Suppose $f$ is continuous and periodic on the reals with period 1. Prove that if $x\in[0,1]$ is an irrational number, then
$$\lim_{N\rightarrow\infty}\frac{1}{N}\sum_{n=1}^N f(nx)=\int_{0}^1f(t)dt$$
Suggestion: First consider $f(t) = e^{2\pi(ikt)}$ where k is an integer.
I can see that this is a limit of a weighted average, but the suggestion throws me off. I've seen the suggestion in fourier transforms but it's not clicking at the moment. Any help would be welcome.
Using the hint you were given, it is easy to verify that $$ \int_0^1 e^{2\pi i k t} \, \mathrm{d}t = \left\{ \begin{array}{lr} 1 & :\ k = 0 \\ 0 & :\ k \neq 0 \end{array} \right. $$ Similarly, for $k \neq 0$ and irrational $x [0,1]$, using geometric series $$ \begin{align*} \lim_{N \to \infty} \frac{1}{N} \left| \sum_{n=1}^N e^{2 \pi i k n x} \right| &= \lim_{N \to \infty} \frac{1}{N} \left| e^{2 \pi i k x} \frac{e^{2 \pi i k N x} - 1}{e^{2 \pi i k x} - 1} \right| \\ &\leq \lim_{N \to \infty} \frac{1}{N} \frac{2}{|e^{2 \pi i k x} - 1|} \\ &= 0 \end{align*} $$ noting that since $x$ is irrational $e^{2 \pi i k x} \neq 1$ for any $k \neq 0$. On the other hand, for $k = 0$ we have $e^0 = 1$, so $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N 1 = 1. $$ It follows that $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N e^{2 \pi i k n x} = \int_0^1 e^{2 \pi i k t} \, \mathrm{d}t. $$
Now for any continuous function $f$ on $\mathbb{R}$ with period $1$, there is a sequence of complex numbers $\{c_k\}_{-\infty}^\infty$ such that $$ f(t) = \sum_{k = -\infty}^\infty c_k e^{2 \pi i k t}. $$ So with a little justification of the interchange between sum and integral, $$ \begin{align*} \int_0^1 f(t) \, \mathrm{d}t &= \int_0^1 \sum_{k = -\infty}^\infty c_k e^{2 \pi i k t} \, \mathrm{d}t \\ &= \sum_{k = -\infty}^\infty c_k \int_0^1 e^{2 \pi i k t} = c_0. \end{align*} $$ And correspondingly, for irrational $x \in [0,1]$, $$ \begin{align*} \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(n x) &= \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N \sum_{k = -\infty}^\infty c_k e^{2 \pi i k n x} \\ &= \sum_{k = -\infty}^\infty c_k \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N e^{2 \pi i k n x} \\ &= c_0. \end{align*} $$
It follows that $$ \lim_{N \to \infty} \frac{1}{N} \sum_{n = 1}^N f(n x) = \int_0^1 f(t)\, \mathrm{d}t. $$