure, a non-negative set function that vanishes on the emptyset, and is countably additive on the algebra E in the sense that

ρ(∪jAj) =∑

j

ρ(Aj),

whenever ∪jAj ∈ E .For example, Lebesgue measure on the line is an extension of the pre-

measure defined by ρ((a, b]) = b−a; E is the algebra of finite disjoint unionsof half-open intervals (a, b].

Theorem 2.2.5 Let µ be the Caratheodory extension of a pre-measure ρdefined on an algebra E. Then every set in E is measurable, and µ(A) = ρ(A)for all A ∈ E.If ρ is σ-finite then the extension is unique.

In the case of Lebesgue measure on Rn, the algebra E is the collectionof finite unions of multi-intervals,

R = (x1, . . . , xn) : aj ≤ x ≤ bj, ρ(R) =n∏

j=1

(bj − aj).

2.3 Borel measures

Let X be a metric space and ρ its metric. The distance between two sets Aand B in X is defined to be

ρ(A,B) = infρ(x, y) : x ∈ A, y ∈ B.A metric outer measure µ∗ on a metric space X is one which satisfies theadditional postulate that

µ∗(A ∪B) = µ∗(A) + µ∗(B), whenever ρ(A,B) > 0.

The importance of this class of measures is that The Borel sets of X arethen µ∗-measurable [1, 2, 4]

An important class of Borel measures on the real line are the Lebesgue-Stieltjes measures. These are extensions of the pre-measure ρF ((a, b]) =F (b) − F (a), where F is a non-decreasing, right continuous function onthe real line, or on an interval. In the following, E denotes the algebra ofsets which can be expressed as finite disjoint unions of half open intervals(aj , bj ], with aj < bj .

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2.3. BOREL MEASURES 11

Lemma 2.3.1 The set function ρF is a pre-measure on E.The proof of this lemma is somewhat more technical then one might expect[2]. The difficulty comes in that any half-open interval can be decomposedin a non-unique fashion:

(a, b] =n⋃

j=1

(cj−1, cj ], c0 = a, cn = b.

Theorem 2.3.2 Let µ be a Lebesgue-Stieltjes measure and let Mµ be itsdomain of measurable sets. Then for any E ∈Mµ we have

µ(E) =

infµ(U) : E ⊂ U, U open

supµ(K) : K ⊂ E, K compact.

Lebesgue measure m on the line is the extension of the pre-measure ρF ,where F (x) = x. The Borel algebra is a sub-class of Lebesgue measurablesets. In the theory of Borel sets, Fσ denotes the class of sets which arecountable unions of closed sets; while Gδ denotes the class of countableintersections of open sets. Any Lebesgue measurable set E can be writtenin the form

E =

V \N1 V ∈ Gδ, µ(N1) = 0;

H ∪N2 H ∈ Fσ, µ(N2) = 0.

A second Lebesgue-Stieltjes measure of interest is that generated by theCantor function, c(x). This function is defined to be 1/2 on the open interval(1/3, 2/3); 1/4 on (1/9, 2/9); 3/4 on (7/9, 8/9), etc.; and then extendingthis to the entire unit interval [0, 1] by continuity. The Cantor function iscontinuous and non-decreasing. Let µc be the Lebesgue-Stieltjes measuregenerated by c. The Cantor set C ⊂ [0, 1] is obtained by removing the openmiddle third (1/3,2/3) of the unit interval, then removing the open middlethirds of the two remaining intervals, and so forth, successively removing theopen middle thirds of the remaining subintervals at each stage. The Cantormeasure of C is one, while the Lebesgue measure of the complement Cc isone. The Cantor measure and Lebesgue measure are examples of mutuallysingular measures: they have their support on complementary subsets of[0, 1].

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12 CHAPTER 2. MEASURES

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Chapter 3

Integration

3.1 Measurable functions

A function f : Rn → R is Lebesgue measurable if f−1(U) is a Lebesguemeasurable set for every open set U . More generally, given any measure µ wesay a real valued function f is measurable with respect to µ if f−1(U) ∈Mwhenever U is open,M being the σ-algebra of µ-measurable sets. In general,we shall simply say f is measurable without reference to the measure µ. Iff is complex-valued, then f is measurable iff its real and imaginary partsare measurable.

If f : X 7→ Y then f−1 is defined by

f−1 : P(Y ) 7→ P(X), f−1(E) = x : x ∈ X, f(x) ∈ E ⊂ Y .It is a simple exercise to show that the inverse map preserves complements,unions, and intersections, that is

f−1(E ∪ F ) = f−1(E) ∪ f−1(F )

f−1(E ∩ F ) = f−1(E) ∩ f−1(F ), f−1(Ec) = f−1(E)c.

It follows that a real-valued function is measurable if f−1(U) is measurablewhenever U is an open or closed interval, or an open ray (a,∞], etc.

A mapping f : X 7→ Y , where X and Y are metric spaces, is Borelmeasurable if f−1(U) is a Borel subset of X for every open set U ⊂ Y . Bythe observations above,

Proposition 3.1.1 Continuous functions are Borel measurable. f is Borelmeasurable iff f−1(B) ∈ BX whenever B ∈ BY . The composition of twoBorel measurable functions on R to itself is Borel measurable.

13

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14 CHAPTER 3. INTEGRATION

On the other hand, the inverse image of a Lebesgue measurable set isnot necessarily Lebesgue measurable, and so Lebesgue measurability is notclosed under composition, [2].

Given two functions f and g we define

f ∨ g = maxf, g, f ∧ g = minf, g;

f+ = f ∨ 0, f− = (−f) ∨ 0.

Proposition 3.1.2 If f and g are measurable, then so are f + g, fg, f ∨ g,f ∧ g, f±, and |f |. The sums, products, and upper and lower envelopes ofany finite number of measurable functions is measurable.

Proof: For any real a we have

x : f + g > a = ∪t∈Qx : f(x) > t ∩ x : g(x) > a− t;

hence f + g is measurable. It is immediate that measurability is invariantunder scalar multiplication; hence f − g is also measurable.

Consider the function f2. If a < 0 then x : f2 > a = X; while fora ≥ 0,

x : f2 > a = x : f(x) >√

a ∪ x : f(x) < −√a.Hence f2 is measurable. The identity

fg =(f + g)2 − f2 − g2

2

then shows that fg is measurable.We have

f+ > a =

X a < 0;

x : f(x) > a a ≥ 0;

hence f+ is measurable. The same argument shows f− is measurable; andtherefore also |f | = f+ + f− is measurable. The identities

f ∨ g =f + g + |f − g|

2, f ∧ g =

f + g − |f − g|2

then show that f ∨ g and f ∧ g are measurable. The result extends to finitecombinations of measurable functions by induction. k. .

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3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 15

Theorem 3.1.3 If fj is a sequence of measurable functions, then thelimits lim supj fj, lim infj fj, as well as limf fj, when it exists, are all mea-surable.

Proof: Let gk(x) = supj≥k fj(x). Then

x : gk(x) > a = ∪j≥kx : fj(x) > a;

hence gk is a sequence of measurable functions. Putting f(x) = infk gk(x),we see that

x : f(x) > a =∞⋂

k=1

x : gk(x) > a,

hence f(x) = lim supj fj(x) is measurable. A similar proof shows thatlim infj fj is measurable. The limit limj fj(x) exists for all x for whichthe limits inferior and superior of the sequence fj(x) coincide; hence thelimj fj is measurable. k. .

3.2 Integration of non-negative functions

A characteristic or indicator function is one which takes on only the values0 and 1; it is generally denoted by χE(x) where E is the set on which thefunction is 1. It is useful to note that

χA∩B = χAχB, χA∪B = χA + χB iff A ∩B = ∅.

A simple function is one which takes on only a finite number of val-ues in the extended real numbers. It is thus a finite linear combination ofcharacteristic functions

ϕ =n∑

j=1

cjχEj , Ej = x : ϕ(x) = cj.

It is clear that a simple function is measurable iff the sets Ej are measurable.Moreover, the class of simple functions is closed under multiplication andaddition.

Theorem 3.2.1 Let f be a measurable real-valued function on a space X.Then there is a sequence of simple functions ϕn such that ϕn → f uniformlyon any set on which f is bounded.

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16 CHAPTER 3. INTEGRATION

Proof: Decompose f = f+ − f−, where f+ = maxf, 0 and f− =max−f, 0. Thus we may assume that f ≥ 0. For n ≥ 0 and 0 ≤ k ≤ 22n−1,let

En,k = f−1((k2−n, (k + 1)2−n)), Fn = f−1((2n,∞]).

Then the sequence of simple functions

ϕn =22n−1∑

k=0

k2−nχEn.k+ 2nχFn

satisfy

0 ≤ · · · ≤ fn ≤ fn+1 · · · ≤ f, 0 ≤ f − ϕn ≤ 2−n, x ∈ F cn.

The convergence is uniform on sets where f is bounded.Let (X,M, µ) be a measure space. The integral of a measurable simple

function ϕ is defined to be

∫ϕ dµ =

n∑

j=1

cjµ(Ej).

We use the convention that 0 · ∞ = 0.

It is clear that the integral of non-negative simple functions is non-negative. Also, since finite linear combinations and finite products of simplefunctions are simple, the integral so defined is a linear functional on thevector space of simple functions.

Lemma 3.2.2 Given a non-negative, measurable simple function ϕ andA ∈ M, define

∫A ϕdµ =

∫ϕχAdµ. Then the set function ν(A) defined

by

ν(A) =∫

A

ϕdµ

is a (countably additive) measure.

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3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 17

Proof: If A is the countable union of disjoint sets in M then

ν(A) =∫

ϕχAdµ =∫ n∑

k=1

ckχEkχAdµ =

∫ n∑

k=1

ckχA∩Ekdµ

=n∑

k=1

ckµ(Ek ∩A) =n∑

k=1

∞∑

j=1

ckµ(Ek ∩Aj) =∞∑

j=1

ckµ(Ek ∩Aj)

=∞∑

j=1

Aj

ϕdµ =∞∑

j=1

ν(Aj).k. .

We now define the integral of any non-negative measurable f by∫

fdµ = supϕ≤f

∫ϕdµ : ϕ simple

.

The reader may check that this definition is consistent with that of the inte-gral of a simple function. If f ≤ g then

∫fdµ is obtained as the supremum

over a smaller set of simple functions than that for g; hence∫

fdµ ≤ ∫gdµ.

The following theorem is fundamental to the development of the theoryof the Lebesgue integral.

Theorem 3.2.3 [Monotone Convergence Theorem] If fn is an increasingsequence of non-negative functions then

∫limn

fn dµ = limn

∫fn dµ.

Proof: Since fn is a nondecreasing sequence of measurable functions,limn fn = f exists and is measurable. It may take on the value infinity.Moreover, the integrals

∫fndµ form a nondecreasing sequence of numbers;

and fn ≤ f implies that∫

fn ≤∫

f , hence

limn

∫fn dµ ≤

∫f dµ.

To prove equality, it suffices to show that for any fixed α, 0 < α < 1,

α

∫f dµ ≤ lim

n

∫fn dµ.

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18 CHAPTER 3. INTEGRATION

Given α and any simple function ϕ, let En = x : fn ≥ αϕ. Then∫

fn dµ ≥∫

En

fn ≥∫

En

αϕ dµ = αν(En).

By Lemma 3.2.2, ν is a countable additive measure. Since En is an in-creasing sequence of sets whose union is X, it follows by Theorem 1.8 c thatlimn ν(En) = ν(X) =

∫ϕ dµ. Thus

limn

∫fn dµ ≥ α

∫ϕdµ.

Taking the supremum over all simple functions ϕ ≤ f we obtain

α

∫f dµ = α sup

ϕ≤f

∫ϕdµ ≤ lim

n

∫fn dµ.

k. .

An immediate consequence of the Monotone Convergence Theorem isthe following characterization of the integral of a general non-negative mea-surable function.

Corollary 3.2.4 Let f be a non-negative measurable function, and let ϕn

be a monotone increasing sequence of measurable functions converging to f .Then ∫

f dµ = limn

∫ϕn dµ.

Using Corollary 3.2.4 we can prove the additivity of the integral on non-negative functions.

Proposition 3.2.5 The integral∫

dµ is finitely additive on non-negativemeasurable functions.

Proof: Let f and g be non-negative measurable functions and let ϕn ↑ f ,ψn ↑ g, where ϕn and ψn are sequences of non-negative simple functions.Then

∫f + g dµ =

∫limn

(ϕn + ψn) dµ

= limn

∫ϕn dµ + lim

n

∫ψn dµ =

∫f dµ +

∫g dµ.

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3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 19

By the same argument, one can show that∫

cf dµ = c∫

f dµ, for any con-stant c. k. .

Applying the monotone convergence theorem to the partial sums of aseries of non-negative functions, we obtain

Corollary 3.2.6 If fn ≥ 0, then∫ ∑

n

fn dµ =∑

n

∫fn dµ,

where both sides may be infinite.

Proposition 3.2.7 If f ≥ 0 then∫

f = 0 iff f = 0 a.e.

Proof: The statement is immediate for simple functions. For general f , iff = 0 a.e. and ϕ ≤ f is simple, then ϕ = 0 a.e. hence

∫f = supϕ≤f

∫ϕ = 0.

Conversely, let

x : f(x) > 0 =⋃

n≥1

En, En = x : f(x) >1n.

If f > 0 on a set of positive measure, then µ(En) > 0 for some n; hence∫

f dµ ≥ 1n

µ(En) > 0.

k. .

An immediate corollary of this proposition is the following strengtheningof the monotone convergence theorem.

Corollary 3.2.8 If 0 ≤ · · · ≤ fn ≤ fn+1 ≤ . . . , and fn ↑ f a.e. thenlimn

∫fn dµ =

∫f dµ.

The following lemma, known as Fatou’s lemma, can also be used as thestarting point in building the convergence theorems of Lebesgue integrationtheory.

Lemma 3.2.9 [Fatou’s Lemma] For any non-negative sequence of measur-able functions fn we have

∫lim inf

nfn dµ ≤ lim inf

n

∫fn dµ.

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20 CHAPTER 3. INTEGRATION

Proof: We have infn≥k fn ≤ fj for j ≥ k, hence∫

infn≥k fn dµ ≤ ∫fj dµ for

j ≥ k; and so ∫infn≥k

fn dµ ≤ infj≥k

∫fj dµ.

Since both sides are increasing in k, we have, in the limit as k →∞,

limk→∞

∫infn≥k

fn dµ ≤ limk→∞

infj≥k

∫fj dµ = sup

k≥1infj≥k

∫fj dµ = lim inf

n

∫fn dµ.

Since the sequence infn≥k fn is non-negative and increasing in k, wehave, by the Monotone Convergence Theorem∫

lim infn

fn dµ =∫

supk≥1

infn≥k

fn dµ =∫

limk→∞

infn≥k

fn dµ = limk→∞

∫infn≥k

fn dµ.

3.3 The Dominated Convergence Theorem

Let (X,M, µ) be a measure space. A real valued function f is said to beintegrable if f is measurable with respect to µ and

∫ |f | dµ < +∞. Since|f±| ≤ |f |, hence

∫f± dµ < +∞ and we can define

∫fdµ =

∫f+dµ−

∫f−dµ.

The class of integrable functions on (X,M, µ) is denoted by L1(X, dµ)or more briefly by L1 when the measure space is clear. The class L1 is closedunder addition and scalar multiplication, hence is a vector space. Moreover,if M contains an infinite number of sets of positive measure, then L1(X, dµis an infinite dimensional vector space. A norm is defined on this vectorspace by

||f ||1 =∫|f |dµ.

It is easily verified that || · ||1 satisfies the triangle inequality and ||cf ||1 =|c|||f ||1. Thus L1 is a metric space with the metric ρ(f, g) = ||f−g||1; and itis natural to ask the question as to whether L1 is a complete metric space.The answer to this question is yes; and not only that, but the completenessis a direct consequence of the introduction of the notion of measure theoryand measurable functions. If one begins with only the continuous functionsin L1, and seeks the completion of this metric space, then one has to includethe measurable functions as well.

We begin by proving:

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3.3. THE DOMINATED CONVERGENCE THEOREM 21

Theorem 3.3.1 [Lebesgue Dominated Convergence Theorem] Let fn bea sequence of functions in L1 such that fn → f a.e. and there exists a non-negative function g ∈ L1 such that |fn| ≤ g, a.e. for all n. Then f ∈ L1 and∫

fdµ = limn

∫fndµ.

Before beginning the proof, let us give an example that shows the neces-sity of the assumption that the sequence fn be dominated by a fixed functiong ∈ L1. Consider the fundamental solution of the heat equation on the line:

G(x, t) =e−x2/4t

√4πt

As t ↓ 0 the function narrows and its maximum value tends to infinity.It is a simple matter to show that

∫ ∞

−∞G(x, t) dx = 1, t > 0,

yet that G(x, t) tends to 0 for all x 6= 0. Thus its pointwise limit is 0 almosteverywhere, and for any sequence tn ↓ 0,

0 =∫ ∞

−∞lim

n→∞G(x, tn) dx < limn→∞

∫ ∞

−∞G(x, tn)dx = 1.

Thus the conclusion of the Dominated Convergence theorem fails while thatof Fatou’s lemma holds with a strict inequality.

−5 −4 −3 −2 −1 0 1 2 3 4 50

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

t=.1

t=1

Figure 3.1: The fundamental solution of the heat equation evaluated att = .1 and t = 1.

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22 CHAPTER 3. INTEGRATION

Proof: Applying Fatou’s lemma to the sequence of functions g + fn ≥ 0we obtain

∫g + fdµ =

∫(g + lim inf

nfn)dµ ≤

∫gdµ + lim inf

n

∫fndµ,

hence∫

fdµ ≤ lim infn∫

fndµ. Applying the same argument to the sequenceg − fn ≥ 0, we obtain,

∫−fdµ ≤ lim inf

n

∫−fndµ = − lim sup

n

∫fndµ.

Combining both inequalities, we have∫

fdµ ≤ lim infn

∫fndµ ≤ lim sup

n

∫fndµ ≤

∫fdµ.

k. ..

Corollary 3.3.2 Let fj be a sequence of integrable functions such that theseries

∑j

∫ |fj |dµ < +∞. Then∑

j fj converges a.e. to a function f ∈ L1

and ∫ ∞∑

j=1

fjdµ =∞∑

j=1

∫fjdµ.

Proof: By the monotone convergence theorem,∫ ∑

j

|fj |dµ =∑

j

∫|fj |dµ < +∞,

hence the function g =∑

j |fj | ∈ L1. Thus g is finite a.e. and for each x forwhich g(x) is finite, the series

∑j fj(x) converges absolutely. The partial

sums of this series are dominated by g, so by the conclusion of the Corollaryfollows by the dominated convergence theorem. k. .

Theorem 3.3.3 The simple functions are dense in L1. The class C0(R) isdense in L1(R, BR, µ), where µ is any Borel measure on R.

Proof: The simple functions are dense in L1(dµ) by Theorem 3.2.1 andthe dominated convergence theorem. To show the continuous functions aredense in L1(R,BR) it is sufficient to show that any simple function can beapproximated by a continuous function in the L1 norm. If I = (a, b) is anopen interval, we may approximate χI by a sequence of continuous, piecewise

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3.4. THE RIEMANN INTEGRAL 23

linear functions Cn,E which take the value 1 on I, 0 on (−∞, a−n−1)∪ (b+n−1,∞). It is a simple calculation to show that ||χE − Cn,E ||1 = n−1.

Next let E be a Borel measurable set in R, with µ(E) = ||χE ||1 < ∞.We show that for any ε > 0, there is a finite union of open intervals A suchthat ∫

|χE − χA|dµ = µ(E∆A) < ε.

To see this, we use the fact that for Borel measures µ on the line,

µ(E) = inf

∞∑

j=1

µ(Ij) : E ⊂∞⋃

j=1

(aj , bj)

.

Thus given any ε > 0 there is a collection of disjoint open intervals Ij =(aj , bj) such that

µ(U\E) <ε

2, U =

∞⋃

j=1

Ij .

Now choose N so that∑∞

N+1 µ(Ij) < ε/2, and put A =⋃N

j=1 Ij , B =U\A. Then E\A = E∩Ac ⊂ (A∪B)∩Ac ⊂ B; hence µ(E\A) < µ(B) < ε/2,and

µ(A∆E) = µ(A\E) + µ(E\A) < µ(U\E) + µ(B) < ε.k. .

3.4 The Riemann Integral

The dominated convergence theorem is a fundamental tool. For example,we may use it to prove that the Lebesgue integral is an extension of the Rie-mann integral; that is, the two integrals give the same result for a Riemann-integrable function. Let [a, b] be a bounded interval. A partition of [a, b]is a set of points P = a = x0 < x1 < · · · < xn = b. Given a boundedreal-valued function f on [a, b], the upper and lower Riemann sums of frelative to a partition P are given by

L(P, f) =n∑

j=1

mj∆xj , U(P, f) =n∑

j=1

Mj∆xj ,

where ∆xj = xj −xj−1 and mj , Mj are the infimum and supremum of f on[xj−1, xj ]. Clearly L(P, f) ≤ U(P, f) for any partition P.

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24 CHAPTER 3. INTEGRATION

Definition 3.4.1 A bounded real-valued function f defined on a bounded in-terval [a, b] is said to be Riemann integrable if supP L(P, f) = infP U(P, f).In that case the Riemann integral of f on [a, b] is defined to be

∫ b

af(x)dx = sup

PL(P, f) = inf

PU(P, f).

The concept of the Riemann integral is clearly an application of themethod of exhaustion, proposed originally by Euxodus of Cnidus (408-355B.C.E.) and developed extensively by Archimedes (287-212 B.C.E.) as atool for computing areas and volumes. It was not until Isaac Barrow, thefirst Lucasian chair of mathematics, and his student and successor, Sir IsaacNewton (1642-1727) that the connection between the method of exhaustionand the notion of anti-derivative was fully established. This is the truecontent of the Fundamental Theorem of Calculus.

Theorem 3.4.2 Let f be a bounded real-valued function on [a, b]. If f isRiemann integrable, then f is Lebesgue measurable and therefore integrableon [a, b]; moreover the Lebesgue and Riemann integrals agree. A function isRiemann integrable iff it is continuous a.e.

Proof: Choose a sequence of partitions Pn such that each partition is arefinement of the previous one and for which maxj ∆xj tends to zero asn → ∞. Then the lower sums L(f,Pn) are increasing and the upper sumsare decreasing. f is Riemann integrable iff these two sequences converge tothe same limit.

The lower and upper sums of a function f are the Lebesgue integrals ofthe simple functions

l(P, f) =n∑

j=1

mjχ(xj−1,xj ], u(P, f) =n∑

j=1

Mjχ(xj−1,xj ].

Let hn = l(Pn, f), and Hn = u(Pn, f). Then hn ≤ f ≤ Hn, and in the limith ≤ f ≤ H, where h = limn hn and H = limn Hn.

If f is Riemann integrable, then the sequences of upper and lower sumsconverge to the same limit. It follows that

limn

[a,b]

hndm =∫ b

af dx = lim

n

[a,b]

Hndm.

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3.5. SOME CONVERGENCE THEOREMS 25

By the Lebesgue dominated convergence theorem,

limn

[a,b]

hndm =∫

[a,b]

hdm =∫

[a,b]

Hdm = limn

[a,b]

Hndm.

Since H ≥ h this implies that h = H a.e. and hence that H = f = h a.e. Itfollows that f is Lebesgue measurable and that

[a,b]

Hdm =∫

[a,b]

fdm =∫ b

afdx.

The proof that the discontinuities of a Riemann integrable function forma null set is left to the reader. k. .

Theorem 3.4.3 Let f(x, t) be a mapping from X × [a, b], where (X,M, µ)is a measure space and suppose that f is differentiable in with respect to tand that for each x ∈ X, |ft(x, t)| ≤ g(x) for all a ≤ t ≤ b, where g ∈L1(X, dµ). Let F (t) =

∫X f(x, t)dµ. Then F is differentiable on a < t < b

and F ′(t) =∫X ft(x, t)dµ.

Proof: For t ∈ (a, b) we apply the Lebesgue dominated convergencetheorem to the sequence

hn(x) =f(x, tn)− f(x, t)

tn − t, tn → t.

By the mean value theorem, the difference quotients hn are bounded byg(x) = supt |ft(x, t)|. k. .

3.5 Some Convergence Theorems

The L1 integral of each term of the following sequences is 1,

1n

χ(0,n), χ(n,n+1), nχ(0,n−1),

yet the sequences themselves tend to 0 uniformly, pointwise, and a.e. re-spectively. In each of these cases the sequences are not dominated by a fixedintegrable function g.

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26 CHAPTER 3. INTEGRATION

On the other hand, there are sequences which tend to 0 in the L1 normbut which do not tend to zero pointwise. For example, the sequence f1 =χ(0,1), f2 = χ(0,1/2), f3 = χ(1/2,1), f4 = χ(0,1/4), f5 = χ(1/4,1/2), etc. and ingeneral fn = χ(j2−k,(j+1)2−k), where n = 2k + j, with 0 ≤ j < 2k. For thissequence, ||fn||1 → 0 as n →∞ but fn(x) does not converge for any x. Thisshows that pointwise convergence and convergence in norm are in generalnot equivalent.

We shall prove here that if ||fn − f ||1 → 0, then some subsequence offn converges to f a.e. In order to do this we introduce the concept ofconvergence in measure.

Definition 3.5.1 A sequence of functions fn is said to converge in measureto f if for every ε > 0,

limn

µ (x : |fn(x)− f(x)| ≥ ε) = 0.

A sequence is said to be Cauchy in measure if for every ε > 0,

µ (x : |fn(x)− fm(x)| ≥ ε) → 0, m, n →∞.

For example, the first, third, and fourth sequences above converge to0 in Lebesgue measure on the line, while the second sequence does not.Convergence in norm implies convergence in measure. We need only provethis for a sequence of functions tending to 0 in the L1 norm. Suppose||fn||1 → 0 and, given ε > 0, put En,ε = x : |fn(x)| > ε. Then for anyfixed ε > 0, ∫

|fn|dµ ≥∫

En,ε

|fn|dµ ≥ εµ(En,ε) → 0.

Thus fn → 0 in measure. k. .

The first and third examples above show that convergence in measuredoes not imply convergence in norm.

Many authors use the nomenclature fn → f [meas], a.e., [unif], or[mean], to signify convergence in measure, pointwise convergence, uniformconvergence, or convergence in the L1 norm.

Theorem 3.5.2 If fn is a Cauchy sequence in measure, then there is ameasurable function f such that fn → f [meas]; and there is a subsequencefnk

such that fnk→ f a.e. Moreover, f is uniquely determined a.e.

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3.5. SOME CONVERGENCE THEOREMS 27

Proof: We may choose a subsequence nk such that for gk = fnk,

µ(Ek) ≤ 2−k, Ek = x : |gk(x)− gk+1(x)| ≥ 2−k.

Putting Fk =⋃

j≥k Ej , we have µ(Fk) ≤∑∞

j≥k 2−j = 21−k, and

|gi(x)− gj(x)| ≤i−1∑

l=j

|gl+1(x)− gl(x)| <i−1∑

l=j

2−l ≤ 21−j , (3.1)

for i ≥ j ≥ k, and x ∈ F ck . This shows that gj(x)j≥k is a Cauchy sequence

for x ∈ F ck .

Let F =⋂∞

1 Fk; then µ(F ) = 0 and gj(x) is a Cauchy sequence onF c. Let f = lim gk on F c and 0 on F . Then f is measurable; gk → f a.e.;and, letting i →∞, in (3.1) we obtain

|f(x)− gj(x)| ≤ 21−j , x ∈ F ck ; j ≥ k.

Hence x : |f(x) − gj(x)| > 21−j ⊂ Fk for j ≥ k. Since µ(Fk) → 0 ask →∞, this shows that gj → f [meas].

But then fn → f [meas] since for any ε > 0,

µ(x : |f(x)− fn(x)| ≥ ε) ≤ µ(x : |f(x)− gj(x)| ≥ ε/2)

+ µ(x : |gj(x)− fn(x)| ≥ ε/2) → 0, j, n →∞.

The proof that the limit in measure of a Cauchy sequence in measure isunique is left as an exercise. k. .

Corollary 3.5.3 If fn → f in L1 then there is a subsequence fnk such

that fnk→ f [a.e.]

Proof: Exercise

The sequence χ[n,n+1] converges pointwise to 0 everywhere, but doesnot converge to 0 in measure. On a finite measure space, however, pointwiseconvergence implies what is called almost uniform convergence. If µ(X) <+∞, a sequence fn is said to converge almost uniformly to f on X ifgiven any ε > 0 there is a set E such that µ(E) < ε and fn → f [unif] onEc. It is easily seen that almost uniform convergence implies convergencein measure.

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28 CHAPTER 3. INTEGRATION

Theorem 3.5.4 [Egorov’s Theorem.] On sets of finite measure, conver-gence almost everywhere implies almost uniform convergence.

Proof: It suffices to show that if fn → 0 everywhere on E, and µ(E) < ∞,then there exists a subset F ⊂ E such that µ(E\F ) < ε and fn → 0uniformly on F . Let Emn = x : |fj(x)| < n−1, j ≥ m. For fixed n, thesequence Emn is increasing, and E =

⋃∞1 Emn. Therefore, by continuity

from below, limm µ(Emn) = µ(E). Hence given ε > 0 and for n ∈ N, thereexists an integer mn such that µ(E\Emnn) < ε2−n.

Let F =⋂

n≥1Emnn. If x ∈ F then for all n ∈ N there is an mn such that

|fj(x)| < n−1 for all j ≥ mn; hence fj → 0 uniformly on F . On the otherhand,

µ(E\F ) = µ

n≥1

E\Emnn

∞∑

n=1

µ(E\Emnn) < ε.k. .

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Chapter 4

Product Spaces

4.1 Fubini’s Theorem

Let Xj ,Mj , µj, j = 1, 2 be two measure spaces, and consider the σ algebraM of subsets of X = X1×X2 generated by rectangles of the form E1×E2,where Ej ∈Mj . This σ-algebra is called the product algebra, and is denotedby M = M1

⊗M2. We shall denote points in X1 and X2 by x and yrespectively.

A product measure on X is constructed as follows. Denote the algebraof sets generated by finite disjoint unions of rectangles by A. Define µ′ onrectangles by µ′(E1 ×E2) = µ1(E1)µ2(E2). Then µ′ is a pre-measure on A.Using µ′, we construct an outer measure µ∗ on X; by Caratheodory’s theo-rem we obtain a complete measure µ on X whose σ-algebra of measurablesets contains the product algebra M = M1

⊗M2. If µj are σ-finite, thenso is µ.

Clearly this construction of product algebras and measures extends toany finite number of factors. Moreover, this definition of product algebrasand measures is associative.

In this section we prove the following theorems of Tonelli and Fubini:

Theorem 4.1.1 Let Xj ,Mj , µj be σ-finite measure spaces, and let theproduct measure space be denoted by X,M, dµ.

Tonelli: Let f ∈ L+(dµ). Then the functions g(x) =∫X2

f(x, y)dµ2(y)

29

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30 CHAPTER 4. PRODUCT SPACES

and h(y) =∫X1

f(x, y)dµ1(x) are in L+(dµ1) and L+(µ2) respectively; and

∫∫

X

fdµ =∫ [∫

f(x, y)dµ2(y)]

dµ1(x)

=∫ [∫

f(x, y)dµ1(x)]

dµ2(y). (4.1)

These three integrals may assume the value +∞.Fubini: Let f ∈ L1(dµ). Then

X2

f(x, y)dµ(y) ∈ L1(dµ1),∫

X1

f(x, y)dµ(x) ∈ L1(dµ2),

and (4.1) holds; that is, the double and each of the iterated integrals areequal.

The strategy of the proof is to begin by proving the result for character-istic functions of rectangles, then simple functions, and ultimately extendthe result to general measurable functions on X.

The x and y sections of a set E ⊂ X are defined to be the subsets

Ex = y : y ∈ X2, (x, y) ∈ E, x ∈ X1

Ey = x : x ∈ X1, (x, y) ∈ E, y ∈ X2.

Similarly, if f : X 7→ R, then the x and y-sections of f are the mappings fx :X2 7→ R and fy : X1 7→ R defined by fx(y) = f(x, y) and fy(x) = f(x, y).

Proposition 4.1.2 If E ∈ M, then the sections Ex and Ey respectivelybelong to M2 for each x ∈ X1, and M1 for each y ∈ X2. If f is measurablewith respect to the product algebra M1

⊗M2, then its sections fx and fy

are measurable with respect to the factors M2 and M1 respectively.

Proof: Let R be the collection of all subsets E ⊂ X such that Ex ∈M2

for all x ∈ X1 and Ey ∈M1 for all y ∈ X2. Then

(A×B)x =

B x ∈ A

∅ x ∈ Ac,

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4.1. FUBINI’S THEOREM 31

and similarly for the section (A × B)y. Hence R contains all rectangles.Moreover, R is a σ algebra, since

(∞⋃

1

Ej

)

x

=∞⋃

1

(Ej)x, (Ex)c = (Ec)x,

and similarly for y-sections. Therefore R ⊃ M1⊗M2. The measurability

of fx and fy follows from the first statement and the relationships

(fx)−1(B) = (f−1(B))x, (fy)−1(B) = (f−1(B))y.k. .

A monotone class on a space X is a collection of subsets C ⊂ P(X)that is closed under countable increasing unions and countable decreasingintersections. Every σ algebra is a monotone class; and the intersection ofmonotone classes is a monotone class. Thus for any collection E ⊂ P(X),there is a unique smallest monotone class containing E , called the monotoneclass generated by E .

Proposition 4.1.3 If A is an algebra of subsets of X then the monotoneclass generated by A coincides with the σ algebra generated by A.

Given a measure µ we denote the non-negative µ measurable functionsby L+(µ).

Lemma 4.1.4 Let Xj ,Mj , µj be σ-finite measure spaces for j = 1, 2,and let M and µ be the product algebras and measures on X = X1 × X2.Given E ∈ M, the sections (χE)x and (χE)y are in L+(dµ2) and L+(dµ1)respectively; and

µ(E) =∫∫

X

χEdµ =∫

X1

X2

(χE)xdµ2(y)

dµ1(x)

=∫

X2

X1

(χE)ydµ1(x)

dµ2(y). (4.2)

Proof: We shall establish the lemma for the case in which the µj arefinite measures; the σ-finite case is left as an exercise. Let C be the class ofsets in M for which the lemma holds. When E is a rectangle, E = A × B

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32 CHAPTER 4. PRODUCT SPACES

then (χE)x(y) = χA(x)χB(y) = (χE)y(x) and all three integrals above areequal to µ1(A)µ2(B). Thus E ⊂ C. By finite additivity of the integral Ccontains finite disjoint unions of rectangles. It therefore suffices to showthat C is a monotone class.

Let E = ∪nEn where En is an increasing sequence of sets. Then thesequence of functions χEn increases monotonically to χE ; and their sectionsincrease monotonically to the corresponding sections of χE ; (4.2) followsfrom the monotone convergence theorem; and E ∈ C.

Similarly, let En be a decreasing sequence of sets in C and put E =∩nEn. The characteristic functions χEn then converge pointwise to χE ; andtheir sections converge pointwise to the corresponding sections of χE . Sinceµ(E1) < ∞ the conclusion follows from the dominated convergence theorem.Thus C is a monotone class and is therefore equal to the σ-algebraM.

k. .

Proof of Theorem 4.1.1: Lemma 4.1.4 establishes (4.1) for characteristicfunctions; and by additivity the result follows for simple functions. Givenf ∈ L+(X, dµ) we approximate f from below by simple functions and passto the limit using the monotone convergence theorem. All three integralsare either finite and equal, or all equal to infinity. Given f ∈ L1(dµ) wedecompose f = f+ − f− and apply Tonelli’s theorem to each of the compo-nents.

4.2 Lebesgue measure in Rn

A specific example of a product measure, obtained by the procedure in §4.1is n dimensional Lebesgue measure on Rn; it is obtained by the premeasuredefined on ”rectangles” in Rn defined by R =

∏j Ij , where the Ij are

intervals (aj , bj) on the real axis. We have

m0(R) = µ

n∏

j=1

Ij

=

n∏

j=1

(bj − aj). (4.3)

Using m0 we obtain an outer measure on Rn and by Caratheodory’s theoremwe obtain a measure, called the Lebesgue measure on Rn. We denote it bym when the dimension is understood.

As on R, Lebesgue measure on Rn has has a number of regularity prop-erties:

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4.2. LEBESGUE MEASURE IN RN 33

Theorem 4.2.1 Suppose E ∈ Ln. Then (a) m(E) = inf m(U) : E ⊂U,U open = sup m(K) : K ⊂ E, K compact ; (b) E = A1∪N1 = A2\N2,where N1 and N2 are sets of measure 0, A1 ∈ Fσ (the class of countableunions of closed sets); and A2 ∈ Gδ (the class of countable intersectionsof open sets); and (c) If m(E) < ∞ then for any ε > 0 there is a finitecollection of disjoint rectangles whose union R satisfies µ(E∆R) < ε.

Moreover,

Theorem 4.2.2 The class of simple functions f =∑

j fjχRj , where Rj =∏k Ijk is a product of intervals Ijk ⊂ R, is dense in L1(Rn). Furthermore

the class of continuous functions with compact support in Rn is dense inL1(Rn).

Proof: The proof is essentially the same as the proof of the correspondingtheorem for Borel measures on R, Theorem 3.3.3.

Theorem 4.2.3 Lebesgue measure on Rn is invariant under the group ofrigid motions.

Proof: It suffices to establish that Lebesgue outer measure is invariantunder the group of rigid motions, that is, under the group of translationsand rotations in Rn. We prove the result in R2 and leave its extension toRn as an exercise.

Lebesgue outer measure was defined in terms of coverings of rectangleswith sides parallel to the axes. Thus the outer measure is invariant undertranslations and reflections in the axes. We therefore have two set functions,area and outer measure, that are both additive and invariant under trans-lations and reflections. Any theorem on area in classical geometry that usesonly these three properties is therefore valid for outer measure.

To show that the outer measure is invariant under rotations, we mustprove that the outer measure of a rectangle in general position is equal tothe area of that rectangle. A rectangle in general position may be writtenas the union of two triangles and a parallelogram (degenerate if one of thediagonals is parallel to the x axis) with bases parallel to the x-axis; so byadditivity of both the area and the outer measure, it suffices to prove thatthe two functionals agree on triangles and parallelograms with base parallelto one of the axes.

Given a parallelogram with base parallel to the x axis, translate one ofits triangular caps parallel to the x axis to the other end, forming a rectanglewith base parallel to the x axis. Both the outer measure and the area of the

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34 CHAPTER 4. PRODUCT SPACES

parallelogram are invariant under this operation; and they coincide on therectangle. Hence the outer measure and area of the parallelogram coincide.

By similar arguments from plane geometry, using only additivity andinvariance under reflections and translations, one can prove that the outermeasure of any triangle is equal to its area.

A second approach is combine a translation of the upper triangle withthe lower triangle to produce a second parallelogram with base parallel tothe x axis, and then apply the previous argument showing the measure of aparallelogram is equal to its area. k. .

Now let T ∈ GL(n,R) and assume that detT 6= 0. We define an invert-ible transformation acting on the Lebesgue measurable functions f ∈ L(Rn)by fT = fT−1. We define an action of GL(n,R) on n dimensional Lebesguemeasure by mT (E) = m(TE), where E is a Lebesgue measurable subset ofRn and TE denotes its image under T . For a rectangle R ∈ Rn we havem(TR) = detTm(R).

Theorem 4.2.4 For f ∈ L1(Rn) and any invertible linear transformationT ∈ GL(n, R) we have

∫∫fT dm =

∫∫f dmT = | det T |

∫∫f dm.

Proof: It is sufficient to prove the result for characteristic functions. Itthen follows for simple functions by additivity, and for the general case bythe dominated convergence theorem.

First, note that for rectangles, A(TR) = detTA(R), hence on rectan-gles, mT (R) = m(TR) = | detT |m(R). By continuity, this identity extendsto all Lebesgue measurable sets E. Next, for any set E we observe thatχE(T−1x) = χTE(x) = (χE)T ; hence

∫∫(χE)T dm =

∫∫χTEdm = m(TE)

= |det T |m(E) = detT

∫∫χE dm =

∫∫χEdmT .

The volume V (E), where E ⊂ Rn, is initially defined on rectangles, andthen extended to polygons by its invariance under translation and reflection.The extension of the volume functional to bounded subsets in Rn is defined

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4.2. LEBESGUE MEASURE IN RN 35

by the method of exhaustion, a reincarnation of which is known as Jordancontent: Let Qk denote the collection of cubes with sides of length 2−k andvertices at the points 2−kZn in Rn. The volume of each cube in Qk is 2−nk,and the collection Qk+1 is obtained by bisecting the sides of Qk.

For E ⊂ Rn we define the inner and outer dyadic approximations of Eto be the open sets

Ik(E) =⋃Q ∈ Qk : Q ⊂ E, Ok(E) =

⋃Q ∈ Qk : Q ∩ E 6= ∅.

For each k, Ik(E) ⊂ E ⊂ Ok(E). The Ik(E) form an increasing sequence ofopen sets, and the Ok(E) a decreasing sequence. Moreover, if E is bounded,then the volumes of these sets are finite and respectively increasing anddecreasing. Denoting their limits respectively by κ(E) and κ(E), we have

κ(E) = limk

V (Ik(E)) ≤ limk

V (Ok(E)) = κ(E).

If these two quantities are equal, then the volume of E, denoted by V (E),is defined to be their common value. In Jordan’s theory, κ(E) and κ(E) areknown as the inner and outer contents of E.

Theorem 4.2.5 If the volume of a set E ⊂ Rn is defined in the abovesense, then E is Lebesgue measurable, and its Lebesgue measure is equal tothe volume.

Proof: Let

A(E) =∞⋃

k=1

Ik(E), A(E) =∞⋂

k=1

Ok(E).

Then A(E) ⊂ E ⊂ A(E), A(E) and A(E) are Borel sets, and by the conti-nuity of Lebesgue measure,

m(A(E)) = κ(E), m(A(E)) = κ(E). k. .

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36 CHAPTER 4. PRODUCT SPACES

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Chapter 5

Differentiation Theory

5.1 Differentiation Theory of Functions

In the following sections we discuss the theory of differentiation of mea-sures; but before going into the general theory of differentiation of measure,we review some basic facts about differentiation of functions of a real vari-able, since these are the ideas constitute the starting point for the theory ofdifferentiation of measures.

In 1904 Lebesgue1 proved the following theorem:

Theorem 5.1.1 Every monotonic function f is differentiable almost every-where.

Lebesgue’s original proof assumed f to be continuous, but that assump-tion may be dropped, since every monotonic function has only jump dis-continuities, and is continuous almost everywhere. The theorem is provedby considering the upper and lower, left and right, derived numbers. Theupper and lower derived numbers are

λ± = lim infh→0±

f(x + h)− f(x)h

, Λ± = lim suph→0±

f(x + h)− f(x)h

.

The derivative of f at x exists if all four derived numbers have the same finitevalue. A detailed proof of the result may be found in the book FunctionalAnalysis, by Riesz & Nagy. The first chapter is devoted to differentiationtheory.

1Lebesgue Lecons sur l’integration et la recherche des fonctions primitive, Paris, 1904

37

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38 CHAPTER 5. DIFFERENTIATION THEORY

Lebesgue’s theorem is easily extended to the larger class of functionsof bounded variation. Let f be a real-valued function on the interval [a, b],where −∞ ≤ a < b ≤ ∞. The total variation function of f on [a, x] is definedto be

Tf (a, x) = supPn

n∑

j=1

|f(xj)− f(xj−1)|, n ∈ N, (5.1)

Pn = a = x0 < x1 < · · · < xn = x.

Clearly Tf (a, x) is a monotone non-decreasing function of x. If Tf (a, b) isfinite, then we say that f is of bounded variation on [a, b]. The class offunctions of bounded variation on [a, b] is denoted by BV [a, b].

Theorem 5.1.2 The functions of bounded variation on [a, b] form a normedlinear space with norm ||f ||var = Tf (a, b). The space BV [a, b] is a completenormed linear space with this norm.

Proof: One must first show that || · ||var satisfies the triangle inequality.There is a minor technical point in that constant functions have total varia-tion zero; so the points in the Banach space are actually equivalence classesof functions which differ by a constant. The proof of completeness requiresthe use of Helly’s theorem.

Theorem 5.1.3 [Jordan Decomposition] Every function of bounded varia-tion can be written as the difference of two non-decreasing functions.

Proof: Let T (x) = T (a, x). It is clear that T is a non-decreasing functionof x, since for x < y, T (y) = T (x) + T (x, y). Noting the identity f =T − (T − f), we see that the theorem will be proved by showing that T − fis non-decreasing. To show that T (x)− f(x) ≤ T (y)− f(y), or equivalentlythat

f(y)− f(x) ≤ T (y)− T (x),

we note that |f(y) − f(x)| is a particular sum of the type given in (5.1);hence |f(y)− f(x)| ≤ Tf (x, y), and the result follows.

Corollary 5.1.4 If f ∈ BV ([a, b]) then the right and left hand limits off(x) exist for all x. The only discontinuities of f are jump discontinuities,and there are at most a countable number of them in [a, b]. Finally, f isdifferentiable almost everywhere.

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5.1. DIFFERENTIATION THEORY OF FUNCTIONS 39

The Cantor function is a continuous, monotonically non-decreasing func-tion whose derivative vanishes almost everywhere.2 Such a function is calleda singular function. The Cantor function cannot be written as the integralof its derivative. Thus, the Cantor function is not an antiderivative.

Definition 5.1.5 A function F on a bounded interval [a, b] is absolutelycontinuous iff given any ε > 0 and any collection (xj , yj), finite or count-ably infinite, of non-overlapping sub-intervals of (a, b), there exists a δ > 0such that

j

F (yj)− F (xj) < ε whenever∑

j

(yj − xj) < δ.

Theorem 5.1.6 Absolute continuity is a necessary and sufficient conditionthat a function be differentiable almost everywhere and equal to the integralof its derivative.

The necessity is not hard to establish. First note that the result is aslam dunk if f is bounded on [a, b]; for in that case

∣∣∣∣∣∣∑

j

F (yj)− F (xj)

∣∣∣∣∣∣≤

j

∫ yj

xj

|f | dm ≤∑

j

∫ yj

xj

C dm

=C∑

j

(yj − xj) = Cδ.

Hence we may choose δ = ε/C.If f is integrable but not bounded, we may decompose f into the sum

of a bounded function g and a function h such that∫ |h|dm < ε/2 where ε

is any positive number. This is so because the bounded functions are densein the integrable functions. It follows that

∣∣∣∣∣∣∑

j

F (yj)− F (xj)

∣∣∣∣∣∣≤

j

∫ yj

xj

|g|+ |h| dm ≤∑

j

∫ yj

xj

C dm +ε

2

<C∑

j

(yj − xj) +ε

2< Cδ +

ε

2.

2In Riesz & Nagy an example is constructed of a strictly increasing function whosederivative vanishes almost everywhere!

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40 CHAPTER 5. DIFFERENTIATION THEORY

The result follows by choosing δ so small that Cδ < ε/2.The proof that absolute continuity implies that F is an indefinite inte-

gral is more involved; the classical proof is given in Riesz & Nagy. Butrather than recount the classical proof here, we shall defer it to the studyof differentiation of measures, and obtain the result as a corollary of theRadon-Nikodym theorem.

Theorem 5.1.7 Every continuous monotone function can be decomposedinto the sum of an absolutely continuous function and a singular function,both monotone. By the Jordan decomposition theorem, the same applies tocontinuous functions of bounded variation.

Proof: Let f(x) be continuous and non-decreasing. Then f ′ exists almosteverywhere, by Lebesgue’s theorem. Now

f(x + h)− f(x)h

≥ 0,

and this ratio tends to f ′(x) a.e. as h → 0. On the other hand, its integralover an interval (a, b) is equal to

1h

∫ b+h

bf(x) dx− 1

h

∫ a+h

af(x) dx;

hence, since f is continuous, its limit as h → 0 is f(b) − f(a). By Fatou’slemma we then have

∫ b

af ′(x) dx ≤ f(b)− f(a).

Since f ′ ≥ 0, it follows that f ′ ∈ l1(dx). Setting g(x) =∫ xa f ′(t) dt and

h(x) = f(x)−g(x), we see that h′(x) = 0 a.e. Moreover, h is non-decreasing,since, for y > x,

h(y)− h(x) = f(y)− f(x)−∫ y

xf ′(t) dt ≥ 0.

Thus h is a non-decreasing singular function. k. .

The result may be extended to general functions of bounded variation,as follows. A saltus function is one which has jump discontinuities at acountable set of point xj. It can be written in the form

s(x) =∑

xj≤x

uj +∑xj<x

vj ,

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5.2. SIGNED MEASURES 41

where the series∑

uj and∑

vj are absolutely convergent. When uj and vj

are positive, the function s is monotone increasing; moreover, by a theoremof Fubini, its derivative is zero almost everywhere.

Every monotone function f can be decomposed into the sum of a con-tinuous function and a saltus function, both of which are monotone. Wesimply define s to be the saltus function whose jumps coincide in locationand size with those of f . Then g = f − s is continuous; and moreover, sinceg′ ≥ 0 a.e., g is monotone by the argument used in the proof of Theorem5.1.7.

Theorem 5.1.8 Every function of bounded variation f can be decomposedinto the sum of an absolutely continuous function, a saltus function, and asingular function.

5.2 Signed Measures

Till now, we have only considered positive measures; we now turn to moregeneral measures, which can take on both positive and negative values, or,more generally, complex values. A real-valued measure µ, defined on theBorel subsets of a topological space X is called a signed (Borel) measure ifit can be written in the form ν = µ1 − µ2, where µ1, µ2 are positive Borelmeasures, at least one of which is finite. A complex Borel measure is onewhich can be decomposed as a sum ν1+iν2 where ν1, ν2 are signed measures.

A signed measure vanishes on the empty set, assumes at most one ofthe values ±∞, and is countably additive: ν(∪jEj) =

∑j ν(Ej), this sum

converging absolutely if ν(∪jEj) is finite. Given a measurable set E, apartition of E is a (finite or countable) disjoint collection Ej such thatE = ∪jEj . The partition is finite if j = 1, . . . , n. The collection of finitepartitions of a set E is denoted by P(E). If ν is a signed measure (real orocmplex), we define the total variation of ν, denoted by |ν|, to be

|ν|(E) = supP(E)

n∑

j=1

|ν(Ej)| .

Theorem 5.2.1 The total variation of a real or complex-valued signed mea-sure ν is a positive, countably additive measure.

Proof: It is clear from its definition that |ν|(∅) = 0; and it is straightfor-ward to show that |ν| is monotone and finitely additive. We thus need to

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42 CHAPTER 5. DIFFERENTIATION THEORY

establish its countable additivity. Suppose Ej is a countable partition ofE. By the monotonicity and finite additivity of |ν|,

|ν|(E) ≥ |ν|

n⋃

j=1

Ej

=

n∑

j=1

|ν|(Ej)

for all n. If the infinite series∑

j |ν(Ej)| diverges, then |ν|(E) = ∞.On the other hand, if the series converges, then |ν|(E) ≥ ∑∞

j=1 |ν|(Ej),and we have to establish the opposite inequality. Let ε > 0 and let E′

i,i = 1, . . . n be a finite partition of E such that

|ν|(E) <n∑

i=1

|ν(E′i)|+ ε.

Then

ν(E′i) =

∞∑

j=1

ν(Eij), Eij = E′i ∩ Ej ,

and it follows that

|ν|(E) <n∑

i=1

|ν(E′i)|+ ε =

n∑

i=1

∞∑

j=1

|ν(Eij)|+ ε

≤∞∑

j=1

n∑

i=1

|ν(Eij |+ ε ≤∞∑

j=1

n∑

i=1

|ν|(Eij |+ ε

=∞∑

j=1

|ν|(Ej) + ε.

Since ε is arbitrarily small, the countable additivity of |ν| follows.In order to prove that |ν| is a Borel measure, we need to show that

|ν|(E) is finite if E is compact. However, |ν|(E) ≤ µ1(E) + µ2(E); so if E

is compact, then µ1(E) and µ2(E), hence |ν|(E), are finite. k. .

If ν is a real signed measure, then

ν± =|ν| ± ν

2

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5.2. SIGNED MEASURES 43

are positive measures, called the positive and negative variations of ν. Wethus have the decompositions:

ν = ν+ − ν−, |ν| = ν+ + ν−.

The first is a Jordan decomposition of the signed measure ν, correspondingto the Jordan decomposition of a function of bounded variation.

The Jordan decomposition of a measure is not unique, since if ν± form aJordan decomposition of ν, then so do ν±+ν ′ for any positive finite measureν ′. However, there is a minimal decomposition in the following sense. Weintroduce an ordering on signed measures by ν1 ≤ ν2 if ν1(E) ≤ ν2(E) forall sets E ∈ X. Then

Proposition 5.2.2 If ν = ν1−ν2 is any Jordan decomposition of the signedmeasure ν, then ν+ ≤ ν1 and ν− ≤ ν2.

Proof: From the inequality

|ν|(E) ≤ |ν(E)| = |(ν1 − ν2)(E)| ≤ (ν1 + ν2)(E), E ∈M,

it follows that |ν| ≤ ν1 + ν2. Hence

ν+ =|ν|+ ν

2≤ (ν1 + ν2) + (ν1 − ν2)

2= ν1,

and similarly for ν−. k. .

Two measures µ and ν on X,M are mutually singular if there exist setsE, F ∈ M such that X = E ∪ F , E ∩ F = ∅, and µ(E) = 0, ν(F ) = 0. Wewrite µ ⊥ ν.

Theorem 5.2.3 [Hahn Decomposition] Let ν = ν+ − ν− be the Jordandecomposition of the signed measure ν. Then ν+ ⊥ ν−, and there existmeasurable sets P and N such that

P ∪N = X, P ∩N = ∅, ν+(N) = ν−(P ) = 0.

Moreover, ν is positive on every subset of P and negative on every subset ofN .

Proof: Assume that ν+(X) < ∞ and let Ej be a sequence of sets suchthat ν(Ej) > ν+(X)− 2−j . Now ν± are monotone, so for A ∈ Ec

j we have

ν(A) ≤ ν+(Ecj ) = ν+(X)− ν+(Ej) ≤ ν+(X)− ν(Ej) ≤ 1

2j.

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44 CHAPTER 5. DIFFERENTIATION THEORY

Since this holds for all A ∈ Ecj we have ν+(Ec

j ) ≤ 2−j . By the Jordandecomposition theorem,

−ν−(Ej) = ν(Ej)− ν+(Ej) ≥ ν(Ej)− ν+(X) ≥ − 12j

.

Hence ν−(Ej) ≤ 2−j .Now let

P = lim supj

Ej , N = P c = lim infj

Ecj .

Then

0 ≤ ν+(N) = lim infj

ν+(Ecj ) ≤ lim inf

j

12j

= 0;

0 ≤ ν−(P ) ≤ ν−(∪j≥kEj) ≤∞∑

j=k

ν−(Ej) ≤∞∑

j=k

12j

=1

2k−1,

for any k. Therefore ν−(P ) = 0.If E ⊂ N then ν(E) = ν+(E) − ν−(E) = −ν−(E) < 0, hence ν is

negative on all subsets of N . Similarly, ν is positive on all subsets of P .k. .

5.3 The Lebesgue-Radon-Nikodym theorem

In the example above, ν±f are mutually singular; and, by the Hahn de-composition theorem, the same holds for the components µ± of the Jordandecomposition of a measure µ. The Borel measure generated by the Cantorfunction and Lebesgue measure are mutually singular with respect to oneanother.

The measure νf has the property that νf (E) = 0 whenever µ(E) = 0.The measure νf is said to be absolutely continuous with respect to µ, andwe write νf ¿ µ. The measure-theoretic analog of Theorem 5.1.7 is thefollowing, known as the Lebesgue decomposition theorem.

Theorem 5.3.1 Let ν be a σ-finite signed measure and µ a positive σ-finitemeasure on (X,M). Then there exists a unique decomposition

ν = ρ + λ, ρ ¿ µ λ ⊥ µ,

where ρ, λ, are σ-finite. Moreover, there is a function f ∈ L1(X, dµ) suchthat dρ = fdµ.

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5.3. THE LEBESGUE-RADON-NIKODYM THEOREM 45

When dρ = fdµ, the density function f is called the Radon-Nikodymderivative of ρ with respect to µ, is denoted by dρ/dµ. We prove Theorem5.3.1 in a sequence of lemmas.

Lemma 5.3.2 Let ν be finite, and µ a positive measure on (X,M), withν. Then ν ¿ µ iff for every ε > 0 there is a δ > 0 such that |ν(E)| < εwhenever µ(E) < δ.

.Proof: By the Jordan decomposition theorem, it suffices to prove the

result when ν is a positive measure. The sufficiency of the ε, δ conditionis immediate; so we have only to establish its necessity. Suppose ν ¿ µbut that the ε, δ condition fails. Then there is an ε > 0 and a sequence ofsets Ej such that ν(Ej) > ε while µ(Ej) < 2−j . Let E = lim supj Ej . ByFatou’s lemma, ν(E) ≥ lim supj ν(Ej) ≥ ε. On the other hand,

µ(E) =µ

k≥1

j≥k

Ej

= lim

k→∞µ

j≥k

Ej

≤ limk→∞

∞∑

j=k

µ(Ej) ≤ limk→∞

21−k = 0.

Thus ν(E) > ε while µ(E) = 0, and so ν is not absolutely continuouswith respect to µ. Therefore the ε, δ condition is a necessary condition forabsolute continuity. k. .

Corollary 5.3.3 The antiderivative of an integrable function is absolutelycontinuous. More generally, for any positive measure µ and f ∈ L1(X,µ)we have that for every ε > 0 there exists a δ > 0 such that

∣∣∫E fdµ

∣∣ < εwhenever µ(E) < δ.

The first statement was proved earlier in §5.1.

Lemma 5.3.4 Let ν ¿ µ, where ν is positive, finite, and not identically 0,and µ is positive. Then there exist ε > 0 and a measurable set E such thatµ(E) > 0 and ν(E′) ≥ εµ(E′) for all E′ ⊂ E.

Proof: Let νj = ν−j−1µ, j = 1, 2, . . . , and let Pj , Nj be the correspond-ing Hahn decompositions of νj : ν−j (Pj) = 0, ν+

j (Nj) = 0. The sequence

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46 CHAPTER 5. DIFFERENTIATION THEORY

of measures νj is increasing; hence the Pj are increasing, and their com-plements therefore decreasing. Setting P = ∪jP and N = ∩jNj we haveP = N c. Since νj is negative on Nj , we have

0 ≤ ν(Nj) ≤ 1jµ(Nj) ≤ 1

jµ(N1) → 0, as j →∞;

and therefore ν(N) = 0. Since ν is not zero, ν(P ) > 0, and therefore ν(Pj) >0 for some j. The lemma is proved by taking E = Pj and ε = j−1 for thisvalue of j. k. .

Lemma 5.3.5 Let ν be finite and positive, and put

F =

f : f ∈ L1(dµ), ν(E) ≥∫

Efdµ, E ∈M

.

Then there is a unique function g ∈ F , finite a.e., such that∫

Xgdµ = sup

F

Xfdµ.

Remark: The class F is nonempty, since one can always take f = ε, thepositive number guaranteed in Lemma 5.3.4.

Proof: Let M denote the supremum on the right and let fj be asequence of functions in F such that

∫fjdµ → M . The sequence gn =

max[f1, f2, . . . , fn], is a monotone increasing sequence of functions; and bythe monotone convergence theorem M = limj

∫gjdµ =

∫limj gjdµ.

Now gj ∈ F for all j. In fact, let Ej = x ∈ E : gj(x) = fj(x). Then

Egndµ =

n∑

j=1

Ej

fjdµ ≤n∑

j=1

ν(Ej) = ν(E).

By the monotone convergence theorem, F is closed under monotone conver-gence, so g = limj gj ∈ F hence

Xgdµ = lim

n

Xgndµ = M.

The proof of uniqueness is left to the reader. k. .

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5.3. THE LEBESGUE-RADON-NIKODYM THEOREM 47

Theorem 5.3.6 [Radon-Nikodym] Let µ and ν be σ-finite measures on aset X, and suppose that ν ¿ µ. Then there is an f ∈ L1(X,M, dµ) suchthat

ν(E) =∫

Efdµ,

for all E ∈M for which ν(E) is finite.

Proof: The proof can be reduced to the case in which ν and µ are real,positive and finite. Let f be the maximal function guaranteed by Lemma5.3.5. Then ν1(E) = ν(E)− ∫

E fdµ satisfies ν1 ≥ 0 and ν1 ¿ µ. If ν1 is notidentically zero, then by Lemma 5.3.4 there is an ε > 0 and a set E′ suchthat

µ(E′) > 0, ν1(E′′) > εµ(E′′), E′′ ⊂ E′.

It follows that

ν(E) ≥∫

Ef + εχE′dµ, E ∈M,

contradicting the maximality of f . k. .

The proof that absolute continuity of a function F is sufficient that it bean antiderivative (Theorem 5.1.6) is an immediate corollary of the Radon-Nikodym theorem. If F is a monotone increasing real valued function onthe line, then the Lebesgue-Stieltjes measure νF generated by F is the Borelmeasure generated by the premeasure ν0((a, b]) = F (b) − F (a). If F is anabsolutely continuous function on the line, then νF is absolutely continuouswith respect to Lebesgue measure. Therefore νF = f dm, where f is theRadon-Nikodym derivative of νF with respect to m. Hence for any interval[a, b],

νF ((a, b]) = F (b)− F (a) =∫

[a,b]f dm =

∫ b

af(x) dx.

It follows that F ′ = f at points of continuity of f .If F is monotone, then νF is a positive measure; and if F is of bounded

variation, then νF is a signed measure with total variation |νF |(I) = TF (I),where I is an interval in R. On the other hand, the Cantor measure, theBorel measure generated by the Cantor function, is singular with respectto Lebesgue measure; and more generally, singular functions generate Borelmeasures which are singular with respect to Lebesgue measure.

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48 CHAPTER 5. DIFFERENTIATION THEORY

Corollary 5.3.7 Let µ and ν satisfy the hypotheses of the Radon-Nikodymtheorem. If E ∈M and if g ∈ L1(E, dν), then gf ∈ L1(X, dµ) and

∫E gdν =∫

E fgdµ.

Proof: When g = χF the result is an immediate consequence of theRadon-Nikodym theorem. The result extends immediately to simple func-tions by additivity, and then to any g ∈ L1(dν) by approximating g bysimple functions and going to the limit. k. .

Lemma 5.3.8 Let µ, ν be positive measures and suppose that if f is anyfunction such that ∫

Efdµ ≤ ν(E), E ∈M

then it follows f = 0 a.e. [µ]. Then µ ⊥ ν.

Proof: Consider the measure µ∧ν = (µ+ν−|µ−ν|)/2, where |µ−ν| denotesthe total variation of µ−ν. For any set E ∈M, |µ−ν|(E) ≥ |µ(E)−ν(E)|,so

(µ ∧ ν)(E) ≤ µ(E) + ν(E)− |µ(E)− ν(E)|2

= min(µ(E), ν(E)) ≤ µ(E).

Consequently µ∧ ν ¿ µ, and so by the Radon-Nikodym theorem, there is afunction g such that

(µ ∧ ν)(E) =∫

Egdµ

for all E for which (µ ∧ ν)(E) is finite.Now also ν ≥ µ ∧ ν, and this implies, by the hypothesis of the lemma,

that g = 0. Hence µ ∧ ν = 0, and it follows that

µ + ν = |µ− ν| = (µ− ν)+ + (µ− ν)−.

Letting X = P ∪ N be the Hahn decomposition relative to µ − ν, we seethat

(µ + ν)(P ) = (µ− ν)+(P ) ≤ µ(P ), (µ + ν)(N) = (µ− ν)−(N) ≤ ν(N);

hence ν(P ) = µ(N) = 0.k. .

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5.4. DIFFERENTIATION ON RN 49

We are now ready to prove the Lebesgue decomposition theorem. Wemay assume that µ and ν are finite. Let f be the maximal element of F asgiven in Lemma 5.3.5. Then

ν(E) ≥∫

Efdµ, E ∈M.

If f = 0 a.e. then ν ⊥ µ and the Lebesgue decomposition is ρ = 0, λ = ν.Otherwise, let ν2(E) = ν(E) − ∫

E fdµ, and suppose g is any function inL1(dµ) such that ν2(E) ≥ ∫

E gdµ for all E ∈ M. Then g = 0 by themaximality of f , hence ν2 ⊥ µ. The Lebesgue decomposition of ν is thusgiven by

ρ(E) =∫

Efdµ, λ(E) = ν(E)−

Efdµ.

If ρ′, λ′ is a second Lebesgue decomposition, then ρ − ρ′ = λ′ − λ. Butthe measure on the left side is absolutely continuous while the right side issingular with respect to µ; thus both sides must be zero. k. .

The Dirac measure is the Borel measure generated by the Heaviside stepfunction; it assigns a measure of 1 to the point 0, and zero measure toall sets not containing 0. It is singular with respect to Lebesgue measure,and therefore does not have a Radon-Nikodym derivative with respect toLebesgue measure. It is the measure-theoretic analog of the saltus functionsin the decomposition of real valued functions of bounded variation (Theorem5.1.8).

The Lebesgue decomposition theorem has a similar extension in the caseof Borel measures on the line. An atom of a measure ν is a point x forwhich ν(x) 6= 0 iff x ∈ A. A measure is called nonatomic if it has no atoms.An absolutely continuous function has no atoms; and if F is of boundedvariation, then the atoms νF are precisely the points of jump discontinuityof F , and so are countable. Let νF be the Borel measure generated byF , where F has finite total variation. By Theorem 5.1.8 we may writeF = F1 + F2 + F3, where F1 is absolutely continuous, F2 is singular, and F3

is a saltus function. It follows immediately that νF = ν1 + ν2 + ν3, whereν1 << m, ν2 ⊥ m and ν3 is atomic.

5.4 Differentiation on Rn

The function class L1loc = L1

loc(Rn, dm) is the class of Borel measurablefunctions on Rn which are Lebesgue integrable over any bounded measurable

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50 CHAPTER 5. DIFFERENTIATION THEORY

set. The measure νf (E) =∫E f dm is absolutely continuous with respect to

Lebesgue measure on Rn, and its Radon-Nikodym derivative is f . In onedimension, when f is continuous, f is obtained as the limit

limr→0

νf (Br(x))m(Br(x))

= limh→0

12h

∫ x+h

x−hfdm = f(x),

where Br(x) is the interval (x− r, x + r).In this section, we examine this relationship in the more general case

when f ∈ L1loc(Rn,m). Let Br(x) denote the ball of radius r centered at x ∈

Rn, and consider the Borel measure on νf on Rn, where f ∈ L1loc(Rn, dm).

It is straightforward to prove that

f(x) = limr→0

νf (Br(x))m(Br(x))

(5.2)

for all x ∈ Rn belonging to the Lebesgue set Lf defined by

Lf =

x : lim

r→0

1m(Br(x))

Br(x)|f(y)− f(x)| dy = 0

.

It is not hard to see that Lebesgue set contains the points of continuity off .

Theorem 5.4.1 [Lebesgue Differentiation Theorem] Let f ∈ L1loc. If x ∈

Lf then the Radon-Nikodym derivative of νf is given by (5.2) for all x in theLebesgue set of f . The Lebesgue set of f contains the points of continuityof f ; its complement has measure 0.

The proof that the complement of the Lebesgue set has measure 0 willbe carried out in a series of lemmas.

Lemma 5.4.2 Let C be a collection of open balls in Rn and let U = ∪B∈CB.For every c < m(U) there exists a finite collection of disjoint sets Bj ∈ C,j = 1, . . . , k, such that c < 3n

∑kj=1 m(Bj).

Proof: Choose a compact set K ⊂ U with m(K) > c. Since K is compactthere is a finite collection of balls A1, . . . , Am which cover K; and we mayorder them in decreasing radii, r1 ≥ r2 ≥ . . . . Choose B1 = A1; cast outall the Ai which have a nonempty intersection with B1; and continue theprocess until all the A′is are exhausted. We have now obtained a finitecollection B1, . . . Bk of disjoint open balls.

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5.4. DIFFERENTIATION ON RN 51

There are two possibilities to consider. If U is originally a disjoint unionof open balls, then Bj = Aj and the result is immediate. Otherwise someof the Ai will be not included in the B′

js, and for each such Ai, there is asmallest integer j such that Ai ∩Bj 6= ∅. By the construction, the radius ofAi will be at most that of Bj ; and so Ai ⊂ B∗

j , the ball concentric with Bj

with three times the radius. Then K ⊂ ∪kj=1B

∗j and

c < m(K) ≤k∑

j=1

m(B∗j ) = 3n

k∑

j=1

m(Bj).k. .

The central method is based on the use of the Hardy-Littlewood maximalfunction, defined by

Hf(x) = supr>0

1m(Br(x))

Br(x)|f | dm, f ∈ L1

loc.

Theorem 5.4.3 [Hardy-Littlewood Maximal Theorem] For f ∈ L1(Rn,m)and α > 0,

m(x : Hf(x) > α) ≤ 3n

α||f ||1.

Proof: Let Uα = x : Hf > α; and let c = m(Uα). For each x ∈ Uα

there is a radius rx such that

αm(Brx(x)) ≤∫∫

Brx (x)

|f | dm.

Since the balls Brx(x) cover Uα, there is a finite disjoint union of them,Bj, for which

c < 3nk∑

j=1

m(Bj) ≤ 3n

α

k∑

j=1

∫∫

Bj

|f | dm ≤ 3n

α

∫∫

Rn

|f | dm.

Since this result is true for all c < m(Uα), the theorem follows, with constantC = 3n. k. .

We also need the Tschebyshev inequality: Given any f ∈ L1(Rn,m),,and all α > 0,

m(Fα) ≤ 1α||f ||1, Fα = x : |f(x)| ≥ α. (5.3)

This simple inequality is well known in Probability theory; its proof is leftto the reader.

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52 CHAPTER 5. DIFFERENTIATION THEORY

Lemma 5.4.4 For x ∈ Rn and r > 0, let

E(f, x, r) =

∫∫Br(x)

f dm

m(Bxr )

. (5.4)

Then if f ∈ L1loc, limr→0 E(f, r, x) = f(x) a.e.

Proof: If f is continuous, the proof is well known, and is the standard methodof proof of the Fundamental Theorem of Calculus. In the general case, weapproximate f by a continuous function g. We may restrict ourselves to acompact ball K ⊂ Rn, since the result is purely local; thus we may truncatef by setting it identically equal to 0 outside K. Thus, given any ε > 0 thereis a continuous function g with support in K such that

∫K |f − g| dm < ε.

Since K is compact, g is uniformly continuous on K; and thus for everyδ > 0 there is an r > 0 such that

|g(x)− g(y)| < δ, |x− y| < r, x, y ∈ K.

Therefore |E(g, x, r)− g(x)| < δ; and limr→0 E(g, x, r) = g(x).Therefore

lim supr→0

|E(f, x, r)− f(x)| =

lim supr→0

|E(f − g, x, r) + (E(g, x, r)− g(x)) + (g(x)− f(x))|

≤ H(f − g)(x) + |f(x)− g(x)|.

Let

Eα = x : lim supr>0

|E(f, x, r)− f(x)| > α, Fα = x : |f − g|(x) > α.

Then by the inequality above, Eα ⊂ Fα/2 ∪ x : H(f − g)(x) > α/2. Bythe Tchebyshev inequality, and the Maximal Theorem,

α

2m(Fα/2) ≤

∫∫

K

|f − g| dm < ε,

α

2m(x : H(f − g)(x) > α/2) < 3n

∫∫

K

|f − g| dm < 3nε;

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5.4. DIFFERENTIATION ON RN 53

hencem(Eα) ≤ 2ε

α(3n + 1).

Since ε is arbitrary, m(Eα) = 0 for all α > 0. Since limr→0 E(f, r, x) = f(x)on the complement of ∪nE1/n, the result follows.

We may now complete the proof of the Lebesgue differentiation theorem.Let

Ez = x : limr→0

E(|f − z|, x, r) = |f(x)− z|.

Then m(Ecz) = 0 by the Lemma above. Let D be a countable dense subset

of C and E = ∪z∈DEz. Then m(Ec) = 0; and for x ∈ E and any ε > 0 thereexists a z ∈ D such that |f(x)− z| < ε/2. Then

limr→0

E(|f − f(x)|, x, r) ≤ limr→0

(E(|f − z|, x, r) + E(|z − f(x)|, x, r))

< limr→0

E(|f − z|, x, r, ) +ε

2= |f(x)− z|+ ε

2

<ε.

Since ε is arbitrary, it follows that limr→0 E(|f − f(x)|, x, r) = 0 for x ∈ E,and the proof of the theorem is complete. k. .

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54 CHAPTER 5. DIFFERENTIATION THEORY

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Chapter 6

Lp spaces

6.1 Banach spaces

Let (X,M, µ) be a measure space and for 1 ≤ p < ∞ define

Lp(X, dµ) = f : ||f ||p < ∞, ||f ||p =(∫

|f |pdµ

)1/p

.

Technically Lp consists of equivalence classes of functions which are equalalmost everywhere; but normally we shall not refer to this matter explicitly.

A Banach space is a complete normed linear vector space. A norm on avector space B is defined to be a positive functional, generally denoted by|| · ||, with the properties

• ||f || ≥ 0; ||f || = 0 iff f = 0.

• ||λf || = |λ|||f ||;• ||f + g|| ≤ ||f ||+ ||g||.

We shall show that the spaces Lp(X, dµ) are Banach spaces for 1 ≤ p < ∞with the norm || · ||p given as above. The case p = ∞ is also included, andwill be discussed below. By complete we mean that any Cauchy sequencefn ∈ B converges to a point f ∈ B.

The first property of the norm is clear, since ||f ||p = 0 implies thatf = 0 a.e. The second property is also easy to prove. Hence the only itemof substance is the triangle inequality.

Theorem 6.1.1 [Holder’s Inequality] For 1 < p < ∞ and p−1 + q−1 = 1,we have ||fg||1 ≤ ||f ||p||g||q for all f ∈ Lp and g ∈ Lq. Equality holds iff fand g are scalar multiples of one another.

55

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56 CHAPTER 6. LP SPACES

Proof: The result is trivial if either f or g is zero, or if ||f ||p or ||g||q is infinite.In the general case we may rescale f and g so that ||f ||p = ||g||q = 1. Then,by the concavity of the logarithm,

ln a1/pb1/q =1p

ln a +1q

ln b ≤ ln(

a

p+

b

q

).

Exponentiating this inequality, and taking a = |f |p, b = |g|q, we obtain

|f ||g| ≤ |f |pp

+|g|qq

.

Integrating this inequality over X we obtain

||fg||1 ≤ 1p||f ||p +

1q||g||q = 1 = ||f ||p||g||q.

The result follows by reintroducing the original scaling. k. .

The indices p and q are called conjugate indices. Unless otherwise statedwe shall take p and q to be conjugate indices in the following.

Theorem 6.1.2 [Minkowski’s Inequality] If 1 ≤ p < ∞ and f, g ∈ Lp then

||f + g||p ≤ ||f ||p + ||g||p.Proof: The result is immediate if p = 1 or f + g = 0 a.e. Otherwise, notethe inequality |f + g|p ≤ (|f |+ |g|)|f + g|p−1; and integrate over X:

||f + g||pp =∫|f + g|pdµ ≤

∫(|f |+ |g|)|f + g|p−1dµ

≤∫|f ||f + g|p−1dµ +

∫|g||f + g|p−1dµ

≤(||f ||p + ||g||p)(∫

|f + g|(p−1)qdµ

)1/q

≤(||f ||p + ||g||p)(∫

|f + g|pdµ

)p/q

=(||f ||p + ||g||p)||f + g||p−1p ,

and the inequality follows by dividing both sides by ||f + g||p−1p .

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6.1. BANACH SPACES 57

Theorem 6.1.3 The Lp spaces are Banach spaces for 1 ≤ p < ∞.

Proof: Let fj be a Cauchy sequence in Lp and choose a subsequence nj

so that ||fnj+1 − fnj ||p < 2−j . Putting Gn =∑n

j=1 |fnj+1 − fnj |, we see that||Gn||p <

∑j 2−j = 1. Since Gn is a monotone increasing sequence, we

have, by the monotone convergence theorem, that∫|G|pdµ = lim

n→∞

∫|Gn|pdµ < 1.

Therefore G =∑∞

j=1 |fnj+1 − fnj | < ∞ a.e. and the series∑

j(fnj+1 − fnj )converges pointwise a.e. The limit f = limj fnj therefore exists and is finitea.e.; and, moreover,

||f − fnk||p = ||

∞∑

j=k+1

fnj+1 − fnj ||p ≤∞∑

j=k+1

||fnj+1 − fnj ||p < 2−k,

||f ||p ≤ ||fnk||p + ||f − fnk

||p.

It follows that fnj → f in the Lp norm, and f ∈ Lp. Moreover, the originalsequence fn → f , since

||f − fn||p ≤ ||f − fnj ||p + ||fnj − fn||p < 2−j + 2−j = 21−j , n > nj .

k. .

We conclude by treating the case p = ∞. We define

||f ||∞ = infa : a ≥ 0, µ(|f | > a) = 0.

The space L∞ = L∞(X,M, µ) is the vector space of all measurable functionsf for which ||f ||∞. The set E = |f | > ||f ||∞ necessarily has measure 0.Putting g = fχE we see that f = g a.e. and ||f ||∞ = sup |g|. Hence || · ||∞is a norm. Moreover, L∞ is a Banach space, since any Cauchy sequence innorm is a uniform Cauchy sequence, and so converges pointwise a.e.

Theorem 6.1.4 If 0 < p < q < r ≤ ∞ then Lp ∩ Lr ⊂ Lq and

||f ||q ≤ ||f ||λp ||f ||1−λr ,

1q

p+

1− λ

r, 0 < λ < 1.

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58 CHAPTER 6. LP SPACES

Proof: For r = ∞, |f |q ≤ ||f ||q−p∞ |f |p; so ||f ||q ≤ ||f ||p/qp ||f ||1−p/q

∞ =||f ||λp ||f ||1−λ

r , where λ = p/q. For r < ∞ we use Holder’s inequality forthe conjugate indices p/qλ, r/q(1− λ) to obtain the inequality

∫|f |qdµ =

∫|f |λq|f |(1−λ)q)dµ ≤ ||f ||λq

p ||f ||(1−λ)qr .

The result follows by taking the qth root of both sides of this inequality.k. .

Theorem 6.1.5 Let 1 ≤ pj , r ≤ ∞ and suppose that fj ∈ Lpj for j =1, . . . , n. Then

||f1 · · · fn||r ≤n∏

j=1

||fj ||pj , where1r

=n∑

j=1

1pj

.

The proof is left as an exercise.

6.2 Duality

A bounded linear functional on a Banach space B is a mapping F : B 7→ Csuch that |F (x)| ≤ C||x||, where || · || is the norm on B. The collection ofall bounded linear functionals on a Banach space B is itself a normed vectorspace with norm

||F || = supx∈B

|F (x)|||x|| = infA : |F (x)| ≤ A||x||.

The space of linear functionals on B is called the dual space and is denotedby B∗.Theorem 6.2.1 B∗ is itself a Banach space.

Proof: We leave as an exercise the proof that the expression ||F || definedabove is space of functionals defined above is in fact a norm. Suppose Fnis a Cauchy sequence of functionals in the dual norm. Then for each x ∈ B,Fn(x) is a Cauchy sequence of complex numbers, hence converges to alimit, denoted by F (x). Since the limit is a linear operation, F (x) is linearin x, hence F is a linear functional. Moreover, |F (x)| ≤ limn |Fn(x)| ≤limn ||Fn||||x||.1 In particular, ||F || ≤ limn ||Fn||.

1Note that | ||Fn||− ||Fm|| | ≤ ||Fn−Fm||; and therefore ||Fn|| is a convergent sequenceof real numbers.

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6.2. DUALITY 59

Given ε > 0, choose N so that ||Fn − Fm|| < ε for all m,n > N. Then

|F (x)− Fn(x)| = limm→∞ |Fm(x)− Fn(x)|

≤ limm→∞ ||Fm − Fn|| ||x|| < ε||x||, for all n ≥ N.

It follows that ||F −Fn|| ≤ ε, hence limn→∞ ||F −Fn|| = 0. This shows thatevery Cauchy sequence in B∗ has a limit in B∗. k. .

Given a function g, measurable on X, consider the functional Fg definedby Fg(f) =

∫fgdµ. By Holder’s inequality, |Fg(f)| ≤ ||f ||p|||g||q; thus if

g ∈ Lq then Fg is a linear functional on Lp, i.e. Fg ∈ Lp ∗. In other words,Lq → (Lp)∗; more precisely, every element q ∈ Lq is mapped into (Lp)∗ viathe mapping g 7→ Fg.

A measure µ is semi-finite if every set B of positive µ-measure has asubset of finite µ measure.

Theorem 6.2.2 For 1 ≤ p < ∞ the injection Lq → (Lp)∗ defined by g 7→Fg is an isometric injection of Lq into (Lp)∗; that is ||Fg|| = ||g||q. The sameholds for p = ∞ when µ is semi-finite.

Proof: We need to show that

||g||q = supf

∫fgdµ, ||f ||p = 1

. (6.1)

For 1 < p < ∞, ||Fg|| ≤ ||g||q by Holder’s inequality. If g 6= 0 let

f =|g|q−1sgn(g)||g||q−1

q

.

Then ||f ||p = 1 and

Fg(f) =∫

fgdµ =∫ |g|q||g||q−1

q

dµ =||g||qq||g||q−1

q

= ||g||q.

This shows that the supremum (6.1) is attained by this choice of f .For p = 1 take f = sgn g. Then ||f ||∞ = 1 while Fg(f) =

∫fgdµ =∫ |g|dµ = ||g||1. Therefore

||g||1 ≤ sup||f ||∞=1

Fg(f) = ||Fg||.

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60 CHAPTER 6. LP SPACES

On the other hand,

|Fg(f)| ≤∣∣∣∣∫

fgdµ

∣∣∣∣ ≤∫|fg|dµ ≤ ||g||1||f ||∞,

so ||Fg|| ≤ ||g||1.Now consider the case p = ∞ and suppose µ is semi-finite. Given ε > 0

choose Aε = x : |g(x)| > ||g||∞ − ε. If µ(Aε) > 0 we choose a subsetB ⊂ Aε with finite measure, and put

f =χB

µ(B)sgn g.

Then ||f ||1 = 1 so

||Fg|| ≥ |Fg(f)| =∣∣∣∣∫

fgdµ

∣∣∣∣ =∫

B

|g|µ(B)

dµ ≥ ||g||∞ − ε.

Since this holds for all ε > 0 we have ||g||∞ ≤ ||Fg||. On the other hand,|Fg(f)| ≤ ∫ |fg|dµ ≤ ||g||∞||f ||1; ||Fg|| ≤ ||g||∞; and therefore ||Fg|| =||g||∞.

Theorem 6.2.3 Let 1 ≤ p < ∞, let q be the conjugate index; and let µ beσ-finite. Then the embedding g 7→ Fg is a surjective mapping of Lq(dµ) ontoLp(dµ)∗.

Proof: Given a functional F ∈ Lp(dµ)∗, let νF (E) = F (χE) for allmeasurable sets E. Then νF is a countably additive σ-finite measure, and isabsolutely continuous with respect to µ. By the Radon-Nikodym theorem,there is a measurable function g ∈ Lp(dµ) such that νF (E) =

∫gdµ; and by

linearity the result also holds for all simple functions φ: F (φ) = νF (φ) =∫φgdµ. Since the simple functions are dense in Lp, the result immediately

extends to all measurable functions f ∈ Lp. k. .

6.3 Distribution functions

Given a real valued function f on a measure space (X,M, µ) its distributionfunction λf is defined by

λf (α) = µ(x : |f(x)| > α).

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6.3. DISTRIBUTION FUNCTIONS 61

Proposition 6.3.1 The distribution function is decreasing and right con-tinuous in α, and increasing in |f |. If fn ↑ f then λfn ↑ λf . Finally,

λf+g(α) ≤ λf (α/2) + λg(α/2).

Proof: The proof is left to the reader.If λf (α) is finite for all α > 0, then the distribution function defines a

negative Borel measure on R+ via the premeasure λf ((a, b]) = λf (b)−λf (a).We denote the associated

Proposition 6.3.2 For any Borel measurable function φ defined on R+ wehave ∫

φ(|f |)dµ = −∫ ∞

0φ(α)dλf (α),

where the integral on the right is the Lebesgue-Stieltjes integral with respectto the measure dλf .

In particular, ∫|f |p = −

∫ ∞

0αpdλf (α).

Proof: The result is immediate for φ = χ(α,∞). The result extends immedi-ately to characteristic functions for any Borel set, then to simple functions,then to integrable functions by the monotone convergence theorem. k. .

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62 CHAPTER 6. LP SPACES

6.4 Linear Transformations

A bounded linear transformation from a Banach space B to a Banach spaceB′ is a linear transformation L : B 7→ B′ such that ||Lu||B′ ≤ C||u||B forsome fixed positive constant C. The best possible such constant is the normof L, denoted by ||L||, and defined by

||L|| = supu∈B

||Lu||B′||u||B .

An important class of linear transformations on Banach spaces is the classof linear integral operators mapping Lp(Y, dµ) to Lq(X, dν) defined by

(Tf)(x) =∫

K(x, y)f(y)dµ(y), (6.2)

where K(x, y), called the kernel is a measurable function on the productspace (X × Y,M⊗N ).

The question of when an integral operator is a bounded transformationfrom Lp to Lq is of fundamental interest in applications; and a variety oftechniques exist. For example, the Fourier transformation

Ff =1√2π

∫e−ikxf(y)dy,

is an isometry from L2(R) to itself, though this fact is not immediatelyobvious.

An interesting class of linear integral operators is the class of convolutionoperators, which typically arise in connection with the Fourier transform andpartial differential equations with constant coefficients. The convolution oftwo functions f and g on the line is defined by

(f ∗ g)(x) =∫

f(x− y)g(y) dy.

By Fubini’s theorem,

||f ∗ g||1 =∫ ∣∣∣∣

∫f(x− y)g(y) dy

∣∣∣∣ dx ≤∫∫

|f(x− y)g(y)| dy dx

=∫|g(y)|

∫|f(x− y)| dx

dy =

∫|g(y)| ||f ||1 dy = ||f ||1||g||1.

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6.4. LINEAR TRANSFORMATIONS 63

This calculation shows that the linear transformation defined by Lgf = f ∗gis a bounded transformation on L1(R) into itself, and that ||Lg|| ≤ ||g||1.Similarly, ||f ∗ g||∞ ≤ ||g||1||f ||∞.

More generally, we have

Theorem 6.4.1 [Generalized Young Inequality] Let f ∈ Lp(R), g ∈ Lq(R).Then

||f ∗ g||r ≤ ||f ||p||g||q, 1p

+1q

= 1 +1r, (6.3)

hence the convolution operator Lg is a bounded linear transformation fromLp to Lr with norm ||Lg|| ≤ ||g||q.

This theorem is easily proved using an interpolation argument.

Theorem 6.4.2 [Riesz-Thorin Interpolation Theorem] Let (X,M, µ) and(Y,N , ν) be measure spaces and 1 ≤ po, p1, q0, q1 ≤ ∞. Suppose T is abounded linear transformation from Lpj (dµ) to Lqj (dν) for j = 0, 1 withrespective norms M0 and M1:

||Tf ||q0 ≤ M0||f ||p0 , ||Tf ||q1 ≤ M1||f ||p1 .

Let

1pλ

=1− λ

p0+

λ

p1,

1qλ

=1− λ

q0+

λ

q1, 0 < λ < 1.

Then T is a bounded linear transformation from Lpλ to Lqλ and

||Tf ||qλ≤ M1−λ

0 Mλ1 ||f ||pλ

.

We apply the interpolation theorem in two stages, first establishing (6.3)for the case q = 1. By the argument above, we know that Lg is a boundedoperator from L1 to L1 and from L∞ to L∞, in both cases with norm ||g||1.We apply the Interpolation theorem with

q0 = p0 = 1, q1 = p1 = ∞

qλ = pλ =1

1− λ, M0 = M1 = ||g||1.

Hence by the interpolation theorem, (p = pλ for 1 < p < ∞),

||Lgf ||p ≤ ||g||1||f ||p, 1 ≤ p ≤ ∞. (6.4)

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64 CHAPTER 6. LP SPACES

This inequality is known as Young’s inequality.Next, suppose 1 < q, let g ∈ Lq, and consider the linear transformation

Lgf = f ∗ g. For p = 1 and r = q we have ||Lgf ||q ≤ ||g||q||f ||1 by Young’sinequality; and for r = ∞ and p = q/(q − 1) we have ||Lgf ||∞ ≤ ||g||q||f ||pby Holder’s inequality. Thus

||Lgf ||r0 ≤ M0||f ||p0 , p0 = 1, r0 = q;

||Lgf ||r1 ≤ M1||f ||p1 , p1 =q

q − 1, r1 = ∞,

and M0 = M1 = ||g||q. By Theorem 6.4.2,

||f ∗ g||rλ= ||Lgf ||rλ

≤ ||g||q||f ||pλ,

where1pλ

=1− λ

p0+

λ

p1,

1rλ

=1− λ

q.

The result then follows from the identity

1pλ

+1q

= 1 +1rλ

.k. .

If vj is a collection of vectors in a Banach space, then by a repeated ap-plication of the triangle inequality, ||∑j vj || ≤

∑j ||vj ||. This result extends

to integrals of a family of vectors v(t). Namely,∣∣∣∣∣∣∣∣∫

v(t)dµ(t)∣∣∣∣∣∣∣∣ ≤

∫||v(t)||dµ(t). (6.5)

This result is established by approximating the integral by integrals of sim-ple, vector valued functions s(t) =

∑j vjχEj (t). In that case the inequality

is a direct consequence of the triangle inequality for the norm.

Theorem 6.4.3 Let (X,M, dµ), (Y,N , dν) be measure spaces and K(x, y)be measurable with respect to M⊗N . Suppose that ||K(·, y)||1 ≤ C and||K(x, ·)||1 ≤ C for some fixed constant C, where ||K(·, y)||1 denotes theL1 norm of K(x, y) in (X, dµ) for fixed y, etc. Then the integral operatorT defined by (6.2) is a bounded linear transformation from Lp(Y, dν) toLp(X, dµ) with norm less than or equal to C.

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6.4. LINEAR TRANSFORMATIONS 65

Proof: For the cases p = 1,∞ the proofs are straightforward and requireonly the hypotheses ||K(x, ·)||1 ≤ C, respectively ||K(·, y)||1 ≤ C.

In the case 1 < p < ∞, let q be the conjugate index and apply Holder’sinequality:∫|K(x, y)f(y)|dν(y) ≤

[∫|K(x, y)|dν(y)

]1/q [∫|K(x, y)||f(y)|pdν(y)

]1/p

≤C1/q

[∫|K(x, y)||f(y)|pdν(y)

]1/p

, a.e.

Hence, by the Tonelli-Fubini theorem∫ [∫

|K(x, y)||f(y)|pdν(y)]p

dµ(x) ≤Cp/q

∫∫|K(x, y)| |f(y)|pdν(y)dµ(x)

≤C(p/q)+1

∫|f(y)|pdν(y).

This implies that the left side is finite; and, taking pth roots, we obtain||Tf ||p ≤ ||f ||p. k. .

Theorem 6.4.4 Let K(x, y) be Lebesgue measurable in R+ × R+ and sup-pose that K(λx, λy) = λ−1K(x, y) for all λ > 0. Suppose also that

∫ ∞

0|K(x, 1)|x−1/p dx = C < ∞, 1 ≤ p < ∞.

Let q be the conjugate index to p. For f ∈ Lp(R+) and g ∈ Lq(R+) let

Tf(x) =∫ ∞

0K(y, x)f(y) dy, Sg(x) =

∫ ∞

0K(x, y)g(y) dy.

Then T and S are bounded linear transformations, with norm less than orequal to C, on Lp(R+) and Lq(R+) respectively.

Proof: We have∫ ∞

0K(y, x)f(y) dy =

∫ ∞

0K(xz, x)f(xz)x dz =

∫ ∞

0K(z, 1)fz(x)dz,

where fz(x) = f(xz). Now ||fz||p = z−1/p||f ||p; so by (6.5),

||Tf ||p ≤∫ ∞

0|K(z, 1)|||fz||pdz = ||f ||p

∫ ∞

0|K(z, 1)|z−1/pdz = C||f ||p.

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66 CHAPTER 6. LP SPACES

The proof of the boundedness of the operator S on Lq is similar; we notethat

∫ ∞

0|K(1, y)|y−1/qdy =

∫ ∞

0|K(u, 1)|u−1−1/qdy

=∫ ∞

0|K(u, 1)|u−1/pdu = C.

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Chapter 7

Exercises

7.1 Set Theory

1. Prove lim infn an ≤ lim supn an for a sequence of real numbers an.Prove that if En is a sequence of sets, then lim infn En ⊂ lim supn En.

2. (a) What is the lexicographic ordering on Rn? Show that Rn is lin-early ordered by the lexicographic ordering. Is the lexicographicordering a well ordering of Rn?

(b) A cone in a real vector space V is a set of vectors K such thatx, y ∈ K implies αx + βy ∈ K for all α, β ≥ 0. Show that therelation x ≥ y iff x− y ∈ K defines an ordering on V .

(c) Let V = R2 and let ≥ be an ordering on V . Define K = x : x ≥0. Is K necessarily a cone? Prove or give a counter example.

(d) Let K be the closed first quadrant in R2 and show that the rela-tion defined by x ≥ y iff x − y ∈ K defines an ordering. Is thisordering linear? Is it a well-ordering?

(e) Let S be the unit square (x1, x2) : 0 ≤ x1, x2 ≤ 1. What arethe maximal linearly ordered subsets of S corresponding to theabove ordering? What are the maximal elements of S? Is therea maximum element of S relative to this ordering?

7.2 Measures

1. Let Ω be the set of events consisting of an infinite sequence of inde-pendent coin tosses. Let Xn be the sequence of independent random

67

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68 CHAPTER 7. EXERCISES

variables with

Xn =

1 nth toss is heads;0 nth toss is tails.

ThenPrXn = 1 = PrXn = 0 = 1

2 .

Using the random variable X given by

X =∞∑

n=1

Xn

2n,

show that Ω can be identified with the unit interval [0, 1] and find thecorresponding probability measure

µ((a, b]) = Pra < X ≤ b, 0 ≤ a < b ≤ 1.

2. Let µ be a measure and En a sequence of measurable sets. Provethat

µ(lim infn→∞ En) ≤ lim inf

n→∞ µ(En) ≤ lim supn→∞

µ(En) ≤ µ(lim supn→∞

En),

provided that µ(∪jEj) < ∞. Hence if En → E and µ(∪jEj) < ∞,then lim

n→∞µ(En) = µ(E).

3. Let µc be the measure generated by the Cantor function, and let C bethe Cantor set, the complement of the union of all open middle thirds.Prove that µc(Cc) = 0, hence that µc(C) = 1; whereas the Lebesguemeasure of Cc equals 1.

7.3 Integration

1. Why is Theorem 3.4.2 important?

2. Let X,M, µ be a finite measure space. Show that the measurablefunctions on this measure space form a metric space under the metric

ρ(f, g) =∫ |f − g|

1 + |f − g|dµ.

Show that convergence in the metric is equivalent to convergence inmeasure. Is this metric space complete? (Identify functions which arealmost everywhere equal.

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7.4. DIFFERENTIATION THEORY 69

3. Show that the conclusion of Egorov’s theorem holds if the hypothesisµ(X) < ∞ is replaced by the condition |fn| ≤ g for all n.

4. Let fn → f [meas] and gn → g [meas]. Prove that fngn → fg [meas]provided µ(X) < ∞, but that the conclusion fails if µ is not finite.

5. Prove Lusin’s theorem: If f : [a, b] → C is Lebesgue measurable, thengiven any ε > 0 there is a compact set E ⊂ [a, b] such that µ(Ec) < εand f restricted to E is continuous.

6. Show the iterated integrals of

x2 − y2

(x2 + y2)2, 0 ≤ x, y ≤ 1,

are unequal.

7. Let f ∈ BV [a, b], and prove the decomposition

f = T+ − T−, T± =12(T ± f).

Show that

T±f (a, b) = supPn

n∑

j=1

[f(xj)− f(xj−1)]±, n ∈ N,

where Pn are the partitions of [a, b] and [x]+ = max(0, x); [x]− =max(0,−x).

8. The indefinite integral F of a function f ∈ L1([a.b],m), where m isLebesgue measure, is of bounded variation, and ||F ||var =

∫ ba |f | dm.

7.4 Differentiation theory

1. Let ν be a signed measure. Show that the positive and negative varia-tions of ν satisfy ν+(E) = ν(E ∩P ), ν−(E) = ν(E ∩N) and are givenby

ν+(E) = supν(A) : A ⊂ E, ν−(E) = − infν(A) : A ⊂ E.

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70 CHAPTER 7. EXERCISES

7.5 Lp Spaces

1. The operators

Tf(x) =1x

∫ x

0f(y) dy, Sf(x) =

∫ ∞

x

1yf(y) dy

are bounded linear tranformations on Lp(R+) and Lq(R+) respectively,for all 1 < p ≤ ∞, 1 ≤ q < ∞; and

||T ||p ≤ p

p− 1, ||S|| ≤ q.

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Bibliography

[1] Caratheodory, C. Vorlesungen uber Reele Funktionen. Teubner, 1918.

[2] Folland, G. Real Analysis: Modern Techniques and Their Applica-tions, second edition. Wiley-Interscience, 1999.

[3] Halmos, P. Naive Set Theory. Van Nostrand, 1960.

[4] Monroe, M.E. Introduction to Measure Theory and Integration.Addison-Wesley, 1953.

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