Question: Fix a universe $\mathcal{U}$. Call a category $\mathcal{U}$-complete if every diagram indexed by a $\mathcal{U}$-small category has a limit, and a functor $\mathcal{U}$-continuous if it preserves $\mathcal{U}$-small limits. Usually, when one fixes a universe, one calls this simply complete and continuous.

Now assume we are given $\mathcal{U}$-complete categories $C,D$ and a $\mathcal{U}$-continuous functor $F : C \to D$. Does then $F$ also preserve limits (which exist in $C$) which are not necessarily $\mathcal{U}$-small?

Answer: 
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.
For the second part of the question: it depends on the mean and variance you have.

Let $X\_t=M\_t+Y\_t$ where $M\_t$ is a martingale and $Y\_t$ is predictable. Let $\mu\_t=\mathbb E[X\_t]=\mathbb E[Y\_t]$ and $\sigma\_t^2=\sum\_{s\le t}\mathbb E[(M\_s-M\_{s-1})^2|M\_1,\ldots,M\_{s-1}]$. By martingale central limit theorem, $M\_t$ is asymptotically normal with mean 0 and variance $\sigma\_t^2$, so a sufficient condition for $X\_t\rightarrow-\infty$ almost surely is $\sigma\_t=o(|\mu\_t|)$ as $t\rightarrow\infty$.

(To be precise, $\sigma\_t$ is a random variable, so you really need to do some fiddling with stopping times as demonstrated in the Wikipedia article on martingale CLT, but the gist of it is that it will converge almost surely if the mean dominates the variance.)