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In general, almost sure convergence and convergence in $L^1$ are independent modes of convergence in the sense that neither implies the other. For martingales, at least one implication, namely that almost sure convergence implies convergence in $L^1$ remains false. How about the other one? Does convergence of a martingale in $L^1$ imply almost sure convergence?

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    $\begingroup$ It's not an answer, but may be interesting: math.stackexchange.com/questions/2200141/… There is a paper sciencedirect.com/science/article/pii/… , so I think that if the answer is positive, the example will be much more complicated, because even an easier question about "Convergence in probability doesn't imply convergence a.s. for martingales" is hard. $\endgroup$ Dec 12, 2020 at 21:21
  • $\begingroup$ @BotnakovN. Thanks for the references, in fact there is a rather simple answer to my question (see below). $\endgroup$
    – n_flanders
    Dec 12, 2020 at 22:12
  • $\begingroup$ I went down the wrong path trying to build a counterexample and didn't succeed. Now I understand, why) $\endgroup$ Dec 13, 2020 at 8:48

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Yes, for a martingale convergence in $L^1$ implies almost sure convergence. Suppose $(M_n) \rightarrow M_\infty$ in $L^1$ where $(M_n)$ is a martingale. Then $\sup_n \mathbb{E}[|M_n|] < \infty$ so by Doob's martingale convergence theorem there exists $N_\infty \in L^1$ such that $(M_n) \rightarrow N_\infty$ almost surely. But $L^1$ convergence implies a.s. convergence along a subsequence so we can also find a subsequence $(M_{n_k})$ such that $(M_{n_k}) \rightarrow M_\infty$ almost surely, so we must have $N_\infty = M_\infty$ a.s.

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  • $\begingroup$ Great, thanks a lot! $\endgroup$
    – n_flanders
    Dec 12, 2020 at 22:11
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Another, somewhat shorter proof: As $L^1$ convergence is equivalent to uniform integrability and convergence in probability, almost sure convergence follows from the convergence theorem for uniformly integrable martingales.

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