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Math Help - Integral proof

  1. #1
    Senior Member Danneedshelp's Avatar
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    Integral proof

    Q: Prove that a bounded function f is integrable on [a,b] if and only if there exists a sequence of partitions (P_{n})_{n=1}^{\infty} satisfying

    lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0.

    Here some of my (incomplete) thoughts:

    (\Leftarrow)
    Let \epsilon>0. Since there exists a sequence of partitions (P_{n})_{n=1}^{\infty} satisfying lim_{n\rightarrow\infty}[U(f,P_{n})-L(f,P_{n})]=0, there exists an N such that, for any n\geq\\N U(f,P_{n})-L(f,P_{n})<\epsilon. Thus, f integrable on [a,b].

    (\Rightarrow)
    For this direction I am not sure how to show there exists a sequence of partitions. Since we are assuming f is integrable, I know that there is a partion P_{\epsilon} of [a,b] such that U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon for any positive \epsilon.
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  2. #2
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    Using a the definition of integral if f on [a,b] then there is a number I such that for any positive \epsilon>0 there is a partition P of [a,b] then \left|\mathcal{S} (P;f)-I\right|<\epsilon .

    Now here is the trick. We can do it twice with \frac{\epsilon}{2}
    \left|\mathcal{U} (P;f)-I\right|<\frac{\epsilon}{2} and \left|\mathcal{L} (P;f)-I\right|<\frac{\epsilon}{2} .
    Now complete it.
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  3. #3
    Senior Member Danneedshelp's Avatar
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    Define U(f)=inf\{U(f,P):P\in\\Q\} and L(f)=sup\{L(f,P):P\in\\Q\}, where Q is the set of all possible partitions.

    We are given f is integrable on [a,b], so there exists a partition P_{\epsilon} of [a,b] such that U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon. Since f is integrable U(f,P_{\epsilon})-L(f,P_{\epsilon})=(U(f,P_{\epsilon})-U(f))+(L(f)-L(f,P_{\epsilon}).

    Let I=U(f)=L(f). Then, |U(f,P_{\epsilon})-I|<\frac{\epsilon}{2} and |L(f,P_{\epsilon})-I|<\frac{\epsilon}{2}.

    So, we have a sequence of paritions (P_{n})_{n=1}^{\infty} such that, there exists an N>0 for which any n\geq\\N implies the upper and lower sums converge to the same value.

    I am not sure how to introduce the limiting variable, n. I understand what is going on, but I am having trouble formalizing the idea of a sequence of partitions.
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  4. #4
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    You cannot do the above for \epsilon=\frac{1}{n}?
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  5. #5
    Senior Member Danneedshelp's Avatar
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    Sorry to bring up an old thread, but I have been away for a while and have not yeat figured out the proof.

    I am unsure how to prove the existance of the sequence of partitions.

    f is integrable on [a,b], so for any \epsilon>0 we have that U(f,P_{\epsilon})-L(f,P_{\epsilon})<\epsilon. Since \epsilon>0 is arbitrary it is certianly true that U(f,P_{\epsilon})-L(f,P_{\epsilon}) can be made less than \epsilon=\frac{1}{n} for each n\in\mathbb{N}. Therefore, there exists a sequenct of paritions P_{n}=\frac{b-a}{n} such that n>N, where N\in\mathbb{N}, implies (P_{n})\rightarrow\\P_{\epsilon}.

    Notice,
    U(f,P_{\epsilon})-L(f,P_{\epsilon})=U(f,P_{\epsilon})-U(f)+L(f)-L(f,P_{\epsilon}) since f is integrable. Thus, U(f,P_{\epsilon})-U(f) can be made less than half the prescribled \epsilon as can L(f)-L(f,P_{\epsilon}). Since U(f)=I=L(f), we have that the lower and upper sums converge to the same limit. Thus,

    U(f,P_{n})-L(f,P_{n})=0 as n\rightarrow\infty.
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