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Thread: Riemann Integrable

  1. #1
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    Riemann Integrable

    Prove the following: Let $\displaystyle I $ be a generalized interval, and let $\displaystyle f $ be a function which is uniformly continuous on $\displaystyle I $. Then $\displaystyle f $ is Riemann integrable.

    We know that $\displaystyle f $ is bounded. We need to show that $\displaystyle \int\limits_{-I} f = \int\limits^{-}_{I} f $ (e.g upper and lower integrals are equal). Let $\displaystyle \varepsilon > 0 $. Then there exists a $\displaystyle \delta > 0 $ such that $\displaystyle |f(x)-f(y)| < \varepsilon $ whenever $\displaystyle |x-y| < \delta $. Skipping some steps, we have $\displaystyle f(x) < f(y) + \varepsilon $ for all $\displaystyle x,y \in J_{k} $ where $\displaystyle J_{k} $ are the subintervals of $\displaystyle I $ with length $\displaystyle |J_{k}| = (b-a)/N $. Then $\displaystyle \sup_{x \in J_{k}} f(x) \leq \inf_{y \in J_{k}} f(y) + \varepsilon $. And so $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon |J_{k}| $ or $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon(b-a) $.

    From here, how would I conclude that $\displaystyle \int\limits^{-}_{I} f = \int\limits_{-I} f $?
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  2. #2
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    Quote Originally Posted by particlejohn View Post

    or $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon(b-a) $.
    the RHS of your inequality is incorrect. the correct one is $\displaystyle \varepsilon (b-a),$ because $\displaystyle \varepsilon \sum_{k=1}^N |J_k|=\varepsilon(b-a).$ so $\displaystyle \forall \epsilon > 0: \ 0 \leq \int_{I}^{-} f - \int_{- \ I} f \leq \varepsilon (b-a).$ letting $\displaystyle \varepsilon \rightarrow 0$ gives us: $\displaystyle \int_{I}^{-}f=\int_{- \ I} f. \ \ \ \square$
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  3. #3
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    I meant $\displaystyle \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon|J_{k}|$.
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