= Solution
For the forward implication, let $(M_n)$ be a <martingale> and let the <stopping time> $T$ satisfy $0\leq T\leq N$ for a deterministic integer $N$. The stopped variable is integrable because it uses only finitely many integrable values. The pathwise identity
$$
M_T=M_0+\sum_{j=1}^{N}(M_j-M_{j-1})\mathbf1_{\{T\geq j\}}
$$
expresses it through <martingale> increments. Since $\{T\geq j\}=\{T>j-1\}\in\mathcal F_{j-1}$, each summand has <expectation> zero by <conditional expectation>. Consequently
$$
\boxed{\mathbb E M_T=\mathbb E M_0.}
$$
This proves the bounded-time case of the <optional stopping theorem> directly; no assumptions about an unbounded stopping time are being used.
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