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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 âŠƒ A 2 âŠƒ A 3 âŠƒ Â·Â·Â· are mea surable subsets of E , then T âˆž i =1 A i is measurable and m * ( T âˆž i =1 A i ) = lim i â†’âˆž m * ( A i ). Proof. Let B i = E \ A i for each i . Then B 1 âŠ‚ B 2 âŠ‚ B 3 âŠ‚ Â·Â·Â· and so, by Corollary 10.9, S âˆž i =1 B i is measurable and m * âˆž [ i =1 B i ! = lim i â†’âˆž m * ( B i ) . Therefore, âˆž i =1 A i = âˆž i =1 ( E \ B i ) = E \ âˆž [ i =1 B i is measurable by Corollary 8.4 and m * âˆž i =1 A i ! = 1 m * âˆž [ i =1 B i ! = 1 lim i â†’âˆž m * ( B i ) . But then m * ( B i ) = 1 m * ( A i ) for each i , so we have m * âˆž i =1 A i ! = 1 lim i â†’âˆž (1 m * ( A i )) = lim i â†’âˆž 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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 Spring '08
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 Topology, Sets, Empty set, Metric space, Lebesgue measure

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