$f$ be a continous function on $\mathbb{R}$ with period $1$ $f$ be a continous function on $\mathbb{R}$ with period $1$, we need to prove that $$\lim_{n\to\infty}\frac{1}{n}\sum_{i=1}^{n} f(n\alpha)=\int_{0}^{1} f(t) dt$$ for every Irrational number $\alpha$
I have no idea how to start this problem, Please help.
 A: I'm trying to solve something similar. Assuming the question was $$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^{N} f(n\alpha) = \int_{0}^{1} f(t) dt$$
for every irrational number $\alpha\in\mathbb{R}$.
Let $f(t)= e^{2\pi i kt}$, then
\begin{align*}
    \sum_{n=1}^Nf(n\alpha) & = \sum_{n=1}^N e^{2\pi i kn\alpha} \\
    & = \sum_{n=1}^N \Big(e^{2\pi i k\alpha}\Big)^n \\
    & = e^{2\pi i k\alpha} \Bigg( \dfrac{e^{2\pi i k\alpha N}-1}{e^{2\pi i k\alpha} -1}\Bigg). 
\end{align*}
Note that since $k\neq 0$ and $\alpha$ is irrational then $e^{2\pi i k\alpha} \neq 1$.  Therefore
$$\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^{N} f(n\alpha) = \lim_{N\to\infty} \frac{1}{N} e^{2\pi i k\alpha} \Bigg( \dfrac{e^{2\pi i k\alpha N}-1}{e^{2\pi i k\alpha} -1}\Bigg)=0 = \int_0^1 e^{2\pi i kt}dt. $$
Now, let $f(t)= \sum_{k=-m}^m c_ke^{2\pi i kt} $ be a trigonometric polynomial. By linearity $\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^{N} f(n\alpha) = \int_{0}^{1} f(t) dt$. Since trigonometric polynomials are dense in the space of continuous functions on the unit circle, we have that $\lim_{N\to\infty} \frac{1}{N} \sum_{n=1}^{N} f(n\alpha) = \int_{0}^{1} f(t) dt$ for every irrational $\alpha \in \mathbb{R}$.
A: Hint : if $\alpha$ irrational, the sequence $(\mathrm e^{2i\pi n \alpha})_{n\in\mathbb N}$ is dense in $S^1$. (As $f$ is $1$-periodic, don't hesitate to consider it of domain $S^1$.)
