Math630-5 - Real Analysis - Math 630 Homework Set #5 -...

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Unformatted text preview: Real Analysis - Math 630 Homework Set #5 - Chapter 4 by Bobby Rohde 10-05-00 Problem 3 Let f be a nonnegative measurable function. Show that that f 0 a.e. Proof BWOC, suppose f 0 a.e. Then f 0 implies some set E with mE > 0 such that f x 0, x E. 1 0 x fx Since x f x n n , we know that the countable sum of the measures of the sets on the righthand side is greater than or equal to the measure of the left hand side which is non1 zero. So N such that if A = x f x N , then mA > 0. Define h x We know nonnegative f F min f x , f sup f F F 1 N , x. Thus h x f f F 1 N F f x, x. Let F A, such that 0 < mF < . g g, and f F f , the latter coming from the fact that since f is with 0. f F and g f, f 0, x. which This implies the f sup f g h mF Thus 0 a.e. QED contradicts assumption that f 0, which implies that f MATH630-5.nb 2 Problem 4 Let f be a nonnegative measurable function. a) Show that an increasing sequence of nonnegative simple functions each of n which vanishes outside a set of finite measure such that f lim n . Construction m fx m with m 22 n 2n and 0, otherwise Thus n assumes exactly 22 n -1 non-zero values over a set of measure 2n. Thus nonnegative simple function and vanishes except on a set of finite measure. m1 2n Define n = , x n, n such that m1 2n n is a x and n as follows. Suppose n x = 0 then clearly Furthermore we can show that n 1 n m1 0, thus suppose n x = 2n for some integer m with 0 < m 22 n . This implies that n1x 2m1 m1 m m1 m 2m fx fx 22 n , then clearly 2n 2n 2n 1 2n 2n 2n 1 , but if m 1 2m1 m1 2 22 n 22 n 1 , so thus we have that n 1 x 2m 1 n x , so its an 2n 1 2n increasing sequence. Now we need only show that f = lim n. Let [x] denote the least natural number -. x, and log2 x = to the logarithim base 2 of x, with log2 0 = Given > 0 and some x0 , we may choose N = max{[ x0 ], f x0 < . We wish to show that n > N, n x0 Since n > x0 non-zero value to x0 , we have that x0 n x0 . log2 , [log2 f x0 ]}. n, n so x0 meets the first condition for assigning a Since n > [log2 f x0 ] > log2 f x0 , we know that 2n 2log2 f x0 f x0 . So m 22 n n , for some m 2 n . Thus x meets the second condition for f x0 2 ,m 2 0 2n 2n assigning a non-zero value to n x0 . Finally since n > log2 ] and n meets the requirements to assign x0 a particular non-zero value, 1 1 we know that n x fx to the spacing of the values of n = 2n < 2 l1 1 = . Thus og fx <. nx n converges pointwise to f x. So f = lim n . 2 MATH630-5.nb 3 Thus we have constructed a b) Show that Proof By definition finite. f h n satisfying all required properties, QED. f. f sup over all simple functions sup f h, where h is a bounded measurable function with m x : h x 0 is We may of course limit ourselves to taking the sup over all h' x such that h' x max h x , 0 . However all such h' x are nonnegative measurable functions. The measurability of h' x deriving from the fact that x : h x 0 and the set x : h x 0 must both be measurable. Hence we may apply part a) and conclude that h' x , an increasing sequence such that n lim n h' x . Thus by Proposition 10 we have that h' n which equals the sup n since it is an increasing sequence. Thus the sup over some subset of all such simple functions must be = h sup f h= f . So f sup . 0. Thus by Proposition However f, and since we are considering a sup we may assume that 8 we have that f . So f sup . f sup , over all simple functions f. QED MATH630-5.nb 4 Problem 5 Let f be a nonnegative integrable function. Show that the function F defined by Fx x f is continuous by using Theorem 10. Proof In order to show continuity we need that F a = limx a F x , x. We need f = limx a 1 f x, x a n Define fn then fn is an increasing sequence with fn f as n x 0, otherwise a). x < a, we have that x x a x f. (- , f a a fN for some N such that x a 1 N a N> a 1 ax . f . But , thus as a Thus limx a f limn fn , and by Theorem 10 we have that limn x a also we know that x such that 0 < a - x < < 1, we have that f fM , x a 0 in the limit x a , we get that limx a f limn fn . So limx a x fn M< 1 f limn x a fn a a f. x Now consider limx a f f limx a a f , by Proposition 12. We know that since f is integrable f < a.e. Since if f = on some set, A, of measure > 0, we would have f mA , which is false since f , by definition of integrable. Thus without loss of generality we may assume that f < everywhere since this won't change it's integral. Now since f is finite over a closed interval we know that f is bounded over that interval. Let N(x) be the x maximum of f y y [a, x]. Hence, a f m[a, x]*N(x), by Proposition 5. However it is clear x that N x1 N x2 , if a x1 x2 . Thus limx a a f limx a m a, x N x = 0. Thus limx a limx a F x . f f limx F is continuous. x a a x a f a f. Hence a f = limx a x f Fa = MATH630-5.nb 5 Problem 6 Let f n be a sequence of nonnegative measurable functions that converge to f, and suppose fn f n. Then f lim fn f lim fn . Also since fn f n, we know that fn f, Proof By Fatou's Lemma we know that n, and thus lim fn f. lim However lim fn f lim fn lim fn fn by definition of lim sup and lim inf. lim fn . QED So by squeezing, Problem 7 a) Show that we may have strict inequality in Fatou's Lemma. Proof Consider the sequence x, since fn defined by fn x = 1 if n 0, n > N. Thus fn 1. Thus 0 x n 1, with fn x f lim 0 otherwise. fn 1, N > x, such that fn x n, we have fn 0 n1 1 n f = 0. So 1 f 0, however lim fn . QED b) Show that the Monotone Convergence Theorem need not hold for decreasing sequences of functions. Proof Let fn x = 0 if x x, lim n and fn x = 1 for x 0, n. f = 0. Hence 0. Hence lim fn f 0. However consider f . QED N > x, such that fn x fn . fn n1 fn n n > N. Thus fn fn 1 1. Hence lim ...
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This note was uploaded on 03/23/2010 for the course MATH 515 taught by Professor Staff during the Spring '08 term at Iowa State.

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