This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 202: Topology and Measure Theory Mark Borgschulte PS10: Measurability, Limits, BorelCantelli Lemma Professor: Justin Holmer November 7, 2007 Exercises (Stein and Shakarchi, Chapter 1, Ex. 5, 7, 8, 11, 16, 26) 5. Suppose E is any given set, and O n is the open set: O n = { x : d ( x,E ) < 1 /n } . Show: (a) If E is compact, then m ( E ) = lim n m ( O n ). (b) However, the conclusion in (a) may be false for E closed and unbounded; or E open and bounded. (a) Pf. Consider Corollary 3.3.ii: If E k & E and m ( E k ) < for some k , then m ( E ) = lim N m ( E N ) . Here, E N = O n (and E = E ). Claim: O n & E . Pf of claim: Each O n contains E , so we need only show that there are no elements in their intersection which are not in E . Let x T n O n , but x 6 E . Since E is compact, hence closed and bounded, we can find an open ball of radius r , disjoint from E , which contains x . Clearly O 1 /r +1 does not contain x . Also because E is compact, we can find r R (different from above r ), s.t. r = the radius of an open ball which covers E , B ( E,r ), and observe that O n are contained within B ( E,r +1) , for all n . Hence, m ( O n ) < , n. This demonstrates the conditions of Cor 3.3.ii, which proves the statement. (b) Closed and Unbounded: Consider N . Clearly, m ( N ) = 0, yet the sum, N N m (( N 1 n ,N + 1 n )) diverges n N , since each natural number makes 2 /n contribution to the sum. Open and Bounded: Consider an enumeration of Q T [0 , 1] , { q i } i =1 . Around each element in the enumeration, place open balls of radius 1 / 2 i +2 . The sum of the measures of these open balls is 1 / 2, an upper bound on the measure of their union. But of course Q is dense in [0 , 1] the sum of the measures of O n = 1 , n ....
View
Full
Document
 Spring '10
 tao/analysis
 Topology, Limits

Click to edit the document details