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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 ....
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 Spring '10
 tao/analysis
 Topology, Limits, Empty set, Topological space, ek, Lebesgue measure

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