# How to show a hitting time is finite almost surely?

A one-dimensional symmetric simple random walk starts at $S_0 = 1$. How to show with probability one it passes $x = 0$ (or I guess equivalently, the stopping time of hitting $x = 0$ at the first time is finite almost surely)?

It's just a new-learner's doubt, not a homework problem, I would appreciate any advise. Thanks in advance!

We have \begin{align} \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right) &= \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\mid X_1=-1\}\right)\mathbb P(X_1=-1)\\ &\quad+ \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\mid X_1=1\}\right)\mathbb P(X_1=1)\\ &= \mathbb P(X_1=-1) + \mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant -1\}\right)\mathbb P(X_1=1)\\ &= \frac12 + \frac12\mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right)^2, \end{align} from which it follows that $$\mathbb P\left( \bigcup_{n=1}^\infty \{S_n\leqslant 0\}\right)=1.$$