Homework 11. Solutions

Transcription

1 Homework 11. Solutions Problem Let f n : R R be 1/n times the characteristic function of the interval (0, n). Show that f n 0 uniformly and f n µ L = 1. Why isn t it a counterexample to the Lebesgue Dominated Convergence theorem. Solution. Uniform convergence. Given ɛ > 0, let N be an integer greater than 1/ɛ. If n > N, then sup f n (x) 1 x R n < ɛ. Thus lim n sup x f n (x) = 0, which is what uniform convergence means. The function f n = (1/n)χ (0,n) is a simple function and 1 n χ (0,n) µ L = 1 n µ L((0, n)) = 1. This example does not contradict the Lebesgue Dominated Convergence theorem because there is no integrable function g such that g f n for all n. Indeed, assume that g f n for all n. Then for each k, the sum s k = k (1/n)χ (n 1,n) is a simple function with g s k 0, because s k = f n on (n 1, n), for all 1 n k. Hence g µ L s k µ L for all k. But the integral s k µ L = k 1 n, which diverges to as k. Hence g µ L =. 1

2 Problem Let (, F, µ) be a measure space. A measurable function f : R is square integrable if f 2 <. Show that if µ() <, then every square-integrable function is integrable. Solution. Let A = { f > 1} and B = { f 1}. Then A and B are disjoint measurable sets with = A B. Then f = f + f because A B = and A B = A B f because f < f 2 on A and f 1 on B A B f because A, B and f 2, 1 0 < by the hypotheses f 2 < and µ() < Problem Let (, F, µ) be a measure space. A sequence, f n, of measurable functions is said to converge in measure to 0 if, for all ɛ > 0, lim µ{ f n > ɛ} = 0 Show that if f n µ 0 as n, then f n converges to 0 in measure, and show that the converse is not true. Solution. By Chebyshev s inequality, µ{ f n > ɛ} 1 ɛ f n µ. Now take the limit as n. For the converse, let = R, µ Lebesgue measure. Take f n = (1/n)χ (0,n). Then lim f n = 1. On the other hand, given ɛ > 0, if n > 1/ɛ, then so that f n converges to 0 in measure. µ{ f n > ɛ} = 0, Problem Let (, F, µ) be a measure space and f a non-negative measurable function on. For A F let µ f (A) = f µ. 2 A

3 Show that µ f is a measure on F. Show that if g 0 is a measurable function, then f µ f = gf µ, for all F. Solution. It is clear that µ f (A) 0 because f 0. To show countable additivity, use Theorem 11, page 67. If A 1, A 2, is a countable collection of disjoint measurable sets, and if A = A n, then µ f (A) = A f µ = A n f µ = µ A (A). For the second part, suppose first that g = χ A, the characteristic function of a measurable set A F. Then χ A µ f = µ f ( A) = fµ = χ A µ. It follows that g µ f = A gfµ for every simple function g 0. In general, if g 0 is measurable, let s n be a sequence of simple functions, 0 s n g, which increase pointwise to g. Then the monotone convergence theorem implies that g µ f = lim s n µ f Also, s n f increases pointwise to gf, so this same theorem implies gf µ = lim s n f µ. The stated identity now follows because for simple nonnegative functions it was just proved that s n µ f = s n f µ. Problem Done in class. 3

4 Problem Let (, M, µ) be a σ-finite measure space, f : R a nonnegative measurable function. Let A f = {(x, t) R 0 t f(x)} Show that A f is a measurable subset of R and that (µ µ L )(A f ) = f µ where µ L is Lebesgue measure on R. Solution. Let F : R R be the map F (x, t) = f(x) t. This map is measurable because it is the composition of the maps G : (x, t) (f(x), t) and H : (z, t) z t. The map H is measurable (xample ), in fact it is continuous. If A B is a product set in R R, then G 1 (A B) = {(x, t) G(x, t) = (f(x), t) A B} = f 1 (A) B is a product set in R. To show that F is measurable it must be shown that if B is a Borel subset of R, then F 1 (B) = {F (x, t) B} is a Borel set. In fact it suffices to consider B an open subset of R. Because F = H G, the inverse image F 1 (B) = (H G) 1 (B) = G 1 (H 1 (B)). Since H is continuous and B is open, the inverse image H 1 (B) can be written as a countable union of open subsets of R R of the form U V, where U, V are open in R. Thus there are countable many open subsets U n, V n R such that Therefore H 1 (B) = F 1 (B) = G 1 ( U n B n ) = U n B n G 1 (U n V n ) = f 1 (U n ) V n. But f 1 (U n ) is measurable and so f 1 (U n ) V n is a product set in R. Hence F 1 (B) is a countable union of product sets, thus an element of the σ-field F B. 4

5 Furthermore, the set A f = {(x, t) 0 t f(x)} = F 1 [0, ) ( [0, )) Let χ be the characteristic function of A f. Then χ 0, so that Fubini 2 applies and gives ( ) (µ µ L )(A f ) = χ (µ µ L ) = χ(x, t) µ L (t) µ(x) R R But for x fixed, the function t R χ(x, t) is the characteristic function of the interval [0, f(x)]. Therefore χ(x, t) µ L = µ L ([0, f(x)]) = f(x) R and the identity follows. 5