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Math Help - Proof of Riemann integrable function

  1. #1
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    Proof of Riemann integrable function

    Let f : [a,b] \rightarrow \mathbb{R} be increasing on the interval [a,b] (that is, f(x) \leq f(y) whenever x < y, so in other words monotone increasing). Prove f is integrable on [a,b].

    By integrable, we're referring to Riemann integrable.

    Any way, uhh help? Would I go about proving this saying that if we can prove its continuous, then its integrable? That's some theorem. But isn't it obvious it's continuous, as it says its increasing
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  2. #2
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    Quote Originally Posted by Auxiliary View Post
    Let f : [a,b] \rightarrow \mathbb{R} be increasing on the interval [a,b] (that is, f(x) \leq f(y) whenever x < y, so in other words monotone increasing). Prove f is integrable on [a,b].
    Note f(a)\leq f(x)\leq f(b) thus it is a bounded function.
    We need to show for any \epsilon > 0 we can choose a partition P of [a,b] such that U(f,P)-L(f,P). Choose a partition P = \{x_0,x_1,...,x_n\} such that \delta = \max_{1\leq k\leq n} (x_k - x_{k-1}) < \frac{\epsilon}{f(b)-f(a)}. Then U(f,P) - L(f,P) = \sum_{k=1}^n [\sup\{ f(x): [x_{k-1},x_k]\} - \inf\{ f(x): [x_{k-1},x_k]\}] (x_k-x_{k-1}) \leq \sum_{k=1}^n [f(x_k)-f(x_{k-1})]\delta  = (f(b)-f(a))\delta < \epsilon
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