Suppose we have a simple random walk $S_n$ and we define a stopping time to be $\tau = min\{n: S_n = A \ \text{or} \ S_n = -B\}$. That is, we stop the first time we hit $A$ or $-B$. With this, I have two questions I was hoping someone might yield some insight into:
The first is why $P(\tau < \infty)$ must be either equal to $1$ or $0$ but cannot be anything in between, say, like $\frac{1}{3}$?
The second is how exactly do we interpret the statement that $P(\tau < \infty) = 1$? I know that formally it can be written as $P(\omega \in \Omega : \tau(\omega) < \infty) = 1$. Is it right to say that this means:
$P(\omega \in \Omega : \tau(\omega) < \infty) = 1 \implies $**"Go and search through all elements in $\Omega$. Take each element $\omega$ and plug it into $\tau(\omega)$. If $\tau(\omega) < \infty$, sound the alerts and note down this specific $\omega$. Then, collect all such $\omega$ that cause $\tau(\omega) < \infty$. If we found at least one $\omega$ such that $\tau(\omega) < \infty$, then we say $P(\tau < \infty) = 1$. If our search was a failure and did not yield any $\omega$ where $\tau(\omega) < \infty$, we say $P(\tau < \infty) = 0$."**
What is confusing to me here is if $P(\omega \in \Omega : \tau(\omega) < \infty)$ means 1) taking all $\omega$ that satisfy $\tau(\omega) < \infty$, then dividing by the total number of all in $\Omega$, or 2) taking all $\omega$ that satisfy $\tau(\omega) < \infty$, then noting that the probability of such a set must be $1$ since we found at least one such $\omega$, hence theres a chance that we may stumble across it?
Thanks so much!