= Solution
Fix deterministic $0\leq s<t$ and $A\in\mathcal F_s$. The <pasting ordered stopping times> argument shows that $U=s\mathbf1_A+t\mathbf1_{A^c}$ is a <bounded stopping time>. The given stopped-expectation property, applied to $U$ and to the deterministic <stopping time> $t$, yields
$$
0=\mathbb E X_U-\mathbb E X_t=\mathbb E[(X_s-X_t)\mathbf1_A].
$$
This holds for every $A\in\mathcal F_s$. Since $X_s$ is $\mathcal F_s$-<measurable> and both time values are integrable, it is exactly the defining test for <conditional expectation>:
$$
\boxed{\mathbb E[X_t\mid\mathcal F_s]=X_s.}
$$
Hence \b[$X$ is a <martingale>]. This proof of the <characterization of a martingale by bounded continuous-time stopped expectations> uses only deterministic two-time pastings; the <càdlàg> assumption and the <usual conditions for a filtration> are stronger than needed for this implication.
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