Solution to Exercise 3

Transcription

1 Spring 207 MATH502 Real Analysis II Solution to Eercise 3 () For a, b > 0, set a sin( b ), 0 < f() = 0, = 0. nπ+π/2 Show that f is in BV [0, ] iff a > b. ( ) b Solution. Put n = for n 0. We claim that f is of bounded variation on [0, 0 ], hence of bounded variation on [0, ]. Now, note that sin n = and sin n sin n =, the total variation of f over [0, 0 ] is f( n ) f( n ) = We have ( nπ + π/2 ) a b ( (n )π + π/2 ) a b + ( ( f( n ) f( n ) 2 nπ + π/2 ) a b (n )π + π/2 It is easy to verify that the partial sum of the series converges iff a > b hence the total variation is bounded iff a > b. We in fact can show that f AC[0, ] if a > b, since for 0 ) a b f () = a a sin( b ) b a b cos( b ) L [0, ]. Now let (0, ] and 0 < ε <, by the fundamental theorem of calculus for smooth function, we have f() f(ε) = ε f (t)dl (t).

2 By the continuity of f at 0, we may let ε 0. Hence f() f(0) = 0 f dl and the absolute continuity of f follows from Theorem 6.8. (2) A function is called Lipschitz continuous on an interval I if M > 0 such that f() f(y) M y. (a) Show that every Lipschitz continuous function is absolutely continuous on I. (b) Show that there are always absolutely continuous functions which are not Lipschitz continuous. Solution. (a) It follows directly from f(i ) f( i) M i i. (b) Put f() = on [0, ]. Then we have f() = 0 2 t dt is an indefinite integral hence is absolutely continuous. But that f() f(0) 0 = as 0 shows that f is not Lipschitz. (3) Assume that < p <, f is absolutely continuous on [a, b], f L p, and α = /q, where q is the eponent conjugate to p. Prove that f Lipα. 2

3 Solution. Since f is absolutely continuous, we have f(y) f() = y f y ( y f f p ) p ( y y q f p ) q by Hölder s inequality. Note. Nowadays the standard notation is f C α [a, b], α = /q. This is a simple version of embedding inequality. (4) Show that the product of two absolutely continuous functions on [a, b] is absolutely continuous. Use this to derive a theorem about integration by parts. Solution. Let f, g be absolutely continuous. Since every absolutely continuous function is bounded on every bounded closed interval, we can suppose that f, g M. f(i )g( i ) f( i)g( i) f( i )g( i ) f( i )g( i) + f( i )g( i) f( i)g( i) M g( i ) g( i) + M f( i ) f( i). This shows that fg is absolutely continuous whenever f, g are absolutely continuous. Integrating (fg) = f g + fg (this is the product rule for differentiable functions, still valid) from a to b gives f(b)g(b) f(a)g(a) since f g is absolutely continuous. b fg = b a a (5) Suppose E [a, b], m(e) = 0. Construct an absolutely continuous monotonic f g 3

4 function f on [a, b] so that f () = at every E. Hint: E V n, V n open, m(v n ) 2 n. Consider the sum of the characteristic functions of these sets. Solution. Since L(E) = 0, there eists a decreasing sequence {V n } of open set with L(V n ) < 2 n such that E V n. Put f() = χ Vn. a n let E. Choosing decreasing δ n > 0 such that W n = ( δ n, + δ n ) V n. If 0 < h < δ N, f( + h) f() = +h χ Vn +h N χ Wn Mh, where M is the number of W n containing ( h, + h). Clearly M as h 0. The same holds for δ N E. < h < 0. This shows that f(+h) f() h at every (6) Let f be in AC[a, b]. Show that the total variation for f of f is also in AC[a, b]. Moreover, T f (b) = b a f (t) dt. Solution. The inequality T f (b) [a,b] f is straightforward, so we work on the other inequality. We first assume f is continuous. For ε > 0,, let A = { : f () > ε}, B = { : 4

5 f () < ε}, and C = { : f () ε}. We have disjoint decompositions A = j (a j, b j ), B = j (c j, d j ). Then b f = f + ( f ) + f B C = f + dj ( f ) + f j a j j c j C f(b j ) f(a j ) + f(d j ) f(c j ) + ε(b a) j j a A bj T f (b) + ε(b a), which implies T f (b) b a f. In general, pick a continuous ϕ so that f ϕ L < ε and define g() = a ϕ. Then f() g() Cε on [a, b] for some constant C. Using T f (b) T g (b) T f g (b) C ε one can show that T f (b) b a f in the general case. (7) Let X and Y be topological spaces having countable bases. (a) Show that X Y has a countable base. (In product topology on X Y, a set G is open if (, y) G, there is some G open in X, G 2 open in Y such that (, y) G G 2 G.) (b) Let µ and ν be Borel measures on X and Y respectively. Show that µ ν is a Borel measure. Solution. (a) Let B X and B Y be the countable bases of X and Y respectively. Put B be the collection of all U V where U B X and V B Y. We know that B is countable. We claim that B forms a base for X Y. Let W be open in 5

6 X Y. Let (, y) W. Then there eists open U, V containing and y respectively. Choose U B X be a subset of U and V B Y be a subset of V. Then U V B and (, y) U V W. (b) As in the proof in (a), every open set in X Y is a countable union of members in B. But each member in B is a measurable rectangle because µ and ν are both Borel. It follows that every open set hence every Borel set is measurable. (8) Let µ be the product measure L L on R n. Show that µ is equal to L n. Solution. Both measures satisfy the following characterization of the Lebesgue measure: (a) translational invariant and (b) the measure of the unit square is. See an eercise in last semester. Note. The definition of the product measure of finitely many measures can be done like the n = 2 case. One can also show that L n L m = L n+m. (9) Fi a = 0 < a 2 < a 3 < < a n and let g n be a continuous function, sptg n (a n, a n+ ), n, g n =. Let f(, y) = Verify that (g n () g n+ ())g n (y). ( ( ) f d dy = 0, but ) f dy d =, and f is L 2 -measurable. Eplain why Fubini s theorem cannot apply. Solution. Since for each (, y), only the N-th term of the sum is non-zero if 6

7 y (a N, a N+ ). Now for each y lies in support of g N, f(, y)d = an+ a N [g N () g N+ ()]g N (y)d = 0 because g n = for all n. Thus f(, y)ddy = 0. If y does not lie in any support of g n, the same conclusion holds. For lies in support of g N, if N > f(, y)dy = g N ()g N (y) + g N ()g N (y)dy = 0 and if N =, If follows that f(, y)dy = a2 a g ()g (y)dy = g (). f(, y)dyd = a2 a f(, y)dyd =. That f is L 2 -measurable follows from the fact that f is a countable sum of continuous functions. Fubini s fails since f(, y) dy = = an g N ()g N (y) + g N ()g N (y) dy a N g N ()g N (y) dy + 2 g N (), an+ a N g N ()g N (y) dy if (a N, a N+ ), which shows that f(, y) dyd = an+ a n 2 g n () d =. 7

8 (0) Let µ and ν be outer measures defined on X and Y respectively and let f be µ-measurable and g ν-measurable with values in (, ]. Is it true that (, y) f()+g(y) measurable in µ ν? How about the map (, y) f()g(y)? Solution. Yes, they are. Since the sum and product of µ ν measurable functions are µ ν measurable. It suffices to check the functions F (, y) := f() and G(, y) := g(y) are µ ν measurable. W.L.O.G. we only consider F. Let U be an open set, then F (U) = f (U) Y is µ ν measurable since it is a measurable rectangle. Hence we have F is µ ν measurable function. () (a) Suppose that f is a real-valued function in R 2 such that each section f is Borel measurable and each section f y is continuous. Prove that f is Borel measurable in R 2. There is a hint given in [R]. (b) Suppose that g is a real-valued function in R n which is continuous in each of the n-variables separately. Prove that g is Borel. Solution. (a) We try to show that f is the pointwise limit of a sequence of Borel measurable functions and so is Borel measurable. We define f n piecewisely in the following way: if a i := i n f n (, y) = < i, for some i Z, then n a i a i a i f(a i, y) + a i a i a i f(a i, y). Obviously f n is Borel measurable by the previous problem and satisfies f n (, y) f(, y) f(a i, y) f(, y) + f(a i, y) f(, y) 0 as n since we have f y is a continuous function. 8

9 (b) We prove the result by doing induction on n. The case for n = is trivial. Suppose it is true for n = k. For n = k +, the map y := ( 2, k+ ) g(,, k+ ) is Borel measurable by induction assmption, i.e. g is Borel measurable. Moreover, by our assumption g y is continuous. g is Borel follows directly by considering the following sequence g m as before, g m (, y) = a i a i a i g(a i, y) + a i a i a i g(a i, y). (2) Suppose that f is real-valued in R 2, f is Lebesgue measurable for each, and f y is continuous for each y. Suppose that g : R R is Lebesgue measurable, and put h(y) = f(g(y), y). Prove that h is Lebesgue measurable on R. Solution. We argue as before to define h n by h n (y) := f n (g(y), y) = a i g(y) a i a i f(a i, y) + g(y) a i a i a i f(a i, y) if g(y) [a i, a i ). Obviously h n is Lebesgue measurable and tends to h pointwisely as n goes to. Therefore, h(y) is Lebesgue measurable. (3) Give an eample of two measurable sets A and B in R 2 but A + B is not measurable. Suggestion: For the two-dimensional case, take A = {0} [0, ] and B = N {0} where N is a non-measurable set in R. Solution. Let N be the non-measurable set in R constructed by picking eactly one element in [0, ] from each equivalence class defined by y iff y Q. Put A = {0} [0, ] and B = N {0}. Then since A and B are of measure 9

10 zero, they are both measurable. But A + B = N [0, ]. We write {r n } is an enumeration of Q in [, ]. We have [0, ] [0, ] (A + B + (r i, 0)) [, 2] [0, ] which shows that it is impossible to have countable additivity for the union above. 0