Tools from Lebesgue integration

Tools from Lebesgue integration E.P. van den Ban Fall 2005 Introduction In these notes we describe some of the basic tools from the theory of Lebesgue integration. Definitions and results will be given with precise formulations, but proofs will often be omitted. The goal is to provide the necessary background for the course on distribution theory. 1 Sets of measure zero By a block in we shall mean a product B = I 1 I n of bounded closed intervals of the form I j = [a j, b j ]. The (Lebesgue) measure µ(b) of such a block is given by n µ(b) = b j a j. j=1 Definition 1 A subset S is said to be of (Lebesgue) measure zero, if for every ε > 0 there exists a countable sequence of blocks (B j ) j N in such that (a) S j N B j ; (b) j=0 µ(b j) < ε. The crucial point in the above definition is that we allow countable sequences of blocks. If instead we had restricted ourselves to finite sequences, the resulting notion would have been the notion of a set of Jordan measure zero. This is the notion that fits well into the theory of Riemann integration. The weaker notion of measure zero given above will allow us to extend the notion of integrability beyond the class of Riemann integrable functions. The usefulness of countable sums is illustrated by the following result. Lemma 2 Let (S k ) k 0 be a countable sequence of sets of measure zero in. Then the union k 0 S k has measure zero as well. Proof: Let ε > 0. Then for each k N there exists a countable sequence (B k,j ) j N of blocks such that S k j N B k,j and such that µ(b k,j ) < ε 2 k 1. j=0 1

It follows that Moreover, S = k N S k (k,j) N 2 B k,j. (k,j) N 2 µ(b k,j ) < Since N 2 is countable, the result follows. ε 2 k 1 = ε. Since obviously a singleton (set consisting of a single point) has measure zero, it follows from the above result that every countable subset of has measure zero. In particular, the set Q [0, 1] has measure zero, Already here we encounter a striking difference with the theory of Riemann integration, as the set Q [0, 1] does not have Jordan measure zero. Definition 3 A function f : C is said to be measurable if there exists a set S of measure zero and a sequence of continuous functions (f k ) C 0 ( ) such that f k f pointwise on \ S. Remark 4 If (f k ) is a sequence of functions on then we say that it converges (pointwise) almost everywhere (abbreviated a.e.) if there exists a set S of measure zero such that lim k f k (x) exists for all x \ S. Thus, according to the above definition, a function f is measurable if and only if there exists a sequence of continuous functions that converges pointwise to it almost everywhere. From the definition it is easily seen that the sum f + g of two measurable functions f, g is measurable again. Similarly, the product of two measurable functions is measurable again. It follows that the set M of measurable functions is an algebra over C; in particular, it is a complex linear space. k=0 We state the following result without proof. Lemma 5 Assume that a sequence of functions (f k ) converges to a function f almost everywhere. If each f k is measurable, then the limit function f is measurable. Definition 6 A subset A is measurable if and only if its characteristic function 1 A is measurable. In particular, a set of measure zero is measurable. One of the very useful facts of Lebesgue theory is the following. Lemma 7 Every countable union of measurable sets is measurable. Proof: If A and B are subsets of, then 1 A B = max(1 A, 1 B ). Assume that A and B are measurable, and let (f j ) be a sequence of continuous functions with f j 1 A almost everywhere. Similarly, let (g j ) be a sequence of continuous functions with g j 1 B almost everywhere (use that the union of two sets of measure zero has measure zero). Then h j = max(f j, g j ) is a sequence of 2

continuous functions with h j max(1 A, 1 B ) almost everywhere. Hence A B is measurable. It follows that every finite union of measurable sets is measurable. Let now A j be a countable sequence of measurable sets and let A be its union. Put U k = A 1... A k. Then it is readily seen that 1 Uk 1 A pointwise on. Since each of the functions 1 Uk is measurable, it follows that 1 A is measurable. Hence A is measurable. Corollary 8 Every open subset of is measurable. Proof: It is readily seen that every open ball is measurable. We now consider the collection B of open balls of the form B(a; r), with a Q n and r Q, r > 0. Then B is a countable collection of measurable sets. We will finish the proof by showing that every open set U is a union of balls from B. Indeed, let x U. Then B(x, R) U for a suitable R > 0. There exists a rational r Q with 0 < 2r < R. Moreover, there exists a point a Q n with x a < r. Now x B(a; r) B(x; R) U. It follows that there exists a B B with x B U. This establishes the claim. Lemma 9 The complement of a measurable set is measurable. The countable intersection of measurable sets is measurable. Every closed subset is measurable. Proof: Let A be measurable. Let B be the complement of A. Then 1 B = 1 1 A, hence B is measurable. The complement of an intersection of sets is the union of the complements of the sets. Thus, the assertion about intersections follows from the similar assertion about unions. Finally, a closed set is the complement of an open set, hence measurable. 2 Integrable functions We now come to the concept of integrability. Definition 10 If f C 0 ( ) is a continuous function with compact support, its L 1 -norm f 1 is defined by the following Riemann-integral f 1 := f(x) dx. A sequence of continuous functions (f k ) k 0 is said to be Cauchy for the L 1 - norm if for every ε > 0 there exists a N N such that p, q N f p f q < ε. Lemma 11 Let (f j ) be a sequence in C 0 ( ) which is Cauchy for the L 1 -norm. Then lim j f j (x) dx exists. If f j 0 almost everywhere, then the above limit equals zero. 3

Proof: Write I j for the integral of f j. Then it follows by the triangle inequality for Riemann integrals that I p I q f p f q 1. This implies that (I k ) is a Cauchy sequence in C, hence converges with a limit I C. The proof of the second assertion is beyond the scope of these notes. Definition 12 A function f : C is said to be integrable if there exists a sequence (f k ) in C 0 ( ) which is Cauchy for the L 1 -norm and converges to f almost everywhere. Lemma 13 Let f be integrable, and let (f k ), (g k ) C 0 ( ) be two approximating sequences as in the above definition. Then lim f k dx = lim g k dx k k Proof: This follows from the previous lemma. Definition 14 integral by Let f be an integrable function. Then we define its Lebesgue f(x) dx = lim f k (x) dx, k where (f k ) is any approximating sequence as in Definition 12. (The limit is independent of the choice of sequence.) The triangle inequality generalizes to integrable functions as follows. Lemma 15 Let f be integrable. Then f is integrable, and f dx f dx. Proof: Let (f j ) C 0 ( ) be an approximating sequence for f. Then f j f pointwise almost everywhere. Moreover, since f p f q f p f q everywhere, it follows that ( f j ) is Cauchy for the L 1 -norm, hence an approximating sequence for f. For every j we have f j dx f j dx by the triangle inequality for Riemann integration. The desired estimate is obtained by passing to the limit. The following result is extremely useful. We omit the proof. A measurable function f is said to be dominated by a non-negative measurable function g if f g almost everywhere. Lemma 16 Let f be a measurable function. Then f is integrable if and only if it is dominated by an integrable function g. Moreover, if this is the case, then f dx g dx. 4

Definition 17 A measurable set A is said to have finite measure if and only if 1 A is integrable. For such a set we put µ(a) := 1 A dx. If A is measurable but 1 A is not integrable, we put µ(a) =. It is not difficult to show that this definition is compatible with the earlier definition of the measure of a block. For a general compact set we have the following. Lemma 18 Let K be compact. Then K is measurable with finite volume. Proof: Since K is closed, K is measurable. By compactness there exists a continuous function f C 0 ( ) with 0 f 1 and f = 1 on K. It follows that 1 K f. Since f is integrable, it follows that 1 K is integrable. Moreover, µ(k) f dx. 3 The space L 1 In this section we denote the space of integrable functions C by L 1 ( ). For f L 1 ( ) we put f 1 = f(x) dx. The following result asserts that in a suitable sense the space L 1 ( ) is complete. Lemma 19 Let (f k ) k N be a sequence of functions in L 1 ( ) that is Cauchy for the L 1 -norm. Then there exists a function f L 1 ( ) such that f k f 1 0. If f is any such function, then f k f almost everywhere. Although we have called 1 a norm, it really is not, as f 1 = 0 need not imply f = 0. However, by the above lemma applied to a constant sequence, if f L 1 and f 1 = 0, then f = 0 almost everywhere. We define N := {f L 1 ( ) f = 0 almost everywhere}. Then N is a linear subspace of L 1. The quotient space L 1 /N is denoted by L 1 ( ). We note that the natural map L 1 ( ) L 1 ( ) is a surjective linear map. Moreover, f, g L 1 ( ) have equal image in L 1 ( ) if and only if they are equal outside a set of measure zero. Thus, passing to L 1 ( ) comes down to identifying functions as soon as the are equal almost everywhere. It follows from the above lemma that 1 factors to a norm on L 1 ( ). Moreover, by the same lemma, we have the followowing corollary. Recall that a Banach space is a linear space equipped with a norm for which it is complete, i.e., every Cauchy sequence has a limit. 5

Corollary 20 The space L 1 ( ), equipped with 1, is a Banach space. We recall from the course on distribution theory that f test (f) defines a linear map from L 1 ( ) into D ( ). The kernel of this linear map equals N. Accordingly, the map factors to a linear embedding f test (f), L 1 ( ) D ( ). 4 Dominated convergence We now come to one of the most important results of the theory known as Lebesgue s dominated convergence theorem. We say that a sequence of measurable functions (f j ) is dominated by a nonnegative measurable function g if each f j is dominated by g. Theorem 21 (Dominated convergence theorem) Let (f j ) be a sequence of integrable functions, dominated by an integrable function g. If f j f almost everywhere, then f is integrable, and f j dx = f dx. lim j The proof of this result is beyond the scope of these notes. However, the result has several important consequences for continuity and differentiation of integrals with a parameter which we shall prove. Theorem 22 such that Let X be a metric space and let f : X C be a function (a) for every x X, the function f(x, ) is integrable; (b) for almost every y, the function x f(x, y) is continuous; (c) there exists an integrable function g : [0, [ such that for almost every y we have f(x, y) g(y), for all x X. Then the function F : X C defined by F (x) = f(x, y) dy is continuous. Proof: Let a X and let (a j ) be any sequence in X with limit a. Then f j := (f(a j, ) defines a sequence dominated by g. Moreover, f j converges almost everywhere, with limit f(a, ). We apply the dominated convergence theorem to conclude that F (a j ) F (a), for every sequence a j ) X with limit a. It follows that F is continuous at a. 6

Another application of the dominated convergence theorem is the following result involving differentiation under the integral sign. Theorem 23 Let I be an interval and f : I C a function such that (a) for almost every y the function f(, y) is differentiable; (b) for every x I the functions f(x, ) and x f(x, ) are integrable; (c) there exists an integrable function g : [0, [ such that for almost every y and every x I, we have the domination x f(x, y) g(y). Then the function F : I C defined by F (x) = f(x, y) dy is differentiable on I. Moreover, for every x I, F (x) = x f(x, y) dy. Proof: Let a I and let a j be a sequence in I such that a j a. We consider the difference quotient F (a j ) F (a) = g j (y) dy, a j a where g j (y) = [f(a j, y) f(a, y)]/(a j a). Let N be a set of measure zero such that f(, y) is differentiable and such that the domination in (c) holds for every y \ N. Then for y \ N, we have that g j (y) = x f(ξ j (y), y) for some ξ j (y) I between a j and a (by the mean value theorem for ordinary differentiation). By the domination in (c) we have g j g almost everywhere. Moreover, g j (y) x f(a, y) for almost every y. By the dominated convergence theorem it follows that F (a j ) F (a) a j a x f(a, y) dy. This holds for every sequence in I with limit a. Hence F is differentiable at a and F F (x) F (a) (a) = lim = x f(a, y) dy. x a x a Corollary 24 Let U R p be open, f : U C a function and k a constant such that (a) for almost every y the function f(, y) is C k ; 7

(b) for every multi-index α N p with α k and every x U the function α x f(x, ) is integrable; (c) there exists an integrable function g : C such that for almost every y and all x U we have the domination α x f(x, y) g(y), ( α k); Then the function F : U C defined by F (x) = f(x, y) dy belongs to C k (U). Moreover, for every α with α k and all x U we have α F (x) dx = x α f(x, y) dy. Finally, the above can be applied to obtain a result about analyticity of integrals with integrand depending complex analytically on a parameter. Theorem 25 such that Let Ω C be an open subset, and f : Ω C a function (a) for every z Ω the function u f(z, u) is integrable; (b) for almost every u, the function f(, u) is complex differentiable on Ω; (c) there exists an integrable function g : [0, [ such that for almost every u and all z Ω, we have the domination f(z, u) g(u). Then the function F : Ω C defined by z f(z, u) du is complex differentiable. Moreover, for every p N the p-th order derivative is given by F (p) (z) = Rn p f(z, u) du. z Proof: Let K be a compact subset of Ω. Then there exists a compact subset K Ω and a constant ε > 0 such that for every α K, we have B(α; ε) K. It follows by application of the Cauchy inequalitities that for every p N there exists a constant C p > 0 such that for every holomorphic function ϕ : Ω C we have ϕ (p) K C p ϕ K ; here K denotes the supnorm over K. It follows from this observation that for almost all u and all z K we have the domination p z f(z, u) C p g(u). By the Cauchy-Riemann equations we have that x = i y = z on all holomorphic functions on Ω. By application of differentiation under the integral sign it now follows that the function F is C 1 and satisfies the Cauchy-Riemann equations, hence is complex differentiable. Moreover, F (z) = x F (z) = x f(z, u) du = z f(z, u) du. By repeated application of this formula we obtain the formula for F (p). 8

5 Convolution with Gaussian functions As an application of the material in the previous section, we consider convolution with the Gaussian function γ(x) = e x 2. Then γ is a real analytic function; in fact it extends to an entire holomorphic function on C n. Lemma 26 analytic. Let f : C be a bounded measurable function. Then γ f is Proof: We consider the function g : C n C defined by g(z, u) = f(u)γ(z u). Let Ω be a bounded open subset of C n. Then there exists a constant R > 0 such that z u u R for all (z, u) Ω. This shows that for every z Ω the function g(z, ) is measurable and dominated by the integrable function h(u) = sup f e ( u R)2. On the other hand, for every u the function g(, u) is complex differentiable with complex partial derivatives given by g (z) = f(u) γ (z u). z j z j It follows that for every z Ω the function g(z, ) is integrable, and by application of Theorerm 25 that the function G(z) = g(z, u) du is complex differentiable on Ω, with respect to each of its variables. Moreover, the complex partial derivatives are given by the formulas G (z) = f(u) γ (z u) du. z j z j By a straightforward application of Theorem 22 it follows that each of the partial derivatives of G is continuous on Ω. This implies that G is holomorphic on Ω. Consequently, G is holomorphic on C n. Its restriction to equals γ f which is therefore real analytic on. 9

6 Repeated integration The following result, known as Fubini s theorem, exposes one of the main advantages of Lebesgue integration beyond Riemann integration. Theorem 27 (Fubini s theorem) Let p, q be positive integers and let f : R p R q C be a function. Then the following two assertions are equivalent. 1. The function f is integrable on R p+q. 2. For almost every x R p the function f(x, ) is integrable on R q ; moreover, the function x f(x, y) dy is integrable on R p. If one of the above conditions is satisfied, then f(z) dz = f(x, y) dy dx. R p+q R p R q The proof of this result lies beyond the scope of these notes. 10