MATH501_HW2 - Serge Ballif MATH 501 Homework 2 October 3...

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Serge Ballif MATH 501 Homework 2 October 3, 2007 In problems 1-3, ( X, M , μ ) denotes a measure space. 1. Let ( E k ) k ⊂ M . Define lim inf E j = [ i =1 \ k = i E k and lim sup E j = \ i =1 [ k = i E k Prove the following. (a) μ (lim inf E j ) lim inf j →∞ μ ( E j ) . (b) If μ ( S j =1 E j ) < , then lim sup j →∞ μ ( E j ) μ (lim sup E j ) . Show the assumption μ ( S j =1 E j ) < is necessary. (c) Borel-Cantelli lemma. If j =1 μ ( E j ) < , then μ (lim sup E j ) = 0 . (a) Recall that lim inf j →∞ x j = lim j →∞ inf k j x k . Claim 1. μ ( S i =1 T k = i E k ) = lim i →∞ μ ( T k = i E k ). Proof. Let F i = T k = i E k . We have the increasing chain of subsets F i F i +1 for all i . Hence, by continuity from below μ ( j =1 F j ) = lim j →∞ μ ( F j ). For each k we know by monotonicity that μ ( T k = j E k ) μ ( E k ). Hence μ ( T k = j E k ) is a lower bound for μ ( E k ). Therefore, μ ( T k = j E k ) inf k j μ ( E k ). Now we take the limit as j → ∞ to get lim j →∞ μ \ k = j E k ! = μ (lim inf E j ) lim inf j →∞ μ ( E j ) . (b) Recall that lim sup j →∞ x j = lim j →∞ sup k j x k . Claim 2. μ T j =1 S k = j E k = lim j →∞ μ S k = j E k . Proof. Let G j = S k = j E k . We have a decreasing chain of subsets G j G j +1 for all j . Also, we know that μ ( G j ) < . Hence, by continuity from above we know that μ ( T j =1 G j ) = lim j →∞ μ ( G j ). For each k we know by monotonicity that μ ( E k ) μ ( S k = j E k ). Hence, sup k>j μ ( E k ) μ ( S k = j E k ). Now we take the limit as j → ∞ to get lim j →∞ μ [ k = j E k ! = μ \ j =1 [ k = j E k ! lim j →∞ sup k j μ ( E k ) . To see that the assumption μ ( S j =1 E j ) < is necessary, we look at the in- tervals E j = ( j, ) where our measure is the Lebesgue measure on . Then lim sup E j = T i =1 S k = i E k = . Then lim sup j →∞ μ ( E j ) = > 0 = μ ( ) = μ (lim sup E j ) Page 1 of 5
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Serge Ballif MATH 501 Homework 2 October 3, 2007 (c) Just as in part ( b ) we let G j = S k = j E k . By additivity we know μ ( G j ) j =1 μ ( E j ). Monotonicity gives us μ ( T j =1 S k = j E k ) μ ( G j ). Combining these
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