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Unformatted text preview: Combining with (5), we proved that f ( b Î· ) f ( a + Î· ) â‰¤ Î¼ * ( I ) + Â² for all Î· > 0. Letting Î· â†“ 0, f ( b ) f ( a +) â‰¤ Î¼ * ( I ) + Â². Since Â² > 0 was arbitrary, we proved the first inequality in (4). Contrary to what happens with Royden, we have to consider also the case of an unbounded I , because it is possible here for I to be unbounded and yet Î¼ * ( I ) < âˆž . This will happen if f ( x ) has a finite limit for x â†’ âˆž or x â†’ âˆž . Assume first that a =âˆž , b finite. We can let J k = ( k,b ) where k < b . Then J k âŠ‚ I and because Î¼ * is an outer measure, Î¼ * ( J k ) â‰¤ Î¼ * ( I ); by what we proved f ( b ) f ( k +) â‰¤ f ( J k ) â‰¤ Î¼ * ( I ) and we can let k â†’ âˆž to get Î¼ * ( I ) â‰¥ f ( b ) f (âˆž ). As similar argument works if a âˆˆ R and b = âˆž , or if a =âˆž ,b = âˆž . Claim 1 is established. Claim 2. If I = ( a,b ) is an open interval, then (6) Î¼ * ( I ) = f ( b ) f ( a +) . In fact, because of (4), it suffices to prove that Î¼ * ( I ) â‰¤ f ( b ) f ( a +). Iâ€™ll assume thatâˆž < a < b < âˆž ; the case in which one or both of a,b is not finite is actually simpler and if you understood what I do in the case I consider, you should have no trouble dealing with the infinite endpoint(s) case. We recall that the set of discontinuities of an increasing function is countable. The complement of a countable set being dense, we can find sequences { c n } , { d n } with the following properties: (a) a < c n +1 < c n , c 1 < d 1 , d n < d n +1 < b for all n âˆˆ N (b) lim n â†’âˆž c n = a , lim n â†’âˆž d n = b . (The sequence { c n } decreases to a , { d n } increases to b . (c) f is continuous at c n ,d n for all n âˆˆ N . 3 Let J n = ( c n +1 ,c n ), K n = ( d n ,d n +1 ) for n âˆˆ N ; H = ( c 1 ,d 1 ). Let also Â² > 0 be given. Because f is continuous at c n , there is Î´ n > 0 such that  x c n  < Î´ n implies  f ( x ) f ( c n )  < Â²/ 2 n +2 . In particular, f ( c n + Î´ n / 2) f ( c n Î´ n / 2) = f ( c n + Î´ n / 2) f ( c n ) + f ( c n ) f ( c n Î´ n / 2) < Â² 2 n +1 . Similarly, there is Î· n > 0 such that  x d n  < Î· n implies  f ( x ) f ( d n )  < Â²/ 2 n +2 . In particular, f ( d n + Î· n / 2) f ( d n Î· n / 2) = f ( d n + Î· n / 2) f ( d n ) + f ( d n ) f ( d n Î· n / 2) < Â² 2 n +1 . Let L n ( c n Î´ n / 2 ,c n + Î´ n / 2), M n ( d n Î· n / 2 ,d n + Î· n / 2) for n âˆˆ N . Since this may look a bit complicated, I will insist that the idea is quite simple and natural; draw a picture! The family of open intervals { J n } âˆª { K n } âˆª { L n } âˆª { M n } âˆª { H } is a countable family covering ( a,b ) and if we add all the Ëœ â€˜ values of these intervals we notice âˆž X n =1 Ëœ â€˜ ( J n ) = lim n â†’âˆž n X k =1 ( f ( c k ) f ( c k +1 )) = f ( c 1 ) lim n â†’âˆž f ( c n +1 ) = f ( c 1 ) f ( a +) ....
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 Spring '11
 Speinklo
 CN, Lebesgue measure, open intervals

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