How to show that this function is integrable

Question

Let $f\colon [0,1] \to \mathbb R $ a function defined by $$ f(x) = \begin{cases} 1, & \text{if $\exists p \in \mathbb N, x = \frac 1p$} \\ 0, & \text{elsewhere} \end{cases} $$ I have to show that $f$ is integrable in $[0,1]$. I tried using Cauchy criterion but no luck so far. any hints are welcome.

Answer 1

hint

Let $\epsilon>0$ be a small real.

$f $ is integrable at $[\frac {\epsilon}{2},1] $ as a stairs function, thus there exist a subdivision $s_1$ of $[\frac {\epsilon}{2},1] $ such that

$U (f,s1)-L (f,s1)<\frac {\epsilon}{2}$

put $s=s_1\cup \{0\} $

we have

$$U (f,s)-L (f,s)\leq $$

$$\frac {\epsilon}{2}+U (f,s_1)-L (f,s_1) <\epsilon$$