Irrational rotation - recurrence times 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.
 A: In the range $n \le jp^{1/2} < n+1$, the number of $j$'s contributing to the sum is $|A|p^{-1/2} +o(p^{-1/2})$. So that portion of the sum satisfies
$$
\sum_{j=np^{-1/2}}^{(n+1)p^{-1/2}} p(1-p)^j \chi_I(j)= (|A|p^{1/2}+o(p^{1/2})) (1-p)^{np^{-1/2}} ;
$$
here, I can treat $(1-p)^j$ as a constant when doing the sum, at the expense of a multiplicative error $(1-p)^{p^{-1/2}}$, which can be absorbed into the other error term, from the ergodic theorem.
So the whole sum equals
$$
|A|p^{1/2}(1+o(1)) \sum_{n\ge 0} (1-p)^{np^{-1/2}}   = (1+o(1))|A| \frac{p^{1/2}}{1-(1-p)^{p^{-1/2}}} \to |A| ,
$$
as desired.
A: 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.
