Consider the answer given here to the question: Integrability of Thomae's Function on $[0,1]$.
What I am looking to know is the following: If $f(x)=\begin{cases}\frac{1}{q} &: x = \frac{p}{q}\in\mathbb{Q}\cap[0,1] \\\ 0 &: x\not\in\mathbb{Q}\cap[0,1]\end{cases}$, $\epsilon > 0$ is given and $N\in\mathbb{N}:\frac{1}{N+1} < \frac{\epsilon}{2}$, $B_N:=\{\frac{k}{l}\mid 1\leq l\leq N, 1\leq k \leq l\}$, $m=\#(B_N)$, $P = (x_0,x_1,\dots,x_n)$ is a partition with $n > m$ and $\max_{j=1,2,\dots,n}\{|x_{j-1} - x_j|\} < \frac{\epsilon}{4m}$ then why are there at most $2m$ subintervals $[x_{j-1}, x_{j}]$ such that $[x_{j-1}, x_{j}]\cap B_N \neq\varnothing$?
I get that the $2m$ can occur from each of element of $B_N$ occurring at the intersection of consequent intervals $[x_{j-2},x_{j-1}],[x_{j-1},x_j]$. But how do we know that the intervals are short enough? That is why
$$\min\{|x - y|\mid x,y\in B_N, x\neq y\} \geq \frac{\epsilon}{4m}$$
Thanks!
