Let $f$ be defined on $\mathbb{Q}\cap [0,1]$ and $$S(f) = \lim_{n \to \infty}\frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})$$Assume that $\{f_n\}$ be a sequence of functions such that for every $n$, $S(f_n)$ exists. If $f_n$ converges uniformly to $f$ on $\mathbb{Q}$, prove that $$\lim_{n \to \infty}S(f_n)$$ exists and $S(f_n) \to S(f)$.
My try: I don't know how to deal with $\mathbb{Q}\cap [0,1]$ because it has zero measure and the summation is the Riemann sum, $$S(f) = \int_{0}^{1}f(x)dx$$ Also is the first $f$, which is defined on $\mathbb{Q}\cap [0,1]$, same as the second $f$? I'm confused about this because I think the second $f$ should be defined on $\mathbb{Q}$ in order to "$f_n$ converges uniformly to $f$ on $\mathbb{Q}$" has meaning.
Edit: I'm taking my first real analysis course and I'm not familiar with Equidistributed sequence and Weyl's ergodic theorem.
The result follows from the inequality
$$\left\vert \frac{1}{n}\sum_{k=1}^{n}f(\frac{k}{n})-\frac{1}{n}\sum_{k=1}^{n}f_m(\frac{k}{n})\right\vert \le \Vert f - f_m\Vert_\infty$$ valid for any $n, m \in \mathbb N$ which implies $$\vert S(f) -S(f_m)\vert \le \Vert f - f_m \Vert _\infty$$ for all $m \in \mathbb N$.
Note: no need to use integration here.