Folland: Real Analysis, Chapter 2 Sébastien Picard

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1 Folland: Real Analysis, Chapter 2 Sébastien Picard Problem 2.3 If {f n } is a sequence of measurable functions on X, then {x : limf n (x) exists} is a measurable set. Define h limsupf n, g liminff n. By Proposition 2.7, h,g are measurable. Let. E g (n, ) h (n, ) n E g (, n) h (, n) It is clear that both E and E are measurable sets. Next, define n { 29 if g h ± w(x) g(x) h(x) else Then w is a measurable function by Exercise 2. Therefore E (w (,) w (, )) c is measurable. Hence the set {x : limf n exists} is measurable since it is equal to {x X : g(x) h(x)} E E E. Problem 2.9 Let f : [,] [,] be the Cantor function (.5), and let g(x) f(x)+x. a. g is a bijection from [,] to [,2] and h g is continuous from [,2] to [,]. b. If C is the Cantor set, m(g(c)). c. By Exercise 29 of Chapter, g(c) contains a Lebesgue nonmeasurable set A. Let B g (A). Then B is Lebesgue measurable but not Borel. d. There exists a Lebesgue measurable function F and a continuous function G on R such that F G is not Lebesgue measurable. (a) We know that f is an increasing function, and therefore f(x)+x is a strictly increasing function. Also, since f is continuous, then f(x) + x is continuous. Since g() and g() 2, and g is continuous and strictly increasing on [,], then g is a bijection from [,] to [,2]. It follows that g exists.

2 Since g is continuous map from a compact set to a Hausdorff space, then g is continuous: indeed, g(k) is compact for any compact set K, and compacts sets are closed in a Hausdorff space, so (g ) (C) is closed for any closed set C in the domain of g. (b) Let [,] C c where E n are countably many disjoint intervals. Since f is constant on a given E n, then g(e n ) is a translate of E n, so m(g(e n )) m(e n ). Therefore n E n. m(g([,] C c )) m( g(e n )) m( E n ) m([,] C c ) m m Since 2 m([,2]) m(g(c))+m(g([,] C c )), we have m(g(c)). (c) Suppose B is a Borel set. Since g is continuous, (g ) (B) A is Borel measurable. But A is not even Lebesgue measurable, which is a contradiction. However, B C. Since C is a null set, B is Lebesgue measurable by completeness of Lebesgue measure. (d) Define G and F as follows: g (x) if x [,2] G(x) x if x if x 2 { x if x B F(x) 69 else Giscontinuous bythepasting Lemma fromelementary topology. F ismeasurable sincef (a, ) is either empty, the whole real line, or a subset of B (which is measurable since B has measure zero). F G is not Lebesgue measurable since G (F (, )) G (B) A. Problem 2.4 If f L +, let λ(e) E fdµ for E M. Then λ is a measure on M, and for any g L+, gdλ fgdµ. (First suppose that g is simple.) We will show that λ is a measure on M. It is clear that λ( ). Let {A k } be a countable sequence of disjoint sets, and define A k A k. 2

3 λ( A k ) fdµ ( A A χ Ak f)dµ k k A χ Ak fdµ λ(a k ). The summation can be taken out from the integral by Theorem 2.5. Let g L + be a simple function. Then g n i a iχ Ai where A i are measurable sets. Therefore n n n gdλ a i λ(a i ) a i fχ Ai dµ a i χ Ai fdµ fgdµ. i i Now take any g L +. By Theorem 2. there exists a sequence {φ n } of simple functions that converges to g pointwise such that φ φ 2 g. Since {φ n } and {fφ n } are monotone increasing, we can apply the monotone convergence theorem twice to obtain gdλ limφ n dλ lim φ n dλ lim φ n fdµ fgdµ. i k Problem 2.6 If f L + and f <, for every ǫ > there exists E M such that µ(e) < and E f > ( f) ǫ. Let f L + such that f <. Let ǫ >. By definition of f, there exists a simple function φ n i a iχ Ei such that φ f and f ǫ < φ. Let E n i E i. Since the E i are in M, then E M. Also, φ f <, so for each E i we have µ(e i ) < and therefore µ(e) <. Hence we have f ǫ < φ f. E E Problem 2.2 (A generalized Dominated Convergence Theorem) If f n, g n, f, g L, f n f and g n g a.e., f n g n, and g n g, then f n f. (Rework the proof of the dominated convergence theorem.) Since f n +g n, we can apply Fatou s lemma: f +g lim(f n +g n ) liminf f n +g n g +liminf f n. The same process can be repeated for g f n : g f lim(g n f n ) liminf g n f n g limsup f n. 3

4 Since limsup f n f liminf f n we can conclude that f n f. Problem 2.2 Suppose f n,f L and f n f a.e. Then f n f iff f n f. (Use Exercise 2.) Suppose f n f. Then limsup f n limsup f n f + f f. On the other hand, we can apply Fatou s lemma to obtain f liminf f n. Therefore limsup f n f liminf f n and hence f n f. Next we prove the converse: suppose f n f. Define g n f n + f. Then f n f g n and g n 2 f L. By applying the generalized Dominated Convergence Theorem (Problem 2) we obtain lim f n f lim f n f. Problem 2.25 Let f(x) x /2 if < x <, f(x) otherwise. Let {r n } be an enumeration of the rationals, and set g(x) 2 n f(x r n ). a. g L (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 <, but g 2 is not integrable on any interval. (a) We need to show g <. Using Theorem 2.5 and the change of variables z x r n, we can compute 4

5 n 2 n f(x r n )dx n 2 n f(x r n )dx 2 n f(z)dz n 2 n z /2 dz n n 2 n 2( 2 n ) 2 <. n (b) Let x R,M >,δ >. It will be shown that there exists a y (x δ,x + δ) such that g(y) > M. This will prove that g is unbounded on each interval and that g is discontinuous at every point. There exists a rational number r n (x δ,x+δ). Furthermore, there exists a y (x δ,x+δ) such that Then < y r n < 2 2n 4M 2. g(y) 2 n f(y r n ) 2 n (2 n (2M)) 2M > M. This proof does not fail after redefining g on any Lebesgue null set, since one can still find an irrational y with the desired properties. (c) Since g < a.e., it follows that g 2 < a.e. However, g 2 2 2n f 2 (x r n )dx n 2 2n f 2 (z)dz 4 n dz z >. Problem 2.27 Let f n (x) ae nax be nbx where < a < b. a. f n (x) dx. b. f n (x)dx. c. f n L ([, ),m), and f n(x)dx log(b/a). 5

6 (a) Since f n is the difference of two exponential functions, we can find a point c R such that f n < on (,c) and f n > on (c, ). In order to find this point c, we solve be nbc ae nac log(b/a) nc(b a) c log(b/a) n(b a) We can now split up the integral in order to integrate f n : f n c o (ae nax be nbx ) dx+ ( e nax ) n e nbx c n 2 ) (e nac e nbc n ( e nbx + n e nax n ( e a b a log(b/a) e b b a log(b/a) ). c ) (ae nax be nbx dx ) 2 n Therefore f n is proportional to (/n), hence f n. c (b) Therefore f n (x)dx. f n (x)dx o e nax n ae nax dx + e nbx n ( +). n be nbx dx (c) f n n ) (ae nax be nbx n ( ) a e nax b ( a n ) e ax a e ax ( b ( n ) e nbx ) e bx b +(b a). e bx 6

7 We can see that n f n is positive on (, ), so to show that n f n L we must compute n f n: o ( a f n n e b ) ax e +(b a) dx bx ( ) log( e ax ) log( e bx )+(b a)x The last step follows from l Hopital s rule: ( ( e ax )(e (b a)x ) ) log e bx ( ( e ax )( e bz )e bx e ax ) lim lim log z x ( e bx )( e az )e bz e az ( ( e bz e ax +e ax+bz )e bx e az ) lim lim log z x ( e az e bx +e bx+az )e bz e ax ( e ax (e ax e bz ax +e bz )e bx e az ) lim lim log z x e bx (e bx e az bx +e az )e bz e ax ( (e ax e bz ax +e bz )e az ) lim lim log z x (e bx e az bx +e az )e bz ( (e bz )e az ) limlog z (e az )e bz ( e bz) limlog z e az log(b/a). lim z e bz be bz lim e az z ae b az a. Problem 2.28 Compute the following limits and justify the calculations: a. lim (+(x/n)) n sin(x/n)dx. b. lim (+nx2 )(+x 2 ) n dx. c. lim nsin(x/n)[x(+x 2 )] dx. d. lim n( + a n2 x 2 ) dx. (The answer depends on whether a >, a, or a <. How does this accord with the various convergence theorems? (a) Denote f n sin(x/n) (+(x/n)) n. First, notice that by the binomial theorem, when n 2, for all x [, ) we have 7

8 (+(x/n)) n +x+ Thus we can bound f n on [, ) by ( ) n x 2 x2 +x+ 2 n2 4. sin(x/n) f n (+(x/n)) n (+(x/n)) n +x+x 2 /4. A rough computation shows that the following integral is bounded: dx +x+x 2 /4 dx +x+x 2 /4 + dx <. dx +x+x 2 /4 dx ++x 2 /4 By the Dominated Convergence Theorem, we can throw the limit under the integral. lim f n dx lim f n dx sin(x/n) dx lim (+(x/n)) n sin()e x dx. (b) Denote By the binomial theorem, we know f n +nx2 (+x 2 ) n. (+x 2 ) n +nx 2. Therefore f n, and <, so by the Dominated Convergence Theorem we can conclude (c) Denote lim f n dx ( lim f n nsin(x/n) x(+x 2 ). +nx 2 ) dx. (+x 2 ) n Since sin(x/n) (x/n) for x >, we can bound f n on (, ) by Next we verify that nsin(x/n) x(+x 2 ) +x 2. 8

9 dx +x arctan(x) π/2 <. 2 By the Dominated Convergence Theorem we can conclude lim f n dx ( lim sin(x/n) x/n (+x2 ) ) dx dx +x 2 π/2. (d) In this case we can evaluate the integral directly: Hence lim a n +n 2 x2dx lim na lim a dy +y lim arctan(y) π/2 lim arctan(na). 2 na n +n 2 x 2dx a > π/2 a π a < This agrees with the fact that onecannot apply the DominatedConvergence Theorem unless a > since there is no way to bound f n (). Problem 2.32 Suppose µ(x) <. If f and g are complex-valued measurable functions on X, define f g ρ(f,g) + f g dµ. Then ρ is a metric on the space of measurable functions if we identify functions that are equal a.e., and f n f with respect to this metric iff f n f in measure. We will first show that ρ is a metric on the space of measurable functions if we identify functions that are equal a.e. It is clear that ρ(f,g) ρ(g,f) and that ρ(f,g). We can also see that f g ρ(f,g) + f g dµ f g a.e. (by Proposition 2.6) + f g f g a.e. It only remains to show the triangle inequality. Let x,y,z X. First, suppose x z x y and z y x y. Then we have 9

10 x y + x y x z z y + + x y + x y x z z y + + x z + z y. On the other hand, suppose x z x y. Then x y + x y x y z y + + x y + z y x z z y + + x z + z y. The above argument can be repeated when z y x y. Hence for all x,y,z X, x y + x y x z z y + + x z + z y. Using basic properties of the integral (Proposition 2.3), we can conclude that ρ(f,g) ρ(f,h)+ ρ(h,g) for any measurable functions f,g,h. This completes the proof that ρ is a metric. Suppose f n f with respect to ρ. Let E n,ǫ {x : f n (x) f(x) ǫ}. It follows that f n f + f n f dµ E n,ǫ µ(e n,ǫ ) +ǫ ǫ f n f + f n f dµ ǫ +ǫ µ(e n,ǫ). f g dµ. + f g Conversely, suppose f n f in measure. Let ǫ >. Choose N N such that for all integers n > N, we have ( { µ x : f n (x) f(x) ǫ } ) < ǫ 2µ(X) 2. Let A {x : f n (x) f(x) ǫ }. Then the following holds for all n > N: 2µ(X) ρ(f n,f) f g + f g dµ A f g + f g dµ+ X\A µ(a)+ ǫ 2µ(X) µ(x\a) < ǫ/2+ǫ/2 ǫ. f g + f g dµ Problem 2.4 In Egoroff s theorem, the hypothesis µ(x) < can be replaced by f n g for all n, where g L (µ). Suppose f,f 2,... and f are measurable complex-valued functions on X such that f n f a.e. and f n g for all n, where g L (µ). We will follow the proof of Theorem 2.33 and make some minor adjustments.

11 Without loss of generality, assume that f n f everywhere on X. For k,n N, let E n (k) {x : f m (x) f(x) k }. mn For fixed k, E n (k) decreases as n increases and ne n (k). To apply continuity of measure from above, we need µ(e ) <. Since f n f 2 g, we observe that We can use the fact that in order to conclude E (k) A(k) : {x : 2 g(x) k }. > 2 g 2 g k µ(a(k)), X A(k) µ(e (k)) µ(a(k)) <. Therefore, by continuity of measure from above, µ(e n (k)) as n. Given ǫ > and k N, there exists a positive integer n k such that µ(e nk (k)) < ǫ2 k. If we define E k E n k (k), then µ(e) < ǫ and f n f uniformly on E c. Problem 2.49 Prove Theorem 2.39 by using Theorem 2.37 and Proposition 2.2 together with the following lemmas. a. If E M N and µ ν(e), then ν(e x ) µ(e y ) for a.e. x and y. b. If f is L-measurable and f λ-a.e., then f x and f y are integrable for a.e. x and y, and fx dν f y dµ for a.e. x and y. (Here the completeness of µ and ν is needed.) (a) Suppose E M N and µ ν(e). Define f χ E. Then f x χ Ex and f y χ E y. Apply Fubini s theorem: ( fd(µ ν) ) ( f x dν(y) dµ(x) ) f y dµ(x) dν(y). It follows that χ Ex dν µ-a.e. and ν(e x ) µ-a.e., and similarly χ E ydµ ν-a.e. and µ(e y ) ν-a.e.. (b) Suppose f is L-measurable and f λ-a.e.. Define A {(x,y) M N : f(x,y) }. Then A E for some E M N such that µ ν(e). By part (a), ν(e x ) and µ(e y ) for a.e. x and y. Since A x E x and A y E y, we have ν(a x ) and µ(a y ). Therefore fx dµ χ Ax f x dµ for µ-a.e. x and f y dν χ A y f y dν for ν-a.e. y.

12 Wenowprove Theorem2.39. Supposef isl-measurableandeither (a)f or(b)f L (λ). By Proposition 2.2, there exists a M N-measurable function g such that f g λ-almost everywhere. By Proposition 2.34, g x is N-measurable and g y is M-measurable. Define h g f. Then h λ-a.e.. By lemma (b), h x is N-measurable for a.e. x and h y is M-measurable for almost every y, hence f x is N-measurable for a.e. x and f y is M-measurable for almost every y In case (b), g L (µ ν), so by lemma (b), h x g x f x is integrable for a.e. x. By Fubini s theorem, g x is integrable for a.e. x, hence f x is integrable for a.e. x. Similarly f y is integrable for a.e. y. By applying lemma (b) on the function h, we can see that for a.e. x we have (g x f x )dν hence g x dν f x dν. Similarly for a.e. y we have g y dµ f y dµ. In case (a), g L + (M N), so by Tonelli s theorem x g x dν f x dν is measurable, and y g y dν f y dν is measurable. In case (b), g L (M N), so by Fubini s theorem x g x dν f x dν is integrable, and y g y dν f y dν is integrable. Since gdλ fdλ, after applying Tonelli s theorem (case a) or Fubini s theorem (case b) and using the fact that g y dν f y dν and g x dµ f x dµ for almost every x and y, we obtain fdλ f(x,y)dµdν f(x, y)dνdµ. Problem 2.55 Let E [,] [,]. Investigate the existence and equality of E fdm2, f(x,y) dy dx for the following f. a. f(x,y) (x 2 y 2 )(x 2 +y 2 ) 2. b. f(x,y) ( xy) a (a > ). c. f(x,y) (x /2) 3 if < y < x /2, f(x,y) otherwise. (a) First, we evaluate x 2 y 2 (x 2 +y 2 ) 2dx. Using the substitution x ytanθ, dx ysec 2 θdθ, we obtain f(x,y) dx dy, and 2

13 x 2 y 2 arctan(/y) (x 2 +y 2 ) 2dx y y y arctan(/y) arctan(/y) arctan(/y) arctan(/y) (tan 2 θ )y 3 sec 2 θdθ (y 2 (tan 2 θ+)) 2 (tan 2 θ )dθ ysec 2 θ (sin 2 θ cos 2 θ)dθ ( 2cos 2 θ)dθ cos 2θdθ 2y sin(2arctan(/y)) y sin(arctan(/y))cos(arctan(/y)) y y +y 2 sin(arctan(/y)) cos(arctan(/y)) tan(arctan(/y)) +tan 2 (arctan(/y)) cos(arctan(/y)) cos(arctan(/y)) Therefore, We observe that Hence x 2 y 2 (x 2 +y 2 ) 2dx x 2 y 2 (x 2 +y 2 ) 2dxdy dy +y 2 π/4. y 2 x 2 (x 2 +y 2 ) 2dx x 2 y 2 (x 2 +y 2 ) 2dy. x 2 y 2 (x 2 +y 2 ) 2dydx By Fubini s theorem, E fdm2 is not defined. dx +x 2 π/4. (b) Since f is non-negative on [,] [,], f L + (E), so by Tonelli s theorem E fdm2 f(x,y)dxdy f(x,y)dydx. The integral may be infinite for some values of a... I haven t had time to do this computation yet. 3

14 (c) First, we compute f(x,y)dydx x.5 The function (x 2 )dx x 2 3 is not integrable on [,]: x 2 dx x 2 3 (x 2 ) 3 dydx dx x. 2 2 (x 2 )dx x 2 3. Therefore, the integral f(x,y)dydx does not exist, and hence E fdm2 does not exist. However, f(x,y)dxdy : f(x,y)dx /2 /2 y /2 χ {y< x /2 } (x 2 ) 3 dx χ { z >y} dz dz z 3 + z 3 /2 y dz z y y2 Problem 2.57 Show that e sx x sinx dx arctan(s ) for s > by integrating e sxy sinx with respect to x and y. (It may be useful to recall that tan(π/2 θ) (tanθ). Cf. Exercise 3d.) We will investigate the integral e sxy sinxdydx. We want to apply Fubini s theorem, so first we verify that Next, we observe that e sxy sinx dydx e sxy sinxdydx s e sxy xdydx e sx s e sx x sinxdx. dx <. This integral can be computed by switching the order of integration. The first step is to use integration by parts to compute 4

15 I e sxy sinxdx ( sy)e sxy cosxdx e sxy cosx ( sy)e sxy cosxdx+ ( s 2 y 2 )e sxy sinxdx sye syx sinx ( s 2 y 2 )e sxy sinxdx+. + Therefore I s 2 y 2 I +. Thus It is given that e sxy sinxdxdy I +s 2 y 2. dy +s 2 y dz 2 s s +z ( π ) 2 s 2 arctan(s). tan( π 2 θ) (tanθ) which implies ( π ) tan 2 arctan(s) s. Therefore π 2 arctan(s) arctan(s ). By Fubini s theorem, we conclude s e sx x sinxdx s arctan(s ). e sx x sinxdx arctan(s ). Problem 2.58 Show that e sx x sin 2 x dx 4 log(+4s 2 ) for s > by integrating e sx sin2xy with respect to x and y. We will investigate the integral 5

16 e sx sin(2xy)dydx. We want to apply Fubini s theorem, so first we verify that Next, we observe that e sx sin(2xy) dydx e sx sin(2xy)dydx e sx x e sxy 2xydydx ( 2 ) 2 cos2x e sx xdx <. e sx sin 2 xdx. x This integral can be computed by switching the order of integration. The first step is to use integration by parts to compute I e sx sin(2xy)dx e sx s e sx 2ycos(2yx)dx e sx s sin(2yx) s 2ycos(2yx)dx (2y) 2 e sx sin(2yx)dx 2y cos(2yx) s 2 s 2e sx (2y) 2 e sx sin(2yx)dx+ 2y s 2 s 2. Therefore I 4y2 I s 2 I + 2y s 2. 2y s 2 +4y 2. e sx sin(2xy)dxdy By Fubini s theorem, this proves that 2y s 2 +4y 2dy 4 log(s2 +4) 4 log(s2 ) 4 log(+4s 2 ). e sx sin 2 xdx x 4 log(+4s 2 ). Problem 2.6 Γ(x)Γ(y)/Γ(x + y) tx ( t) y dt for x,y >. (Recall that Γ was defined in 2.3. Write Γ(x)Γ(y) as a double integral and use the argument of the exponential as a new variable of integration.) 6

17 By definition, we have Γ(x)Γ(y) t x s y e t s dsdt. We perform the change of variables s u uv, t uv. The Jacobian is (s, t) (u,v) s t u v s t ( v)u ( uv) u. v u The change of variables formula for multiple integrals yields Γ(x)Γ(y) (uv) x (u uv) y e u ududv By definition of Γ(x+y) and Fubini s theorem, this can be rewritten as This proves that ( Γ(x)Γ(y) Γ(x)Γ(y) Γ(x+y) ) v x ( v) y dv Γ(x+y). t x ( t) y dt. v x ( v) y e u u x+y dudv. Problem 2.63 The technique used to prove Proposition 2.54 can also be used to integrate any polynomial over S n. In fact, suppose f(x) n xα j j (α j N {}) is a monomial. Then fdσ if any α j is odd, and if all α j s are even, fdσ 2Γ(β ) Γ(β n ), where β j α j +. Γ(β + +β n 2 By Theorem 2.49, we know that R n e x 2 n j x α j j dx First we compute the right-hand side: S n e r2 j S n e r2 n j ( xj ) αjr αj r n dσdr. x j ( n ( xj ) αjr )( ) αj r n dσdr fdσ e r2 r n + α j dr x j S n ( )( ) e s αj n fdσ S n 2 s 2 s 2 s 2 ds ( ) fdσ S 2 Γ( α j + ). 2 n 7

18 If any α i is odd, then e x2 i x α i i dx i by symmetry, hence then left-hand side is zero: n n x α j j dx e x2 α j x j j dx j. Therefore R n e x 2 j If every α i is even, then by symmetry then R n e x 2 n j x α j j dx 2n n j j S n fdσ. e x2 i x α i e x2 j x α j j dx j i dx i 2 e x2 i x α i i dx i. The left-hand side is n j e s s α j n 2 2 ds By combining the identities of the left-hand side and right-hand side, we obtain fdσ 2Γ(β ) Γ(β n ) S Γ(β n + +β n ) where β j α j+ 2. j Γ( α j + ). 2 8