Real Analysis: Homework # 12 Fall Professor: Sinan Gunturk Fall Term 2008

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1 Eduardo Corona eal Analysis: Homework # 2 Fall 2008 Professor: Sinan Gunturk Fall Term 2008 #3 (p.298) Let X be the set of rational numbers and A the algebra of nite unions of intervals of the form (a; b] with (a; b] = (a < b) and (?) = 0: The extension of to the smallest -algebra containing A is not unique. Proposition (9) Let C be a semialgebra of sets and a nonnegative set function de ned on C with? = 0 (if? 2 C). Then has a unique extension to a measure on the algebra A generated by C if the following conditions are satis ed: (i) If a set B in C is the union of a nite disjoint collection fb i g k of sets in C, then B = P k B i. (ii) If a set B in C is the union of a countable disjoint collection fb i g of sets in C; then B P B i. #4 (p.298) Prove Proposition 9 by showing: (a) Condition (i) implies that if A is the union of each of two disjoint collections fc i g and fd j g of sets in C; then P C i = P D j (b) Condition (ii) implies that is countably additive on A (for nite additivity and monotonicity already imply the reverse inequality) #5a (p.298) Let C be a semialgebra of sets and A the smallest algebra of sets containing C. (a) Show that A is comprised of sets of the form A = S C i with C i 2 C. #7 (p.298) Let be a nite measure on an algebra A, and the induced outer measure. Show that a set E is measurable if and only if for each " > 0 there is a set A 2 A ; A E; such that (AE) < ". #2 (p.30) Let X = Y = [0; ]; and let = be the Lebesgue measure. Show that each open set in X Y is measurable, and hence each Borel set in X Y is measurable. #28b (p.32) If f is measurable with respect to A B; then f x is measurable with respect to B for each x 2 X. #29 (p.32) Let X = Y = and = =Lebesgue measure. Then is the two dimensional Lebesgue measure on X Y = 2. (a) For each measurable subset E of, let: (E) = f(x; y) : x y 2 Eg Show that (E) is a measurable subset of 2 [Hint: consider rst cases when E is open, G, E of measure zero and then E measurable] (b) If f is a measurable function on ; the function F de ned by F (x; y) = f(x y) is a measurable function on 2. (c) If f and g are integrable functions on ; then for almost all x the function given by (y) = f(x y)g(y) is integrable. If we denote its integral h(x); then h is integrable and jhj jfj jgj #30 (p.32) Let f and g be functions in L (), and de ne f g to be the function h de ned by h(y) = f(y x)g(x)dx: (a) Show that f g = g f (b) Show that (f g) h = f (g h) (c) For f 2 L ; de ne b f by b f(s) = e ist f(t)dt: Then b f is a bounded complex function and [f g = b fbg

2 Extra Show that for each Lebesgue measurable function f, there is a Borel measurable function F such that f = F Lebesgue almost everywhere. (Hint: Take a sequence of simple functions that converge to f and modify them appropriately.) Hence when f is transformed by an integral operator, we may assume that f is Borel measurable. Use this to give an alternative proof of #29c that avoids showing that f(x y)g(y) is Lebesgue measurable in X Y. Solution 2 (#3) First of all, we check that A is indeed an algebra:? = (q; q]; X = ( ; ]; (a; b] [ (c; d] is either a disjoint union or (min(a; c); max(b; d)] and nally (a; b] c = ( ; a] [ (b; ]: Now, as de ned is a pre-measure on this algebra (since it assigns 0 to? and to everything else), and it is certainly not - nite (since the only element of nite measure in the algebra is?). Hence, we can t guarantee that the extension to (A) is unique. Now, we come up with two valid extensions: One possible extension certainly is the Caratheodory extension. This extension is highly trivial:? = 0; and for all E 6=?, since all covers of E by sets in A contain a nonempty set A i ; it follows that E =. The measurable sets are P (X); and restricted to (A) is an extension to by the Caratheodory extension theorem. Now, we consider the counting measure in (A). Because of the properties of cardinality, it is easy to see the counting measure de ned on any -algebra of sets is a measure. In this case, (?) = 0; and (a; b] = for a < b (an interval contains an in nite amount of points). Hence, restricted to a coincides with and hence is an extension of on (A). However, and are very di erent: for any q 2 Q; the singleton fqg = T (q =n; q] 2 (A), (fqg) = and (fqg) =. Also, this shows is not - nite, while is (Q = S q2q fqg). Solution 3 (#4) (a) So, for A 2 A(C) we have A = S n C i = S m j= D j: This implies that C i = S m j= C i \ D j 8i and D j = S n C i \ D j 8j (both disjoint unions, and C i \ D j 2 C by de nition). The key is that even if A is not in C; we know these two disjoint unions of sets in C give the same set. Applying (i): nx (C i ) = nx mx mx nx mx (C i \ D j ) = (C i \ D j ) = (D j ) j= j= Hence, assigning (A) = P n (C i) is a well de ned set function on A(C), and its values are uniquely determined by the values of for sets in C. Finite additivity follows directly from this de nition: if we have any two A and B disjoint and in A(C); A = S n C i and B = S m j= D j disjoint unions of sets in C, n[ m[ (A [ B) = ( C i [ D j ) = j= j= nx mx (C i ) + (D j ) = (A) + (B) (b) Now, let A = S k= A k disjoint union of sets in A(C) such that A 2 A(C) as well. By de nition of the generated algebra, each A k in turn is a disjoint union of elements in C: A k = S n k C k;i. Therefore, A = [ [ n k k= And by (ii) (applied to C k;i ) and (i) (for each A k by de nition of for nite disjoint unions of elements in C) this implies: Xn k A (C k;i ) = (A k ) k= Now, to prove the reverse inequality, it su ces to note that is monotone: A = S n C i and B = S m disjoint unions of sets in C, B A: 2 3 mx nx mx nx mx B = (D j ) = (C i \ D j ) 4 (C i \ D j ) 5 + (C i \ \ nx Dj) c = (C i ) = A j= j= j= C k;i k= j= j= D j 2

3 So, now for the countable, disjoint union of sets in A(C); by monotonicity and nite additivity, we have: ( n[ A k ) = k= nx (A k ) (A) 8n 2 N k= (A k ) (A) k= Hence, is the unique pre-measure in A(C) induced by the values of in C. Solution 4 (#5a) To prove this, it su ces to show that the set A = fa = S n C i j C i 2 C disjointg is an algebra of sets that contains C: (i)? 2 A (by convention, we take this as an empty union) and X 2 A, since for any element D of C; D c 2 A (by de nition of semialgebra) and X = D [ D c. (ii) A; B 2 A and A \ B =? then we have A = S n C i and B = S m j= D j disjoint unions of elements in C; then A [ B is also a union of disjoint sets of C; and therefore A [ B 2 A. So, this implies A is an algebra of sets that contains C; and so by de nition of the generated algebra, A(C) A. Also by de nition of A(C); it is clear that 8A 2 A; A 2 A(C) (since it is a nite union of sets in A(C)). Thus, A(C) = A. Solution 5 (#7) (=)) Lets call A the set of measurable sets. If E 2 A then E c 2 A and by de nition of the outer measure, 9(A n ) A such that E c S A n and: [ (E c ) = (E) = inff (A n ) : (A n ) A; E A n g (A n ) < (E c ) + " < Now, we consider A " = ( S A n) c = T Ac n 2 A (countable intersection of A c n 2 A). We have that A " E (since E c A c ") and: (EA " ) = (E) (A " ) We can do this, since is nite and both E and A " are in A. Also, we have that (E) = (X) (E c ) and (A " ) = (X) (A c "). Thus: (E) (A " ) = (A c ") (E c ) < " ((=) Now, let B X: As usual, we want to prove: (B) (B \ E) + (B \ E c ) Now, 8" > 0; let A " 2 A such that A " E and (EA " ) < ". Then (since A " 2 A ): (B \ E) = (B \ E \ A " ) + (B \ E \ A c ") (B \ A " ) + (EA " ) (B \ E c ) (B \ A c ") And combining these two, we get: (B \ E) + (B \ E c ) (B \ A " ) + (B \ A c ") + (EA " ) < (B) + " Since this is true for any " > 0; we obtain the inequality and that E 2 A. 3

4 Solution 6 (#2) As we know, the algebra A() with which we construct is the one generated by rectangles A B 2 with A; B Lebesgue measurable sets in [0; ]. Here it su ces to prove that S = [a; b] [c; d] [0; ] 2 (which is an actual open rectangle in [0; ] 2 ) is measurable, since then all open squares in this space are a basis for the Borel topology. So, 8B [0; ] 2 we need to prove: (B) (B \ S) + (B \ S c ) Now, by de nition, since (B) < ; 8" > 0 9 ( n = A n B n ) such that B S A n B n and P ( n) (B) + ": Now, we observe that S and S c 2 A() (S is a rectangle and S c = ([0; a) [0; ] [ (b; ] [0; ]) [ ([a; b] [0; a) [ (b; ]) = [ 2 so: Finally, we have: n = ( n \ S) [ ( n \ S c ) = ( n \ S) [ ( n \ ) [ ( n \ 2 ) ( n ) = ( n \ S) + ( n \ ) + ( n \ 2 ) [ [ (B \ S) + (B \ S c ) ( ( n \ S)) + ( ( n \ S c )) = = ( n \ S) + ( n \ ) + ( n \ 2 ) ( n ) (B) + " Hence, S is measurable. Then, since sets of this form are a basis for the Borel topology and the set of all measurable sets A is a -algebra, it follows that all open sets in [0; ] 2 are measurable (since they can be written as unions of sets like S). Finally, the Borel -algebra ("Borel sets") is the -algebra generated by the open sets, and again using that A is a -algebra that contains sets like S; it follows that all Borel sets are measurable (by de nition of the generated -algebra, the Borel sets are contained in A ). Solution 7 (#28b) f measurable with respect to A B means that f ((; )) 2 A B 8 2. Now, for a given x 2 X; f x = f(x; y) and fx ((; )) = fy 2 Y : (x; y) 2 f ((; ))g = f ((; )) x. Finally, by 28a; if f ((; )) 2 A B then fx ((; )) = f ((; )) x 2 B. Thus, f x is measurable with respect to B 8x 2 X. Solution 8 (#29) (a) So, let E be an open set in. Then, we can show (E) = f(x; y) : x y 2 Eg is also open in 2 : given (x ; y ) 2 (E), (E) y = E + fy g and (E) x = E + fx g open sets in (translation and re ection of an open set is open). Then, since x 2 (E) y and y 2 (E) x ; there exist open intervals (x ; x + ) (E) y and (y "; y + ") (E) x. If = minf; "g; we claim that the open ball B () (x ; y ) = f(x; y) : jx x j + jy y j < g (E). For (x; y) 2 B () (x ; y ); j(x y) (x y )j = j(x x ) + (y y )j jx x j + jy y j < And since x y 2 E and our choice of intervals implies that (x y ; x y + ) 2 E. Thus, (E) is open in 2 ; and in particular, it is measurable. Alternatively, we can consider (x; y) = x y; which is a continuous function. Then (E) = (E) is open if E is open. This also tells us that E 2 G =) (E) 2 G 2 by properties of the inverse image. For (G n ) sequence of open sets, E = T G n:! \ \ G n = (G n ) 2 G 2 And so, (E) is measurable. (We know this, since continuous =) Borel measurable). Finally, we prove that if E is of measure 0; then 2 ((E)) = 0. Now, we know that, 8n 2 N 9(I (n) m ) m= 4

5 a sequence of disjoint intervals such that E S m= I(n) m and P m= I m (n) < 2n. Now, if we de ne 2 n = f(x; y) 2 (E) : jyj < ng, it is clear that ( n ) is a monotonely increasing sequence of sets in 2 ; and S n = (E). Also, P n = f(x; y) : (x; y) 2 S m= I(n) m ; jyj < ng is a union of disjoint parallelograms, and we can deduce (by computing the area of each parallelogram, which is 2n ji n j and should coincide with the Lebesgue measure on 2 ) that: n 2 P n and 2 (P n ) < n =) 2( n ) < n Since, for a xed n; n n+p 8p 2 N, this means 2( n ) < n+p 8p 2 N =) 2( n ) = 0. Finally, since S n = (E); we have that: 2((E)) = 0 Since all 2 null sets are measurable, and a countable union of null sets is null (the set of null sets is a -ring), it follows that (E) is null, and thus, measurable. Alternatively, we could have covered our measure zero set E by a G set with zero measure, and then used Fubini to prove the measure of of this G set is zero. Finally, let E be a general measurable set in. We know that there exists a set H 2 G such that E H and (HE) = 0. But then 2 ( (HE)) = 0 (since we have proved the inverse image of a null set is null) and this implies (H) (E) is measurable. Since (H) is measurable, this in turn means (E) = (E) is measurable. (b) If f is a measurable function in, then the function de ned as F (x; y) = f(x y) is a measurable function on 2 : This is immediate from (a): clearly F = f, and so F ((; )) = (f (; )): Since f is measurable, f (; ) is a Lebesgue measurable set, and by (a); (f (; )) is Lebesgue measurable. (c) Now, we need to prove that (y) = f(x since f(x y) is 2 measurable (by b). Now, j(y)j dy = y)g(y) is integrable for almost every x 2. is measurable, jf(x y)j jg(y)j dy Using the Theorem for kernel functions, we know that since: jf(x y)j dy kfk jf(x y)j dx kfk By Tonelli, 0 h(x)dx = jf(x y)j jg(y)j dydx = jf(x So, this proves the result for h; and that f(x L () for almost every x: y)j dxa dy kfk kgk y)g(y) 2 L ( 2 ; d 2 ). Then, by Fubini, we have that is in Solution 9 (#30) (a) (Commutativity) By translation invariance of the Lebesgue measure, we can "substitute" y = x z and obtain: f g(x) = f(x y)g(y)dy = f(z)g(x z)dz = g f(x) (b) (Associativity) Using the de nitions of convolution, Fubini s Theorem and #29, we have that: f g(z)h(x z)dz = f(y)g(z y)dy h(x z)dz = f(y)g(z y)h(x z)dydz = f(y) g(z y)h(x z)dz dy 5

6 Now, by the translation invariance of the Lebesgue measure, it follows that: g(z y)h(x z)dz = g((z + y) y)h(x (z + y))dz = g(x)h((x y) z)dz g h(x y) So, we can conclude that: f g(z)h(x z)dz = f(y)g h(x y)dy (c) (Fourier Transform) First, we prove that f() b is bounded (using basic results of complex integration): f() b e = e it f(t)dt it jf(t)j dt = kfk 8 2 =) f b kfk Now, we compute the Fourier Transform of a convolution f g of L functions. First, we apply Fubini to change the order of integration, and we split the exponential: e it f(t y)g(y)dydt = e iy g(y) e i(t y) f(t y)dt dy Then, by translation invariance of the Lebesgue measure, it follows that: = e iy g(y) f()dy b = bg() f() b Solution 0 (Extra) Let f Lebesgue measurable function, and f n g a sequence of simple functions such that n!. These functions are n = P k n Ai ; with A i Lebesgue measurable sets. Then, for each c(n) i A i ; we take a F i 2 F such that F i A i and (A i F i ) = 0; and de ne n = P k n simple functions are such that n = n A:E: (since they only di er in k n [ c(n) i Fi. These modi ed A i F i ): They are also borel measurable, since they are now de ned on borel sets. So, if we de ne F (x) = lim n! n (x); it follows that f = F A:E: and that, being a limit of borel measurable functions, F is also borel measurable. Hence, we can replace f in #29 by F borel measurable, and we know that f g(x) = F g(x) for A:E: x 2. However, we don t need to check measurability of F (x y)g(y); since F (x y) = F ; and we know the composition of a borel measurable function and a continuous function is borel measurable. 6