Let $(\Omega,\mathcal A,\operatorname P)$ be a probability space, $\tau$ be a measurable map on $(\Omega,\mathcal A)$ with $\operatorname P\circ\:\tau^{-1}=\operatorname P$, $X:\Omega\to[-\infty,\infty)$ be $\mathcal A$-measurable with $\operatorname E[X^+]<\infty$ and $$\Omega_1:=\left\{\limsup_{n\to\infty}\frac{X\circ\tau^n}n\le0\right\}.$$ It's easy to see that $$\sum_{n\in\mathbb N}\operatorname P\left[\frac{X\circ\tau^{n-1}}n>\varepsilon\right]\le\frac{\operatorname E[X^+]}\varepsilon<\infty\tag1\;\;\;\text{for all }\varepsilon>0.$$
Why can we conclude that $\operatorname P[\Omega_1^c]=0$?
This is obviously an application of the Borel-Cantelli lemma, shouldn't it only yield $\operatorname P\left[\limsup_{n\to\infty}\left\{\frac{X\circ\tau^{n-1}}n>\varepsilon\right\}\right]=0$ for all $\varepsilon>0$? Why does this imply the desired claim?
The application of the Borel-Cantelli lemma shows that for all $\varepsilon\gt 0$, there exists a set $\Omega(\varepsilon)$ of probability one such that for all $\omega\in\Omega(\varepsilon)$, there exists $N=N(\omega,\varepsilon)$ for which $n\geqslant N$ implies $\frac{X\circ\varphi^{n-1}(\omega)}n\leqslant\varepsilon$. Therefore, for all $\omega\in \Omega(\varepsilon)$, $\limsup_{n\to+\infty}\frac{X\circ\varphi^{n-1}(\omega)}n\leqslant\varepsilon$. Now, let $\Omega':= \bigcap_{k\geqslant 1}\Omega(1/k)$: this set has probability one and for all $\omega\in\Omega'$, $\limsup_{n\to+\infty}\frac{X\circ\varphi^{n-1}(\omega)}n\leqslant0$. This show that $\Omega'\subset\Omega_1$; since $\Omega'$ has probability one, so has $\Omega_1$.