# Proving that a non-negative submartingale converges if the compensator has a finite limit

I'm stuck on the following problem:

Let $X_n$ be a non-negative submartingale with Doob-Meyer decomposition $X_n = X_0 + M_n + A_n$, where $M_n$ is the martingale part and $A_n$ is a strictly increasing predictable process. Prove that on the event $A_\infty = \lim_{n \to \infty} A_n < \infty$, $X_n$ converges to a finite limit almost surely.

Since $X_n$ is a non-negative martingale we can show by Doob's upcrossing/downcrossing lemma that it converges to some (possibly infinite) value as $n \to \infty$, which forces $M_n$ to converge to some (against possibly infinite) value. I'm sure that an important fact to consider here is that we must have $M_n^- \le A_\infty + X_0$ to preserve non-negativity of $X_n$, but I'm not sure how to infer from this that $M_n$ won't blow up to infinity.

-
$A_n \in F_{n-1}$ so $T = inf (n: A_{n+1} > K)$ is a stopping time, and $K + M_{min(n,T)}$ is a positive martingale, which therefore converges, and its covergence implies that of $M_n$ on $T = \infty$ – mike May 2 '12 at 23:04
Oh right, I see, silly me. Thanks! – Dominic May 3 '12 at 0:50