Let $f[0,1]\to [0,1]$ given by $$f(x):=\begin{cases} 0& x\notin Q\\ 0& x=0\\ \dfrac{1}{q+1}& x=\dfrac{q}{p}, x\in \mathbb{Q}\setminus \{0\}, \gcd(p,q)=1 \end{cases}$$ Is $f$ Riemann integrable?
I know Thomae function is Riemann integrable but This fuction is some different. I ve'spent a lot of time but I can't solve this function is Riemann integrable.
It is. Let's proof that it's continuous in any irrational point.
For simplicity, choose any irrational $x > 0$ (negative case is similar).
If $q < n$ and $p > \frac{2n}{\varepsilon}$, then $\frac{q}{p} < \frac{\varepsilon}{2}$. So for any $n$ there are only finitely many points $\frac{q}{p}$ with $q < n$ in $\frac{\varepsilon}{2}$-neighborhood of $x$. None of this points coincide with $x$ (as it's irrational), so some neighborhood of $x$ contains no such points. And $f$ is at most $\frac{1}{n}$ in this neighborhood.
Thus, for any $n$, $x$ has neighborhood $U_n(x)$ s.t. $|f(U_n(x))| < \frac{1}{n}$. It means $\lim\limits_{y \to x} f(y) = 0$. And as $f(x) = 0$ too, it means $f$ is continuous in $x$.
As $f$ is continuous in all irrational points, it's continuous almost everywhere, and thus Riemann integrable by Lebesgue criteria (as it's also bounded).