# Riemann Integral

• Jan 8th 2012, 04:34 AM
younhock
Riemann Integral
Suppose i have $\displaystyle f(x)=0$ for$\displaystyle x \neq \frac{1}{n},n=1,2,3,.....$ and $\displaystyle f(\frac{1}{n})=1$.
how to show that $\displaystyle f$ is Riemann Integrable and the riemann sum is 0 by using definition?
• Jan 8th 2012, 05:23 AM
girdav
Re: Riemann Integral
Fix $\displaystyle \varepsilon>0$, and $\displaystyle n_0$ such that $\displaystyle n_0^{-1}\leq \varepsilon$. Put $\displaystyle s_2(x):=\begin{cases} 1&\mbox{ if } x\leq n_0^{-1}\\0&\mbox{ otherwise}\end{cases}$ and $\displaystyle s_1(x)=0$. Choose a good subdivision and you are done.
• Jan 8th 2012, 08:59 AM
younhock
Re: Riemann Integral
Quote:

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

Why we construct $\displaystyle s_2 (x)$ and $\displaystyle s_1 (x)?$ Can it related to $\displaystyle f?$
• Jan 8th 2012, 09:21 AM
girdav
Re: Riemann Integral
It's just a definition. $\displaystyle s_1$ and $\displaystyle s_2$ are step functions. Using the subdivision $\displaystyle 0<n_0^{-1}<(n_0-1)^{-1}<\ldots< 2^{-1}<1$, you can check the definition of Riemann integrable.