Birkhoff theorem for irrational rotation Lately, I have come across this problem, that I was not sure exactly how to tackle. 

Let $\alpha$ be an irrational number, and let $0 < a < b < 1$. Prove that
  $$\lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(x))$$
  exists for all $x \in [0,1)$.

Here's my attempt of the solution: By Birkhoff ergodic theorem, we know that the limit exists almost everywhere. Hence, the set where it does exist is dense. Choosing $x \in [0,1)$, we can pick $y$ such that the limit at $y$ exists and $|x - y| < \eta$ for $\eta$ as small as we like. Now I’d like to prove that $\frac{1}{n} \sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(x))$ is a Cauchy sequence. I break it up into 3 pieces:
$$
\begin{align*}
\left| \frac{1}{n} \sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(x)) - \frac{1}{m} \sum_{k =0}^{m-1} \chi_{[a,b]}(R_{\alpha}^k(x)) \right| 
&\leq \left| \frac{1}{n} \sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(x)) - \frac{1}{n} \sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(y)) \right| \\
&\qquad + \left| \frac{1}{n}\sum_{k =0}^{n-1} \chi_{[a,b]}(R_{\alpha}^k(y)) - \frac{1}{m} \sum_{k =0}^{m-1} \chi_{[a,b]}(R_{\alpha}^k(y)) \right| \\
&\qquad + \left|\frac{1}{m} \sum_{k =0}^{m-1} \chi_{[a,b]}(R_{\alpha}^k(y)) - \frac{1}{m} \sum_{k =0}^{m-1} \chi_{[a,b]}(R_{\alpha}^k(x))\right|.
\end{align*}
$$
Now, the middle part is pretty easy to estimate since the limit exists at $y$. The problem I'm having is with the other two expressions. Could anyone give me any hints on how to proceed?
 A: This is a result of weyl's criterion for equi-distribution.
This is a result of weyl's criterion for equi-distribution:
for any sequence $(x_k) \subset [0,1]$,
$\lim_n (1/n)\sum^n_{k=1} \chi_{[a,b]}(x_k) =b-a, (\forall0\leq a <b \leq1) \iff  \lim_n (1/n)\sum^n_{k=1} e^{imx_k}=0, (\forall m \in \mathbb{N})$
for a bit more "direct" proof, it is enough to find one single point $x_0 \in [0,1)$ for which
$\lim_n (1/n)\sum^n_{k=1} \chi_{[a,b]}(R^k_{\alpha}(x_0)) =b-a, \space \space \space (\forall0\leq a <b \leq1) \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space \space (1)$
since for any $x \in [0,1)$ and $0\leq a <b \leq1 $ it holds that
$\chi_{[a,b]}(R^k_{\alpha}(x))=\chi_{[a-(x-x_0),b-(x-x_0)] \space (mod \space 1)}(R^k_{\alpha}(x_0))$, 
hence 
$\lim_n (1/n)\sum^n_{k=1} \chi_{[a,b]}(R^k_{\alpha}(x))=\lim_n (1/n)\sum^n_{k=1} \chi_{[a-(x-x_0),b-(x-x_0)] \space (mod \space 1)}(R^k_{\alpha}(x_0))$
(clearly, the set $[a-(x-x_0),b-(x-x_0)]$ is a finite disjoint union of intervals, hence the limit in the left side exists).
we will find such $x_0$:
for any interval $[a,b]$ with $a,b \in \mathbb{Q}$, equation (1) holds for allmost all x in [0,1), hence by countability of $\mathbb{Q}$ there exist some $x_0$ for which equation (1) holds for all  $a,b \in \mathbb{Q}$. (in fact, for almost all x it  holds.)
now fix some $a,b \in [0,1)$, and take some sequence of intervals with sides in $\mathbb{Q}$,   $(I_n)$ and $(J_n)$,  for which $I_n \subset I_{n+1}, J_{n+1} \subset J_{n},\space \cup I_n=(a,b),\space \cap J_n=[a,b]$, then
$(1/n)\sum^n_{k=1} \chi_{I_N}(R^k_{\alpha}(x_0)) \leq (1/n)\sum^n_{k=1} \chi_{[a,b]}(R^k_{\alpha}(x_0)) \leq (1/n)\sum^n_{k=1} \chi_{J_N}(R^k_{\alpha}(x_0))$,
for all $N$, hence taking the limit over $n$ and than taking the limit over $N$, we conclude that equation (1) holds for all $a,b \in [0,1)$, and we are done.
