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Thread: Analysis - c-translate of an integrable function is integrable

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    Analysis: c-translate of an integrable function is integrable

    I have difficulty proving this simple result: If f\in \mathcal{R}[a,b], then g(y):=f(y-c)\in \mathcal{R}[a+c,b+c] for c\in \Bbb R.
    I want to use Cauchy condition to prove this. (If you know other simple proofs, please tell me.)

    My attempt: Let \varepsilon>0. Since f\in \mathcal{R}[a,b], there exists a partition of [a,b], P=\{a=x_0,x_1,\cdots,x_n=b\} such that |U(f,P)-L(f,P)|<\varepsilon. Pick P_c=\{a+c=x_0+c,x_1+c,\cdots,x_n+c=b+c\}. Then |U(g,P_c)-L(g,P_c)|=|U(f,P_c)-L(f,P_c)|. I have no idea what to do now. How can I compare this with |U(f,P)-L(f,P)|? Those are completely different partition points.

    Can someone also explain the arguments in this alternative proof?
    Last edited by ach4124; Sep 17th 2016 at 12:14 AM.
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