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?