 Solution

For the forward implication, let $(M_n)$ be a martingale and let the stopping time $T$ satisfy $0\le T\le N$ for a deterministic integer $N$. The stopped variable is integrable because it uses only finitely many integrable values. The pathwise identityexpresses it through martingale increments. Since $\{T\ge j\}=\{T>j-1\}\in {\mathcal{F}}_{j-1}$, each summand has expectation zero by conditional expectation. ConsequentlyThis proves the bounded-time case of the optional stopping theorem directly; no assumptions about an unbounded stopping time are being used.