Prove this function is Riemann integrable (similar to Thomae’s function)

Question

Let $f:[0,1]\rightarrow \mathbb{R}$ be defined by \begin{equation*} f(x) = \begin{cases} 1- (\frac 1q)^2 & \text{if } x\in \mathbb{Q} \text{ and } x=\frac pq \text{ in lowest terms,}\\ 1 & \text{otherwise} \end{cases} \end{equation*} Prove that $f$ is Riemann Integrable.

I assume this has a similar proof to Thomae’s function being Riemann integrable. I can see that $U(f,P) = 1$ for all partitions as the irrationals are dense in $\mathbb{R}$ . I want to show for all $\epsilon > 0$ there is a partition P such that $U(f,P) - L(f,P) < \epsilon$ which means $1 - L(f,P) < \epsilon$. However I am unsure how to do this.

Answer 1

If $t$ is Thomae's function, then $f=1-t^2$. Therefore, since $t$ is Riemann-integrable and $x\mapsto x^2$ is continuous, $f$ is Riemann-integrable.