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](https://www.researchgate.net/publication/285380172_Les_algorithmes_stochastiques_contournent-ils_les_pieges) 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.



Does a sequence that verifies the assumptions of a square integrable martingale on some event need to be convergent on this event?
I think that $e\_m - E(e\_m | \mathcal F\_{m-1}) $ is a square summable, martingale difference sequence, so $\Sigma e\_m - E(e\_m | \mathcal F\_{m-1}) $ converges a.s. If that is true, then $\Sigma e\_m $ converges on any set where $ E(e\_m | \mathcal F\_{m-1})$ is eventually 0.


