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Combining with(5 we proved that f b ? f a ? ≤ μ i

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Unformatted text preview: 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 { 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 ∈ N (b) lim n →∞ c n = a , lim n →∞ 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 ∈ N . 3 Let J n = ( c n +1 ,c n ), K n = ( d n ,d n +1 ) for n ∈ N ; H = ( c 1 ,d 1 ). Let also ² > 0 be given. Because f is continuous at c n , there is δ n > 0 such that | x- c n | < δ n implies | f ( x )- f ( c n ) | < ²/ 2 n +2 . In particular, f ( c n + δ n / 2)- f ( c n- δ n / 2) = f ( c n + δ n / 2)- f ( c n ) + f ( c n )- f ( c n- δ n / 2) < ² 2 n +1 . Similarly, there is η n > 0 such that | x- d n | < η n implies | f ( x )- f ( d n ) | < ²/ 2 n +2 . In particular, f ( d n + η n / 2)- f ( d n- η n / 2) = f ( d n + η n / 2)- f ( d n ) + f ( d n )- f ( d n- η n / 2) < ² 2 n +1 . Let L n ( c n- δ n / 2 ,c n + δ n / 2), M n ( d n- η n / 2 ,d n + η n / 2) for n ∈ 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 } ∪ { K n } ∪ { L n } ∪ { M n } ∪ { H } is a countable family covering ( a,b ) and if we add all the ˜ ‘ values of these intervals we notice ∞ X n =1 ˜ ‘ ( J n ) = lim n →∞ n X k =1 ( f ( c k )- f ( c k +1 )) = f ( c 1 )- lim n →∞ f ( c n +1 ) = f ( c 1 )- f ( a +) ....
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Combining with(5 we proved that f b f a ≤ μ I ² for...

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