= Solution
For the converse in the <characterization of a martingale by stopped expectations>, fix $n\geq0$ and $A\in\mathcal F_n$. Define the <bounded stopping time>
$$
T_A=\begin{cases}n+1,&\omega\in A,\\ n,&\omega\notin A.\end{cases}
$$
Indeed, its only nontrivial sublevel <event> is $\{T_A\leq n\}=A^c\in\mathcal F_n$. Applying the assumed stopped-expectation equality to $T_A$ and to the deterministic stopping time $n$ gives
$$
0=\mathbb E(M_{T_A}-M_n)=\mathbb E\bigl[\mathbf1_A(M_{n+1}-M_n)\bigr].
$$
This holds for every $A\in\mathcal F_n$. Since $M_n$ is <adapted> and both variables are integrable, the defining test-event property of <conditional expectation> yields
$$
\boxed{\mathbb E[M_{n+1}\mid\mathcal F_n]=M_n.}
$$
Together with the given adaptation and integrability, this is exactly the <martingale> property.
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