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Math Help - Riemann Integrable

  1. #1
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    Riemann Integrable

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

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

    From here, how would I conclude that  \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  \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 \varepsilon (b-a), because \varepsilon \sum_{k=1}^N |J_k|=\varepsilon(b-a). so \forall \epsilon > 0: \ 0 \leq \int_{I}^{-} f - \int_{- \ I} f \leq \varepsilon (b-a). letting \varepsilon \rightarrow 0 gives us: \int_{I}^{-}f=\int_{- \ I} f. \ \ \ \square
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  3. #3
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    I meant  \int\limits^{-}_{I} f - \int\limits_{-I} f \leq \sum_{k=1}^{N} \varepsilon|J_{k}|.
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