# Thread: Riemann Integral

1. ## 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?

2. ## 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.

3. ## Re: Riemann Integral

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 $\displaystyle f?$

4. ## 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.