I came across this claim by reading some literature on stochastic approximation.

Let $(\Omega, \mathcal{A}, \mathbb{P}$) be a probability space, $(\mathcal{F}_n)$ a filtration on it. Let $(\epsilon_{n})$ be a sequence adapted to $(\mathcal{F}_n)$ such that $\mathbb{E}[\epsilon_{n+1} | \mathcal{F}_n] = 0$ and $$ \sum_{i=0}^{+\infty}\mathbb{E}[||\epsilon_{n}||^2] < +\infty \, . $$

We know that this implies that $M_n = \sum_{i=0}^n \epsilon_{n}$ is a square integrable martingale, and, hence, $M_n$ converges almost surely.

**Question:** Now let $A \in \mathcal{A}$ and assume that we still have the last equation but only $\mathbb{1}_A\mathbb{E}[\epsilon_{n+1} | \mathcal{F}_n] = 0$ is verified. Does the sequence $M_n$ converges almost surely on $A$?.

The authors of Les algorithms stochastiques évitent-ils les pièges claim that on $A$, $M_n$ is some sort of square integrable martingale difference sequence, but since the event $A$ is not necessarily measurable with respect of any of the algebras $\mathcal{F}_n$ I can't see why. My thought process was to look at the trace filtration with respect to $A$, and try to show that $M_n$ is a martingale with respect to this filtration; but i didn't really manage to make it work.

Am I missing something simple? Any help would be greatly appreciated.

**Edit**: It may be important that the event $A$ is $\mathcal{F}_{+\infty}$-measurable.