REAL ANALYSIS I HOMEWORK 4 CİHAN BAHRAN The questions are from Stein and Shakarchi s text, Chapter 2.. Given a collection of sets E, E 2,..., E n, construct another collection E, E 2,..., E N, with N = 2 n, so that n k= E k = N j= E j ; the collection {E j } is disjoint; also E k = E j E k E j, for every k. Observe that it suffices to construct such a collection with N 2 n because in case N < 2 n we can pad empty sets to get to exactly 2 n sets and the conclusions will still hold. Let C = {E,..., E n }. Let D be the collection of all subsets of X = n k= E k which have expressions of the type E E n where E i is either E i or E c i for each i =,..., n. The function f : 2 C D E E E E is surjective by the definition of D, hence D 2 C = 2 n. Note that if E =, then f(e) = E c E c n = X c =. So we can restrict f as f : 2 C { } D { }. E / E E c First, we show that X = f(e) = D. E 2 C { } D D { } ( ) The second equality directly follows from f being surjective. For the first equality, let x X. If we let E x = {E C : x E}, we have x f(e x ) by the definition of f. Second, we show disjointness. Say x f(e). Then for every E E, we have x E. Hence E E x. Conversely whenever E / E, we have x / E. Thus E = E x. From here we deduce that Third, for E C we claim that f(e) f(e ) E = E. E = E 2 C { } f(e) E f(e) = D D { } D E Again the second equality is immediate. For the first equality the part is trivial. Let x E, so E E x, equivalently E E x, and x f(e x ) so we are done. Thus writing D { } = {E,..., E N}, we have N 2 n with the desired properties. D. 2. In analogy to Proposition 2.5, prove that if f is integrable on R d and δ > 0, then f(δx) converges to f(x) in the L -norm as δ.
REAL ANALYSIS I HOMEWORK 4 2 Since the set of continuous functions with compact support is dense in L (R d ), given ε > 0 there exists such a function g such that f g ε. Writing f δ for the function x f(δx), we have f δ g δ = (f g) δ = δ d f g. Since g is uniformly continuous on K := supp g, there exists λ > 0 such that g(x) g(y) ε/m(k) whenever x y λ and x, y K. Since K is compact, there exists M > 0 such that x K implies x M. Choosing δ close enough to, we can guarantee that δ λ/m. Therefore for every x K we get Thus for every x K Thus g g δ = K g g δ ε. Hence x δx = δ x λ. g(x) g(δx) ε/m(k). f f δ f g + g g δ + f δ g δ = (δ d + ) f g + g g δ (δ d + )ε + ε = ε(δ d + 2) and given δ even closer to we can make sure that δ d < 2. So given ε > 0, we can pick η > 0 such that δ < η implies This yields f f δ 4ε. lim f f δ = 0. δ 5. Suppose F is a closed set in R, whose complement has finite measure, and let δ(x) denote the distance from x to F, that is, Consider δ(x) = d(x, F ) = inf{ x y : y F }. I(x) = R x y 2 dy. (a) Prove that δ is continuous, by showing that it satisfies the Lipschitz condition δ(x) x y. (b) Show that I(x) = for each x / F. (c) Show that I(x) < for a.e. x F. This may be surprising in view of the fact that the Lipschitz condition cancels only one power of x y in the integrand of I. [Hint: For the last part, investigate F I(x)dx.] (a) For every z F we have x z x y + y z.
Hence REAL ANALYSIS I HOMEWORK 4 3 δ(x) = inf{z F : x z } inf{z F : x y + y z } x y + inf{z F : y z } = x y + so δ(x) x y. Changing the roles of x and y, we get δ(x) y x = x y. Thus δ(x) x y. (b) Note that x / F means that x has a ball around it disjoint from F since F is closed and hence δ := d(x, F ) = δ(x) > 0. Thus x y δ/2 implies So δ/2. Thus x+δ/2 x δ/2 δ(x) δ/2. x+δ/2 x y dy δ/2 δ/2 2 x δ/2 x y dy = 2 δ/2 (c) Consider the function f : F R R (x, y) x y 2. δ/2 δ/2 u du = δ du 2 0 u = δ =. 2 Note that f(x, x) = for every x F. f is continuous since δ is continuous. Also clearly f is non-negative. Hence by Tonelli s theorem, we have I(x)dx = F R F x y dxdy 2 Ç å = R F x y dx dy 2 Ç å = F c F x y dx dy. 2 Observe that for fixed y F c, for every x F we have x y and hence and so F {x R : x y } F = {x R : x y } {x R : x y } = [ + y, ) (, y ] x y 2 dx = = y+ y+ = 2 = 2 lim L y x y dy + 2 y (x y) dy + 2 u du + 2 du = 2 lim u2 L Ç å L u 2 du L u = 2. x y 2 dy (x y) 2 dy
Thus F REAL ANALYSIS I HOMEWORK 4 4 I(x)dx 2 F c dy = 2m(F c ) < since m(f c ) < by assumption. So we showed that F I(x)dx is finite hence I(x) is finite almost everywhere for x F. 7. Let Γ R d R, Γ = {(x, y) R d R : y = f(x)}, and assume f is measurable on R d. Show that Γ is a measurable subset of R d+, and m(γ) = 0. Consider the function Given α R, the set F : R d R R (x, y) f(x). F ((α, )) = f ((α, )) R is measurable in R d+ ; hence G is a measurable function. Also the projection R d R R (x, y) y is measurable (in fact continuous). Thus their difference G : R d R R (x, y) f(x) y is a measurable function, thus G ({0}) = Γ is a measurable subset of R d+. And by Corollary 2.3.3, m(γ) = m(γ x )dx = m({f(x)})dx = 0dx = 0. R d R d 0. Suppose f 0, and let E 2 k = {x : f(x) > 2 k } and F k = {x : 2 k < f(x) 2 k+ }. If f is finite almost everywhere, then F k = {f(x) > 0}, and the sets F k are disjoint. Prove that f is integrable if and only if 2 k m(f k ) <, if and only if 2 k m(e 2 k) <. Use this result to verify the following assertions. Let x a if x, x b if x >, f(x) = and g(x) = 0 otherwise, 0 otherwise. Then f is integrable on R d if and only if a < d; also g is integrable on R d if and only if b > d.
REAL ANALYSIS I HOMEWORK 4 5 Note that F k s are disjoint so letting E = {x : f(x) > 0}, as f = 0 outside E by non-negativity we have f = f = and by the definition of F k, we have 2 k m(f k ) f F k so f is finite if and only if Observe that for every k, we have hence Now since 2 k m(e 2 k) = 2 k+ m(e 2 k+) = = 2 so we get the second if and only if. E 2 k m(f k ) is finite. E 2 k = E 2 k+ F k 2 k m(f k ) = 2 F k f 2 k+ m(f k ) = 2 2 k m(e 2 k+) + 2 k+ m(e 2 k+) + 2 k m(e 2 k), we obtain 2 k m(f k ) 2 k m(e 2 k) 2 k m(f k ) 2 k m(f k ). Consider the given f and E 2 k s defined by it. If a < 0 then f is clearly integrable. So we may assume a > 0. Then we have E 2 k = {x R d : x a > 2 k, x } = {x R d : x a < 2 k, x } = {x R d : x < min{, 2 k/a }} and since 2 k/a if and only if k/a 0 if and only if k 0, we get B E 2 k = R d() if k < 0 B R d(2 k/a ) if k 0 where B R d(r) denotes the ball centered at the origin of radius r in R d. Thus 2 k m(e 2 k) = 2 k m(b R d()) + 2 k m(b R d(2 k/a )) Observe that a ball of radius r in R d contains a cube of side-length r and be contained in a cube of side-length 2r in R d. Thus r d m(b R d(r)) (2r) d = 2 d r d
REAL ANALYSIS I HOMEWORK 4 6 therefore Ñ é 2 k + 2 k (2 k/a ) d k= 2 k m(e k ) 2 d 2 k + 2 k (2 k/a ) d so as 2 k =, it is enough to determine when integrability of f. And 2 k (2 k/a ) d = 2 k (2 k/a ) d converges for the 2 k kd/a = 2 k( d/a) = (2 ( d/a) ) k converges iff 2 d/a < iff d/a < 0 iff d/a > iff a < d. Now consider the given g and E 2 k s defined by it. We may define g to be when x which does not affect the integrability of g. If b < 0 then g is clearly not integrable. So we may assume b > 0. In this case we have E 2 k = {x R d : x b > 2 k, x > } {x R d : > 2 k, x } = {x R d : x b < 2 k, x > } {x R d : > 2 k, x } = {x R d : < x < 2 k/b } {x R d : > 2 k, x }. Note that since 2 k/b > iff k/b > 0 iff k < 0; so E 2 k = if k 0. And for k < 0 we have E 2 k = {x R d : < x < 2 k/b } {x R : x } = {x R d : x < 2 k/b }. Because E 2 k = B R d(2 k/b ), we have 2 k (2 k/b ) d 2 k m(e k ) 2 d 2 k (2 k/b ) d. Since 2 k (2 k/b ) d = 2 k( d/b) = 2 (d/b )k converges iff 2 (d/b ) < iff b > d, k= the series 2 k m(e 2 k) = 2 k m(e 2 k) also converges iff b > d as desired. 3. Give an example of two measurable sets A and B such that A + B is not measurable. [Hint: In R 2 take A = {0} [0, ] and B = N {0}.] Let A, B be as in the hint where N R is a nonmeasurable set. Then A + B = {(0, y) + (x, 0) : y [0, ], x N } = N [0, ] By Lemma 3.5, both A and B are null sets, hence are measurable. But A + B cannot be measurable, because since m([0, ]) = > 0, the measurability of A+B would imply the measurability of N by Proposition 3.4.
REAL ANALYSIS I HOMEWORK 4 7 5. Consider the function defined over R by x /2 if 0 < x <, f(x) = 0 otherwise. For a fixed enumeration {r n } n= of the rationals Q, let F (x) = 2 n f(x r n ). n= Prove that F is integrable, hence the series defining F converges for almost every x R. However, observe that the series is unbounded on every interval, and in fact, any function F that agrees with F a.e. is unbounded in any interval. Note that fχ (/n,) = /n x /2 dx = 2 x /n = 2 2» /n so by monotone convergence theorem f = fχ (0,) = n lim (2 2» /n) = 2. Note that by the translation invariance of the Lebesgue integral, we have f(x)dx = f(x r)dx for any r R. Thus N N N 2 n f(x r n )dx = 2 n f(x)dx = 2 n+ N = 2 n n= n= n= n=0 and again by the monotone convergence theorem F = 2 n = 2. n=0 Let I be a interval. Let r N be a rational number in I. Then for every M >, whenever x (r N, r N + 2 2N /M 2 ) we have 0 < x r N < 2 2N /M 2 < so since f is decreasing f(x r N ) > f(2 2N /M 2 ) = 2 N M. But (r N, r N +2 2N /M 2 ) intersects I in a nonempty interval which has positive measure. So on a set of positive measure contained in I, we have F (x) = 2 n f(x r n ) 2 N f(x r N ) > M. n= So if F is a.e. equal to F then F also has to be larger than M in a set of positive measure zero contained in I. Since M was arbitrary, we get the desired conclusion. 8. Let f be a measurable finite-valued function on [0, ], and suppose that f(x) f(y) is integrable on [0, ] [0, ]. Show that f(x) is integrable on [0, ]. We are assuming that the function F : [0, ] [0, ] R (x, y) f(x) f(y)
REAL ANALYSIS I HOMEWORK 4 8 is integrable. So by a part of Fubini s theorem, for almost every y [0, ] the function [0, ] R x f(x) f(y) is in L ([0, ]). So pick and fix such a y. Then the function g : [0, ] R x f(x) f(y) is in L ([0, ]), because g is. Now since f(y) is finite, the constant function c : [0, ] R x f(y) is also in L ([0, ]) (its integral is f(y)). Thus g + c = f L ([0, ]). 9. Suppose f is integrable on R d. For each α > 0, let E α = {x : f(x) > α}. Prove that f(x) dx = m(e α )dα. R d 0 Consider the function F : (0, ) R d R if f(x) > α, (α, x) 0 otherwise. Since F is non-negative and measurable (since F = χ f(x) >α and f is measurable). So by Tonelli s theorem we have F (α, x)dxdα = F (α, x)dαdx 0 R d R d 0 Å ã Å ã χ Eα (x)dx dα = χ (0, f(x) ) (α)dα dx 0 R d R d 0 m(e α )dα = f(x) dx. 0 R d