For a martingale $M_n$ with bounded increment, then, almost surely :

• either $M_n$ converges to a finite limite.

• either $\limsup M_n=\infty$ and $\liminf M_n=-\infty$.

Sketch of the proof. One can assume that $M_0=0$. Let $T$ be the first time at which the martingale goes below $-A$. Then $M_{n \wedge N}$ is a martingale bounded from below, therefore it converges. Then $M_n$ converges if $N$ is infinite etc.

For your problem consider the Doob decomposition.

For a martingale $M_n$ with bounded increment, then, almost surely :

• either $M_n$ converges to a finite limite.

• or $\limsup M_n=\infty$ and $\liminf M_n=-\infty$.

Sketch of the proof. One can assume that $M_0=0$. Let $T$ be the first time at which the martingale goes below $-A$. Then $M_{n \wedge N}$ is a martingale bounded from below, therefore it converges. Then $M_n$ converges if $N$ is infinite etc.

For your problem consider the Doob decomposition.