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1、MEASURE AND INTEGRATION: LECTURE 1 Preliminaries. We need to know how to measure the “size” or “vol ume” of subsets of a space X before we can integrate functions f : X R or f : X C. Were familiar with volume in Rn . What about more general spaces X? We need a measure function : subsets of X 0,. For
2、 technical reasons, a measure will not be defi ned on all subsets of X, but instead a certain collection of subsets of X called a algebra, a collection of subsets of X (i.e., a collection M P(X) that is a subset of the power set of X) satisfying the following: .(1) X M c (2) If A M, then A. X A M (3
3、) If Ai M (i = 1,2,.), then . i=1 M Constrast with a topology P(X), which satisfi es (1) and X . n (2) If Ui (i = 1,.,n), then i=1Ui . (3) If U ( I) is an arbitrary collection in , then I U . Remarks on algebras: (a) By (1), X M, so by (2), M. c (b) Ai = (Ai )c countable intersections are in M. i=1i
4、=1 (c) A,B M A B M (since A B = A Bc). Let (X,X ) and (Y,Y ) be a topological spaces. Then f : XY is continuous if f1(U) X for all U Y . “Inverse images of open sets are open.” Let (X,M) be a measure space (i.e., M is a algebra for the space X). Then f : X Y is measurable if f1(U) M for all U Y . “I
5、nverse images of open sets are measurable.” Basic properties of measurable functions. Proposition 0.1. Let X,Y,Z be topological spaces such that X f Y g Z. (1) If f and g are continuous, then g f is continuous. Proof. (g f)1(U) = f1(g1(U) = f1(open) = open. Date: September 4, 2003. 1 2 MEASURE AND I
6、NTEGRATION: LECTURE 1 (2) If f is measurable and g is continuous, then g f is measurable. Proof. (g f)1(U) = f1(g1(U) = f1(open) = open. Theorem 0.2. Let u: X R, v : X R, and : R R Y . Set h(x) = (u(x), v(x) : XY . If u and v are measurable and is continuous, then h: XY is measurable. Proof. Defi ne
7、 f : X = R2 by f(x) = u(x) v(x). Then R R h = f. We just need to show (NTS) that f is measurable. Let R R2 be a rectangle of the form I1 I2 where each Ii R(i = 1, 2) is an open interval. Then f1(R) = u1(I1) v1(I2). Let x f1(R) so that f(x) R. Then u(x) I1 and v(x) I2. Since u is measurable, u1(I1) M
8、, and since v is measurable, v1(I2) M. Since M is a algebra, u1(I1) v1(I2) M. Thus f1(R) M for any rectangle R. Finally, any open set U = (rectangle around points with i=1Ri rational coordinates). So f1(U) = f1 (Ri) = f1(Ri). Each i=1i=1 term in the union is in M, so since countable unions of elemen
9、ts in M are in M. , f1(U) M Examples. (a) Let f : X C with f = u + iv and u, v real measurable func tions. Then f is complex measurable. (b) If f = u + iv is complex measurable on X, then u, v, and |f| are real measurable. Take to be z Re z, z Im z, and z , respectively. z | | (c) If f, g are real m
10、easurable, then so are f + g and fg. (Also holds for complex measurable functions.) (d) If E X is measurable (i.e., E M), then the characteristic function of E, E (x) = 1 if x E; 0 otherwise. Proposition 0.3. Let F be any collection of subsets of X. Then there exists a smallest algebra M such that F
11、 M. We call M the algebra generated by F. Proof. Let = the set of all algebras containing F. The power set P(X) = the set of all subsets of X is a algebra, so is not empty. Defi ne M = . Since F M for all M , we have F M.MM If M is a algebra containing F, then M M by defi nition. Claim: is a algebra
12、. If A M, take M . M is a algebra and M . Thus, Ac M, and so Ac since M . If Ai MA M M M 3 MEASURE AND INTEGRATION: LECTURE 1 for each i = 1,2,., then A M, and so iAi . It follows that M . iAi M Borel Sets. By the previous proposition, if X is a topological space, then there exists a smallest algebr
13、a B containing the open sets. Ele ments of B are called Borel sets. If f : (X,B) (Y,) and f1(U) B for all U , then f is called Borel measurable. In particular, continuous functions are Borel measurable. Terminology: F (“Fsigma”) = countable union of closed sets. G (“Gdelta”) = countable intersection
14、 of open sets. MEASURE AND INTEGRATION: LECTURE 2 Proposition 0.1. Let M be a algebra on X, let Y be a topological space, and let f : XY . (a) Let be a collection of sets E Y such that f1(E) M. Then is a algebra on Y . (b) If f is measurable and E Y is Borel, then f1(E) M. (c) If Y = (with open sets
15、 along with ,a) and (b, with a,b R) and f1(,) M for all , then , f is measurable. Proof. (a) Since f1(Y ) = X M, we have Y . Also f1(Ec) = Lastly, (f1(E)c M Ec M. f1(Ei) = f1 i=1i=1 (Ei) M. (b) Because f is measurable, all open sets are in . Since is a algebra, we have B . (c) Recall = E Given R, choose n 1 n are both measurable. Proof. NTS g1(,) M for all . We have g
