Consider the Thomae function $$ f(x) = \begin{cases} \frac1q, & \text{if }x\text{ is a rational number}\frac pq\text{, in lowest terms, with }q>0, \\ 0, & \text{if }x\text{ is irrational.} \end{cases} $$
How do I prove the following (where U stands for upper Riemann sum)?
For each positive number $\epsilon$, there is a partition P of $[0,1]$ such that $$U(f,P)<\epsilon.$$
(This would then imply that $\int_0^1f=0$.)
