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.
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