We will assume first that the interval i of endpoints

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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
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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 ) . Combining with (5), we proved that f ( b - η ) - f ( a + η ) μ * ( I ) + ² for all η > 0. Letting η 0, f ( b - ) - f ( a +) μ * ( I ) + ². Since ² > 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 μ * ( I ) < . This will happen if f ( x ) has a finite limit for x → -∞ or x → ∞ . Assume first that a = -∞ , b finite. We can let J k = ( - k, b ) where - k < b . Then J k I and because μ * is an outer measure, μ * ( J k ) μ * ( I ); by what we proved f ( b - ) - f ( - k +) f ( J k ) μ * ( I ) and we can let k → ∞ to get μ * ( I ) f ( b - ) - f ( -∞ ). As similar argument works if a R and b = , or if a = -∞ , b = . Claim 1 is established. Claim 2. If I = ( a, b ) is an open interval, then (6) μ * ( I ) = f ( b - ) - f ( a +) . In fact, because of (4), it suffices to prove that μ * ( I ) f ( b - ) - f ( a +). I’ll assume that -∞ < a < b < ; 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
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