Unformatted text preview: Real Analysis  Math 630
Homework Set #11  Chapter 6 & 11
by Bobby Rohde 112500 Problem 6.21 a) Let g be an integrable function on 0, 1 . Show that 0 and measurable function f such that f
f g g
1 a bounded f Proof
Case Let
g
1 0 sgn g x , where sgn 0 0 ess sup f(x) = 1. g= g , 0, 0 . 0 Since g
1 fx 0, that implies that m x:g x Hence Case f g = g sgn g =
g
1 1= g 1 f . QED 0 0,
1 Let f x f g= g x. Then f f = 0. QED. . f g= g g 0 finite . Thus MATH63011.nb 2 b) Let g be a bounded measurable function. Show that integrable function f such that
f g g f
1 >0 an Proof
By definition ess sup g(x) = inf M : m t : f E x:g x g . Thus mE > 0. Let g by definition of E. Hence E g f 1 . Thus f g g f tM 0. Given > fx Then f g = E. we have this = mE g 1 . QED. 0.
E Let g = Problem 6.22 Find a representation for the bounded linear functionals on l p, 1 Construction p . Problem 6.24 Show that the element g in L p given by Theorem 13 is unique. Proof MATH63011.nb 3 Problem 11.7 Prove Proposition 4: If X , , is a measure space, then we can find a complete measure space X , 0, 0 such that i. ii. E iii. E Proof
First we must show that the 0 defined by (iii) is in fact a algebra. Thus we need that E1 , E2 E1 E2 0, 0 and E1 0 and the union of any countable collection of sets in is in 0 .
0 0. 0 E E A 0 E B where B and A C, C ,C 0. Clearly E1 E1 A1 B1 where B1 and A1 C1 , C1 , C1 0. And similarly for 0 E2 . Thus E1 E2 = A1 A2 B1 B2 . And we know that B1 B2 , since is a C1 C2 . C1 C2 , since is a algebra and algebra. Furthermore A1 A2 0 C1 C2 C1 C2 0. So C1 C2 0. E1 = hence A1 B1 B1 = A1 B1 = C1 C1 . Also C1 A1 A1 B1 B1 B1 C1 . However B1 C1 , and C1 . Hence E1 has the appropriate form to be in 0 . Thus we need only show that the union of a countable collection of sets in 0 is in 0 . Consider An Bn = Bn An , we know that Bn , so we only need 0 n 1 En , En that An is the subset of some set of measure 0. However An Cn , with Cn 0, and the union of countable collection of things of meaurse 0 is still measure 0. Hence it suffices to consider the union of the Cn and we are done. Thus 0 is a algebra. A B as above so that A is a subset of a set of measure 0. We wish to Consider E 0 with E show that B is not dependant on the choice of B . Given two such representations A1 B1 A2 B2 . Let A1 C1 , with C1 0 and A2 C2 , with C2 0. Thus, B1 A1 A2 B2 B1 C2 B2 = B2 , however by symmetry of argument, we have B2 C1 B1 = B1 , and hence B1 B2 . Thus we may define 0 E B, without concern that the value will change depending on our choice of B. So that 0 A 0 if A C, C 0. MATH63011.nb 4 It then follows immediately that 0 is a measure from the fact that is a measure and that we have only added sets which have measure 0 and the sets generated by their unions with sets of , which will neccesarily preserve the additivity of the measure. Thus we have created a complete measure with the neccesary properties. Problem 11.10 Prove Proposition 7: Let f be a nonnegative measurable function . Then a sequence of simple functions with n 1 lim n such that f n at each point of X. If f is defined on a finite measurable space, then we may choose the fucntions n so that each vanishes outside a set of finite measure. Construction
Define En,k x:k 2 n pair of integers n, k . I wish to show that
n n fx k 1 2 n and n 2 n 22 n k 0k En,k , with respect to each thus defined has the neccesary properties.
2n 2 . Thus m k 1 En,k , and so Since f is finite, for a given n, k , n 0, k 0, mEn,k 0, except on that set which is finite. So each function vanishes off a set of finite measure. n = Furthermore it should be clear that for a given n, the sets En,k are mutually exclusive and that if for a given n and k, f x is bracketed by k0 2 n , k0 1 2 n then for the next n, we will have f x bracketed by either 2 k0 2 n 1 , 2 k0 1 2 n 1 or 2 k0 1 2 n 1 , 2 k0 2 2 n 1 . And the lower bound of the interval is the same as the value returned by the sum, it is then clear that n 1 n. Furthermore, since the width of the interval between k 2 n and k 1 2 n goes to 0 as n 0, while the range goes to both 0 and as k 0 and k 22 n . This allows that for any value of f x , it will eventually be in some interval and the value can never exceed f x . Thus since the interval successively approximates f x with an width equal to that of the interval hence the limit must go to f x . MATH63011.nb 5 Problem 11.17 Prove Proposition 15: If f and g are integrable functions and E is a measurable set, then i. E c1 f c2 g c1 E f c2 E g ii. If h f and h is measurable then h is integrable. iii. If f g a.e., then f g. Proof
E c1 f c2 g c1 E f c2 Eg Simply break f and g into + and  parts and apply Proposition 13. If h
f and h is measurable then h is integrable. If we can first show part (iii), then by applying that result we can conclude that f h f , and thus h in integrable. So let us proceed to part (iii). If f
g a.e., then f g. If we consider
n f inf
f , then clearly a sequence of monotone increasing simple functions which converge to g which will be less than f a.e. By montone convergence theorem lim
n g inf
f f . Hence f g. QED. ...
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 Spring '08
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 Math, Lebesgue measure, Lebesgue integration, integrable, Borel measure, C1 C2 C1

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