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The definition of Thomae function is $$g(x) = \begin{cases} \dfrac{1}{q}, & \text{if $x=\dfrac{p}{q}$} \\[2ex] 0, & \text{if $x$ is irrational} \end{cases}$$ where $p\in \mathbb{Z}, q\in \mathbb{Z}^+$ and $\gcd(p,q)=1.$

I know how to prove $g$ is Riemann integrable on [−1, 1], but how to prove this integral $\int_{-1}^1 g(x)dx$ is 0?

My attempt: for any partition, the smallest Riemann sum is 0, simply because we can always choose representative point to be irrational. But how to complete the proof $\int_{-1}^1 g(x)dx=0$?

Mariana
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  • Your attempt basically finishes the proof. Since the Riemann integral is monotonic, the integral must be non-negative. And you've shown that the upper integral of $g$ is $0$. – Rushabh Mehta Jul 03 '21 at 16:41
  • @DonThousand I don't know that "the Riemann integral is monotonic". I think what I show in the above is the lower integral of g is 0. – Mariana Jul 03 '21 at 16:47
  • @Mariana That's an equally valid approach, although you haven't done that explicitly. What you need to show (explicitly) is that both the upper and lower integrals are $0$. I just used monotonicity since it was easier for me. – Rushabh Mehta Jul 03 '21 at 16:48
  • @DonThousand yes, but how to show the upper integral is 0? – Mariana Jul 03 '21 at 16:51
  • Let $S_n=\left{\frac pq:\Big\vert:q< n\wedge|p|\leq|q|\right}$ for $n\in\mathbb N$. Let $P_n$ be a partition such that each $s\in S_n$ is in an interval of length $\leq\frac1{n\cdot|S_n|}$ (you can think about how to construct this). Thus, the upper sum of this partition is bounded from above by $\frac2n+\frac1n=\frac3n$. Since $n$ was arbitrary, we are done. – Rushabh Mehta Jul 03 '21 at 17:10

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