= Solution
The required condition is <uniform integrability>, written entirely in terms of the given <probability laws>:
$$
\boxed{\lim_{R\to\infty}\sup_n\int_{\{|x|>R\}}|x|\,\mu_n(dx)=0.}
$$
Together with the given <almost sure convergence>, this implies <L1 convergence> to an integrable $M_\infty$. Passing to the limit in $\mathbb E[M_m\mid\mathcal F_n]=M_n$, using the contraction property of <conditional expectation> in <L1 space>, gives $M_n=\mathbb E[M_\infty\mid\mathcal F_n]$.
For every bounded <stopping time> $S$, the same identity holds with $\mathcal F_S$: split an event of its <stopping-time sigma-algebra> according to $\{S=k\}$ and use the deterministic-time identity. In particular $M_{T\wedge n}=\mathbb E[M_\infty\mid\mathcal F_{T\wedge n}]$. These <conditional expectations> are <uniformly integrable>: for an integrable $Y$ and $Z=\mathbb E[Y\mid\mathcal G]$, the event $A=\{|Z|>K\}$ satisfies $\mathbb P(A)\leq\mathbb E|Y|/K$, and
$$
\mathbb E[|Z|;A]\leq\mathbb E[|Y|;|Y|>R]+R\mathbb E|Y|/K.
$$
First choose $R$, then $K$, uniformly in $\mathcal G$. Define $M_T=M_\infty$ on $\{T=\infty\}$. Now $M_{T\wedge n}\to M_T$ <almost surely> and in <L1 space>, so the bounded-time mean identities imply \b[$\mathbb E M_T=\mathbb E M_0$ even when $T$ can be infinite].
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