 Solution

The required condition is uniform integrability, written entirely in terms of the given probability laws: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 $\mathbb{P}(A)\le \mathbb{E}|Y|/K$, andFirst 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 $\mathbb{E}{M}_T=\mathbb{E}{M}_0$ even when $T$ can be infinite.