Let $(X_{k})_{k}$ be IID random variables on $(\Omega, \mathcal{F}, P)$ where $0 < \mathbb E[|X_{1}|]<\infty$ and $S_{n}:=\sum_{k=1}^{n}X_{k}$
Show that $(S_{n})_{n}$ leaves the compact set about $0$ almost surely.
I am given the hint:
For a fixed $ l > 0$ define $\tau(\omega):=\inf \{n \in \mathbb N_{0}: S_{n}(\omega)\notin [-l,l]\}, \omega \in \Omega$ and use Borel-Cantelli Lemma to show that $\tau < \infty$ almost surely.
I am not sure how I can use Borel-Cantelli on the random variable $\tau$ described above. I believe I have to construct a sequence of random variables $(\tau_{m})_{m}$ but I am unsure how to define them.
One idea would be
$\tau_{m}(\omega):=\inf\{m\in\{1,...,n\}:S_{m}(\omega)\in[-l,l]\}$ and let $\epsilon > 0$.
and then $\sum_{m \in \mathbb N} P(|\tau_{m}|>\epsilon)$ but honestly I do not know whether I am on the right path on how to even evaluate $\sum_{m \in \mathbb N} P(|\tau_{m}|>\epsilon)$
Let us pick fixed number of steps $n$ and number $\epsilon > 0$ such that $n \epsilon > 2l$ and $P(X_i > \epsilon) > \beta^{1/n}$ with some constant $\beta$. Then event $\{S_n > n\epsilon = 2l \}$ also has probability greater than zero.
From independence, it means that for any starting $n_0, \lvert S_{n_0} \rvert < 2l$, $X_{n_0} + X_{n_0+1} + ... + X_{n_0+n} > 2l$
Then events $A_k = \{X_i > \epsilon\ \text{for i from nk to n(k+1)}\}$ are independent. If we manage to show that $P(\bigcup A_k) = 1$, then we are done, since anytime $A_k$ happens, we make a jump of length $2l$. But $P(A_k)$ is fixed number, $P(A_k) \geq \beta$, so it follows immediately that $\sum P(A_k)$ diverges. Then we apply second BC lemma and we are done.
In a qualitative sense, this means that there exist infinite number of subsequences $\{X_i\}_{nk}^{n(k+1)}$, such that $S_{n(k+1)} - S_{nk} > 2l$. This means that, whatever happens earlier (and we cannot assume anything about it, since we need independence for BC lemma), we can always make a jump of length $2l$, and subsequently, leave desired compact set. This means that stopping time is a.s. finite.