# Show that $\lim_{n \rightarrow \infty} \frac{|f(T^n(x))|}{n}=0$

Let $T: (X, \mathcal{A},\mu) \rightarrow (X, \mathcal{A},\mu)$ be ergodic wrt a measure $\mu$ on $(X,\mathcal{A})$. Show that for any $f \in L^1(X,\mathcal{A})$ and $\mu$-almost every $x \in X$ we have

$$\lim_{n \rightarrow \infty} \frac{|f(T^n(x))|}{n}=0 .$$

This question screams out for the use of the Birkhoff Ergodic Theorem for ergodic transformations (which is essentially $\displaystyle \lim_{n \rightarrow \infty} \frac{1}{n} \sum_{k=0}^{n-1}f(T^k(x)) = \int f d\mu$).

How can you make this work? Or am I on the wrong track?

Hints: Show successively that:

• $\displaystyle\frac{1}{n} \sum_{k=0}^{n-1}f(T^k(x))\to \int f d\mu$

• $\displaystyle\frac{1}{n+1} \sum_{k=0}^{n}f(T^k(x))\to \int f d\mu$

• $\displaystyle\frac{1}{n} \sum_{k=0}^{n}f(T^k(x))\to \int f d\mu$

• $\displaystyle\frac{f(T^nx)}n=\frac{1}{n} \sum_{k=0}^{n}f(T^k(x))-\frac{1}{n} \sum_{k=0}^{n-1}f(T^k(x))\to$ $______$ $-$ $______$ $=0$

Birkhoff's ergodic theorem does the job. An alternative way is to use the Borel-Cantelli lemma: for a fixed $\varepsilon$, define $A_n :=\{|f\circ T^n| /n\gt\varepsilon \}$. Then using the fact that $T$ preservers the measure $\mu$, we get $\mu(A_n) =\mu\left(\{|f|/\varepsilon\gt n\}\right)$.

If $g$ is an integrable random variable, then we have $$\sum_{ n=1}^\infty \mu\left(\{|g\gt n\}\right) =\sum_{ n=1}^\infty \sum_{j\geqslant n} \mu\left(\{j\leqslant |g|\lt j+1 \}\right) =\sum_{j=1}^\infty j\mu\left(\{j\leqslant |g|\lt j+1 \}\right)\leqslant \sum_{j=1}^\infty \int_{\{j\leqslant |g|\lt j+1 \}} |g|\mathrm d\mu.$$