TITLE: Trying to prove that $\lim_{N \rightarrow \infty} \frac{1}{N} \Sigma_{n=1}^N f(n\alpha) = \int_0^1 f(x) dx$
QUESTION [5 upvotes]: Let $\alpha$ be an irrational number. Let $f: \mathbb{R} \rightarrow \mathbb{C}$ be a continuous periodic function with period 1. Show that $\lim_{N \rightarrow \infty} \frac{1}{N} \Sigma_{n=1}^N f(n\alpha) = \int_0^1 f(x)\,dx$
The beginning (but probably not the end) of my confusion with this problem has to do with the irrational inputs. Why would that be necessary? Any help is appreciated!

REPLY [2 votes]: You can use the equidistribution theorem to conclude that the sequence $\{n\alpha\},n=1,2,\dots$ is equidistributed on the unit interval $[0,1]$ (note that we need the condition that $\alpha$ is irrational), where $\{n\alpha\}:=n\alpha-\lfloor n\alpha\rfloor$ is the fraction part of $n\alpha$. Then apply the Riemann integral criterion for equidistribution (see here).