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I set d n = b for all n otherwise set d n = b 1/n now

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Unformatted text preview: I , set d n = b for all n ; otherwise set d n = b +1 /n . Now I ⊂ J n (which can be interpreted as I ⊂ J n ∪∅∪∅∪··· ) so that by the definition of μ * we see that μ * ( I ) ≤ ˜ ‘ ( J n ) = f ( d n )- f ( c n ) for all n ∈ N . Letting n → ∞ we get μ * ( I ) ≤ f ( b +)- f ( a- ) , the second inequality in (4). To prove the first inequality in (4), we may assume that μ * ( I ) < ∞ , otherwise the inequality is obvious. Given then ² > 0, there is a countable collection { J n } of open intervals such that I ⊂ ∞ [ n =1 J n and such that if J n = ( c n ,d n ) for n ∈ N , (5) ∞ X n =1 ˜ ‘ ( J n ) = ∞ X n =1 ( f ( d n )- f ( c n )) ≤ μ * ( I ) + ². We will assume first that the interval I of endpoints a,b is bounded; i.e.,-∞ < a < b < ∞ . Let η > 0 be given; assume also η < ( b- a ) / 2 so that [ a + η,b- η ] makes sense as an interval. Now [ a + η,b- η ] is compact and contained in ∞ [ n =1 J n , thus there is N ∈ N such that [ a + η,b- η ] ⊂ of S N n =1 J n . We will select indices n 1 ,...,n k ∈ { 1 ,...,N } as follows. There is then n , 1 ≤ n ≤ N such that a + η ∈ J n . Of all possible such n ’s, Let n 1 be one with a maximal d n . So n 1 has the properties: 1 ≤ n 1 ≤ N , c n 1 < a + η < d n 1 and if 1 ≤ n ≤ N , c n < a + η < d n , then d n ≤ d n 1 . If b- η < d n 1 2 we are done; k = 1. If not, d n 1 ∈ I and there is n , 1 ≤ n ≤ 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 ≥ a + η so c n 1 < c n < d n 1 < d n . Of all such n ’s, let n 2 be one for which d n is maximal so that now we have 1 ≤ n 2 ≤ N , c n 1 < c n 2 < d n 1 < d n 2 and if n is such that 1 ≤ n ≤ N , c n 1 < c n < d n 1 < d n , then d n ≤ d n 2 . Of course, there is no reason to assume that n 2 > n 1 . If d n 2 > b- η , 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 + η,b- η ] ⊂ k [ j =1 ( c n j ,d n j ) and c n 1 < a + η , d n k > b- η c n 1 < c n 2 < d n 1 < c n 3 < ··· < c n k < d n k- 1 < d n k . Because f is increasing, f ( a + η ) ≥ f ( c n 1 ), f ( b- η ) ≤ f ( d n k ) and thus f ( b- η )- f ( a + η ) ≤ 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 ) + ··· + 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!! Don’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 ) ≤ f ( d n j )- f ( c n j- 1 ) and we see that f ( b- η )- f ( a + η ) ≤ k X j =1 f ( d n j )- f ( c n j- 1 ) = k X j =1 ˜ ‘ ( J n k ) ≤ ∞ X n =1 ˜ ‘ ( J n ) ....
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I set d n = b for all n otherwise set d n = b 1/n Now I ⊂...

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