= Solution
Use the standard definition of a <local martingale>, including an integrable initial value, and let $T_n\uparrow\infty$ be a <localizing sequence>. If $X\geq0$, the <Fatou lemma> gives $\mathbb EX_t\leq\liminf_n\mathbb EX_{t\wedge T_n}=\mathbb EX_0<\infty$. For $s\leq t$, the <Conditional Fatou lemma> gives
$$
\mathbb E(X_t\mid\mathcal F_s)\leq\liminf_n\mathbb E(X_{t\wedge T_n}\mid\mathcal F_s)=\liminf_nX_{s\wedge T_n}=X_s.
$$
For clarity, the <Conditional Fatou lemma> itself follows by setting $U_m=\inf_{n\geq m}U_n$ for nonnegative $U_n$, applying the conditional version of the <monotone convergence theorem> to $U_m\uparrow\liminf U_n$, and observing $\mathbb E(U_m\mid\mathcal F_s)\leq\inf_{n\geq m}\mathbb E(U_n\mid\mathcal F_s)$. Hence \b[a nonnegative <local martingale> is a <supermartingale>.]
If $|X_t|\leq C$ simultaneously for all $t$, with a deterministic $C<\infty$, the <dominated convergence theorem> applied to the stopped <martingale> identity gives, for each $A\in\mathcal F_s$,
$$
\mathbb E(\mathbf1_AX_t)=\lim_n\mathbb E(\mathbf1_AX_{t\wedge T_n})=\lim_n\mathbb E(\mathbf1_AX_{s\wedge T_n})=\mathbb E(\mathbf1_AX_s).
$$
This is exactly the <conditional expectation> identity, so \b[a bounded <local martingale> is a true <martingale>.] The same proof works for deterministic bounds on every finite time interval; merely having a finite random pathwise bound supplies no dominating integrable <random variable> and is a different convention.
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