Measure Theory & Integration

Transcription

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

2 Contents 1 Preliminaries 1 2 Measures Area and Measure Carathéodory s Theorem Borel measures Integration Measurable functions Integration of non-negative functions The Dominated Convergence Theorem The Riemann Integral Some Convergence Theorems Product Spaces Fubini s Theorem Lebesgue measure in R n Differentiation Theory Differentiation Theory of Functions Signed Measures The Lebesgue-Radon-Nikodym theorem Differentiation on R n L p spaces Banach spaces Duality Distribution functions Linear Transformations i

3 ii CONTENTS 7 Exercises Set Theory Measures Integration Differentiation theory L p Spaces

4 List of Figures 3.1 Fundamental solution of the heat equation iii

5 iv LIST OF FIGURES

6 Chapter 1 Preliminaries Let {a n } be a sequence of real numbers. The limit superior and limit inferior of the sequence are defined as lim sup a n = inf n sup k 1 n k a n, lim inf n a n = sup inf a n. n k k 1 The following is left as an exercise: lim inf n a n lim sup a n. (1.1) n Let {E n } be a countable collection of sets. We define Ē = lim sup E n = n E = lim inf n E n = k=1 n=k k=1 n=k E n = {x : x E n for infinitely many n}. E n = {x : x E n for all but finitely many n}. It is a simple exercise to show that E = lim inf n E n lim sup n E n = Ē. If E = Ē we say the limit of the sequence of sets exists and is equal to this common set. A partial ordering of a set X is a relation which satisfies the following axioms: Given x, y, z X, x y, y z implies x z. x y, y x implies x = y. x x. 1

7 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 real line is linearly ordered; the space of continuous real-valued functions on a topological 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 set X, called the power set of X and denoted by P(X) is partially ordered by the inclusion relation. A maximal element of a partially ordered set X is an element M such that if x X and x M, then x = M. A minimal element is defined similarly. A maximal or minimal element may or may not exist, and in general is not unique. The Hausdorff Maximal Principle states that very partially ordered set has a maximal linearly ordered subset. Zorn s Lemma states that if X is a partially ordered set for which every linearly ordered subset has an upper bound, then X has a maximal element. The Cartesian product of a collection of sets {X α } α A, denoted by α X α is defined to be the set of all maps f : A α X α such that f(α) X α. The Axiom of Choice states that if {X α } is a non-empty collection of non-empty sets, then α X α is non-empty. That is, it is possible to construct a set with one element from each of the X α. The Hausdorff Maximal principle, Zorn s Lemma and the Axiom of Choice are all equivalent. A well-known introduction to set theory is Paul Halmos Naive Set Theory [3]

8 Chapter 2 Measures 2.1 Area and Measure The modern notion of measure, developed in the late Nineteenth and early Twentieth centuries, is an extension of the notion of area, developed by the Greeks. The area of a planar rectangle is defined to be the product of its length and width. By elementary geometric arguments, the area of a triangle, and hence the area of any polygon, can be determined. To determine the area of a circle, the Greeks constructed sequences of inscribed and circumscribed regular polygons, with the number of sides tending to infinity, to obtain inner and outer approximations to the circle. This gives a sequence of lower and upper estimates of the area of the circle, and the area of the circle is defined to be the common limit as the number of sides tends to infinity. This procedure, known as the Method of Exhaustion (You could Google it), was formulated by Euxodus of Cnidos ( B.C.E.), and was developed systematically by Archimedes ( B.C.E.). Archimedes developed systematic algorithms for calculating not only the areas, but the volumes of a wide variety of geometric figures, and laid the foundations of integration theory that are still relevant today. In fact, the Riemann integral is really an application of the method of exhaustion and the principles of Archimedes. What properties do we want of a notion of Area? Suppose A denotes the area of a set E. The most obvious properties to require of the set function 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 Euclidean geometry, 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

9 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 ( ) E j = µ(e j ), if E j E k =, j k. The sequence of sets {E j } is said to disjoint. Definition Let X be a set and let M P(X) be a collection of subsets of X. A set function µ defined on M is said to be a measure if µ is non-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 nonmeasurable sets that is, that µ is not defined everywhere on P(X). We demonstrate the existence of non-measurable sets for any measure µ on the unit interval [0, 1] which is invariant under translations (this happens to be Lebesgue measure, as we shall see). Define an equivalence relation on [0, 1] by x y if x y is rational. This partitions the unit interval into equivalence classes. By the Axiom of Choice, there is a set N consisting of one element from each equivalence class. For each rational number r, let N r = [N+r] mod1, that is, the translate of N by r with wrapped around around 1. (Or, do the construction on the unit circle). Now clearly [0, 1] = r N r and the sets N r are disjoint. Hence by countable additivity, and translation invariance, µ([0, 1]) = r µ(n r) = r µ(n). This sum must be infinity or zero, depending on whether µ(n) is positive or zero. Thus µ cannot be defined on N. We now turn our attention to the collection M, which are called measurable sets. In view of our requirement of countable additivity of µ it is natural to require that M be closed under countable disjoint unions. In addition, it is natural to require that the complement of any measurable set also be measurable. Then if A and B are measurable, A B = (A c B c ) c is measurable. Definition A σ-algebra M is a collection of subsets of which is closed under countable unions and complementations. The verification of the following proposition is left to the reader.

10 2.1. AREA AND MEASURE 5 Proposition Any countable union of sets may be replaced by a countable disjoint union of sets. The intersection of any family of σ-algebras is again 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 the σ-algebra generated by E. For example, if X is a topological space, the open sets on X generate a σ-algebra, called the Borel σ-algebra, which is denoted by B X. The Borel algebra of sets on R is generated by the open intervals, closed intervals, half-open intervals, or by rays (a, ), (, a), or by 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 Borel algebra B X is called a Borel measure. Borel measures play a pre-eminent role in measure theory on R n. Borel measures on the line are constructed by specifying the measure of each 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-open interval (a, b] is defined by µ 0 ((a, b]) = F (b) F (a). The extension of µ 0 to B R 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 variable X, defined by F (x) = Pr(X x), generates a probability measure, and Pr(a < X b) = F (b) F (a) = µ F ((a, b]). The Dirac measure is obtained by taking F equal to the Heaviside step function: H(x) = 0 for x < 0 and H(x) = 1 for x 1. The Dirac measure assigns the measure 1 to any set containing the origin, and 0 to all other sets. The Cantor function F generates a measure µ F which assigns the measure 1 to the Cantor set, which has Lebesgue measure 0; and 0 the the union of open middle thirds, which has total length 1. Theorem 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 E j M, µ( j E j ) j µ(e j);

11 6 CHAPTER 2. MEASURES continuous from below: If E j E j+1 is an increasing sequence of measurable sets, then µ ( E j ) = lim j µ(e j ); µ is continuous from above: If E j E j+1 and µ(e 1 ) <, then µ ( E j ) = lim j µ(e j ). Proof: If E F then F = E (F \E) so µ(f ) = µ(e) + µ(f \E) µ(e). If {E j } is a sequence of sets with non-empty intersections, then for all n, µ ( n ) ( ) n n E j = µ n F j = µ(f j ) µ(e j ), where ( ) F 1 = E 1, F k = E k \ k 1 E k. Then the F j are disjoint and F j E j ; letting n we obtain the second statement in the theorem. To prove continuity from below, let E 0 = and note that µ ( ) E j = µ(e j \E j 1 ) = lim n n µ(e j \E j 1 ) = lim n µ(e n). Finally, to prove continuity from above, put F j = E 1 \E j. Then the sequence {F j } is increasing, and Therefore by the previous result, Since µ(e 1 ) < it follows that E 1 = ( F j ) ( E j ). µ(e 1 ) =µ ( F j ) + µ ( E j ) = lim j µ(f j ) + µ ( E j ) = lim j µ(e 1 ) µ(e j ) + µ ( E j ). µ ( E j ) = lim j µ(e j ).

12 2.2. CARATHÉODORY S THEOREM 7 A null set of a measure space (X, M, µ) is a set N for which µ(n) = 0. By subadditivity, countable unions of null sets are null sets. If P is a proposition which is true except on a set of measure zero, we say that P holds almost everywhere (a.e.) If the σ-algebra M includes all null sets, we say that the measure µ is complete. Any measure can be extended to be complete. 2.2 Carathéodory s Theorem In this section we describe a general constructive procedure for obtaining complete measures. Recall that the area enclosed by a circle is obtained as the limit of the areas of a sequence of circumscribed polygons. In fact, the area of a disk is by definition the infimum of the areas of all circumscribed polygons. This leads to the general notion of outer measure, as follows. Definition An outer measure on a non-empty set X is a set function µ defined on P(X) which is non-negative, monotone, and countably subadditive. In practice, an outer measure is obtained by starting with an elementary concept of measure, called a pre-measure, which is defined on a restricted class of sets E P(X) and approximating general subsets X from the outside. We assume that E is an algebra, that is, that it is closed under complementation and finite unions. For example, Lebesgue measure in the plane is constructed beginning with the notion of area of a rectangle, extending it to polygons, and defining the outer area of a plane figure by taking the infimum of the areas of all circumscribed polygons. Proposition Let E P(X) be an algebra of sets and ρ a nonnegative function defined on E, such that ρ( ) = 0. For any A X define µ (A) = inf ρ(e j ) : A E j, E j E. Then µ is an outer measure. Proof: By its definition, µ vanishes on the empty set. If A B then any covering of B is also a covering of A; hence the infimum for A is taken over a larger collection of coverings, and therefore smaller. Hence µ is monotone.

13 8 CHAPTER 2. MEASURES It remains to demonstrate countable additivity. Let A = j A j X, and let ε > 0. For each j there is a covering of A j by sets E jk E such that hence ρ(e jk ) < µ (A j ) + 2 j ε; k=1 µ (A) < j,k=1 ρ(e jk ) < µ (A j ) + ε. Since ε is arbitrary, µ (A) j µ (A j ), and the countable subadditivity of µ.. is established. A measure is obtained by restricting the outer measure to a subclass of sets, called measurable sets, on which it is countably additive. Definition Let µ be an outer measure on a set X. A subset A X is said to be µ -measurable if µ (E) = µ (E A) + µ (E A c ), E X. By subadditivity µ (E) µ (E A) + µ (E A c ) for all A and E; hence A is measurable iff µ (E) µ (E A) + µ (E A c ), E X. This definition of measurability which is due to the Twentieth century Greek mathematician Carathéodory [1]. Theorem [Carathéodory] If µ is an outer measure on X, then the collection M of measurable sets is a σ-algebra, and the restriction of µ to M is a complete measure. Proof: First note that M is closed under complementation, since the definition of measurability is symmetric in A and A c. If A, B M and E X, then µ (E) = µ (E A) + µ (E A c ) = µ (E A B)+ µ (E A B c ) + µ (E A c B) + µ (E A c B c )

14 2.2. CARATHÉODORY S THEOREM 9 Since A B = (A B) (A B c ) (A c B) we have, by subadditivity, µ (E A B) + µ (E A B c ) + µ (E A c B) µ (E (A B)). Therefore µ (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 disjoint sets in M then µ (A B) = µ ((A B) A) + µ ((A B) A c ) = µ (A) + µ (B), so µ is finitely additive on M. Now suppose that {A j } is a sequence of disjoint measurable sets, and let B n = n 1 A j, with B = 1 A j. Then for any E X Therefore µ (E B n ) =µ (E B n A n ) + µ (E B n A c n) =µ (E A n ) + µ (E B n 1 ) =... = n µ (E A j ). µ (E) = µ (E B n ) + µ (E B c n) n µ (E A j ) + µ (E B c ). Letting n and using the countable subadditivity of µ, we obtain µ (E) µ (E A j ) + µ (E B c ) µ (E B) + µ (E B c µ (E). Thus B M; and, taking E = B we obtain also that µ is countably additive on M. Finally, if µ (A) = 0, then µ (E) µ (E A) + µ (E A c ) = µ (E A c ) µ (E), hence A is measurable, and µ restricted to M is a complete measure. This completes the proof of Theorem

15 10 CHAPTER 2. MEASURES The Carathéodory extension theorem is a primary method of constructing measures. Typically, the set function ρ of Proposition is taken to be a pre-measure, a non-negative set function that vanishes on the empty set, and is countably additive on the algebra E in the sense that ρ( j A j ) = j ρ(a j ), whenever j A j E. For example, Lebesgue measure on the line is an extension of the premeasure defined by ρ((a, b]) = b a; E is the algebra of finite disjoint unions of half-open intervals (a, b]. Theorem Let µ be the Carathéodory extension of a pre-measure ρ defined 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 R n, the algebra E is the collection of finite unions of multi-intervals, R = {(x 1,..., x n ) : a j x b j }, ρ(r) = 2.3 Borel measures n (b j a j ). Let X be a metric space and ρ its metric. The distance between two sets A and B in X is defined to be ρ(a, B) = inf{ρ(x, y) : x A, y B}. A metric outer measure µ on a metric space X is one which satisfies the additional 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 are then 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 ρ F ((a, b]) = F (b) F (a), where F is a non-decreasing, right continuous function on the real line, or on an interval. In the following, E denotes the algebra of sets which can be expressed as finite disjoint unions of half open intervals {(a j, b j ]}, with a j < b j.

16 2.3. BOREL MEASURES 11 Lemma The set function ρ F 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 decomposed in a non-unique fashion: (a, b] = n (c j 1, c j ], c 0 = a, c n = b. Theorem Let µ be a Lebesgue-Stieltjes measure and let M µ be its domain of measurable sets. Then for any E M µ we have inf{µ(u) : E U, U open} µ(e) = sup{µ(k) : K E, K compact}. Lebesgue measure m on the line is the extension of the pre-measure ρ F, where F (x) = x. The Borel algebra is a sub-class of Lebesgue measurable sets. In the theory of Borel sets, F σ denotes the class of sets which are countable unions of closed sets; while G δ denotes the class of countable intersections of open sets. Any Lebesgue measurable set E can be written in the form V \N 1 V G δ, µ(n 1 ) = 0; E = H N 2 H F σ, µ(n 2 ) = 0. A second Lebesgue-Stieltjes measure of interest is that generated by the Cantor 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 extending this to the entire unit interval [0, 1] by continuity. The Cantor function is continuous and non-decreasing. Let µ c be the Lebesgue-Stieltjes measure generated by c. The Cantor set C [0, 1] is obtained by removing the open middle third (1/3,2/3) of the unit interval, then removing the open middle thirds of the two remaining intervals, and so forth, successively removing the open middle thirds of the remaining subintervals at each stage. The Cantor measure of C is one, while the Lebesgue measure of the complement C c is one. The Cantor measure and Lebesgue measure are examples of mutually singular measures: they have their support on complementary subsets of [0, 1].

17 12 CHAPTER 2. MEASURES

18 Chapter 3 Integration 3.1 Measurable functions A function f : R n R is Lebesgue measurable if f 1 (U) is a Lebesgue measurable set for every open set U. More generally, given any measure µ we say a real valued function f is measurable with respect to µ if f 1 (U) M whenever 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. If f is complex-valued, then f is measurable iff its real and imaginary parts are measurable. If f : X Y then f 1 is defined by f 1 : P(Y ) 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 (E c ) = f 1 (E) c. It follows that a real-valued function is measurable if f 1 (U) is measurable whenever U is an open or closed interval, or an open ray (a, ], etc. A mapping f : X Y, where X and Y are metric spaces, is Borel measurable if f 1 (U) is a Borel subset of X for every open set U Y. By the observations above, Proposition Continuous functions are Borel measurable. f is Borel measurable iff f 1 (B) B X whenever B B Y. The composition of two Borel measurable functions on R to itself is Borel measurable. 13

19 14 CHAPTER 3. INTEGRATION On the other hand, the inverse image of a Lebesgue measurable set is not necessarily Lebesgue measurable, and so Lebesgue measurability is not closed under composition, [2]. Given two functions f and g we define f g = max{f, g}, f g = min{f, g}; f + = f 0, f = ( f) 0. Proposition 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 of any finite number of measurable functions is measurable. Proof: For any real a we have {x : f + g > a} = t Q {x : f(x) > t} {x : g(x) > a t}; hence f + g is measurable. It is immediate that measurability is invariant under scalar multiplication; hence f g is also measurable. Consider the function f 2. If a < 0 then {x : f 2 > a} = X; while for a 0, {x : f 2 > a} = {x : f(x) > a} {x : f(x) < a}. Hence f 2 is measurable. The identity fg = (f + g)2 f 2 g 2 2 then shows that fg is measurable. We have X a < 0; {f + > a} = {x : f(x) > a} a 0; hence f + is measurable. The same argument shows f is measurable; and therefore also f = f + + f is measurable. The identities f g = f + g + f g f + g f g, f g = 2 2 then show that f g and f g are measurable. The result extends to finite.. combinations of measurable functions by induction.

20 3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 15 Theorem If {f j } is a sequence of measurable functions, then the limits lim sup j f j, lim inf j f j, as well as lim f f j, when it exists, are all measurable. Proof: Let g k (x) = sup j k f j (x). Then {x : g k (x) > a} = j k {x : f j (x) > a}; hence {g k } is a sequence of measurable functions. Putting f(x) = inf k g k (x), we see that {x : f(x) > a} = {x : g k (x) > a}, k=1 hence f(x) = lim sup j f j (x) is measurable. A similar proof shows that lim inf j f j is measurable. The limit lim j f j (x) exists for all x for which the limits inferior and superior of the sequence {f j (x)} coincide; hence the.. lim j f j is measurable. 3.2 Integration of non-negative functions A characteristic or indicator function is one which takes on only the values 0 and 1; it is generally denoted by χ E (x) where E is the set on which the function is 1. It is useful to note that χ A B = χ A χ B, χ A B = χ A + χ B iff A B =. A simple function is one which takes on only a finite number of values in the extended real numbers. It is thus a finite linear combination of characteristic functions ϕ = n c j χ Ej, E j = {x : ϕ(x) = c j }. It is clear that a simple function is measurable iff the sets E j are measurable. Moreover, the class of simple functions is closed under multiplication and addition. Theorem Let f be a measurable real-valued function on a space X. Then there is a sequence of simple functions ϕ n such that ϕ n f uniformly on any set on which f is bounded.

21 16 CHAPTER 3. INTEGRATION Proof: Decompose f = f + f, where f + = max{f, 0} and f = max{ f, 0}. Thus we may assume that f 0. For n 0 and 0 k 2 2n 1, let E n,k = f 1 ((k2 n, (k + 1)2 n )), F n = f 1 ((2 n, ]). Then the sequence of simple functions satisfy ϕ n = 2 2n 1 k=0 k2 n χ En.k + 2 n χ Fn 0 f n f n+1 f, 0 f ϕ n 2 n, x F c n. 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 n ϕ dµ = c j µ(e j ). We use the convention that 0 = 0. It is clear that the integral of non-negative simple functions is nonnegative. Also, since finite linear combinations and finite products of simple functions are simple, the integral so defined is a linear functional on the vector space of simple functions. Lemma Given a non-negative, measurable simple function ϕ and A M, define A ϕdµ = ϕχ A dµ. Then the set function ν(a) defined by ν(a) = ϕdµ is a (countably additive) measure. A

22 3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 17 Proof: If A is the countable union of disjoint sets in M then ν(a) = = ϕχ A dµ = n c k χ Ek χ A dµ = k=1 n c k µ(e k A) = k=1 n k=1 n c k χ A Ek dµ k=1 c k µ(e k A j ) = c k µ(e k A j ) = ϕdµ = A j ν(a j )... We now define the integral of any non-negative measurable f by fdµ = sup ϕ f { } ϕdµ : ϕ simple. The reader may check that this definition is consistent with that of the integral 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 theory of the Lebesgue integral. Theorem [Monotone Convergence Theorem] If {f n } is an increasing sequence of non-negative functions then lim f n dµ = lim f n dµ. n n Proof: Since {f n } is a nondecreasing sequence of measurable functions, lim n f n = f exists and is measurable. It may take on the value infinity. Moreover, the integrals f n dµ form a nondecreasing sequence of numbers; and f n f implies that f n f, hence lim n f n dµ f dµ. To prove equality, it suffices to show that for any fixed α, 0 < α < 1, α f dµ lim f n dµ. n

23 18 CHAPTER 3. INTEGRATION Given α and any simple function ϕ, let E n = {x : f n αϕ}. Then f n dµ f n αϕ dµ = αν(e n ). E n E n By Lemma 3.2.2, ν is a countable additive measure. Since {E n } is an increasing sequence of sets whose union is X, it follows by Theorem 1.8 c that lim n ν(e n ) = ν(x) = ϕ dµ. Thus lim f n dµ α ϕ dµ. n Taking the supremum over all simple functions ϕ f we obtain.. α f dµ = α sup ϕ dµ lim f n dµ. ϕ f n An immediate consequence of the Monotone Convergence Theorem is the following characterization of the integral of a general non-negative measurable function. Corollary Let f be a non-negative measurable function, and let ϕ n be a monotone increasing sequence of measurable functions converging to f. Then f dµ = lim ϕ n dµ. n Using Corollary we can prove the additivity of the integral on nonnegative functions. Proposition The integral dµ is finitely additive on non-negative measurable functions. Proof: Let f and g be non-negative measurable functions and let ϕ n f, ψ n g, where ϕ n and ψ n are sequences of non-negative simple functions. Then f + g dµ = lim(ϕ n + ψ n ) dµ n = lim n ϕ n dµ + lim n ψ n dµ = f dµ + g dµ.

24 3.2. INTEGRATION OF NON-NEGATIVE FUNCTIONS 19 By the same argument, one can show that cf dµ = c f dµ, for any constant c... Applying the monotone convergence theorem to the partial sums of a series of non-negative functions, we obtain Corollary If f n 0, then f n dµ = n n where both sides may be infinite. f n dµ, Proposition If f 0 then f = 0 iff f = 0 a.e. Proof: The statement is immediate for simple functions. For general f, if f = 0 a.e. and ϕ f is simple, then ϕ = 0 a.e. hence f = sup ϕ f ϕ = 0. Conversely, let {x : f(x) > 0} = n 1 E n, E n = {x : f(x) > 1 n }. If f > 0 on a set of positive measure, then µ(e n ) > 0 for some n; hence f dµ 1 n µ(e n) > 0... An immediate corollary of this proposition is the following strengthening of the monotone convergence theorem. Corollary If 0 f n f n+1..., and f n f a.e. then lim n fn dµ = f dµ. The following lemma, known as Fatou s lemma, can also be used as the starting point in building the convergence theorems of Lebesgue integration theory. Lemma [Fatou s Lemma] For any non-negative sequence of measurable functions {f n } we have lim inf f n dµ lim inf f n dµ. n n

25 20 CHAPTER 3. INTEGRATION Proof: We have inf n k f n f j for j k, hence inf n k f n dµ f j dµ for j k; and so inf f n dµ inf f j dµ. n k j k Since both sides are increasing in k, we have, in the limit as k, lim inf f n dµ lim inf f j dµ = sup inf f j dµ = lim inf f n dµ. k n k k j k j k n Since the sequence {inf n k f n } is non-negative and increasing in k, we have, by the Monotone Convergence Theorem lim inf f n dµ = sup inf f n dµ = lim inf f n dµ = lim inf f n dµ. n n k k n k k n k k The Dominated Convergence Theorem Let (X, M, µ) be a measure space. A real valued function f is said to be integrable 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 L 1 (X, dµ) or more briefly by L 1 when the measure space is clear. The class L 1 is closed under addition and scalar multiplication, hence is a vector space. Moreover, if M contains an infinite number of sets of positive measure, then L 1 (X, dµ is an infinite dimensional vector space. A norm is defined on this vector space by f 1 = f dµ. It is easily verified that 1 satisfies the triangle inequality and cf 1 = c f 1. Thus L 1 is a metric space with the metric ρ(f, g) = f g 1 ; and it is natural to ask the question as to whether L 1 is a complete metric space. The answer to this question is yes; and not only that, but the completeness is a direct consequence of the introduction of the notion of measure theory and measurable functions. If one begins with only the continuous functions in L 1, and seeks the completion of this metric space, then one has to include the measurable functions as well. We begin by proving: k 1

26 3.3. THE DOMINATED CONVERGENCE THEOREM 21 Theorem [Lebesgue Dominated Convergence Theorem] Let {f n } be a sequence of functions in L 1 such that f n f a.e. and there exists a nonnegative function g L 1 such that f n g, a.e. for all n. Then f L 1 and fdµ = limn fn dµ. Before beginning the proof, let us give an example that shows the necessity of the assumption that the sequence f n be dominated by a fixed function g L 1. 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 0. Thus its pointwise limit is 0 almost everywhere, and for any sequence t n 0, 0 = lim G(x, t n) dx < lim n n G(x, t n )dx = 1. Thus the conclusion of the Dominated Convergence theorem fails while that of Fatou s lemma holds with a strict inequality t= t= Figure 3.1: The fundamental solution of the heat equation evaluated at t =.1 and t = 1.

27 22 CHAPTER 3. INTEGRATION Proof: Applying Fatou s lemma to the sequence of functions g + f n 0 we obtain g + fdµ = f n )dµ f n dµ, (g + lim inf n gdµ + lim inf n hence fdµ lim inf n fn dµ. Applying the same argument to the sequence g f n 0, we obtain, fdµ lim inf f n dµ = lim sup f n dµ. n n Combining both inequalities, we have fdµ lim inf f n dµ lim sup n n f n dµ fdµ.... Corollary Let {f j } be a sequence of integrable functions such that the series j fj dµ < +. Then j f j converges a.e. to a function f L 1 and f j dµ = f j dµ. Proof: By the monotone convergence theorem, f j dµ = f j dµ < +, j j hence the function g = j f j L 1. Thus g is finite a.e. and for each x for which g(x) is finite, the series j f j(x) converges absolutely. The partial sums of this series are dominated by g, so by the conclusion of the Corollary.. follows by the dominated convergence theorem. Theorem The simple functions are dense in L 1. The class C 0 (R) is dense in L 1 (R, B R, µ), where µ is any Borel measure on R. Proof: The simple functions are dense in L 1 (dµ) by Theorem and the dominated convergence theorem. To show the continuous functions are dense in L 1 (R, B R ) it is sufficient to show that any simple function can be approximated by a continuous function in the L 1 norm. If I = (a, b) is an open interval, we may approximate χ I by a sequence of continuous, piecewise

28 3.4. THE RIEMANN INTEGRAL 23 linear functions C n,e which take the value 1 on I, 0 on (, a n 1 ) (b + n 1, ). It is a simple calculation to show that χ E C n,e 1 = n 1. Next let E be a Borel measurable set in R, with µ(e) = χ E 1 <. We show that for any ε > 0, there is a finite union of open intervals A such that χ E χ A dµ = µ(e A) < ε. To see this, we use the fact that for Borel measures µ on the line, µ(e) = inf µ(i j ) : E (a j, b j ). Thus given any ε > 0 there is a collection of disjoint open intervals I j = (a j, b j ) such that µ(u\e) < ε 2, U = I j. Now choose N so that N+1 µ(i j) < ε/2, and put A = N I j, B = U\A. Then E\A = E A c (A B) A c 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 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 Riemann integral; that is, the two integrals give the same result for a Riemannintegrable function. Let [a, b] be a bounded interval. A partition of [a, b] is a set of points P = {a = x 0 < x 1 < < x n = b}. Given a bounded real-valued function f on [a, b], the upper and lower Riemann sums of f relative to a partition P are given by L(P, f) = n m j x j, U(P, f) = n M j x j, where x j = x j x j 1 and m j, M j are the infimum and supremum of f on [x j 1, x j ]. Clearly L(P, f) U(P, f) for any partition P.

29 24 CHAPTER 3. INTEGRATION Definition A bounded real-valued function f defined on a bounded interval [a, b] is said to be Riemann integrable if sup P L(P, f) = inf P U(P, f). In that case the Riemann integral of f on [a, b] is defined to be b a f(x)dx = sup L(P, f) = inf U(P, f). P P The concept of the Riemann integral is clearly an application of the method of exhaustion, proposed originally by Euxodus of Cnidus ( B.C.E.) and developed extensively by Archimedes ( B.C.E.) as a tool for computing areas and volumes. It was not until Isaac Barrow, the first Lucasian chair of mathematics, and his student and successor, Sir Isaac Newton ( ) that the connection between the method of exhaustion and the notion of anti-derivative was fully established. This is the true content of the Fundamental Theorem of Calculus. Theorem Let f be a bounded real-valued function on [a, b]. If f is Riemann integrable, then f is Lebesgue measurable and therefore integrable on [a, b]; moreover the Lebesgue and Riemann integrals agree. A function is Riemann integrable iff it is continuous a.e. Proof: Choose a sequence of partitions P n such that each partition is a refinement of the previous one and for which max j x j tends to zero as n. Then the lower sums L(f, P n ) are increasing and the upper sums are decreasing. f is Riemann integrable iff these two sequences converge to the same limit. The lower and upper sums of a function f are the Lebesgue integrals of the simple functions l(p, f) = n m j χ (xj 1,x j ], u(p, f) = n M j χ (xj 1,x j ]. Let h n = l(p n, f), and H n = u(p n, f). Then h n f H n, and in the limit h f H, where h = lim n h n and H = lim n H n. If f is Riemann integrable, then the sequences of upper and lower sums converge to the same limit. It follows that lim n [a,b] h n dm = b a f dx = lim n [a,b] H n dm.

30 3.5. SOME CONVERGENCE THEOREMS 25 By the Lebesgue dominated convergence theorem, lim h n dm = hdm = Hdm = lim n n H n dm. [a,b] [a,b] [a,b] [a,b] Since H h this implies that h = H a.e. and hence that H = f = h a.e. It follows that f is Lebesgue measurable and that [a,b] Hdm = [a,b] fdm = b a fdx. The proof that the discontinuities of a Riemann integrable function form.. a null set is left to the reader. Theorem 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 t and that for each x X, f t (x, t) g(x) for all a t b, where g L 1 (X, dµ). Let F (t) = X f(x, t)dµ. Then F is differentiable on a < t < b and F (t) = X f t(x, t)dµ. Proof: For t (a, b) we apply the Lebesgue dominated convergence theorem to the sequence h n (x) = f(x, t n) f(x, t), t n t. t n t By the mean value theorem, the difference quotients h n are bounded by.. g(x) = sup t f t (x, t). 3.5 Some Convergence Theorems The L 1 integral of each term of the following sequences is 1, 1 n χ (0,n), χ (n,n+1), nχ (0,n 1 ), yet the sequences themselves tend to 0 uniformly, pointwise, and a.e. respectively. In each of these cases the sequences are not dominated by a fixed integrable function g.

31 26 CHAPTER 3. INTEGRATION On the other hand, there are sequences which tend to 0 in the L 1 norm but which do not tend to zero pointwise. For example, the sequence f 1 = χ (0,1), f 2 = χ (0,1/2), f 3 = χ (1/2,1), f 4 = χ (0,1/4), f 5 = χ (1/4,1/2), etc. and in general f n = χ (j2 k,(j+1)2 k ), where n = 2 k + j, with 0 j < 2 k. For this sequence, f n 1 0 as n but f n (x) does not converge for any x. This shows that pointwise convergence and convergence in norm are in general not equivalent. We shall prove here that if f n f 1 0, then some subsequence of f n converges to f a.e. In order to do this we introduce the concept of convergence in measure. Definition A sequence of functions f n is said to converge in measure to f if for every ε > 0, lim n µ ({x : f n (x) f(x) ε}) = 0. A sequence is said to be Cauchy in measure if for every ε > 0, µ ({x : f n (x) f m (x) ε}) 0, m, n. For example, the first, third, and fourth sequences above converge to 0 in Lebesgue measure on the line, while the second sequence does not. Convergence in norm implies convergence in measure. We need only prove this for a sequence of functions tending to 0 in the L 1 norm. Suppose f n 1 0 and, given ε > 0, put E n,ε = {x : f n (x) > ε}. Then for any fixed ε > 0, f n dµ f n dµ εµ(e n,ε ) 0. E n,ε Thus f n 0 in measure... The first and third examples above show that convergence in measure does not imply convergence in norm. Many authors use the nomenclature f n f [meas], a.e., [unif], or [mean], to signify convergence in measure, pointwise convergence, uniform convergence, or convergence in the L 1 norm. Theorem If {f n } is a Cauchy sequence in measure, then there is a measurable function f such that f n f [meas]; and there is a subsequence f nk such that f nk f a.e. Moreover, f is uniquely determined a.e.

32 3.5. SOME CONVERGENCE THEOREMS 27 Proof: We may choose a subsequence n k such that for g k = f nk, µ(e k ) 2 k, E k = {x : g k (x) g k+1 (x) 2 k }. Putting F k = j k E j, we have µ(f k ) j k 2 j = 2 1 k, and i 1 i 1 g i (x) g j (x) g l+1 (x) g l (x) < 2 l 2 1 j, (3.1) l=j for i j k, and x Fk c. This shows that {g j(x)} j k is a Cauchy sequence for x Fk c. Let F = 1 F k; then µ(f ) = 0 and {g j (x)} is a Cauchy sequence on F c. Let f = lim g k on F c and 0 on F. Then f is measurable; g k f a.e.; and, letting i, in (3.1) we obtain l=j f(x) g j (x) 2 1 j, x F c k ; j k. Hence {x : f(x) g j (x) > 2 1 j } F k for j k. Since µ(f k ) 0 as k, this shows that g j f [meas]. But then f n f [meas] since for any ε > 0, µ({x : f(x) f n (x) ε}) µ({x : f(x) g j (x) ε/2}) + µ({x : g j (x) f n (x) ε/2}) 0, j, n. The proof that the limit in measure of a Cauchy sequence in measure is.. unique is left as an exercise. Corollary If f n f in L 1 then there is a subsequence {f nk } such that f nk f [a.e.] Proof: Exercise The sequence {χ [n,n+1] } converges pointwise to 0 everywhere, but does not converge to 0 in measure. On a finite measure space, however, pointwise convergence implies what is called almost uniform convergence. If µ(x) < +, a sequence {f n } is said to converge almost uniformly to f on X if given any ε > 0 there is a set E such that µ(e) < ε and f n f [unif] on E c. It is easily seen that almost uniform convergence implies convergence in measure.

33 28 CHAPTER 3. INTEGRATION Theorem [Egorov s Theorem.] On sets of finite measure, convergence almost everywhere implies almost uniform convergence. Proof: It suffices to show that if f n 0 everywhere on E, and µ(e) <, then there exists a subset F E such that µ(e\f ) < ε and f n 0 uniformly on F. Let E mn = {x : f j (x) < n 1, j m}. For fixed n, the sequence {E mn } is increasing, and E = 1 E mn. Therefore, by continuity from below, lim m µ(e mn ) = µ(e). Hence given ε > 0 and for n N, there exists an integer m n such that µ(e\e mn n) < ε2 n. Let F = E mnn. If x F then for all n N there is an m n such that n 1 f j (x) < n 1 for all j m n ; hence f j 0 uniformly on F. On the other hand, µ(e\f ) = µ E\E mnn.. µ(e\e mnn) < ε. n 1 n=1

34 Chapter 4 Product Spaces 4.1 Fubini s Theorem Let {X j, M j, µ j }, j = 1, 2 be two measure spaces, and consider the σ algebra M of subsets of X = X 1 X 2 generated by rectangles of the form E 1 E 2, where E j M j. This σ-algebra is called the product algebra, and is denoted by M = M 1 M2. We shall denote points in X 1 and X 2 by x and y respectively. A product measure on X is constructed as follows. Denote the algebra of sets generated by finite disjoint unions of rectangles by A. Define µ on rectangles by µ (E 1 E 2 ) = µ 1 (E 1 )µ 2 (E 2 ). Then µ is a pre-measure on A. Using µ, we construct an outer measure µ on X; by Carathéodory s theorem we obtain a complete measure µ on X whose σ-algebra of measurable sets contains the product algebra M = M 1 M2. If µ j are σ-finite, then so is µ. Clearly this construction of product algebras and measures extends to any finite number of factors. Moreover, this definition of product algebras and measures is associative. In this section we prove the following theorems of Tonelli and Fubini: Theorem Let {X j, M j, µ j } be σ-finite measure spaces, and let the product measure space be denoted by {X, M, dµ}. Tonelli: Let f L + (dµ). Then the functions g(x) = X 2 f(x, y)dµ 2 (y) 29

35 30 CHAPTER 4. PRODUCT SPACES and h(y) = X 1 f(x, y)dµ 1 (x) are in L + (dµ 1 ) and L + (µ 2 ) respectively; and X [ fdµ = ] f(x, y)dµ 2 (y) dµ 1 (x) [ = ] f(x, y)dµ 1 (x) dµ 2 (y). (4.1) These three integrals may assume the value +. Fubini: Let f L 1 (dµ). Then X 2 f(x, y)dµ(y) L 1 (dµ 1 ), X 1 f(x, y)dµ(x) L 1 (dµ 2 ), and (4.1) holds; that is, the double and each of the iterated integrals are equal. The strategy of the proof is to begin by proving the result for characteristic functions of rectangles, then simple functions, and ultimately extend the result to general measurable functions on X. The x and y sections of a set E X are defined to be the subsets E x = {y : y X 2, (x, y) E}, x X 1 E y = {x : x X 1, (x, y) E}, y X 2. Similarly, if f : X R, then the x and y-sections of f are the mappings f x : X 2 R and f y : X 1 R defined by f x (y) = f(x, y) and f y (x) = f(x, y). Proposition If E M, then the sections E x and E y respectively belong to M 2 for each x X 1, and M 1 for each y X 2. If f is measurable with respect to the product algebra M 1 M2, then its sections f x and f y are measurable with respect to the factors M 2 and M 1 respectively. Proof: Let R be the collection of all subsets E X such that E x M 2 for all x X 1 and E y M 1 for all y X 2. Then (A B) x = { B x A x A c,

36 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 ( ) E j = (E j ) x, (E x ) c = (E c ) x, 1 x 1 and similarly for y-sections. Therefore R M 1 M2. The measurability of f x and f y follows from the first statement and the relationships (f x ) 1 (B) = (f 1 (B)) x, (f y ) 1 (B) = (f 1 (B)) y... A monotone class on a space X is a collection of subsets C P(X) that is closed under countable increasing unions and countable decreasing intersections. Every σ algebra is a monotone class; and the intersection of monotone classes is a monotone class. Thus for any collection E P(X), there is a unique smallest monotone class containing E, called the monotone class generated by E. Proposition If A is an algebra of subsets of X then the monotone class generated by A coincides with the σ algebra generated by A. Given a measure µ we denote the non-negative µ measurable functions by L + (µ). Lemma Let {X j, M j, µ j } be σ-finite measure spaces for j = 1, 2, and let M and µ be the product algebras and measures on X = X 1 X 2. Given E M, the sections (χ E ) x and (χ E ) y are in L + (dµ 2 ) and L + (dµ 1 ) respectively; and µ(e) = X χ E dµ = X 1 (χ E ) x dµ 2 (y) dµ 1(x) X 2 = X 2 (χ E ) y dµ 1 (x) dµ 2(y). (4.2) X 1 Proof: We shall establish the lemma for the case in which the µ j are finite measures; the σ-finite case is left as an exercise. Let C be the class of sets in M for which the lemma holds. When E is a rectangle, E = A B

37 32 CHAPTER 4. PRODUCT SPACES then (χ E ) x (y) = χ A (x)χ B (y) = (χ E ) y (x) and all three integrals above are equal to µ 1 (A)µ 2 (B). Thus E C. By finite additivity of the integral C contains finite disjoint unions of rectangles. It therefore suffices to show that C is a monotone class. Let E = n E n where {E n } is an increasing sequence of sets. Then the sequence of functions χ En increases monotonically to χ E ; and their sections increase monotonically to the corresponding sections of χ E ; (4.2) follows from the monotone convergence theorem; and E C. Similarly, let {E n } be a decreasing sequence of sets in C and put E = n E n. The characteristic functions χ En then converge pointwise to χ E ; and their sections converge pointwise to the corresponding sections of χ E. Since µ(e 1 ) < the conclusion follows from the dominated convergence theorem... Thus C is a monotone class and is therefore equal to the σ-algebra M. Proof of Theorem 4.1.1: Lemma establishes (4.1) for characteristic functions; and by additivity the result follows for simple functions. Given f L + (X, dµ) we approximate f from below by simple functions and pass to the limit using the monotone convergence theorem. All three integrals are either finite and equal, or all equal to infinity. Given f L 1 (dµ) we decompose f = f + f and apply Tonelli s theorem to each of the components. 4.2 Lebesgue measure in R n A specific example of a product measure, obtained by the procedure in 4.1 is n dimensional Lebesgue measure on R n ; it is obtained by the premeasure defined on rectangles in R n defined by R = j I j, where the {I j } are intervals (a j, b j ) on the real axis. We have n m 0 (R) = µ I j = n (b j a j ). (4.3) Using m 0 we obtain an outer measure on R n and by Carathéodory s theorem we obtain a measure, called the Lebesgue measure on R n. We denote it by m when the dimension is understood. As on R, Lebesgue measure on R n has has a number of regularity properties:

38 4.2. LEBESGUE MEASURE IN R N 33 Theorem Suppose E L n. Then (a) m(e) = inf {m(u) : E U, U open } = sup {m(k) : K E, K compact }; (b) E = A 1 N 1 = A 2 \N 2, where N 1 and N 2 are sets of measure 0, A 1 F σ (the class of countable unions of closed sets); and A 2 G δ (the class of countable intersections of open sets); and (c) If m(e) < then for any ε > 0 there is a finite collection of disjoint rectangles whose union R satisfies µ(e R) < ε. Moreover, Theorem The class of simple functions f = j f jχ Rj, where R j = k I jk is a product of intervals I jk R, is dense in L 1 (R n ). Furthermore the class of continuous functions with compact support in R n is dense in L 1 (R n ). Proof: The proof is essentially the same as the proof of the corresponding theorem for Borel measures on R, Theorem Theorem Lebesgue measure on R n is invariant under the group of rigid motions. Proof: It suffices to establish that Lebesgue outer measure is invariant under the group of rigid motions, that is, under the group of translations and rotations in R n. We prove the result in R 2 and leave its extension to R n as an exercise. Lebesgue outer measure was defined in terms of coverings of rectangles with sides parallel to the axes. Thus the outer measure is invariant under translations and reflections in the axes. We therefore have two set functions, area and outer measure, that are both additive and invariant under translations and reflections. Any theorem on area in classical geometry that uses only these three properties is therefore valid for outer measure. To show that the outer measure is invariant under rotations, we must prove that the outer measure of a rectangle in general position is equal to the area of that rectangle. A rectangle in general position may be written as the union of two triangles and a parallelogram (degenerate if one of the diagonals is parallel to the x axis) with bases parallel to the x-axis; so by additivity of both the area and the outer measure, it suffices to prove that the two functionals agree on triangles and parallelograms with base parallel to one of the axes. Given a parallelogram with base parallel to the x axis, translate one of its triangular caps parallel to the x axis to the other end, forming a rectangle with base parallel to the x axis. Both the outer measure and the area of the

39 34 CHAPTER 4. PRODUCT SPACES parallelogram are invariant under this operation; and they coincide on the rectangle. Hence the outer measure and area of the parallelogram coincide. By similar arguments from plane geometry, using only additivity and invariance under reflections and translations, one can prove that the outer measure of any triangle is equal to its area. A second approach is combine a translation of the upper triangle with the lower triangle to produce a second parallelogram with base parallel to the x axis, and then apply the previous argument showing the measure of a.. parallelogram is equal to its area. Now let T GL(n, R) and assume that det T 0. We define an invertible transformation acting on the Lebesgue measurable functions f L(R n ) by f T = f T 1. We define an action of GL(n, R) on n dimensional Lebesgue measure by m T (E) = m(t E), where E is a Lebesgue measurable subset of R n and T E denotes its image under T. For a rectangle R R n we have m(t R) = det T m(r). Theorem For f L 1 (R n ) and any invertible linear transformation T GL(n, R) we have f T dm = f dm T = det T f dm. Proof: It is sufficient to prove the result for characteristic functions. It then follows for simple functions by additivity, and for the general case by the dominated convergence theorem. First, note that for rectangles, A(T R) = det T A(R), hence on rectangles, m T (R) = m(t R) = det T m(r). By continuity, this identity extends to all Lebesgue measurable sets E. Next, for any set E we observe that χ E (T 1 x) = χ T E (x) = (χ E ) T ; hence (χ E ) T dm = χ T E dm = m(t E) = det T m(e) = det T χ E dm = χ E dm T. The volume V (E), where E R n, is initially defined on rectangles, and then extended to polygons by its invariance under translation and reflection. The extension of the volume functional to bounded subsets in R n is defined