# Does a sequence that verifies the assumptions of a square integrable martingale on some event need to be convergent on this event?

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.

• Is $A$ necessarily $\mathcal F_{\infty}$-measurable? Jun 17, 2021 at 23:00
• @NateRiver actually yes it is. Jun 18, 2021 at 0:32

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.