Let $\alpha$ be an irrational number. Then the sequence $\{\{n\alpha\}\}$ is equidistributed.
I am using the following definition of equidistribution. A sequence $\{a_i\}$ is equidistributed if $\frac1n\sum\limits_{i=1}^n f(a_i) \to \lambda(f):=\int_0^1 f(x)\,dx$ for all continuous $f:[0,1]\to \mathbb{R}_{\ge 0}$.
In the proof I used the fact that if $T_g(x)=(g+x) \pmod 1$, then $\lambda(f\circ T_g)=\lambda(f)$ for all continuous $f:[0,1]\to \mathbb{R}_{\ge 0}$. Hence $\lambda$ is the unique Haar probability measure.
Now suppose for $\theta=e^{2i\pi\alpha}$ I have proved $$\frac1n\sum\limits_{i=1}^n g(\theta^i) \to \hat{\lambda}(g)$$ for all continuous $g:T\to \mathbb{R}_{\ge 0}$ where $\hat{\lambda}$ is the unique haar measure in $T$ and $T$ is the circle group.
Then how to deduce the fact that $n\alpha \pmod 1$ is equidistributed. Note that if we define the map $G: [0,1)\to T$ as $G(x)=e^{2i\pi x}$. Then $f=g\circ G$ does not covers all continuous functions.