Real Variables, Fall 2014 Problem set 3 Solution suggestions xercise 1. Let f be a nonnegative measurable function. Show that f = sup ϕ, where ϕ is taken over all simple functions with ϕ f. For each n N we divide [0, n) to disjoint intervals I k = [ k 1 2 n, k 2 n ), where k = 1,..., n2 n. We then define f n for all n N so that f n (x) = k 1 2 n when x f 1 (I k ) and f n (x) = n when x f 1 [n, ]. In other words, f n (x) = 2 n n k=1 k 1 2 n χ f 1 (I k ) + nχ f 1 [n, ]. Since f was measurable then f n is a simple function. Now f n f n+1 by construction and f n (x) f(x) for all x. Indeed if x is so that f(x) < then there exists m N so that f(x) < m. When n m then k 1 f(x) < k for some k = 1,..., n2 n. Thus 2 n 2 k f n (x) = k 1 2 n f(x) < k 2 n = f n(x) + 1 2 n, which implies that f(x) 1 < f 2 n n (x) f(x). Letting n it follows that lim f n (x) = f(x). If on the other hand x is such that f(x) = then f n (x) = n for all n N and thus f n (x) = f(x). We have thus constructed a nondecreasing sequence of bounded simple functions (f n ) n=1 so that f n f. To truncate the supports, we define ϕ n = f n χ B(0,n) for all n N, where B(0, n) is the open n-radius ball around 0. Note that (ϕ n ) n=1 is still a nondecreasing sequence of bounded simple functions converging pointwise to f. In addition 1
m({x : ϕ n (x) 0}) m(b(0, n)) = 2n < for all n N. Now using monotone convergence theorem we conclude that sup ϕ lim ϕ n = lim ϕ n = f. And the converse inequality sup ϕ f holds by assumptions since the supremum is taken over all simple functions with ϕ f. So the claim follows. xercise 2. Let (f n ) n=1 be a sequence of nonnegative measurable functions on (, ) such that f n f almost everywhere, and suppose f n f <. Then for each measurable set we have f n f. Let be a measurable set. Note that since f n f almost everywhere then f n χ fχ almost everywhere. Thus by Fatou s lemma f = fχ = lim inf f nχ lim inf f n χ = lim inf f n. Similarly f lim inf c c f n. Now by assumption and the above inequalities we have f = f + f lim inf f n + lim inf c = lim f n = f, c f n lim ( f n + so there in fact holds an equality everywhere. Since f < then it follows that lim inf f n + lim inf c f n f = 0. In other words, by writing f = f + f, we have c ( ) ( ) lim inf f n f + lim inf f n f = 0. c c 2 c f n )
Since both terms are non-negative and their sum is equal to zero, they must both be zero. In other words lim inf f n f = 0 and lim inf Now the first equation gives lim inf f n f = 0 c c f n = and using the second equation we obtain ( f = lim inf f n = lim inf f n c c = lim inf f n lim sup f n = f lim sup f n. f f n ) Above we used the fact that lim inf (a n + b n ) = a + lim inf b n for sequences (a n ) n=1 and (b n ) n=1 for which lim a n exists and a n a. Thus lim sup f n = f f = f. c So we conclude that lim sup f n = f = lim inf which finally implies that the sequence ( f n) n=1 converges and its limit is f. In other words, f n f. f n, xercise 3. Let (f n ) n=1 be a sequence of integrable functions such that f n f almost everywhere and f is integrable. Show that f n f 0 if and only if f n f. 3
: Assume first that f n f 0. Note that by the reverse triangle inequality we have f n f = ( f n f ) f n f f n f. Letting n it follows that f n f. : Assume then the converse that f n f holds. Define g n = f + f n f f n for all n N, which is non-negative by the triangle inequality. Note that g n 2 f almost everywhere. Thus by Fatou s lemma 2 f = 2 f = lim n lim inf g n = lim inf f + lim inf f n lim sup f f n = 2 f lim sup f f n. Above we used the fact that lim inf (a n + b n ) = a + lim inf b n for sequences (a n ) n=1 and (b n ) n=1 for which lim a n exists and a n a. Since f is integrable then this implies that 0 lim inf f f n lim sup f f n 0, so there in fact holds an equality everywhere and in particular f f n = 0. lim xercise 4. Prove the Riemann-Lebesgue Theorem: If f is integrable on R then f(x) cos(nx) dx 0 as n. The general idea is to approximate f by step functions with bounded support so we prove the statement in multiple steps: (a) Assume first that f = χ Ik is an indicator function of some bounded interval I k with endpoints a k and b k. Then bk f(x) cos(nx) dx = χ Ik cos(nx) dx = cos(nx) dx = 1 n sin(nb k) sin(na k ) 2 n 0 a k 4
as n. So the claim holds for indicator functions of bounded intervals. (b) Assume then that f is a step function with bounded support. In other words, f = k a i χ Ai i=1 for some real a i and bounded intervals A i. By part (a) and linearity of the integral we have f(x) cos(nx) dx = = ( k ) a i χ Ai cos(nx) dx i=1 k a i χ Ai cos(nx) dx 0 i=1 as n, as each of the terms in the sum go to zero. Hence the statement holds for all step functions with bounded support. (c) We then assume that f is integrable and nonnegative, and let ε > 0 be fixed. By xercises 1. and 3. of this problem set we find a simple function ϕ f with bounded support so that f ϕ < ε. Let p be the measure of a compact interval K containing the support of ϕ. We may assume that p 1 by bounding the support of ϕ by a larger interval if necessary. Since ϕ is a simple function with bounded support then M := sup ϕ <. Again, we can assume that M 1 by replacing M by max{1, sup ϕ } if necessary. We then use xercise 4 of problem set 1 to find a step function g so that ϕ g < ε except on a pm set of measure less than ε, and 0 g M. Call the set where pm the bound fails by. Now by triangle inequality, the fact that cos(nx) 1 for all n N and x R, and the above estimates, 5
we have = f(x) cos(nx) dx ε + (f(x) + ϕ(x) ϕ(x) + g(x) g(x)) cos(nx) dx f ϕ + ε pm p + ϕ g + ϕ g + g(x) cos(nx) dx g(x) cos(nx) dx ε + ε ε(sup ϕ + sup g ) + + g(x) cos(nx) dx M pm ε + ε M + ε2m pm + g(x) cos(nx) dx 4ε + g(x) cos(nx) dx. Since the choice of ε > 0 was independent of n N, we can let n in the above inequality and note that part (b) implies lim sup f(x) cos(nx) dx 4ε since g was a step function with bounded support. Since the choice of ε > 0 was arbitrary then which implies that lim sup lim f(x) cos(nx) dx = 0, f(x) cos(nx) dx = 0. Hence the statement is true for any nonnegative integrable function. (d) To lastly verify the statement for the most general case, assume that f is an integrable function. Compose f to its positive and negative parts, i.e. f = f + f, where f + and f are integrable 6
and nonnegative. Now by part (c) we have f(x) cos(nx) dx = = 0 (f + (x) f (x)) cos(nx) dx f + (x) cos(nx) dx f (x) cos(nx) dx as n since f + and f were integrable and nonnegative. So lim f(x) cos(nx) dx = 0. Hence the statement is true for all integrable functions f. xercise 5. Let g be a bounded measurable function and let f be integrable on R. Show that lim g(x)(f(x) f(x + t)) dx = 0. t 0 We first establish the following lemma: Lemma (a) Let f be integrable. Then for any ε > 0 there exists a continuous function h with compact support so that f h ε. Proof: The proof follows the same pattern as the proof of the previous exercise. Assume first that f is nonnegative. Then by xercises 1. and 3. of this problem set we find a simple function ϕ f with bounded support so that f ϕ < ε. Let p be the measure of a compact interval K 4 containing the support of ϕ. We may assume that p 1 by bounding the support of ϕ by a larger interval if necessary. Since ϕ is a simple function then M := sup ϕ <. Again, we can assume that M 1 by replacing M by max{1, sup ϕ } if necessary. We then use xercise 4 of problem set 1 to find a continuous function h on K so that ϕ h < ε 4pM ε except on a set of measure less than, and 0 h M. Call 4pM the set where the bound fails by. Technically the given h is only continuous in K. This does not however pose any problems as we can choose one such h and extend it continuously to the whole real line by truncating it by hχ [a+δ,b δ] for small δ > 0, where K = [a, b], and then 7
linearly interpolating the endpoints to zero and making it zero outside of K. The original h can be chosen so that the new h obtained by this truncation satisfies the bounds that we want. Since h and ϕ vanish outside K, then by triangle inequality and the above bounds we have f h = ε 4 + f ϕ + f ϕ + K\ ϕ h ϕ h + K ε 4pM p + 2Mε 4pM = ε 4 + ε 4M + ε 2p ε 4 + ε 4 + ε 2 = ε. ϕ h So the claim is true when we assume that f is nonnegative. If f is not nonnegative, then compose f to its positive and negative parts f = f + f and apply the previous reasoning to the nonnegative integrable functions f + and f, i.e. find continuous h 1 and h 2 with compact support so that f + h 1 ε and f h 2 2 ε. 2 Now h 1 h 2 is a continuous function with compact support and f h 1 + h 2 = f + f h 1 + h 2 f + h 1 + ε 2 + ε 2 = ε. So the proof of the lemma is done. f h 2 We then return to the exercise. Denote M := sup g, which is finite since g is bounded. Now since f is integrable then by the above Lemma (a) we find a compactly supported continuous function h so that f h ε. Since h is compactly supported there exists 2M n N so that the support of h is contained in B(0, n), where B(0, n) is as usual the open n-radius ball around 0 and the overline denotes closure. Now for all 0 t < 1 we have h(x) = h(x + t) = 0 if x R \ B(0, n + 1). Thus for all 0 t < 1 we have by triangle 8
inequality and the above estimates that g(x)(f(x) f(x + t)) dx = B(0,n+1) g(x)(f(x) f(x + t) h(x) + h(x) h(x + t) + h(x + t)) dx g(x) h(x) h(x + t) dx + + g(x) h(x + t) f(x + t) dx M h(x) h(x + t) dx + 2M B(0,n+1) M h(x) h(x + t) dx + 2M ε B(0,n+1) 2M = M h(x) h(x + t) dx + ε. B(0,n+1) g(x) f(x) h(x) dx f h Now as a continuous function, h is uniformly continuous on the compact set B(0, n + 1). So it follows that h(x) h(x + t) 0 as t 0 uniformly. Hence by letting t 0 we get that lim t 0 g(x)(f(x) f(x + t)) dx ε. Since the choice of ε > 0 was arbitrary then lim g(x)(f(x) f(x + t)) dx = 0. t 0 xercise 6. Let f be integrable over a measurable set. Show that for any ε > 0 there is a continuous function g supported on a finite measure set such that f g ε. Apply Lemma (a) from xercise 5. to the integrable function fχ to find a compactly supported continuous function g so that fχ g ε. Now f g = f g χ = fχ gχ fχ g ε. So g satisfies the desired properties. 9