Riemann Integral

**younhock** · January 8th 2012, 04:34 AM

Suppose i have $f(x)=0$ for $x \neq \frac{1}{n},n=1,2,3,.....$ and $f(\frac{1}{n})=1$ .
how to show that $f$ is Riemann Integrable and the riemann sum is 0 by using definition?

**girdav** · January 8th 2012, 05:23 AM

Fix $\varepsilon>0$ , and $n_0$ such that $n_0^{-1}\leq \varepsilon$ . Put $s_2(x):=\begin{cases} 1&\mbox{ if } x\leq n_0^{-1}\\0&\mbox{ otherwise}\end{cases}$ and $s_1(x)=0$ . Choose a good subdivision and you are done.

**younhock** · January 8th 2012, 08:59 AM

Originally Posted by girdav

Fix $\varepsilon>0$ , and $n_0$ such that $n_0^{-1}\leq \varepsilon$ . Put $s_2(x):=\begin{cases} 1&\mbox{ if } x\leq n_0^{-1}\\0&\mbox{ otherwise}\end{cases}$ and $s_1(x)=0$ . Choose a good subdivision and you are done.

Why we construct $s_2 (x)$ and $s_1 (x)?$ Can it related to $f?$

**girdav** · January 8th 2012, 09:21 AM

It's just a definition. $s_1$ and $s_2$ are step functions. Using the subdivision $0<n_0^{-1}<(n_0-1)^{-1}<\ldots< 2^{-1}<1$ , you can check the definition of Riemann integrable.

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