I consider the irrational rotation
$T\_\alpha(x) = x + \alpha \text{ mod } 1$ for given irrational $\alpha \in [0,1]$. For a given open interval $A \subset [0,1]$ with length $|A|>0$, I consider the recurrence times $I = \{n\in \mathbb{N}: T^n(0) \in A \}$. I want to show that
$\sum\_{i \in I} p\cdot(1-p)^i \to |A|$ as $p \to 0$.


My very informal motivation for this is that the above sum should be equal to $\sum\_{n\in \mathbb{N}} p \cdot (1-p)^{n\cdot\frac{1}{|A|}}$ "give or take" "a few" $(1-p)$-factors (which tend to $1$ as $p \to 0$), and the latter sum can be shown to converge to $|A|$ as $p \to 0$.


I have obtained a somewhat similar (but obviously not identical) result for the case of rational $\alpha$ (happy to add details, but I'm not sure if it's useful), and tried to derive the above using a rational series converging to $\alpha$, but wasn't successful.


Unfortunately, I have virtually no background in ergodic theory, numbers theory and similar subjects which apparently treat irrational rotations, and thus don't quite know where to start.



Irrational rotation - recurrence times
You can recover this result in two steps:


* a variation on Birkhoff's ergodic theorem yields the Cesàro convergence of the sums;
* Cesàro convergence implies Abel convergence.


First step: $T\_\alpha$ preserves the Lebesgue measure and is uniquely ergodic. Hence, for all $f \in \mathcal{C} (\mathbb{S}\_1, \mathbb{R})$,


$$\lim\_{n \to + \infty} \frac{1}{n} \sum\_{k=0}^{n-1} f \circ T\_\alpha^k = \int\_{\mathbb{S}\_1} f(x) \ dx,$$


where the limit is uniform. By approximating $\mathbf{1}\_A$ from above and from below by continuous functions, we get that:


$$\lim\_{n \to + \infty} \frac{1}{n} \sum\_{k=0}^{n-1} \mathbf{1}\_A \circ T\_\alpha^k = |A|,$$


where the limit is again uniform. I had to use uniform ergodicity to be able to tell something about a specific starting point ($0$) instead of a generic one.


Second step: let $p \in (0,1)$. Let $(a\_k)\_{k \geq 0}$ be a bounded real sequence. Then:


$$\sum\_{n=0}^{+ \infty} (1-p)p^n a\_n 
= \sum\_{n=1}^{+ \infty} \frac{n(1-p)^2 p^n}{p} \frac{1}{n} \sum\_{k=0}^{n-1} a\_k$$


Now, putting $b\_n^p := n(1-p)^2p^{n-1}$, for all $p \in (0,1)$, the sequence $(b\_n^p)\_{n \geq 1}$ defines a probability measure on the positive integers. In addition, for all $n$, $b\_n^p$ converges to $0$ as $p$ goes to $1$. Hence, 


$$\lim\_{n \to + \infty} \frac{1}{n} \sum\_{k=0}^{n-1} a\_k = \ell \quad \implies \quad \lim\_{p \to 1} \sum\_{n=0}^{+ \infty} (1-p)p^n a\_n = \ell.$$


Finally, take $a\_n := \mathbf{1}\_A \circ T\_\alpha^n (0)$, and apply the first step.


