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Thread: Indicator rational

  1. #1
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    Indicator rational

    Suppose we have the indicator function $\displaystyle I_{\mathbb{Q}}: [0,1] \to \mathbb{R} $ of rationals. To show that it is not Riemann integrable (without looking at upper and lower sums) I said that the Riemann sum can take on three possible general values:

    $\displaystyle \mathcal{R}(f,P) = \begin{cases} 0 \ \ \ \ \ \ \ \ \ \ \ \ \text{if} \ x_{i}^{*} \notin \mathbb{Q} \ \text{for any} \ i \\ x_{i}-x_{i-1} \ \ \text{if} \ x_{i}^{*} \in \mathbb{Q} \ \text{for exactly one} \ i \\ \geq ||P||, \ \ \ \text{if} \ x_{i}^{*} \in \mathbb{Q} \ \text{for some} \ i \end{cases} $


    Here $\displaystyle ||P|| $ is the mesh of the partition. Because we know that $\displaystyle \mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1}) $. So we can't bound $\displaystyle |\mathcal{R}(f,P)| $? Because either $\displaystyle \mathcal{R}(f,P) = 1 $ or $\displaystyle \mathcal{R}(f,P) = 0 $.
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  2. #2
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    Because in general you want to show that $\displaystyle |\mathcal{R}(f,P)| < \epsilon $ and $\displaystyle |\mathcal{R}(f,P)-1| < \epsilon $.
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