Could someone give me a hint on how to prove this equality?

Question

Please just a hint, I'd prefer to solve this on my own, but I can't see what to do. Here is the question:

Given that $c \notin 2\pi\mathbb{Q}$, $f$ is everywhere continuous and $f$ is $2\pi$ periodic, prove that $$\lim_{n\to\infty} \frac{1}{n}\sum_{j=0}^{n-1} f(x-cj) = \frac{1}{2\pi}\int_0^{2\pi} f(t)dt.$$

Answer 1

$\frac{1}{2\pi}\int_0^{2\pi} f(t)dt$ is the average value of $f(x)$ over one period.

Since $f(x)$ is periodic with period $2\pi$

$f(x-jc) = f(a)$ with $a\in [0,2\pi]$

And since Since $c\ne 2\pi\mathbb Q$

Every $x - jc$ is equivalent to a different value of $a$

And with a large enough value of $n$ this set we expect to be uniformly distributed over $[0,2\pi]$

Answer 2

Hint: It's enough to show

$$\lim_{n\to\infty} \frac{1}{n}\sum_{j=0}^n f(cj) = \frac{1}{2\pi}\int_0^{2\pi} f(t)dt.$$

Prove this first for $f(t) = e^{imt}, m \in \mathbb Z.$ Think about Stone-Weierstrass ...