 Solution

For $A>0$, the first time $T_A=\inf \{n\ge 0:X_n<-A\}$ is a stopping time. On $\{T_A\ge 1\}$, the bounded increments assumption controls the overshoot:Before that time, $X_n\ge -A$. Allowing the possibility $T_A=0$, the processis a nonnegative martingale. Indeed, the stopped martingale in discrete time is a martingale, and the integrable ${\mathcal{F}}_0$-measurable variable $X_0^{-}$ is a constant-in-time martingale. If $T_A=0$, nonnegativity follows from $X_0+X_0^{-}=X_0^{+}$. Its expectations are constant and finite, so the Martingale convergence theorem makes $Y_n$ converge to a finite value almost surely.

On $\{T_A=\infty \}$ this proves finite convergence of $X_n$. Taking the countable union over positive integer $A$, we conclude that $X_n$ converges finitely on the event $\{{\inf }_n{X}_n>-\infty \}$. Apply the same reasoning to the martingale $-X$ to obtain finite convergence on $\{{\sup }_n{X}_n<\infty \}$.

Outside the event of finite convergence, both the infimum and the supremum of the sequence must therefore be infinite in the respective directions. Removing any finite initial segment cannot change this, because each of its values is finite. HenceThis is the bounded-increment martingale convergence-or-oscillation dichotomy. The $X_0^{-}$ term ensures the proof also covers an unbounded integrable initial value.