# Birkhoff's ergodic theorem holds everywhere

Let $$\mu$$ be the Lebesgue measure on $$(0,1]$$. Let $$\theta(x) = x + \alpha \mod 1$$ for an irrational number $$\alpha$$. Consider the set $$A = (a,b]$$ with $$0 < a < b < 1$$, and write $$S_n(\mathbb1_A)=\frac{1}{n}\sum_{i=0}^{n-1} \mathbb{1}_A \circ \theta^{i}$$ I am asked to prove that $$S_n(\mathbb1_A) \to \mu(A)$$ everywhere.
Using Birkhoff's almost everywhere ergodic theorem, it is easy to prove almost everywhere convergence. I was given the hint, to consider this result on the sets $$A_k = (a + k^{-1}, b - k^{-1}]$$ for all sufficiently large $$k$$. However, even with this I can't see why it is not possible that there is a point $$x \in (0, 1]$$ so that $$S_n(\mathbb{1}_{A_k}) \not \to \mu(A_k)$$ for infinitely many $$k$$, which would contradict convergence.

• I assume $a$ in the rotation should be $\alpha$ and this is not the same as $a$ in the interval, right?
– User
Commented May 14 at 20:03
• Yes, sorry, let me correct. Commented May 14 at 20:08

$$\alpha$$ being irrational implies that the sequence $$(n\alpha)_{n\in \mathbb{N}}$$ is equidistributed mod1. This means that $$\lim_{n\to \infty} \frac{|\{1\leq i \leq n: \{i\alpha\} \in (c,d] \} |}{n}=d-c,$$ for any $$0\leq c\leq d \leq 1$$, where $$\{i\alpha\}=i\alpha-[i\alpha]$$ denotes the fractional part of $$i\alpha$$.

Then simply observe that $$\sum_{i=1}^{n} f(\theta^{i}x)= \sum_{i=1}^{n} \mathbb{1}_{A}(\{x+i\alpha\})=| \{1\leq i \leq n: \{i\alpha\} \in (a-x,c-x] \} |,$$ where $$(a-x,c-x]$$ is considered mod1.

Second answer, building on Birkhoff’s ergodic theorem which is much heavier machinery than, say, Weyl’s criterion, which is the most that is needed for the previous proof.

Step 1: For each rational interval $$(r_1,r_2]$$ with $$0\leq r_1 \leq r_2 \leq 1$$ and $$r_1,r_2 \in \mathbb{Q}$$ it easily follows, as you note, by Birkhoff's PET, that there is a set of full measure of points $$x\in (0,1]$$ such that $$\lim_{N\to \infty} \frac{1}{N} \sum_{n=1}^N \mathbb{1}_{(r_1,r_2]}({\theta^n(x)})=r_2-r_1. \ \ \ (*)$$

Step 2: Extract a set of full measure of points $$x\in (0,1]$$ satisfying $$(*)$$ for all rational intervals simultaneously. This can be done because there are countably many such intervals.

Step 3: A point arising from Step 2 will satisfy $$(*)$$ for all intervals $$(a,b]$$ by a standard approximation argument using the density of the rational numbers.

Step 4: If you have one point $$x$$ satisfying $$(*)$$ for all intervals then this holds for any point $$y\in (0,1]$$. Indeed, just observe that for any interval $$A\subset (0,1]$$, $$\{y+n\alpha\} \in A \iff \{x+n\alpha\} \in A-(y-x),$$ where $$A-(y-x)$$ is another (shifted) interval.

• Thanks, but I am not sure if this is the solution I am looking for. Among others, this presumes that the sequence is in fact equidistributed, which is not part of the course I'm taking, and I doubt that the solution would involve relying on that Commented May 15 at 6:57
• So, you haven't seen Weyl's criterion?
– User
Commented May 15 at 8:38
• No, I have not. Commented May 15 at 9:04
• Weird, but fair enough. Have you heard of unique ergodicity? In particular, do you know that $\theta$ is a uniquely ergodic transformation?
– User
Commented May 15 at 10:16
• No. My course only covered enough ergodic theory so that the strong law of large numbers got within reach. So essentially what I have available is the Maximal Ergodic lemma, Birkhoff's a.e. thm, the von Neumann Lp ergodic thm, apart from general measure theory. Commented May 15 at 10:29

So I think that there is an argument which uses Birkhoff's theorem and some version of the trick.

Fix $$0 and write $$A = [a,b]$$. Let $$\varepsilon > 0$$ such that $$2 \varepsilon < b-a$$.

By Birkhoff's theorem, for almost every $$x$$,

$$\lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a+\varepsilon, b-\varepsilon]} (\theta^k x) = b-a-2\varepsilon \ \ (*).$$

Let $$x \in (0,1]$$. Since almost every point satisfies $$(*)$$, there exists a point $$y$$ such that $$|x-y| < \varepsilon$$ and with this property.

Now, if $$\theta^k y \in [a+\varepsilon, b-\varepsilon]$$, then $$\theta^k x \in [a,b]$$. Hence,

$$\liminf_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a,b]} (\theta^k x) \ge \lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a+\varepsilon, b-\varepsilon]} (\theta^k y) = b-a-2\varepsilon.$$

Now, if $$|x-y| < \varepsilon$$ and $$\theta^k x \in [a, b]$$, then $$\theta^k y \in [a-\varepsilon,b+\varepsilon]$$. By the same reasoning, for a generic $$y$$,

$$\limsup_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a,b]} (\theta^k x) \le \lim_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a-\varepsilon, b+\varepsilon]} (\theta^k y) = b-a+2\varepsilon.$$

Summarizing:

$$b-a-2\varepsilon \le \liminf_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a,b]} (\theta^k x) \le \limsup_{n \to + \infty} \frac{1}{n} \sum_{k=0}^{n-1} \mathbf{1}_{[a,b]} (\theta^k x) \le b-a+2\varepsilon.$$

Since this holds for all $$\varepsilon > 0$$, we get the claim.