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Unformatted text preview: I , set d n = b for all n ; otherwise set d n = b +1 /n . Now I âŠ‚ J n (which can be interpreted as I âŠ‚ J n âˆªâˆ…âˆªâˆ…âˆªÂ·Â·Â· ) so that by the definition of Î¼ * we see that Î¼ * ( I ) â‰¤ Ëœ â€˜ ( J n ) = f ( d n ) f ( c n ) for all n âˆˆ N . Letting n â†’ âˆž we get Î¼ * ( I ) â‰¤ f ( b +) f ( a ) , the second inequality in (4). To prove the first inequality in (4), we may assume that Î¼ * ( I ) < âˆž , otherwise the inequality is obvious. Given then Â² > 0, there is a countable collection { J n } of open intervals such that I âŠ‚ âˆž [ n =1 J n and such that if J n = ( c n ,d n ) for n âˆˆ N , (5) âˆž X n =1 Ëœ â€˜ ( J n ) = âˆž X n =1 ( f ( d n ) f ( c n )) â‰¤ Î¼ * ( I ) + Â². We will assume first that the interval I of endpoints a,b is bounded; i.e.,âˆž < a < b < âˆž . Let Î· > 0 be given; assume also Î· < ( b a ) / 2 so that [ a + Î·,b Î· ] makes sense as an interval. Now [ a + Î·,b Î· ] is compact and contained in âˆž [ n =1 J n , thus there is N âˆˆ N such that [ a + Î·,b Î· ] âŠ‚ of S N n =1 J n . We will select indices n 1 ,...,n k âˆˆ { 1 ,...,N } as follows. There is then n , 1 â‰¤ n â‰¤ N such that a + Î· âˆˆ J n . Of all possible such n â€™s, Let n 1 be one with a maximal d n . So n 1 has the properties: 1 â‰¤ n 1 â‰¤ N , c n 1 < a + Î· < d n 1 and if 1 â‰¤ n â‰¤ N , c n < a + Î· < d n , then d n â‰¤ d n 1 . If b Î· < d n 1 2 we are done; k = 1. If not, d n 1 âˆˆ I and there is n , 1 â‰¤ n â‰¤ N such that c n < d n 1 < d n . The way we selected n 1 , this forces any such n to satisfy c n â‰¥ a + Î· so c n 1 < c n < d n 1 < d n . Of all such n â€™s, let n 2 be one for which d n is maximal so that now we have 1 â‰¤ n 2 â‰¤ N , c n 1 < c n 2 < d n 1 < d n 2 and if n is such that 1 â‰¤ n â‰¤ N , c n 1 < c n < d n 1 < d n , then d n â‰¤ d n 2 . Of course, there is no reason to assume that n 2 > n 1 . If d n 2 > b Î· , we are done; k = 2. If not ..., I hope you get the drift. In this way we get a subset { n 1 ,...,n k } of { 1 ,..., N } such that [ a + Î·,b Î· ] âŠ‚ k [ j =1 ( c n j ,d n j ) and c n 1 < a + Î· , d n k > b Î· c n 1 < c n 2 < d n 1 < c n 3 < Â·Â·Â· < c n k < d n k 1 < d n k . Because f is increasing, f ( a + Î· ) â‰¥ f ( c n 1 ), f ( b Î· ) â‰¤ f ( d n k ) and thus f ( b Î· ) f ( a + Î· ) â‰¤ f ( d n k ) f ( c n k ) = f ( d n k ) f ( d n k 1 ) + f ( d n k 1 ) f ( d n k 2 ) + Â·Â·Â· + f ( d n 2 ) f ( d n 1 ) + f ( d n 1 ) f ( c n 1 ) = k X j =2 f ( d n j ) f ( d n j 1 ) + f ( d n 1 ) f ( c n 1 ) . The best way to avoid being intimidated by a long string of calculations is by doing them oneself. This is not hard!! Donâ€™t let it get you down. Anyway, we use once more the fact that f is increasing, thus f ( d n j ) f ( d n j 1 ) â‰¤ f ( d n j ) f ( c n j 1 ) and we see that f ( b Î· ) f ( a + Î· ) â‰¤ k X j =1 f ( d n j ) f ( c n j 1 ) = k X j =1 Ëœ â€˜ ( J n k ) â‰¤ âˆž X n =1 Ëœ â€˜ ( J n ) ....
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 Spring '11
 Speinklo
 CN, Lebesgue measure, open intervals

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