Task:
Let $f_n: [0, 1] \rightarrow \Bbb R$, $n \in \Bbb N$, be a sequence of functions with
$f_n(x) := 1$ for $n!x \in \Bbb Z$, $0$ otherwise.
Show that $f_n$ is Riemann integrable.
Question:
We didn't do anything with Lebesgue, so I think this might work by induction, but I don't see where to start.
We need to prove that the number of points, where $f_n(x)=1$ is finite.
$f_n(x)=1$, when $n!x\in\mathbb{N}$.
$n!x\in\mathbb{N}$, when $x\in\mathbb{Q}$ and $x=\frac{a}{b}: a, b \in\mathbb{N}, n!\equiv 0\pmod b$.
It is obvious that there is only finite number of $\frac{a}{b}\in[0;1]$, which satisfy these conditions. So, $f_n(x)$ is continious at all but finitely many points, thus, integrable.
