Show that $f$ is Riemann-Integrable Let $f$ be a function defined on $[0,1]$ by: $f(x)=1$ if $x=\frac{1}{n}, n \in \mathbf{N}^*$ and $f(x)=0$ if not. Show that $f$ is Riemann integrable on $[0,1]$.
I know that there is already a post on this function and this problem might not be difficult, but I didn't really find an answer to my question. I'm not really good at proving if a function is integrable, so I would like to understand how it works correctly and become good x)
To show that $f$ is Riemann intergable on $[a,b]$, we consider the upper and lower Darboux sum. By definition we have to show the following: $\exists$ a subdivision $\sigma$: $\overline{S}_{\sigma}(f)\le \underline{S}_{\sigma}(f)+\epsilon \ \forall \epsilon >0$. By densition of irrational numbers, we have that $\underline{S}_{\sigma}(f)=0 \ \forall \sigma$. So, we have to show that $\overline{S}_{\sigma}(f)\le\epsilon$.
I remark that there exists $\epsilon$ such that there is a finite number of discontinuities on the interval $[\epsilon,1]$, so probably I can bound the things in the sum quite "easily". My problem is how to choose my subdibision correctly (and to be sure that it's correct) and how to work correctly on the interval $[0,\epsilon]$ as more we approach to 0, more there are of discontinuities (because of density of irrational numbers). I ask this question not only for this case, but in general how it works. If someone could explain these things in details, I would really appreciate it. Thank you very much in advance!
 A: Hint
For $0 \lt \epsilon \lt 1$, consider the subdivision
$$\begin{aligned}\sigma \equiv x_ 0 &= 0 \lt x_1 = \frac{\epsilon}{2}\\&\lt x_2 = \frac{1}{N} -\frac{\epsilon}{2 2^{N+1}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{N+1}}\\
&\dots\\
&\lt x_{N} = \frac{1}{2} -\frac{\epsilon}{2 2^{2}} \lt x_3 = \frac{1}{N} +\frac{\epsilon}{2 2^{2}}\\
&\lt x_{N+2} = \frac{1}{2^0} -\frac{\epsilon}{2 2^{1}} \lt x_{N+3} = 1
\end{aligned}$$
Where $N$ is the largest integer such that $\frac{1}{N}\gt \epsilon / 2$.
To understand why that works, I suggest that you take an example, i.e. $\epsilon = 1/10$, and make a drawing of the subdivision vs. the function. From there, you can compute the Darboux lower and upper sum to complete the proof. You'll essentially see that Darboux lower sum is equal to zero while Darboux upper sum is less than
$$\frac{\epsilon}{2} + \frac{\epsilon}{2}\left(\frac{1}{2} + \frac{1}{2^2} + \dots + \frac{1}{2^{N+1}}\right) \lt \epsilon.$$
The idea, as you have seen is to isolate zero on one side from the other discontinuities of the function.
