Let $T = \inf\{ n : S_n = a \text{ or } S_n = -b\}$ be a stopping time, where $S_n = X_1 + \dots +X_n$ and each $X_n$ is a martingale. I am looking at a proof which shows that $T < \infty$ almost surely. They state:
$$P(T = \infty) \leq P(T > n) \leq P(|S_n| \leq \text{max}\{a,b\})$$
Could someone explain these inequalities for me? The first one holds for all $n$ which I can somewhat see, but I have no idea about the second one. Surely $|S_n| \leq \text{max}\{a,b\}$ doesn't make sense, as if $S_n = a$ or $-b$ then $ T\not > n$?
To have $T=\infty$ you must have $T>n$ for all $n$, so in particular you must have $T>n$ for a fixed $n$. To have $T>n$ for a fixed $n$, $S_k$ can't be $a$ or $-b$ for any $k \leq n$. But this would have to happen at some point if $|S_k|$ ever became larger than both $a$ and $b$. (The process doesn't let you skip over $a$ or $-b$ on the way to the outside of the interval $(-b,a)$.) So $\max_{k \leq n} |S_n|$ has to be less than $\max \{ a,b \}$.
So $\{ T=\infty \} \subseteq \{ T>n \} \subseteq \{ \max_{k \leq n} |S_n| \leq \max \{ a,b \} \}$. Then use monotonicity of $P$ to get the result.
By the way, in the terminology that I learned, a "Markov time" is what you are referring to and a "stopping time" is an a.s. finite Markov time. I'm not sure how universal this terminology is, but I find it a useful distinction.