Μ i 1 n μ i 2 n f b n f a f a f a n f b n f a n as

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μ * ( I 1 n ) + μ * ( I 2 n ) = f ( b n - ) - f ( a +) + f ( a +) - f ( a n +) = f ( b n - ) - f ( a n +) as asserted. Now f ( b n - ) - f ( a n +) f ( b n ) - f ( a n ) ˜ ( I n ) so that (3) and (8) imply μ * ( A 1 ) + μ * ( A 2 ) X n =1 ˜ ( I n ) , completing the proof of (2). This concludes the treatment of part (b), except for a some comments. Borel sets are measurable. Not every countable or even finite set has measure zero in this model. For example, if f has a jump at c , then from μ * (( c - ², c ]) = f ( c +) - f ( c - ² +) it is quite easy to deduce that μ * ( { c } ) = f ( c +) - f ( c - ) . 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 ! lim inf n →∞ m ( A n ) lim sup 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 ! < lim inf n →∞ m ( A n ) lim sup 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 ) = lim inf n →∞ m ( A n ) . This proves the first inequality. 5
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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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