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to Solutions homework 5 18. Let A P(X) be an algebra, the collection of countable unions of sets in A, and A the collection of countable intersections of sets in A . Let 0 be a premeasure on A and the induced outer measure. a. For any E X and > 0 there exists A A with E A and (A) (E) + . Proof. Let > 0. Since (E) = inf { 0 (Ai ) : Ai A, E Ai }, we can take Ai A such that the Ai s cover E and 0 (Ai ) < (E) + . Put A = Ai . Then A A , and (A) (Ai ) = 0 (Ai ) < (E) + . b. If (E) < , then E is -measurable i there exists B A with E B and (B\E) = 0. Proof. Let E X with (E) < . To each n there is An A such that E An and 1 (An ) < (E) + n . Put B = An . Then B A , and E B, which implies (E) (B). On the other hand, for 1 any n, we have (B) (An ) (E) + n . Therefore (E) = (B). First suppose E is -measurable; we need to show (B\E) = 0. But (B) = (B E) + (B E c ) = (E) + (B\E). The result follows from subtracting (B) = (E). (This is where we use (E) < .) Now suppose (B\E) = 0. Let C X; we need to show that (C) (C E) + (C E c ). Since B M(A), B is -measurable. Therefore. (C) = (C B) + (C B c ) (C E) + (C B c ). Note that (C E c ) (C B c ) + (C E c B c ) (C B c ) + (B\E) = (C B c ). Therefore (C) (C E) + (C E c ). Since C was arbitrary, E is -measurable. 1 c. If 0 is - nite, the restriction (E) < in (b) is super uous. Proof. Write X = Xi , where Xi A and 0 (Xi ) < . Without loss of generality we can take the Xi s to be disjoint. Let E X and put Ei = E Xi . Then each Ei is of nite measure, following so the construction in part (b) there is for each i, n a set Ain in A such that 1 Ei Ain Xi and (E) (Ain ) < (E) + i . 2n Then putting Bi = n Ain , we get (Bi \Ei ) = 0 by the argument in part (b). Then putting B = Bi , we get E B and (B\E) = ( Bi \Ei ) (Bi \Ei ) = 0. The Bi s are in A , so B A . However, the disjointness of the Xi s, together with Bin Xi allows us to collapse Bi into a A set. If x j n Bjn , then there is a j such that for all n, x Bjn . This implies that for all n, there is an i such that x Bin : namely i = j. Therefore Bjn j n n j Bjn . In general, however, the implication does not go the other way. In this case, though, it does: Suppose x n i Bin . Then for all n there is an i such that x Bin . But this i can only be the unique i for which x Xi . Hence it is the same i for each n; for that i we have x n Bin . Then x is in the union over all i, so Bjn n j j n i Bij ; Bjn . and hence B is Therefore B = A . n i Bij . Since the Bij are in A , so are the 19. Let be an outer measure on X induced from a nite premeasure 0 . If E X, de ne the inner measure of E to be (E) = 0 (X) (E c ). Then E is -measurable i (E) = (E). Proof. If E is -measurable, then (X) = (E) + (E c ). Since (E) is nite we can subtract from both sides: (E) = (X) (E c ) = (E). Now suppose (E) = (E). Choose B A such that E B and (B) = (E), using the construction in the rst two paragraphs for 18(b), above. By problem 18(b), it su ces to show that (B\E) = 0. Note that B is -measurable, so (B) = (B). (X) = (E) + (E c ) (E) + (B\E) + (B c ) (E) + (B\E) + (X) (B) Cancelling the (X) and using (E) = (B) = (B) gives the result. 2

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