The complement of a countable set being dense we can

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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
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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 +) . X n =1 ˜ ( K n ) = lim n →∞ n X k =1 ( f ( d k +1 ) - f ( d k )) = lim n →∞ f ( d n +1 ) - f ( d 1 ) = f ( b - ) - f ( d 1 ) . ˜ ( H ) = f ( d 1 ) - f ( c 1 ) . X n =1 ˜ ( L n ) = X n =1 f c n + δ n 2 - f c n - δ n 2 ¶¶ < X n =1 e 2 n +1 = ² 2 . X n =1 ˜ ( M n ) = X n =1 f d n + η n 2 · - f d n - η n 2 ·· < X n =1 e 2 n +1 = ² 2 . Because we have a countable cover of I , the sum of the expressions on the left of the first of all these equali- ties/inequalitites dominates μ * ( I ); adding up we thus get μ * ( I ) < f ( b - ) - f ( a +) + ². Since ² > 0 is arbitrary, μ * ( I ) f ( b - ) - f ( a +) follows, establishing Claim 2. Claim 3. If I = ( a, b ], -∞ ≤ a < b < , then (7) μ * ( I ) = f ( b +) - f ( a +) Assume first that a > . We can proceed exactly as we did in proving the first inequality in (4), with two difference. The countable cover { J n } is defined in the same way in terms of ² , but then we use the fact that [ a + η, b ] is compact to extract a finite subcover of [ a + η, b ]. That is the first difference; that b - η does not appear; it is replaced by b . The second difference is that where we used f ( b - η ) f ( d n k ), now we use that f ( b +) f ( d n k ). Everything else is the same, and we end with μ * ( I ) f ( b +) - f ( a +). For the converse of this inequality, select any c ( a, b ) at which f is continuous. For a small δ consider the open intervals ( a, c ) , ( c - δ, c + d ) , ( c, b + δ ). These intervals cover I ; because μ * is an outer measure and by Claim 2: μ * ( I ) μ * (( a, c ))+ μ * (( c - δ, c + δ ))+ μ * (( c, b + δ )) = f ( c - ) - f ( a +)+ f (( c + δ ) - ) - f (( c - δ )+)+ f (( b + d ) - ) - f ( c +) . Using the continuity of f at c , letting δ 0, we get μ * ( I ) f ( b +) - f ( a +), establishing Claim 3 . With this we can return to where we were on hold, namely at (3). With the notation as defined there, we have (8) μ * ( I 1 n ) + μ * ( I 2 n ) = μ * ( I n ) = f ( b n - ) - f ( a n +) 4
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for all n N . (The last equality is, of course, just claim 2). Let I n = ( a n , b n ). If one of I 1 n , I 2 n is empty, then the other one is I n and (8) is trivially true. Otherwise, a n < a < b n and I 1 n = ( a, b n ), I 2 n = ( a n , a ] and by claims 2, 3
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