Andrew van Herick Math 710 Dr. Alex Schuster Nov. 2, 2005 Assignment 8 Section 3: 20, 24, 25, 30, 31 20. Show that the sum and product of two simple functions are simple. Show that A\B = A B A[B = A + B A B A c = 1 A Let E R and suppose f; g : E! R are two simple functions. By Proposition 19 (p. 67, Royden) both f g and f + g are measurable. Because f and g are simple f (E) and g (E) are nite sets. It follows that both fuv 2 R : u 2 f (E) ; v 2 g (E)g and fu + v 2 R : u 2 f (E) ; v 2 g (E)g each have cardinality which is at most jf (E)j jg (E)j ; in other words, nite. see that It is easy to (f g) (E) fuv 2 R : u 2 f (E) ; v 2 g (E)g and (f + g) (E) fu + v 2 R : u 2 f (E) ; v 2 g (E)g Hence it follows that (f g) (E) and (f + g) (E) must be nite as well. f g; f + g are simple functions. By de nition Let A; B R: First we ll show that A\B = A B : Let x 2 R: If x 2 A \ B then x 2 A and x 2 B; so A (x) = 1 and B = 1: Hence A\B (x) = 1 = 1 1 = A (x) B (x) = A B (x) : If x =2 A \ B; then A\B = 0: By de Morgan s laws x 2 A c [ B c ; i.e. x =2 A or x =2 B: Either way A (x) B (x) = 0. Hence A\B (x) = A B (x) : This covers all cases of x and establishes that A\B (x) = A B (x) for all x 2 R: By de nition A\B = A B :
2 Now we ll show that A c = 1 A : Let y 2 R: If x 2 A then A c (y) = 0 and A (y) = 1 by de nition, so A c (y) = 0 = 1 1 = 1 A (y) = (1 A ) (y) : If x 2 A c ; then then A c (y) = 1 and A (y) = 0; so A c (y) = 1 = 1 0 = 1 A (y) = (1 A ) (y) : This covers all cases of y and establishes that A c (y) = (1 A c = 1 A : A ) (y) for all y 2 R: Hence To show that A[B = A B. By de Morgan s laws A [ B = A c \ B c : By what we proved above A[B 1 = (A[B) c = A c \B c = A c B c = ( 1 + A ) (1 B ) = A B + B + A 1 It follows immediately that A[B = A + B A B ; as desired.
Andrew van Herick 3 24. Let f be measurable and B a Borel set. Then f 1 [B] is a measurable set. Let f be measurable and B a Borel set. Let E := E R : f 1 [E] is measurable show that E contains B; the collection of Borel sets. : We ll First we ll show that E contains every open interval. Let I R be an open interval. Then I = (a; b) for some (possibly extended) real numbers a; b such that a < b: Note that I = ( 1; b) \ (a; 1) : As f is measurable f 1 (( 1; b)) = fx : f (x) < bg is measurable by de nition. Similarly f 1 ((a; 1)) is also measurable. By the fact that the measurable sets form a -algebra f 1 (I) = f 1 (( 1; b) \ (a; 1)) = f 1 (( 1; b)) \ f 1 ((a; 1)) is also measurable. Hence I 2 E: This establishes that I 2 E for every open interval I R: Now we ll show that E is a -algebra. Let A 0 be a countable index set and fe g 2A0 be a collection of sets from E: Then by the fact that the measurable sets form a -algebra 0 1 f 1 @ [ E A = [ f 1 (E ) 2A 0 2A 0 is measurable, since f 1 (E ) 2A0 is a countable collection of measurable sets. Now let E 2 E: By the same reasoning f 1 (E c ) is measurable since f 1 (E c ) = f 1 (E) c : This establishes that E is a -algebra. Thus E is a -algebra containing every open interval. We proved in Exercise 10.4.5. (p. 311, Wade) that every open set in R is a countable union of open intervals. It follows immediately that every open subset of R belongs to E: This establishes that E is a -algebra containing every open subset of R: As B is the smallest -algebra containing all the open sets, it follows that B E: B 2 B; it follows immediately that B 2 E: Equivalently B is measurable. As
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Andrew van Herick 5 25. Show that if f is a measurable real valued function and g a continuous function de ned on ( 1; 1) ; then g f is measurable. Let f be a measurable real valued function and g a continuous function de ned on ( 1; 1) : As of g f is R; an open set the domain of g f is measurable. Let 2 R and E = fx : g f (x) > g : It is easy to see that E = fx : g f (x) 2 (a; 1)g = (g f) 1 ((a; 1)) : By Lemma 1 below E = f 1 g 1 ((; 1)) : Because g : R! R is continuous and (; 1) is open, by Theorem 10.58 (p. 317, Wade) g 1 ((a; 1)) is is open in R: By de nition g 1 ((a; 1)) is a Borel set. Problem 24 above E = f 1 g 1 ((; 1)) is a measurable set. Lemma 1 Let E; F; G be sets and suppose f : E! F and g : F! G: (g f) 1 (A) = f 1 g 1 (A) : It A G; then Let x 2 (g f) 1 (A) : By de nition (g f) (x) = g (f (x)) 2 A: By the de nition of inverse images we get f (x) 2 g 1 (A) and x 2 f 1 g 1 (A) : Now let y 2 f 1 g 1 (A) : Then f (y) 2 g 1 (A) : By de nition g (f (y)) 2 A. As g (f (y)) = (g f) (y) we have (g f) (y) 2 A: By de nition y 2 (g f) (A) : This establishes that (g f) 1 (A) = f 1 g 1 (A) :
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Andrew van Herick 7 30. Prove Egoro s Theorem: If hf n i is a sequence of measurable functions that converge to a real-valued function f a.e. on a measurable set E of nite measure, then given > 0; there is a subset A E with ma < such that f n converges uniformly on EnA: [Hint: Apply Proposition 24 repeatedly with " n = 1 n and n = 2 n.] Let > 0: For each k 2 N de ne " k = 1 k ; k = 2 k. Clearly " k ; k > 0: Using Proposition 24 (p. 73, Royden) for each k 2 N choose A k E such that m (A k ) < k, and choose N k 2 N such that for all x =2 A k and all n N k ; jf n (x) f (x)j < " k : Let A := S k2n A k: Because each A k E it follows that A E: Note that each A k is measurable. By Proposition 13 (p. 62, Royden) 1X 1X 1X m (A) = m (A k ) < k = 2 k = k=1 k=1 k=1 This establishes that m (A) < : Now we ll show that f n converges uniformly on EnA: Let " > 0: Then there exists M 2 N such that 1 M < ": Choose N 2 N such that N = N M: Let n N and x 2 EnA: As x 2 A c and A c = T k2n (A k) c ; we have x 2 (A M ) c : From our construction of A M above jf n (x) f (x)j < " M = 1 M < ": This establishes that for all " > 0 there is a N 2 N such that n N implies jf n (x) f (x)j < " for all x 2 EnA: By de nition f n is uniformly convergent on EnA: This establishes the existence of a subset A E with ma < such that f n converges uniformly on EnA:
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Andrew van Herick 9 31. Prove Lusin s Theorem: Let f be a measurable real-valued function on an interval [a; b] : Then given > 0; there is a continuous function ' on [a; b] such that such that m fx : f (x) 6= ' (x)g < : [Hint: Use Egoro s Theorem, Propositions 15 and 22, and Problem 2.40.] Let f be a measurable real-valued function on an interval [a; b] : Let > 0: As f is real valued f takes on the values 1 on ;; a set of measure zero. Let k = 1 k : By Proposition 22 (p. 71, Royden) for each k 2 N choose a continuous function h k on [a; b] such that jf h k j < k : It follows almost immediately from this construction that h k (x)! f (x) for every x 2 [a; b] : By Egoro s Theorem there is a subset A [a; b] with m (A) < 2 such that h k converges to f uniformly on [a; b] na: By Proposition 15 (p. 63, Royden) there exists a closed set F [a; b] na such that m ([a; b] nanf ) < 2 : Because all of the h k s are continuous on [a; b] ; they are also continuous on F [a; b] : Hence f is continuous on F by Exercise 7.7.1(a) (p. 191, Wade). By Problem 2.40 (p. 49, Royden) there is a continuous function g : R! R such that f (x) = g (x) for each x 2 F: Let ' = gj [a; b] : Then ' is clearly continuous on [a; b] : Clearly fx 2 [a; b] : f (x) 6= ' (x)g ([a; b] nf ) [ A: Because ([a; b] nf ) [ A = ([a; b] nanf ) [ A; it follows that m (fx : f (x) 6= ' (x)g) m (([a; b] nf ) [ A) = m (([a; b] nanf ) [ A) < 2 + 2 = This establishes the existence of a continuous function ' : [a; b]! R such that m fx : f (x) 6= ' (x)g < as desired.