Bounded increments ($|X_{n+1}-X_n|\le C(n\ge 0)$)

A discrete-time stochastic process has bounded increments if a single deterministic $C<\infty$ satisfies $|X_{n+1}-X_n|\le C$ almost surely for every $n$. The constant is uniform in time and sample outcome; bounding each increment by a different constant does not suffice. This property applies to differences, so it does not require the process itself to stay bounded.

A discrete-time martingale with a common deterministic bound on all increments either converges finitely or has $\mathrm{liminf}{X}_n=-\infty$ and $\mathrm{limsup}{X}_n=+\infty$, almost surely. Stopping on first crossing below $-A$ gives a lower bound $-A-M$ on the overshoot. Adding $A+M+X_0^{-}$ produces a nonnegative martingale, so the martingale convergence theorem forces finite convergence on the event of never crossing. Apply the same argument to $-X$.

Let $S_n={\sum }_{j=1}^n{X}_j$, where the increments are nonconstant independent and identically distributed random variables with zero expected value and $|X_j|\le c$. Then the random walk oscillates between arbitrarily large positive and negative levels almost surely. For $T_a=\inf \{n:S_n\ge a\}$, the stopped martingale $a+c-S_{n\wedge T_a}$ is nonnegative. The martingale convergence theorem implies that $S_n$ converges finitely on $\{T_a=\infty \}$. However, some $\delta >0$ has $\mathbb{P}(|X_1|>\delta )>0$, and the Borel-Cantelli lemmas imply $|X_n|>\delta$ infinitely often. Hence finite convergence is impossible. Apply this to every positive integer $a$ and to $-S_n$.

The bounded increments also imply infinitely many visits to $[-c/2,c/2]$: a path cannot cross from above $c/2$ to below $-c/2$ without entering this interval. Identical distribution cannot simply be dropped. For independent symmetric signs ${\epsilon }_j$, the increments $X_j=c{4}^{-(j-1)}{\epsilon }_j$ satisfy $|S_n|\ge 2c/3$ for every $n$, even though they are nonzero, independent, symmetric and uniformly bounded.