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Unformatted text preview: We only evaluated the exact outer measure of open intervals and of half open intervals open on the left. But it is easy to see that if I = [ a,b ], then μ * ( I ) = f ( b +) f ( a ) and if I = [ a,b ), then μ * ( I ) = f ( b ) f ( a ). (c) This is not much different from part (b). You might notice first that μ * of an interval works out to the same value as in part (b). Once that is established, all goes in (essentially) the same way. 2. Let { A n } be a sequence of measurable subsets of R . Assume they are all contained in some measurable set E with m ( E ) < ∞ . Prove: m ˆ ∞ [ n =1 ∞ \ m = n A m ! ≤ liminf n →∞ m ( A n ) ≤ limsup n →∞ m ( A n ) ≤ m ˆ ∞ \ n =1 ∞ [ m = n A m ! . Conclude that if S ∞ n =1 T ∞ m = n A n = T ∞ n =1 S ∞ m = n A n =: lim n A n , then m ‡ lim n A n · = lim n →∞ m ( A n ) . By considering the case of A n = A when n is even, A n = B when n is odd, where A,B are measurable, A ∩ B = ∅ , m ( A ) > 0, m ( B ) > 0, show that it is possible to have m ˆ ∞ [ n =1 ∞ \ m = n A n ! < liminf n →∞ m ( A n ) ≤ limsup n →∞ m ( A n ) < m ˆ ∞ \ n =1 ∞ [ m = n A n ! . Solution. Let E be a measurable subset of R with m ( E ) < ∞ and let A n be a measurable subset of E for all n ∈ N . For the first inequality to be proved, set B n = T ∞ m = n A m . Then B 1 ⊂ B 2 ⊂ ··· (and it goes without saying, so I won’t say it, that the B n ’s are measurable) and m ˆ ∞ [ n =1 ∞ \ m = n A n ! = m ˆ ∞ [ n =1 B n ! = lim n →∞ m ( B n ) . On the other hand, B n ⊂ A m for all m ≥ n , thus m ( B n ) ≤ m ( A m ) for all m ≥ n , hence m ( B n ) ≤ inf m ≥ n m ( A m ). We know from our studies of liminf and limsup that the sequence { inf m ≥ n m ( A m ) } is increasing and thus has a limit, and we know what this limit is. We can return to our calculations above and complete them as follows: m ˆ ∞ [ n =1 ∞ \ m = n A m ! = m ˆ ∞ [ n =1 B n ! = lim n →∞ m ( B n ) ≤ lim n →∞ inf m ≥ n m ( A m ) = liminf n →∞ m ( A n ) . This proves the first inequality. 5 Note. The symbol m is unfortunately overloaded; I hope it is clear from the context when it is a subindex, when it represents Lebesgue measure. The basic unwritten rule not followed by Royden is NOT to use a letter that is commonly used for a variable to represent a fixed object. That is why one introduced symbols like R , N , etc., to avoid preempting the use of R,N , etc. I’ll also use this opportunity to say that there was an unfortunate typo in the statement of the inequalities; two subindices of n that should have been m . Since I had stated the inequalities already in class, and because the typos are obvious, I do expect people to have corrected them....
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 Spring '11
 Speinklo
 CN, Lebesgue measure, open intervals

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