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.

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

2 Answers 2


First answer:

$\alpha$ 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.

  • $\begingroup$ 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 $\endgroup$
    – blomp
    Commented May 15 at 6:57
  • $\begingroup$ So, you haven't seen Weyl's criterion? $\endgroup$
    – User
    Commented May 15 at 8:38
  • $\begingroup$ No, I have not. $\endgroup$
    – blomp
    Commented May 15 at 9:04
  • $\begingroup$ Weird, but fair enough. Have you heard of unique ergodicity? In particular, do you know that $\theta$ is a uniquely ergodic transformation? $\endgroup$
    – User
    Commented May 15 at 10:16
  • $\begingroup$ 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. $\endgroup$
    – blomp
    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 $0<a<b<1$ 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 $y$,

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


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

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


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