# Show that the function is Riemann integrable

Let $N>0$ and $f:(0,1)\to [0,1]$ be denoted by $f(x)=\frac{1}{i}$ is $x=\frac{1}{i}$ for some integer $i<N$, and $f(x)=0$ elsewhere. Show that $f$ is Riemann integrable.

My approach: So the points where $f(x)\neq 0$ are {$\frac{1}{2}$, $\frac{1}{3}$$\ldots}. Now say we take first n of this points. We define intervals {x_1,x_2},{x_3,x_4}\ldots{x_{2n-1},x_{2n}} of equal length to enclose all the points where f(x)\neq 0. Length of each interval be \Delta x= \frac{k}{n} (for some k\gt0), as we have observed distance between two consecutive point decreases as we increase n. Say at each interval m_i and M_i respectively denotes infimum and supremum of f(x) in that interval. Now the lower sum s=\sum_{i}{m_i\Delta x} =0. The upper sum S=\sum_{i=1}^{n}{M_i\Delta x} = \Delta x(\frac{1}{2}+\frac{1}{3}+\cdots$$\frac{1}{n+1}$)$\le$n$\Delta x$$= k$. (for all $n$)

So as we keep choosing smaller value of $k$, the $S$ shrinks and as $S \ge 0$, we conclude inf $S$=$0$.

as sup $s$=inf $S$=$0$, we conclude the given function is Riemann integrable.

I want to know whether my is approach right. I am just a beginner in Riemann Integration so just want it to be checked and corrected.

• That's all right... – Mostafa Ayaz Jan 4 '18 at 10:25

## 1 Answer

Discontinuities of the function happen in ${1,{1\over 2},{1\over 3},...,{1\over N-1}}$ and they are zero-set. So the function is Reimann integrable