Can a martingale converge in $L^1$ but not almost surely?

Question

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?

Answer 1

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.

Answer 2

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.