Indicator rational

**manjohn12** · March 3rd 2009, 09:10 PM

Suppose we have the indicator function $I_{\mathbb{Q}}: [0,1] \to \mathbb{R}$ 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:

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

Here $||P||$ is the mesh of the partition. Because we know that $\mathcal{R}(f,P) = \sum_{i=1}^{n} f(x_{i}^{*})(x_{i}-x_{i-1})$ . So we can't bound $|\mathcal{R}(f,P)|$ ? Because either $\mathcal{R}(f,P) = 1$ or $\mathcal{R}(f,P) = 0$ .

**manjohn12** · March 4th 2009, 09:02 AM

Because in general you want to show that $|\mathcal{R}(f,P)| < \epsilon$ and $|\mathcal{R}(f,P)-1| < \epsilon$ .

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