[proofplan]
We apply the [Characterization of Martingales Among Local Martingales](/theorems/2078): $X$ is a martingale if and only if it is a local martingale and the families $\chi_t = \{X_T : T \leq t\}$ are uniformly integrable for every $t$. The domination hypothesis $|X_t| \leq Z$ with $Z \in L^1$ immediately gives uniform integrability of each $\chi_t$, since a family dominated by a single integrable random variable is uniformly integrable by the de la Vallee-Poussin criterion.
[/proofplan]
[step:Verify that the stopped families $\chi_t$ are uniformly integrable via domination]
Fix $t \geq 0$. For any stopping time $T$ with $T \leq t$, we have
\begin{align*}
|X_T(\omega)| = |X_{T(\omega)}(\omega)| \leq Z(\omega) \quad \text{a.s.}
\end{align*}
since $|X_s| \leq Z$ for all $s \geq 0$ by hypothesis. Therefore every element of $\chi_t = \{X_T : T \text{ stopping time}, T \leq t\}$ is bounded in absolute value by the integrable random variable $Z$.
A family of random variables dominated by a single $L^1$ random variable is uniformly integrable: for any $c > 0$,
\begin{align*}
\sup_{T \leq t} \mathbb{E}\!\left[|X_T| \cdot \mathbb{1}_{\{|X_T| > c\}}\right] \leq \mathbb{E}\!\left[Z \cdot \mathbb{1}_{\{Z > c\}}\right] \to 0 \quad \text{as } c \to \infty,
\end{align*}
where the limit holds because $Z \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ and the dominated convergence theorem applied to $Z \cdot \mathbb{1}_{\{Z > c\}} \to 0$ a.s. with dominator $Z$.
[guided]
Fix $t \geq 0$. We need to show that $\chi_t = \{X_T : T \leq t\}$ is uniformly integrable. Recall the definition: a family $(Y_\alpha)$ is uniformly integrable if
\begin{align*}
\lim_{c \to \infty} \sup_\alpha \mathbb{E}\!\left[|Y_\alpha| \cdot \mathbb{1}_{\{|Y_\alpha| > c\}}\right] = 0.
\end{align*}
For any stopping time $T$ with $T \leq t$, since $|X_s| \leq Z$ for all $s$ (a pointwise-in-time hypothesis applied at the random time $T(\omega)$), we get $|X_T| \leq Z$ a.s. Therefore
\begin{align*}
|X_T| \cdot \mathbb{1}_{\{|X_T| > c\}} \leq Z \cdot \mathbb{1}_{\{Z > c\}}
\end{align*}
for every $c > 0$, since $|X_T| > c$ implies $Z \geq |X_T| > c$. Taking expectations and the supremum over all stopping times $T \leq t$:
\begin{align*}
\sup_{T \leq t} \mathbb{E}\!\left[|X_T| \cdot \mathbb{1}_{\{|X_T| > c\}}\right] \leq \mathbb{E}\!\left[Z \cdot \mathbb{1}_{\{Z > c\}}\right].
\end{align*}
The right-hand side does not depend on $T$ and tends to $0$ as $c \to \infty$ by the dominated convergence theorem (the functions $Z \cdot \mathbb{1}_{\{Z > c\}}$ decrease to $0$ pointwise and are dominated by $Z \in L^1$). This establishes uniform integrability of $\chi_t$.
[/guided]
[/step]
[step:Conclude by invoking the characterization theorem]
Since $X$ is a local martingale (by hypothesis) and the families $\chi_t$ are uniformly integrable for every $t \geq 0$ (by the previous step), the [Characterization of Martingales Among Local Martingales](/theorems/2078) gives that $X$ is a martingale.
For the "in particular" statement: if $X$ is a bounded local martingale with $|X_t| \leq C$ for some constant $C > 0$ and all $t$, take $Z = C \in L^1(\Omega, \mathcal{F}, \mathbb{P})$ (since $\mathbb{E}[C] = C < \infty$). Then $|X_t| \leq Z$ for all $t$, and the preceding argument applies.
[/step]
