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# The second inequality relating liminf and limsup is

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The second inequality relating liminf and limsup is standard, there is nothing to prove at this level of the course. Notice that so far we did NOT have to assume that thee sets have finite measure. But now, for the last inequality, we will set B n = S m = n A m and now we’ll need to use that m ( B 1 ) < to conclude that the measure of the intersection is the limit of the measures. Because B 1 B 2 B 3 ⊃ · · · and because m ( B n ) m ( A m ) for all m n , hence m ( B n ) sup m n m ( A m ), we get m ˆ \ n =1 [ m = n A m ! = m ˆ \ n =1 B n ! = lim n →∞ m ( B n ) lim n →∞ sup m n m ( A m ) = lim sup n →∞ m ( A n ) . The only non-trivial thing that remains to be done is to see that the first and last inequalities cannot be strengthened to equalities. Assume A n = A for n even, A n = B for n odd, where A, B are sets of positive measure with empty intersection. Then one see that [ n =1 \ m = n A n = , \ n =1 [ m = n A n = A B so that m ˆ [ n =1 \ m = n A m ! = 0 < min( m ( A ) , m ( B )) = lim inf n →∞ m ( A n ) lim sup n →∞ m ( A n ) = max( m ( A ) , m ( B )) < m ( A ) + m ( B ) = m ( A B ) = m ˆ \ n =1 [ m = n A m ! . 3. Prove; For every β [0 , 1) there exists a compact subset E of [0 , 1] such that m ( E ) = β and [0 , 1] \ E is dense in [0 , 1]. (Cantor would know what to do.) Solution. This is Exercise 14b of Chapter 3 of Royden. Assume β (0 , 1) and let α = 1 / 3. Here is a way to proceed that isn’t too messy. We follow Royden’s construction. We let E = \ n =0 C n , where C 0 , C 1 , . . . are constructed inductively as follows. C 0 = [0 , 1]. The interval C n +1 will be constructed from the interval C n by removing from the middle of each interval of C n an open interval of length ( α/ 3 n +1 ). For this of course we need to verify that each interval of C n is larger than α/ 3 n +1 A bit of experimenting, perhaps, shows that a correct inductive approach is the following: We will construct inductively for each n N a set C n consisting of the union of 2 n pairwise disjoint intervals, say C n = S 2 n k =1 I ( n ) k , I ( n ) k I ( n ) j = if k 6 = j and each I ( n ) k is a closed interval of length γ n = (1 - α )2 - n + α 3 - n . We already started the process by setting C 0 = [0 , 1], so assume we have constructed C n for some n 0. First of all, let us notice that α 3 n +1 < γ n . In fact γ n - α 3 n +1 = 1 - α 2 n + α 3 n - α 3 n +1 = 1 - α 2 n + 2 α 3 n +1 > 0 . This calculation also shows that if we remove an open interval of length α/ 3 n +1 from the center of one of these intervals I ( n ) k , what remains will be two intervals whose lengths add up to 1 - α 2 n + 2 α 3 n +1 so that each one will be of length 6

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one half that number; i.e., of length 1 - α 2 n +1 + α 3 n +1 = γ n +1 . If we let C n +1 be the set obtained by deleting an open interval of length α/ 3 n +1 we see that C n +1 satisfies the same properties as C n , but with n replaced by n + 1. Moreover C n +1 C n .
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