SOLUTIONS OF SELECTED PROBLEMS

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1 SOLUTIONS OF SELECTED PROBLEMS Problem 36, p. 63 If µ(e n < and χ En f in L, then f is a.e. equal to a characteristic function of a measurable set. Solution: By Corollary.3, there esists a subsequence χ Enj f a.e.. Let A = {x : χ Enj (x f(x}. For x A, {χ Enj (x} is a sequence of s and s, so its limit can be either or. Let E = {x A : f(x = }. Then f = χ E a.e.. Problem 38-b, p. 63 Suppose that µ(x <, f n, g n are measurable functions, and f n f, g n g in measure. Prove that f n g n fg in measure. Show that it may not be the case if µ(x =. Solution:. Notice that n {x : f(x > n} =. lim n µ({x : f(x > n} = (here, we have used µ(x <. the same is true for the function g. we conclude that for every η > there exists a number M such that ( µ({x : f(x > M} < η and µ({x : g(x > M} < η.. For sufficiently large values of n, one has µ({x : f(x f n (x } < η. Notice that {x : f n (x > M + } {x : f(x > M} {x : f(x f n (x }. The same is true if one replaces f by g. there exists N such that ( µ({x : f n (x > M +} < η and µ({x : g n (x > M +} < η, n > N. 3. Fix a number ɛ >. Notice that Then, and {x : f n (xg n (x f(xg(x ɛ} {x : g n (x(f n (x f(x ɛ/} {x : f(x(g n (x g(x ɛ/}. {x : g n (x(f n (x f(x ɛ/} {x : g n (x > M + } {x : f n (x f(x ɛ/((m + } {x : f(x(g n (x g(x ɛ/} {x : f(x > M} {x : g n (x g(x ɛ/(m}. Typeset by AMS-TEX

2 SOLUTIONS OF SELECTED PROBLEMS µ({x : f n (xg n (x f(xg(x ɛ} 4η + δ n, n > N where δ n = µ({x : f n (x f(x ɛ/((m +}+µ({x : g n (x g(x ɛ/(m} when n. Hence, µ({x : f n (xg n (x f(xg(x ɛ} 5η when n is large enough. This completes the proof (η was an arbitrary number. 4. Let us drop the condition µ(x <. Take X = R, µ = m- the Lebesgue measure, f n (x = g n (x = x + /n, f(x = g(x = x. Clearly, f n f and g n g in measure. On the other hand, (x + /n x ɛ when x nɛ/. Therefore µ({x : f n (xg n (x f(xg(x ɛ} = for every ɛ > and for every n. Remark. The assumption µ(x < was used in the proof when we derived (. The same conclusion holds in the case when both functions f(x and g(x are integrable. Then ( is a consequence of the Chebyshev inequality µ({x : f(x > M} M X f(x dµ. If f(x and g(x are integrable functions then the assumption µ(x < can be dropped. Problem 43-b, p. 63 Suppose that µ(x < and f : X [, ] C is a function such that f(, y is measurable for each y and f(x, is continuous for each x. Then for every ɛ > there exists a measurable set E X with µ(e < ɛ and f(, y converges to f(, uniformly on E c as y. The solution is a modification of the proof of Egoroff s Theorem. Let E n (k = <r</n,r Q {x : f(x, r f(x, k. The sets E n (k are measurable: they are countable unions of measurable sets. For a fixed value of k, the sets E n (k are decreasing as n, and their intersection is empty. Therefore lim µ(e n(k =. n Choose n k in such a way that µ(e nk (k < k ɛ, and take E = k= E n k (k.

3 SOLUTIONS OF SELECTED PROBLEMS 3 Problem 44, p. 64 If f : [a, b] C is Lebesgue measurable and ɛ >, there is a compact set E [a, b] such that m(e c < ɛ and f E is continuous. (Lusin s Theorem Solution: As in the problem 38-b, there exists M such that m({x : f(x > M} < ɛ/3. We denote A = {x : f(x > M}. Let { f(x, when f(x M f M (x = M, when f(x > M. Then f M (x L, and, by Theorem.6, there exists a sequence of continuous functions that converges to f M in L. By Corollary.3, a subsequence of that sequence converges to f M a.e.. We conclude that there exists a sequence of continuous functions f n (x on [a, b] that converges to f M (x a.e.. By Egoroff s Theorem, there exists a set A such that m(a < ɛ/3 and f n converges to f uniformly on A c. Take a compact set E (A A c such that m((a A c \ E < ɛ/3. Notice that f(x = f M (x when x E. The restriction f E is the limit of uniformly convergent sequence of continuous functions f n E, so it is continuous. Clearly, m(e c < ɛ. Remark. Lusin s theorem does not say that f(x is continuous on E. It says that the restriction of f to E is continuous. For example, a function {, when x is rational D(x =, when x is irrational is not continuous at any point. However, the restriction of D(x to the set of irrational numbers (and to any subset of the set of irrational numbers is continuous. One can prove that a continuous function on a compact set can be extended to a continuous function on the whole interval (this is a special case of the Tietze extension theorem. Lusin s theorem implies the following fact: for every measurable function f(x on [a, b] and for every ɛ > there exists a continuous function f ɛ on [a, b] such that m({x : f(x f ɛ (x} < ɛ. In words, every measurable function can be changed on a set of arbitrarily small measure to get a continuous function. Problem 46, p. 68 (a part Let X = Y = [, ], M = N = B [,], µ Lebesgue measure, ν-counting measure. If D = {(x, x} [, ] is the diagonal, what is µ ν(d? Solution: Let D j=a j B j, A j, B j B [,] be a covering of D by rectangular sets. Let J = {j : ν(b j < }, let B = j J B j, and let E = [, ] \ B. Notice that the set B is either finite or countable; therefore µ(e =. On the other hand, E j J A j, so there exists j J such that µ(a j >. We conclude that µ(a j ν(b j =, and µ(a j ν(b j = j= for every covering of D by rectangles. µ ν(d =.

4 4 SOLUTIONS OF SELECTED PROBLEMS Problem 55, p. 77 Let E = [, ] [, ]. Investigate the existence and equality of fdm, f(x, ydxdy, and E f(x, ydydx for the following f. a f(x, y = (x y (x + y. Solution: Let us investigate substitution x = yz to get f(x, ydxdy first. For y >, one makes a f(x, ydx = y The substitution z = /w leads to /y z ( + z dz. so ( One notice that z ( + z dz = w ( + w dw, f(x, ydx = y /y /y z ( + z dz >, z ( + z dz. so The equality f(x, y = f(y, x implies f(x, ydydx = f(x, ydxdy f(x, ydxdy <. f(x, ydxdy >. f(x, ydydx, and the function f(x, y is not integrable on E. Remark. Though it not necessary to do for solving the problem, one can actually evaluate f(x, ydxdy. The function (z /[y(z + ] is positive in the domain {(z, y : < y <, z /y}, so one can use Tonelli s theorem to obtain (see ( f(x, ydxdy = z ( + z dz dy /z y = z ( + z ln zdz = I.

5 A substitution w = /z leads to SOLUTIONS OF SELECTED PROBLEMS 5 z w ( + z ln zdz = ( + w ln wdw. I = x ( + x ln xdx. Now, let us take the branch of the function ln z in the complex plane with negative imaginary half-axis removed that equals ln x on the positive real half-axis; ln z = ln z + i arg z, where π/ < arg z < 3π/. Then ln( x = ln x + iπ for x >, and In other words, z x ( + z ln zdz = 4I + iπ ( + x dx = 4I I = 4 z ( + z ln zdz. The last integral equals lim R,ɛ I(R, ɛ where z (3 I(R, ɛ = C R,ɛ ( + z ln zdz; here C R,ɛ = [ R, ɛ] {z : z = ɛ, Imz } [ɛ, R] {z : z = R, Imz }. When R > and ɛ <, the integrand in (3 has one pole, z = i, that lies inside the contour. The residue at this pole equals d dz ((z (z + i ln z z=i = i. 4I = (πi i = π. Finally, f(x, ydxdy = I = π 4. b f(x, y = ( xy a, a >. Solution: The function f(x, y is non-negative, so by Tonelli s theorem the integral over E and the iterated integrals are equal to each other. To find out, under what conditions they are finite, consider, say, f(x, ydxdy. The substitution z = xy leads to g(y = f(x, ydx = ( y a ( ay, when a ; ln( y y, when a =.

6 6 SOLUTIONS OF SELECTED PROBLEMS The function g(y is continuous on [,, and it goes to as y when a. The integral g(ydy is finite when a >, i.e. a <. We conclude that the function f(x, y is integrable when < a <. c ( 3 x f(x, y =, when < y < x (/ ;, otherwise. Solution: One has f((/ + z, y = f((/ z, y, therefore f(x, ydx =, and f(x, ydxdy =. On the other hand, h(x = and h(x dx = ( f(x, ydy = x 3 x, ( x dx =. the function h(x is not integrable, and the function f(x, y is not integrable as well. Answer: f(x, ydxdy =, integrals f(x, ydydx = and f(x, ydm E do not exist.