Birkhoff ergodic theorem

Question

I am reading a paper, and am struggling to understand the following. I'll cut it down to the bare basics.

Suppose we have an ergodic measure preserving system $(X,T,\mu)$. Suppose that $f\in\mathcal{L}^1(X)$, where $f$ takes values in $\mathbb{N}$ (not sure if this is important, I will just state it). The paper says that by the ergodic theorem, $f\circ T^n=o (n)$ a.e. I can't seem to see how this holds.

We know by the ergodic theorem that for a.e. $x$, $\frac{1}{n}\sum\limits_{k=0}^{n-1}f(T^kx)\to \int_X fd\mu$ a.e.

I can't see how this implies $\frac{f\circ T^n}{n}\to 0$ a.e.

At first I thought it could follow from the null-sequence test, but I am not sure if this can be applied since we have the $\frac{1}{n}$ term lying around. Any help will be appreciated!

Thanks.

Answer 1

Hint: If we set

$$a_n := \frac{1}{n} \sum_{k=0}^{n-1} f(T^k x)$$

then $$\frac{1}{n} f(T^n x) = \frac{n+1}{n} a_{n+1} - a_{n} = (a_{n+1}-a_{n}) + \left(\frac{n+1}{n}-1 \right) a_{n+1}$$

Now use the convergence of the sequence $a_n$ to show that the right-hand side converges to $0$ as $n \to \infty$.

Answer 2

Assume it is not, since $f$ take positive values you can take $c>0$ such that for infinitely many $n$ we have $f\circ T^n >cn$, hence $$\frac{1}{N} \sum_{n=0}^N f\circ T^n\geq\frac{1}{N}\sum_{N<n\in I} f\circ T^n\geq\sum_{N<n\in I} c\rightarrow\infty$$ where $I$ is the set of those $n$ for which $f\circ T^n>cn$ by contradiction.