 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 $X\ge 0$, the Fatou lemma gives $\mathbb{E}{X}_t\le {\mathrm{liminf}}_n\mathbb{E}{X}_{t\wedge T_n}=\mathbb{E}{X}_0<\infty$. For $s\le t$, the Conditional Fatou lemma givesFor clarity, the Conditional Fatou lemma itself follows by setting $U_m={\inf }_{n\ge m}{U}_n$ for nonnegative $U_n$, applying the conditional version of the monotone convergence theorem to $U_m\uparrow \mathrm{liminf}{U}_n$, and observing $\mathbb{E}(U_m\mid {\mathcal{F}}_s)\le {\inf }_{n\ge m}\mathbb{E}(U_n\mid {\mathcal{F}}_s)$. Hence a nonnegative local martingale is a supermartingale.

If $|X_t|\le C$ simultaneously for all $t$, with a deterministic $C<\infty$, the dominated convergence theorem applied to the stopped martingale identity gives, for each $A\in {\mathcal{F}}_s$,This is exactly the conditional expectation identity, so 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.