 Solution

The bounded optional stopping theorem gives $\mathbb{E}{X}_{T\wedge n}=\mathbb{E}{X}_0$. Decompose the difference from the terminal value:ConsequentlyThe first term tends to zero by hypothesis. The second tends to zero by the dominated convergence theorem, because $X_T$ is integrable and $T$ is finite with probability one. Thus the stopped martingale converges to $X_T$ with convergence in L1, which permits passage of expectations to the limit:The explicit tail condition supplies exactly the missing control for an unbounded stopping time.