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Unformatted text preview: ∞ 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 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 μ * ( I 1 n ) + μ * ( I 2 n ) = f ( b n ) f ( a +) + f ( a +) f ( a n +) = f ( b n ) f ( a n +) as asserted. Now f ( b n ) f ( a n +) ≤ f ( b n ) f ( a n ) ˜ ‘ ( I n ) so that (3) and (8) imply μ * ( A 1 ) + μ * ( A 2 ) ≤ ∞ X n =1 ˜ ‘ ( I n ) , completing the proof of (2). This concludes the treatment of part (b), except for a some comments. Borel sets are measurable. Not every countable or even finite set has measure zero in this model. For example, if f has a jump at c , then from μ * (( c ²,c ]) = f ( c +) f ( c ² +) it is quite easy to deduce that μ * ( { c } ) = f ( c +) f ( c ) ....
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 Spring '11
 Speinklo
 CN, Lebesgue measure, open intervals

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