MAT205a, Fall 2019 Part II: Integration Lecture 4, Following Folland, ch 2.1, part of 2.4 1. Measurable functions 1.1. Definition. Suppose that (X; M) and (Y; N ) are two measurable spaces, i.e., M and N are σ-algebras of subsets of X and Y respectively. Consider a function f : X ! Y and subsets of X of the form ff −1(E): E 2 N g. These subsets form a σ-algebra. Definition 1.1. A function f : X ! Y is called (M; N )-measurable if the pre- image of any measurable set is measurable, i.e., f −1(E) 2 M for any E 2 N . A function f : X ! R is called M-measurable if for any Borel set E ⊂ R its preimage, f −1(E), is in M. It is easy to see that fE ⊂ Y : f −1(E) 2 Mg is a σ-algebra. Thus f : X ! R is measurable if and only if fx 2 X : f(x) > ag are measurable for all a 2 R. We will denote this set by ff > ag. Suppose that A 2 M, we define the characteristic function of A by χA(x) = 1 when x 2 A and χA(x) = 0 when x 62 A. Then χA is measurable. Clearly for any −1 c E ⊂ R we have χA (E) 2 f;; A; A ;Xg. We remark that if X is a topological space with the Borel σ-algebra BX and f : X ! R is continuous then f is BX -measurable. By the definition of continuity, the preimage of any open interval I ⊂ R is open. Thus the preimage of any Borel set is a Borel set in X. 1.2. Properties of measurable functions. We show that the class of measurable functions is closed under the standard operation. Proposition 1.1. If f is a measurable function, then for any c 2 R the function cf is also measurable. If f; g 2 X ! R are measurable, then maxff; gg; minff; gg; f + g and fg are mea- surable. −1 −1 −1 Proof. First, clearly if c 6= 0 then (cf) (E) = f (c E) and if E 2 BR then −1 c E 2 BR, thus cf is measurable. Now, fmaxff; gg > ag = ff > ag \ fg > ag and, similarly, fminff; gg > ag = ff > ag [ fg > ag. Thus maxff; gg; minff; gg are measurable. For the sum we have fx : f + g > ag = [r(ff > rg \ fg > a − rg), where the union is taken over all rational numbers r. It is a countable union of measurable sets, thus it is measurable. To show that fg is measurable it is enough to show that for any measurable function f the function f 2 is also measurable and 1 2 1 2 2 2 2 then write fg = 2 ((f + g) − f − g ). But ff > ag = X when a < 0 and 2 p p 2 ff > ag = ff > ag [ ff < − ag when a ≥ 0. Therefore f is measurable. For any measurable function f we define f + = maxff; 0g and f − = max{−f; 0g, those functions are non-negative and measurable, f = f + − f −. Definition 1.2. A finite linear combination of characteristic functions of measurable sets is called a simple function. It follows from the previous proposition that simple functions are measurable. We will show that any measurable function can be approximated by simple ones. Proposition 1.2. If ffng is a sequence of measurable functions then the follow- ing functions are measurable: g1(x) = supn fn(x), g2(x) = infn fn(x), g3(x) = lim supn fn(x), and g4 = lim infn fn(x). Proof. Clearly, fg1 > ag = [nffn > ag is measurable as the countable union of measurable sets. Similarly, for g2, fg2 ≥ ag = \nffn ≥ ag is measurable. Further- more, g3 = lim supn fn = infn supk≥n fk is measurable, and so is g4 = lim infn = supn infk≥n fk. The proposition implies that if there is a limit, then it is equal to both the upper and the lower limits and is measurable. Corollary 1.1. Suppose that ffng is a sequence of measurable functions on (X; M), and for any x 2 X there is the limit f(x) = limn!1 fn(x) (i.e., f is the point-wise limit of fn). Then f is also measurable. 1.3. Approximation of measurable functions. We denote by L+(M) the family of all non-negative measurable functions on (X; M). As we have seen above, for any measurable function f there are functions f +; f − 2 L+ such that f = f + − f −. Proposition 1.3. Let (X; M) be a measurable space and f 2 L+(M) be a non- negative measurable function. Then there exists a sequence of simple functions fφjg such that φ1 ≤ φ2 ≤ ::: and f(x) = limj!1 φj(x) for any x 2 X. Proof. For each n define the sets −n −n n En;k = ff 2 [k2 ; (k + 1)2 )g;En;1 = ff ≥ 2 g for 0 ≤ k ≤ 4n. Clearly, those are measurable sets, now we set X −n n φn = k2 χEn;k + 2 χEn;1 : k Then it is easy to check that φn ≤ φn+1 ≤ f and f(x) = limn!1 φn(x) for each x 2 X. 3 Now assume that X = R and µ is a Lebesgue-Stieltjes measure on R. We want to approximate a measurable function by continuous ones almost everywhere. Proposition 1.4. If µ is a Lebesgue-Stieltjes measure on (R; M) and f is M- measurable, then there is a sequence of continuous functions fgng which converges to f almost everywhere. Proof. It is enough to prove the proposition for the case of a non-negative function. Let En;k be sets defined in the proof of the previous proposition. Since µ is a Lebesgue- Stieltjes measure there exists an open set Un;k such hat En;k ⊂ Un;k and µ(Un;k n −n−k En;k) ≤ 2 . Let also n;k be a continuous function with support on Un;k such that −n−k 0 ≤ n;k ≤ 1 and µ(Un;k n f n;k < 1g) < 2 . Now let Cn = [k(Un;k n En;k) [ (Un;k n f n;k 6= 1)g 1 and C = \j [n+j Cn. Then C is the set of x which belong to infinitely many sets Cn and µ(C) = 0. We define X −n gn = k2 n;k: k Clearly gn is a continuous function. Now if x 2 RnC then there exists n(x) such that n(x) x 62 Cn for n > n(x) and f(x) < 2 . Then for n > n(x) we have jgn(x) − f(x)j < n c 1=2 . Therefore gn converges to f on C . 2. Modes of convergence 2.1. Point-wise and uniform convergence. Let (X; M; µ) be a measure space. The statement "the sequence ffng converges to f" can be understood in a number of ways. First we discuss the following three types of convergence. (1) We say that ffng converges to f uniformly on X if for any " > 0 there exists m such that jf(x) − fn(x)j < " for every x 2 X and n ≥ m. (2) We say that ffng converges to f point-wise if for any x 2 X and " > 0 there exists m = m(x) such that jf(x) − fn(x)j < " for n ≥ m. (3) If the property described in (2) holds for x 2 E such that µ(Ec) = 0, we say that ffng converges to f almost everywhere. Clearly, uniform convergence implies point-wise convergence, which implies con- vergence almost everywhere. It is easy to construct examples showing that inverse implications are wrong. If a sequence of measurable functions converges to a func- tion almost everywhere and the measure is complete, then the limit function is also measurable. Theorem 2.1 (Egorov). Suppose that (X; M; µ) is a measure space with a finite measure, µ(X) < 1. If a sequence of measurable functions ffng converges to a 4 measurable function f almost everywhere then for any " > 0 there exists E 2 M such that µ(X n E) < " and fn converges to f uniformly on E. Proof. First, let Y be the set where ffng converges to f, µ(X n Y ) = 0. For each n and k we define An;k = [m≥nfjf − fmj > 1=kg: c Then A1;k ⊃ A2;k ⊃ :: for each k and \nAn;k ⊂ Y . Since the measure is finite, −k we have limn!1 µ(An;k) = 0. We choose n(k) such that µ(An(k);k) < "2 . And let A = [kAn(k);k, so that µ(A) ≤ ". On the complement of A we have jfn − fj < 1=k c when n ≥ n(k). Thus fn converges to f uniformly on A . 2.2. Lusin theorem. Now we show that any measurable function is continuous on some compact subset of almost full measure. Theorem 2.2 (Lunsin). Let µ be a Lebesgue-Stieltjes measure on (R; M). Suppose that f : R ! R is a M-measurable function and A ⊂ R is such that µ(A) < 1. Then for any " > 0 there exists a compact set K ⊂ A such that µ(A n K) < " and f is continuous on K. Proof. By Proposition 1.4 there is a sequence of continuous functions gn that con- verges to f almost everywhere on A. By Egorov's theorem there exists E ⊂ A such that µ(AnE) < "=2 and gn converges to f uniformly on E.