Let $f(x)=\frac{1}{n}$ if $x=\frac{1}{n}$ and $f(x)=0$ if not. Show that $f$ is Riemann integrable on $[0,1]$. I would like to have a feedback on my proof and to know if it holds.
My attempt is to pass by Darboux upper and lower sums.
To show that $f$ is Riemann integrable, we have to show the following:$\forall \epsilon>0$ $\exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\underline{S}_{\sigma}(f)+\epsilon$.
We remark that $\underline{S}_{\sigma}(f)$ is always equal to $0$, whatever the partition (by density of rational and irrational numbers). So, we need to show that $\forall \epsilon>0 \ \exists$ a partition $\sigma$: $\overline{S}_{\sigma}(f)<\epsilon$. Let $0<\epsilon<1$ and consider the following intervals: $[0,\epsilon/2]$ and $(\epsilon/2,1]$. We remark that on the second interval there is clearly a finite number of discontinuities. Suppose $N$ such that $\frac{1}{N}>\frac{\epsilon}{2}$ (so $\frac{1}{N}\in (\epsilon/2,1]$). We see as well that the number of numbers of the form $1/n$ on the interval $(\epsilon/2,1]$ is at most $\frac{2}{\epsilon}$. So, to "simplify" the sum we can fix the distance between these numbers as $\frac{\epsilon^2}{4}$, which is possible (we can check that $\frac{1}{N-1}-\frac{1}{n}>\frac{\epsilon^2}{4}$). Thus,
$\overline{S}_{\sigma}(f)=M_i(x_{i+1}-x_i)=\frac{1}{n}\cdot\frac{\epsilon}{2}+\frac{2}{\epsilon}\cdot\frac{\epsilon^2}{4}\cdot\frac{1}{n}=\frac{1}{n}\cdot\frac{\epsilon}{2}+\frac{1}{n}\cdot\frac{\epsilon}{2}=\frac{\epsilon}{n}\le \epsilon \ \forall n\in\mathbf{N}^*$
Therefore, $f$ is integrable on $[0,1]$