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Unformatted text preview: Math 318 HW #6 Solutions 1. (a) Exercise 16.7. Prove Corollary 10.10, which says that if A 1 A 2 A 3 are mea surable subsets of E , then T i =1 A i is measurable and m * ( T i =1 A i ) = lim i m * ( A i ). Proof. Let B i = E \ A i for each i . Then B 1 B 2 B 3 and so, by Corollary 10.9, S i =1 B i is measurable and m * [ i =1 B i ! = lim i m * ( B i ) . Therefore, i =1 A i = i =1 ( E \ B i ) = E \ [ i =1 B i is measurable by Corollary 8.4 and m * i =1 A i ! = 1 m * [ i =1 B i ! = 1 lim i m * ( B i ) . But then m * ( B i ) = 1 m * ( A i ) for each i , so we have m * i =1 A i ! = 1 lim i (1 m * ( A i )) = lim i m * ( A i ) , as desired. (b) Exercise 16.36. Show that Corollary 10.10 is false for unbounded sets A i , i = 1 , 2 ,... . Where does the proof of Corollary 10.10 break down in this case?...
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This note was uploaded on 08/20/2011 for the course MATH 318 taught by Professor Staff during the Spring '08 term at Haverford.
 Spring '08
 Staff
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