[proofplan]
We prove both implications separately. For $(1) \Rightarrow (2)$: if $X$ is a martingale, then it is a fortiori a local martingale. Uniform integrability of the family $\chi_t = \{X_T : T \text{ stopping time}, T \leq t\}$ follows from the optional stopping theorem, which represents each $X_T$ as $\mathbb{E}[X_t \mid \mathcal{F}_T]$; the family of conditional expectations of a single integrable random variable is always uniformly integrable. For $(2) \Rightarrow (1)$: given a reducing sequence $(T_n)$, the martingale property of $X^{T_n}$ gives $\mathbb{E}[X_{T \wedge T_n}] = \mathbb{E}[X_0]$ for every bounded stopping time $T$. As $n \to \infty$, $X_{T \wedge T_n} \to X_T$ a.s., and uniform integrability of $\chi_t$ upgrades this to $L^1$ convergence via the Vitali convergence theorem, yielding $\mathbb{E}[X_T] = \mathbb{E}[X_0]$.
[/proofplan]
[step:Prove $(1) \Rightarrow (2)$: a martingale has uniformly integrable stopped families]
Assume $X$ is a martingale. Then $X$ is a local martingale (take the reducing sequence $T_n = n$). It remains to show that $\chi_t = \{X_T : T \text{ stopping time}, T \leq t\}$ is uniformly integrable for every $t \geq 0$.
Fix $t \geq 0$. For any stopping time $T$ with $T \leq t$, the [Optional Stopping Theorem](/theorems/2109) applied to the martingale $X$ gives
\begin{align*}
X_T = \mathbb{E}[X_t \mid \mathcal{F}_T] \quad \text{a.s.}
\end{align*}
The family $\{X_T : T \leq t\}$ is therefore a subset of $\{\mathbb{E}[X_t \mid \mathcal{G}] : \mathcal{G} \text{ sub-$\sigma$-algebra of } \mathcal{F}\}$. The conditional expectations of a single $L^1$ random variable form a uniformly integrable family (since $\mathbb{E}[|X_T|] \leq \mathbb{E}[|X_t|] < \infty$ and for any $c > 0$, $\mathbb{E}[|X_T| \mathbb{1}_{\{|X_T| > c\}}] \leq \mathbb{E}[|X_t| \mathbb{1}_{\{|X_t| > c/2\}}] + \mathbb{E}[|X_t|] \cdot \mathbb{P}(|X_T| > c)$, which can be made uniformly small). Hence $\chi_t$ is uniformly integrable.
[guided]
Assume $X$ is a martingale. Then $X$ is automatically a local martingale — the constant sequence $T_n = n$ serves as a reducing sequence, since the stopped process $X^n$ restricted to $[0, n]$ is a martingale and $T_n = n \to \infty$.
The substantive content is uniform integrability of $\chi_t$. Fix $t \geq 0$. For any stopping time $T$ with $T \leq t$, the optional stopping theorem applies because $T$ is bounded by $t$ and $X$ is a martingale. The conclusion is
\begin{align*}
X_T = \mathbb{E}[X_t \mid \mathcal{F}_T] \quad \text{a.s.}
\end{align*}
Why does this imply uniform integrability? We use a standard fact from measure theory: if $Z \in L^1(\Omega, \mathcal{F}, \mathbb{P})$, then the family $\{\mathbb{E}[Z \mid \mathcal{G}] : \mathcal{G} \subseteq \mathcal{F}\}$ is uniformly integrable. To verify this, note that for any $c > 0$,
\begin{align*}
\mathbb{E}\!\left[|\mathbb{E}[Z \mid \mathcal{G}]| \cdot \mathbb{1}_{\{|\mathbb{E}[Z \mid \mathcal{G}]| > c\}}\right] \leq \mathbb{E}\!\left[\mathbb{E}[|Z| \mid \mathcal{G}] \cdot \mathbb{1}_{\{\mathbb{E}[|Z| \mid \mathcal{G}] > c\}}\right],
\end{align*}
using $|\mathbb{E}[Z \mid \mathcal{G}]| \leq \mathbb{E}[|Z| \mid \mathcal{G}]$ a.s. (conditional Jensen). Since $\mathbb{E}[|Z| \mid \mathcal{G}]$ is a non-negative submartingale indexed by the directed set of sub-$\sigma$-algebras, Doob's maximal inequality and the de la Vallee-Poussin criterion give uniform integrability.
In our case $Z = X_t \in L^1$ (since $X$ is a martingale), and each $X_T = \mathbb{E}[X_t \mid \mathcal{F}_T]$ is a conditional expectation of $Z$. Therefore $\chi_t \subseteq \{\mathbb{E}[X_t \mid \mathcal{G}]\}$, which is uniformly integrable.
[/guided]
[/step]
[step:Prove $(2) \Rightarrow (1)$: reduce to showing $\mathbb{E}[X_T] = \mathbb{E}[X_0]$ for bounded stopping times]
Assume $X$ is a local martingale and all families $\chi_t$ are uniformly integrable. We show $X$ is a martingale.
It suffices to prove: for every deterministic $0 \leq s \leq t$,
\begin{align*}
\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s \quad \text{a.s.}
\end{align*}
We will establish the equivalent condition: $\mathbb{E}[X_T] = \mathbb{E}[X_0]$ for every bounded stopping time $T$. The martingale property then follows from applying this to stopping times of the form $T = s \cdot \mathbb{1}_A + t \cdot \mathbb{1}_{A^c}$ for $A \in \mathcal{F}_s$, which gives $\mathbb{E}[X_s \mathbb{1}_A] + \mathbb{E}[X_t \mathbb{1}_{A^c}] = \mathbb{E}[X_0]$ and by comparison with $T' = s$ (giving $\mathbb{E}[X_s] = \mathbb{E}[X_0]$), we deduce $\mathbb{E}[X_t \mathbb{1}_A] = \mathbb{E}[X_s \mathbb{1}_A]$ for all $A \in \mathcal{F}_s$, i.e., $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$.
[/step]
[step:Use the reducing sequence and Vitali convergence to pass to the limit]
Let $(T_n)_{n \geq 1}$ be a reducing sequence for $X$. Fix a bounded stopping time $T$ with $T \leq t$ for some $t \geq 0$.
Since $X^{T_n}$ is a martingale for each $n$, and $T \wedge T_n \leq T \leq t$ is a bounded stopping time, the optional stopping theorem gives
\begin{align*}
\mathbb{E}[X_{T \wedge T_n}] = \mathbb{E}[X^{T_n}_T] = \mathbb{E}[X^{T_n}_0] = \mathbb{E}[X_0]
\end{align*}
for each $n$, where the last equality uses $T_n > 0$ a.s. for large $n$, so $X^{T_n}_0 = X_{0 \wedge T_n} = X_0$.
As $n \to \infty$, $T_n \to \infty$ a.s., so $T \wedge T_n \to T$ a.s., and the càdlàg property of $X$ gives $X_{T \wedge T_n} \to X_T$ a.s. Since $T \wedge T_n \leq t$ is a stopping time bounded by $t$, we have $X_{T \wedge T_n} \in \chi_t$ for each $n$. By hypothesis, $\chi_t$ is uniformly integrable. The [Vitali convergence theorem](/theorems/2111) states that if a sequence converges a.s. and is uniformly integrable, then it converges in $L^1$. Therefore $X_{T \wedge T_n} \to X_T$ in $L^1(\mathbb{P})$, which gives
\begin{align*}
\mathbb{E}[X_T] = \lim_{n \to \infty} \mathbb{E}[X_{T \wedge T_n}] = \mathbb{E}[X_0].
\end{align*}
Since $T$ was an arbitrary bounded stopping time, the previous step yields $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$ for all $0 \leq s \leq t$. Combined with adaptedness (inherited from the local martingale) and integrability ($X_t \in L^1$ since $X_t \in \chi_t$ and $\chi_t$ is uniformly integrable), $X$ is a martingale.
[guided]
Let $(T_n)_{n \geq 1}$ be a reducing sequence for $X$. Fix a bounded stopping time $T$ with $T \leq t$.
Since $X^{T_n}$ is a martingale for each $n$, and $T \wedge T_n$ is a stopping time bounded by $t$, the optional stopping theorem for the martingale $X^{T_n}$ gives
\begin{align*}
\mathbb{E}[X_{T \wedge T_n}] = \mathbb{E}[X^{T_n}_T] = \mathbb{E}[X^{T_n}_0] = \mathbb{E}[X_0].
\end{align*}
We want to take $n \to \infty$. Since $T_n \to \infty$ a.s., we have $T \wedge T_n \to T$ a.s. By the càdlàg property of $X$, this gives $X_{T \wedge T_n}(\omega) \to X_{T(\omega)}(\omega)$ for a.e. $\omega$. So we have pointwise convergence. Can we pass the limit through the expectation?
This is where uniform integrability enters. Each $T \wedge T_n$ is a stopping time with $T \wedge T_n \leq T \leq t$, so $X_{T \wedge T_n} \in \chi_t$. By hypothesis, $\chi_t$ is uniformly integrable. The Vitali convergence theorem states: if $(Y_n)$ is a sequence of random variables such that $Y_n \to Y$ a.s. and the family $\{Y_n\}$ is uniformly integrable, then $Y_n \to Y$ in $L^1$ and in particular $\mathbb{E}[Y_n] \to \mathbb{E}[Y]$.
We verify the hypotheses of Vitali: (i) $X_{T \wedge T_n} \to X_T$ a.s. (established above); (ii) the sequence $(X_{T \wedge T_n})_{n \geq 1}$ is uniformly integrable, since it is a subset of the uniformly integrable family $\chi_t$. Therefore
\begin{align*}
\mathbb{E}[X_T] = \lim_{n \to \infty} \mathbb{E}[X_{T \wedge T_n}] = \mathbb{E}[X_0].
\end{align*}
Since this holds for every bounded stopping time $T$, the argument in the previous step (using stopping times of the form $T = s \cdot \mathbb{1}_A + t \cdot \mathbb{1}_{A^c}$) gives $\mathbb{E}[X_t \mid \mathcal{F}_s] = X_s$ for all $0 \leq s \leq t$.
Integrability of $X_t$ is automatic here: $X_t \in \chi_t$ (take the stopping time $T = t$), and $\chi_t$ is uniformly integrable, which implies $\sup_{T \leq t} \mathbb{E}[|X_T|] < \infty$. In particular $\mathbb{E}[|X_t|] < \infty$.
Together with adaptedness (which $X$ inherits from being a local martingale), this establishes that $X$ is a martingale.
[/guided]
[/step]
