Math 2210 Real Analysis Problem Set 3 Solutions I. Minevich (small corrections by R. Kenyon) November 27, 2009 p. 48 # 3. If {fn } is a sequence of measurable functions on X, then {x : lim fn (x) exists} is a measurable set. Solution. {x : lim fn (x) exists} = {x : ∀ > 0 ∃N ∀n, m > N |fn (x) − fm (x)| < since the real (or complex) numbers are complete, so a sequence (namely (fn (x))∞ n=1 ) converges iff it is Cauchy. Thus {x : lim fn (x) exists} = ∞ [ ∞ \ ∞ \ ∞ \ k=1 N =1 m=N n=N {x : fn (x)−fm (x) < 1 1 }∩{x : fm (x)−fn (x) < }. k k Since fm and fn are measurable, so is ±(fm − fn ), so this is a countable intersection of a countable union of a countable intersection of measurable sets, which is measurable because the measurable sets form a σ-algebra. p. 48 # 5. If X = A ∪ B where A, B ∈ M, a function f on X is measurable iff f is measurable on A and on B. Solution. If f is measurable on A and on B and f : (X, M) → (Y, N ) then ∀E ∈ N , f −1 (E) = (f −1 (E) ∩ A) ∪ (f −1 (E) ∩ B) = (f |A )−1 (E) ∪ (f |B )−1 (E): a union of two measurable sets, hence measurable. If f is measurable on A ∪ B then (f |A )−1 (E) = f −1 (E) ∩ A: an intersection of two measurable sets, hence measurable; similarly for f |B . 1 p. 48 # 9. Let f : [0, 1] → [0, 1] be the Cantor function (§1.5), and let g(x) = f (x) + x. a. g is a bijection from [0, 1] to [0, 2], and h = g −1 is continuous from [0, 2] to [0, 1]. b. If C is the Cantor set, m(g(C)) = 1. c. By Exercise 29 of Chapter 1, g(C) contains a Lebesgue nonmeasurable set A. Let B = g −1 (A). Then B is Lebesgue measurable but not Borel. d. There exist a Lebesgue measurable function F and a continuous function G on R such that F ◦ G is not Lebesgue measurable. Solution. a. Since f (x) is monotone increasing, if y > x then f (y) ≥ f (x), so g(y) = f (y) + y > f (x) + x = g(x), which shows g is injective. It is surjective by the intermediate value theorem because it is continuous (being the sum of two continuous functions), g(0) = 0, and g(1) = 2. Since g is monotone and bijective, g((a, b)) = (g(a), g(b)), so g is also an open map, which implies h is continuous. b. Since g is surjective, g([0, 1] \ C) t g(C) = [0, 2], so we just have to show m(g([0, 1] \ C)) = 1 (we are using the fact that C is measurable). But [0, 1] \ C is a disjoint union of open intervals (a, b) on which f (x) is constant, so g([0, 1] \ C) is a disjoint union of open intervals (g(a), g(a) + b − a), hence m(g([0, 1] \ C)) = the length of the union of these intervals, which is 1. Therefore m(g(C)) = 2 − 1 = 1. c. Since A ⊂ g(C), we have B ⊂ C, so m(B) ≤ m(C) = 0, and any null set is measurable. Now if B is Borel then the continuity of h ensures h(B) = A is Borel: a contradiction since A is not even in the Lebesgue σalgebra. d. Let F = χB , so F (x) = 1 if x ∈ B and F (x) = 0 if x ∈ / B, and G = h. −1 Then F is Lebesgue measurable since F (a, ∞) = ∅, B, or R, all of which are Lebesgue measurable. And (F ◦ G)−1 (1/2, ∞) = G−1 ◦ F −1 (1/2, ∞) = 2 G−1 (B) = g(B) = A, which is not Lebesgue measurable, so F ◦ G is not Lebesgue measurable. p. 59 # 22. Let µ be counting measure on N. Interpret Fatou’s Lemma and the monotone and dominated convergence theorems as statements about infinite series. Solution. R P P k k By setting fk = ∞ a χ , we have fk = ∞ {n} n=1 n n=1 an . In each P statement k ∞ we consider a sequence of fk ’s, so we have a sequence of series { ∞ n=1 an }k=1 . Fatou’s lemma states ∞ X n=1 lim inf k akn ≤ lim inf k ∞ X akn . n=1 The monotone convergence theorem says that if ∀n ∀k akn ≤ ak+1 n , and k ∀n an = limk→∞ an , then ∞ X n=1 an = lim k→∞ ∞ X akn . n=1 P k The dominated convergence theorem says that if ∞ n=1 an is an absolutely convergent akn → an and (b) there exists a P∞ series for all k such that (a)P∀n ∞ series n=1 bn with bn ≥ 0 for all n and n=1 bn < ∞ such that ∀n ∀k |akn | ≤ bn , then ∞ ∞ ∞ X X X |an | < ∞ and an = lim akn . n=1 n=1 k→∞ n=1 p. 59 # 25. Let f (x) = x−1/2 if 0 < x < 1, f (x)P= 0 otherwise. let {rn } be −n an enumeration of the rationals, and set g(x) = ∞ f (x − rn ). 1 2 a. g ∈ L1 (m), and in particular g < ∞ a.e. b. g is discontinuous at every point and unbounded on every interval, and it remains so after any modification on a Lebesgue null set. c. g 2 < ∞ a.e., but g 2 is not integrable on any interval. 3 Solution. a. By the monotone convergence theorem, Z ∞ Z ∞ Z ∞ X −n |g(x)|dx = g(x)dx = 2 −∞ −∞ rn n=1 1 x −1/2 dx = 0 n=1 f (x − rn )dx. −∞ n=1 But this equals Z Z rn +1 ∞ ∞ X X −1/2 −n −n (x−rn ) dx = 2 2 ∞ ∞ X n=1 21−n x1/2 |10 = ∞ X 21−n = 2. n=1 if ∃E ∈ M, m(E) = > 0, such that g = ∞ on E then RIn particular, R g ≥ E g ≥ · ∞ = ∞ > 2: a contradiction. R b. g is unbounded on any interval because any interval (a, b) contains a rational rn for some n, and for any large M ∈ R and any x ∈ (rn , b) ∩ (rn , r + 1/(22n M 2 )) we have f (x − r)2−n > M , and g(x) ≥ f (x − rn )2−n since g(x) is a sum of nonnegative terms, hence we can pick x ∈ (a, b) so that g(x) is as large as we want. Any modification on a null set will still leave uncountably many such x ∈ (rn , b) ∩ (rn , r + 1/(22n M 2 )), so g will remain unbounded. Now suppose g is continuous at a point x0 . Then there exists δ such that |x − x0 | < δ ⇒ |g(x) − g(x0 )| < 1. But g is unbounded on (x0 − δ, x0 + δ): a contradiction. c. Because g < ∞ a.e., g 2 < ∞ a.e. since g(x) = ∞ ⇔ g(x)2 = ∞. Again using the monotone convergence theorem, !2 Z Z b X Z bX ∞ ∞ 2 −n 2−2n (f (x − rn ))2 dx |g| = 2 f (x − rn ) dx ≥ (a,b) a a n=1 n=1 = ∞ X f (x − rn )2 dx 2 (1) a n=1 Let rk ∈ (a, b). Then Z b Z −2k 2 −2k (1) ≥ 2 f (x−rk ) dx = 2 a b Z −2n min{b,rk +1} rk 1 dx = 2−2k (ln(min{b, 1})−ln(0)) = ∞. x − rk p. 63 # 32. Suppose µ(X) < ∞. If f and g are complex-valued measurable functions on X, define Z |f − g| ρ(f, g) = dµ. 1 + |f − g| 4 Then ρ is a metric on the space of measurable functions if we identify functions that are equal a.e., and fn → f with respect to this metric iff fn → f in measure. Solution. To prove ρ is a metric, first we check that ρ(f, g) = 0 ⇔ f = g. Obviously, ρ(f, f ) = 0. If ρ(f, g) = 0 but f 6= g then ∃E ∈ M|m(E) > 0 and ∀x ∈ R |f −g| |f −g| > 0 on E, so ρ(f, g) ≥ > 0. Ef (x) 6= g(x), which implies 1+|f −g| E 1+|f −g| Second, obviously ρ(f, g) = ρ(g, f ). Third, we check the triangle inequality: |f − h| 1 1 =1− ≤1− 1 + |f − h| 1 + |f − h| 1 + |f − g| + |g − h| = |f − g| |g − h| |f − g| |g − h| + ≤ + 1 + |f − g| + |g − h| 1 + |f − g| + |g − h| 1 + |f − g| 1 + |g − h| Now suppose fn → f in this metric but not in measure. Then ∃ > 0, δ > 0 ∀N ∃n > N µ({x : |fn (x) − f (x)| ≥ }) ≥ δ. This implies Z |fn − f | ≥ δ > 0, ρ(fn , f ) ≥ 1+ {x:|fn (x)−f (x)|≥} 1 + |fn − f | a contradiction to convergence in the metric. Finally, suppose R f|fnn −f→| f in measure but not in the metric. Then ∃ > 0 ∀N ∃n > N 1+|f ≥ . Without loss of generality we may take < n −f | |fn −f | 2µ(X). Letting M = µ(X), g = 1+|f , and E = {x : g(x) ≥ /(2M )}, we n −f | have µ(E) ≥ /2: otherwise, since g ≤ 1 we have Z Z Z ≤ g= g+ g < µ(E) + µ(X \ E) < + M = , 2M 2 2M X E X\E a contradiction. But the fact that µ(E) ≥ /2 means exactly that µ({x : |fn (x) − f (x)| ≥ /(2M ) > 0}) ≥ , 1 − /(2M ) 2 a contradiction to convergence in measure. 5 p. 63 # 33. If fn ≥ 0 and fn → f in measure, then R f ≤ lim inf R fn . Solution. By theorem 2.30, there is a subsequence fnk that converges to f a.e., so lim inf fn (x) ≤ f (x) a.e. If there exists a measurable S∞ set E, µ(E) S∞ > 0, such that ∀x ∈ E lim inf fn (x) < f (x) then E = k=1 Ek = k=1 {x : lim inf fn (x) ≤ f (x) − k1 }, so ∃k|µ(Ek ) = δ > 0. But fn → f in measure, so ∃N ∀n > N µ({x : |fn (x) − f (x)| ≥ k1 }) R < δ: R a contradiction, hence R lim inf fn (x) = f (x) a.e. By Fatou’s lemma, f = lim inf fn ≤ lim inf fn . p. 63 # 34. Suppose |fn | ≤ g ∈ L1 and fn → f in measure. R R a. f = lim fn . b. fn → f in L1 . Solution. a. By theorem 2.30 there exists a subsequence fnk → f a.e., so by Proposition 2.11 f is measurable (up to a null set). Looking at the imaginary and real parts of fn and f separately, we can assume the fn and g are all real-valued, so g +fn ≥ 0 a.e. and g −fn ≥ 0 a.e. By definition of convergence in measure, g + fn → g + f and g − fn → g − f in measure, so by exercise 33, Z Z Z Z Z Z g + f = g + f ≤ lim inf (g + fn ) = g + lim inf fn so R R f ≤ lim inf fn , and Z Z Z Z Z Z g − f = g − f ≤ lim inf (g − fn ) = g − lim sup fn so R R f ≥ lim sup fn . Together, Z Z Z Z lim inf fn ≥ f ≥ lim sup fn ≥ lim inf fn andR we have equality everywhere: lim fn . R f = lim inf R R fn = lim sup fn = b. Since |fn − f | → 0 in measure and |fn − f | ≤R 2g ∈ L1R, we can apply part (a) to the sequence (|fn − f |)∞ 0 = lim |fn − f |. n=1 to get 0 = p. 63 # 38. Suppose fn → f in measure and gn → g in measure. 6 a. fn + gn → f + g in measure.. b. fn gn → f g in measure if µ(X) < ∞, but not necessarily if µ(X) = ∞. Solution. a. Given > 0, δ > 0, ∃N1 , N2 ∀n > N1 µ({x : |fn (x) − f (x)| ≥ /2}) < δ/2 and ∀n > N2 µ({x : |gn (x) − g(x)| ≥ /2}) < δ/2. Letting N = max{N1 , N2 } we have ∀n > N µ({x : |fn (x) + gn (x) − (f (x) + g(x))| ≥ }) < δ since ≤ |fn (x) + gn (x) − (f (x) + g(x))| ≤ |fn (x) − f (x)| + |gn (x) − g(x)| implies |fn (x) − f (x)| ≥ /2 or |gn (x) − g(x)| ≥ /2. b. Note first that if fn → f in measure then f must be finite a.e.; similarly for g. Let An = {x : f (x) > n} and Bn = {x : g(x) > n}. Since ∞ µ(X) < ∞, and X = ∪∞ n=1 An = ∪n=1 Bn , given > 0, there exists N1 such that for n > N1 , µ(An ), µ(Bn ) < . There exists N2 such that for n ≥ N2 , µ{x : |fn (x) − f (x)| > /n} < and µ{x : |gn (x) − g(x)| > /n} < . Let Cn = {x : |fn (x) − f (x)| < /n} and Dn = {x : |gn (x) − g(x)| < /n}. Now |fn (x)gn (x) − f (x)g(x)| ≤ |fn (x)||gn (x) − g(x)| + |g(x)||fn (x) − f (x)| and if n ≥ max{N1 , N2 } then the right-hand side at most 2 on An ∩ Bn ∩ Cn ∩ Dn , a set whose complement has small measure. For a counterexample when µ(X) = ∞, let f (x) = g(x) = x and fn (x) = gn (x) = x + 1 χ[n,n+1) . n Clearly fn → f and gn → g in measure. However for x ∈ [n, n + 1), |fn (x)gn (x) − f (x)g(x)| = (x + 1/n)2 − x2 = 2x/n + 1/n2 > 2. So fn gn 9 f g in measure. p. 63 # 42. Let µ be counting measure on N. Then fn → f in measure iff fn → f uniformly. Solution. If fn → f in measure then, given > 0, ∃N ∀n > N µ({x : |fn (x) − f (x)| ≥ }) < 1/2, i.e. {x : |fn (x) − f (x)| ≥ } = ∅, so ∀x|fn (x) − f (x)| < , which is uniform convergence. 7 If fn → f uniformly then, given δ > 0 and > 0, ∃N ∀n > N ∀x|fn (x) − f (x)| < , so µ({x : |fn (x) − f (x)| ≥ }) = 0 < δ, which is convergence in measure. p. R77 # 55. the existance and equality R1 R 1 E = [0, 1] × [0, 1]. R 1 Investigate R 1 Let 2 of E f dm , 0 0 f (x, y)dxdy, and 0 0 f (x, y)dydx for the following f . a. f (x, y) = (x2 − y 2 )(x2 + y 2 )−2 . b. f (x, y) = (1 − xy)−a (a > 0). c. f (x, y) = (x − 1/2)−3 if 0 < y < |x − 1/2|, f (x, y) = 0 otherwise. Solution. The strategy is to first integrate |f |. If the integral exists and is finite, then f ∈ L1 so by Fubini’s theorem, all three integrals are equal. a. 1 1 1 1 |x2 − y 2 | |f |dm = |f (x, y)|dxdy = dxdy 2 2 2 0 0 0 0 (x + y ) y 1 ! Z 1 Z y 2 Z 1 2 Z 1 y − x2 x − y2 x x − dy = dx + dx dy = 2 2 2 2 2 2 x2 + y 2 0 x2 + y 2 y 0 0 (x + y ) y (x + y ) 0 Z 1 1 1 = − dy = ln y|10 − tan−1 y|10 = ∞ 2 y 1+y 0 R so f dm2 does not exist. But Z 1 Z 0 1 Z 0 and Z 1Z 0 0 Z 2 x2 − y 2 dxdy = (x2 + y 2 )2 1 Z 1 Z x2 − y 2 dydx = (x2 + y 2 )2 Z 0 Z 1 0 Z 1 Z 1 −1 −x π −1 1 dy = dy = − tan y| = − 0 2 x2 + y 2 0 4 0 1+y 1 Z 1 y 1 π −1 1 dx = dx = tan x| = 0 2 x2 + y 2 0 4 0 1+x b. Case 1. a < 2, a 6= 1: Z 2 Z 1 Z |f |dm = 1 −a Z (1−xy) dxdy = 0 0 0 8 1 1 Z 1 (1 − xy)1−a 1 (1 − y)1−a 1 dy = − dy y(a − 1) 0 a−1 0 y y Z 1/2 = 0 (1 − y)1−a 1 − y y Z 1 dy + 1/2 (1 − y)1−a 1 − y y dy Note that the function is bounded in the first integral since (1 − y)1−a 1 − = lim (1 − a)(1x )−a = 1 − a lim y→0 y→0 y y And the denominator in the second integral is ≥ 1/2 so the integral is Z 1 (1/2)2−a ≤2 ((1 − y)1−a − 1)dy = −1 2−a 1/2 Thus, by Fubini’s theorem, all three integrals are equal. Case 2. a = 1: Z Z Z 1Z 1 −1 2 (1−xy) dxdy = |f |dm = 0 1 Z 1 ln(1 − xy) ln(1 − y) dy = − dy −y y 0 0 1 0 0 1/2 Z =− 0 ln(1 − y) dy − y Z 1 1/2 ln(1 − y) dy < ∞ y because ln(1 − y) −1 = lim = −1 y→0 y→0 1 − y y so the function is bounded from 0 to 1/2, which makes the first integral finite, and Z 1 Z 1 ln(2) 1 ln(1 − y) − dy ≤ −2 ln(1 − y)dy = 2( + ) < ∞. y 2 2 1/2 1/2 lim Again, by Fubini’s theorem, all three integrals are equal. Case 3. a ≥ R2: From case 1, we can see that the second integral will be 1 infinite since 1/2 (1 − y)−b = ∞ when b > 1. Since the function is symmetric in x and y, integration with respect to dxdy is the same as with respect to dydx, so none of the integrals exist. c. Z 2 Z 1/2 Z |f |dm = 0 0 1/2−x 1 −x 2 −3 Z 1 Z dydx + 1/2 9 0 x−1/2 −3 1 x− dydx 2 Z 1/2 = 0 R 1 −x 2 −2 −2 −1 12 −1 1 1 1 1 x− dx = −x − x− = ∞+∞ = ∞ 1 2 2 2 1/2 Z dx+ 1 0 2 2 so f dm does not exist. Also, −3 −3 Z 1 Z x−1/2 Z 1Z 1 Z 1/2 Z 1/2−x 1 1 x− dydx+ dydx = ∞−∞ f (x, y)dydx = x− 2 2 1/2 0 0 0 0 0 R1R1 so 0 0 f (x, y)dydx does not exist. And −3 −3 Z 1Z 1 Z 1/2 Z 1/2−y Z 1/2 Z 1 1 1 x− f (x, y)dxdy = dxdy+ x− dxdy 2 2 0 0 0 0 0 y−1/2 "Z # Z 1/2 1/2 1 1 1 1 1 − − =− dy + dy = −∞ 2 0 y2 4 4 (y − 1)2 0 p. 77 # 59. Let f (x) = x−1 sin x. R∞ a. Show that 0 |f (x)|dx = ∞. Rb b. Show that limb→∞ 0 f (x)dx = π/2 by integrating e−xy sin x with respect to x and y. (In view of part (a), some care is needed in passing to the limit as b → ∞.) Solution. a. Z ∞ ∞ Z (k+1)π ∞ X X sin x | sin x| 1 1 dx ≥ √ dx ≥ x x 4 2 k=1 k + 0 k=1 kπ 1 4 1 + k+ 3 4 =∞ since this is the harmonic series. In more detail, by trapezoidal rule, Z (k+1)π | sin x| π 1 1 1 1 3 dx ≥ · | sin((k + )π)| + | sin((k + )π)| x 2 2 (k + 14 )π 4 4 (k + 43 )π kπ 1 1 1 = √ + . k + 43 4 2 k + 41 b. On the one hand, Z ∞Z ∞ Z −xy e sin xdydx = 0 0 0 ∞ ∞ Z ∞ −1 −xy sin x e sin x dx = dx. x x 0 0 10 On the other hand, Z Z ∞Z ∞ −xy e sin xdxdy = 0 0 0 ∞ ∞ −e−xy cos x − ye−xy sin x dy y2 + 1 0 Noting that | cos x| ≤ 1 and | sin x| ≤ 1, the limits at ∞ are 0: Z ∞ 1 π dy = tan−1 y|∞ = 0 = 2 y +1 2 0 11