Let $(M, \mathcal{A}, \mu)$ to be a measure space, and $f: M\to M$ measurable function, and $\mu$ a finite invariant measure. If $\varphi: M \to \mathcal{R}$ is integrable then $\lim \limits_{n\to \infty} \frac{\varphi(f^{n}(x))}{n} = 0$, $\mu-a.e. x \in M$
proof:
Fix $\varepsilon > 0$. Define $E_{n} = \{ x\in M | |\varphi(f^{n}(x))| \geq n \varepsilon \}$. So,
$\mu (E_{n}) = \mu(\{ x\in M | |\varphi(x)| \geq n \varepsilon \})= \sum_{k=n}^{\infty} \mu \{ x\in M | k \leq \frac{|\varphi(x)|}{\varepsilon} < k+1 \}$. We obtain that:
$\sum_{n=1}^{\infty} \mu (E_{n}) = \sum_{n=1}^{\infty} \sum_{k=n}^{\infty} \mu \{ x\in M | k \leq \frac{|\varphi(x)|}{\varepsilon} < k+1 \}=$
$=\sum_{k=1}^{\infty} k \mu \{ x\in M | k \leq \frac{|\varphi(x)|}{\varepsilon} < k+1 \} $.
But $\sum_{k=1}^{\infty} k \mu \{ x\in M | k \leq \frac{|\varphi(x)|}{\varepsilon} < k+1 \} \leq \int \frac{|\varphi|}{\varepsilon}d\mu $ (*),
and $\varphi: M \to \mathcal{R}$ is integrable, then $\sum_{n=1}^{\infty} \mu (E_{n}) < \infty$. By Borel-Cantelli, we have done.
This proof can be found in ON THE SUBADDITIVE ERGODIC THEOREM, AVILA & BOCHI, Lemma 2.
How can I show this inequality:
$\sum_{k=1}^{\infty} k \mu \{ x\in M | k \leq \frac{|\varphi(x)|}{\varepsilon} < k+1 \} \leq \int \frac{|\varphi|}{\varepsilon}d\mu $ ?
Thanks
Let $f \geq 0$ be a measurable function, then
$$k \cdot \mu \left( \left\{x; k \leq f(x) < k+1 \right\} \right) = \int 1_{[k,k+1)}(f(x)) \cdot \underbrace{k}_{\leq f(x)} \, \mu(dx) \leq \int 1_{[k,k+1)}(f(x)) \cdot f(x) \, \mu(dx).$$
Summing over $k \geq 1$ we obtain
$$\sum_{k \geq 1} k \mu \left( \left\{x; k \leq f(x) < k+1 \right\} \right) \leq \int f(x) \, \mu(dx).$$
Applying this inequality for $f(x) := |\varphi(x)|/\epsilon$ proves the desired inequality.