[proofplan] We prove the equivalence by a cyclic chain (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i). For (i) $\Rightarrow$ (ii), uniform integrability supplies the $L^1$-boundedness needed to invoke the [Almost Sure Martingale Convergence Theorem](/theorems/1157), and the [Uniform Integrability and $L^1$ Convergence](/theorems/1162) criterion then upgrades almost sure convergence to $L^1$ convergence. For (ii) $\Rightarrow$ (iii), we [set](/page/Set) $Z = X_\infty$ and use the $L^1$ contraction property of conditional expectation together with the martingale property to show $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. For (iii) $\Rightarrow$ (i), the representation $X_n = \mathbb{E}[Z \mid \mathcal{F}_n]$ places $(X_n)$ inside a family of conditional expectations of a single integrable random variable, which is uniformly integrable by the [Conditional Expectations are Uniformly Integrable](/theorems/1161) theorem. [/proofplan] [step:Show that uniform integrability implies $L^1$-boundedness] A uniformly integrable family is $L^1$-bounded. To see this, fix $\alpha > 0$ such that $\sup_n \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}] \le 1$ (such $\alpha$ exists by the definition of uniform integrability). Then for every $n \ge 0$, \begin{align*} \mathbb{E}[|X_n|] &= \mathbb{E}[|X_n| \mathbb{1}_{|X_n| \le \alpha}] + \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}] \\ &\le \alpha + 1 < \infty. \end{align*} Therefore $\sup_n \mathbb{E}[|X_n|] \le \alpha + 1 < \infty$. [guided] Uniform integrability controls the tails $\mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}]$ uniformly in $n$, but we also need a uniform bound on the full $L^1$ norms. The idea is to split at the threshold $\alpha$: the part where $|X_n| \le \alpha$ is bounded by $\alpha$ pointwise, and the tail is bounded by the uniform integrability condition. Concretely, choose $\alpha > 0$ so that $\sup_n \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}] \le 1$ (this is possible since the definition of uniform integrability requires this supremum to tend to $0$ as $\alpha \to \infty$). For each $n \ge 0$, decompose the [integral](/page/Integral): \begin{align*} \mathbb{E}[|X_n|] &= \mathbb{E}[|X_n| \mathbb{1}_{|X_n| \le \alpha}] + \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}] \\ &\le \alpha \cdot \mathbb{P}(\Omega) + 1 = \alpha + 1. \end{align*} Since this bound is independent of $n$, we conclude $\sup_n \mathbb{E}[|X_n|] \le \alpha + 1 < \infty$, as required for the [Almost Sure Martingale Convergence Theorem](/theorems/1157). [/guided] [/step] [step:Apply the [Almost Sure Martingale Convergence Theorem](/theorems/1157) to obtain $X_n \to X_\infty$ a.s.] Since $(X_n)_{n \ge 0}$ is a martingale (hence a supermartingale) and $\sup_n \mathbb{E}[|X_n|] < \infty$ by the preceding step, the hypotheses of the [Almost Sure Martingale Convergence Theorem](/theorems/1157) are satisfied. We conclude that there exists $X_\infty \in L^1(\Omega, \mathcal{F}_\infty, \mathbb{P})$, where $\mathcal{F}_\infty = \sigma(\mathcal{F}_n : n \ge 0)$, such that $X_n \to X_\infty$ a.s. as $n \to \infty$. [/step] [step:Upgrade to $L^1$ convergence via the Vitali criterion to complete (i) $\Rightarrow$ (ii)] We now have $X_n \to X_\infty$ a.s. and $(X_n)_{n \ge 0}$ is uniformly integrable. These are precisely the two hypotheses of the [Uniform Integrability and $L^1$ Convergence](/theorems/1162) criterion. Applying it, we conclude $X_n \to X_\infty$ in $L^1(\Omega, \mathcal{F}, \mathbb{P})$. This completes the proof that (i) implies (ii). [guided] The a.s. convergence from the previous step does not by itself yield $L^1$ convergence — a.s. convergence is strictly weaker, and Fatou's lemma only gives the inequality $\mathbb{E}[|X_\infty|] \le \liminf_n \mathbb{E}[|X_n|]$, not convergence of the norms. Uniform integrability is the additional condition that bridges this gap. The [Uniform Integrability and $L^1$ Convergence](/theorems/1162) criterion states: if $X_n \to X$ a.s. and $(X_n)$ is uniformly integrable, then $X_n \to X$ in $L^1$. We have verified both hypotheses — a.s. convergence in the preceding step, and uniform integrability by assumption (i). Therefore $\mathbb{E}[|X_n - X_\infty|] \to 0$. [/guided] [/step] [step:Prove the representation $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. to establish (ii) $\Rightarrow$ (iii)] Set $Z := X_\infty$. Fix $n \ge 0$. For every $m \ge n$, the martingale property gives $X_n = \mathbb{E}[X_m \mid \mathcal{F}_n]$ a.s. We estimate the distance between $X_n$ and $\mathbb{E}[X_\infty \mid \mathcal{F}_n]$: \begin{align*} \|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1} &= \|\mathbb{E}[X_m \mid \mathcal{F}_n] - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1} \\ &= \|\mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n]\|_{L^1} \\ &\le \|X_m - X_\infty\|_{L^1}, \end{align*} where the second equality uses linearity of conditional expectation ([Basic Properties of Conditional Expectation](/theorems/1148), part (v)), and the inequality is the $L^1$ contraction property of conditional expectation ([Conditional Convergence Theorems](/theorems/1149), conditional Jensen with $\varphi(t) = |t|$). Since $X_m \to X_\infty$ in $L^1$ by hypothesis (ii), the right-hand side tends to $0$ as $m \to \infty$. But the left-hand side is independent of $m$, so it must equal $0$. Therefore $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s., and setting $Z = X_\infty$ gives (iii). [guided] We must show that $X_n$ coincides a.s. with the conditional expectation $\mathbb{E}[X_\infty \mid \mathcal{F}_n]$. The natural candidate is $Z = X_\infty$, the a.s. and $L^1$ [limit](/page/Limit). The key idea is to exploit the martingale property at finite times and then pass to the limit. Fix $n \ge 0$ and let $m \ge n$. Since $(X_k)$ is a martingale, the iterated conditioning property gives $X_n = \mathbb{E}[X_m \mid \mathcal{F}_n]$ a.s. — this is the defining property of a martingale composed with the [Tower Property of Conditional Expectation](/theorems/1150). Now consider the difference: \begin{align*} \|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1} &= \|\mathbb{E}[X_m \mid \mathcal{F}_n] - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1}. \end{align*} By linearity of conditional expectation ([Basic Properties of Conditional Expectation](/theorems/1148), part (v)), we may combine the two terms: \begin{align*} \|\mathbb{E}[X_m \mid \mathcal{F}_n] - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1} = \|\mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n]\|_{L^1}. \end{align*} The $L^1$ contraction property of conditional expectation — which follows from applying conditional Jensen ([Conditional Convergence Theorems](/theorems/1149), part (iv)) with the convex [function](/page/Function) $\varphi(t) = |t|$ and then taking expectations — gives \begin{align*} \|\mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n]\|_{L^1} \le \|X_m - X_\infty\|_{L^1}. \end{align*} By hypothesis (ii), $X_m \to X_\infty$ in $L^1$, so the right-hand side tends to $0$ as $m \to \infty$. The left-hand side $\|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_{L^1}$ does not depend on $m$ at all, so a non-negative constant bounded above by a [sequence](/page/Sequence) tending to zero must itself be zero. Hence $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. [/guided] [/step] [step:Deduce uniform integrability from the closed representation to complete (iii) $\Rightarrow$ (i)] Assume $X_n = \mathbb{E}[Z \mid \mathcal{F}_n]$ a.s. for all $n \ge 0$, with $Z \in L^1(\Omega, \mathcal{F}, \mathbb{P})$. The family $(X_n)_{n \ge 0}$ is a subfamily of $\{\mathbb{E}[Z \mid \mathcal{G}] : \mathcal{G} \subset \mathcal{F}\}$, obtained by restricting to the sub-$\sigma$-algebras $\mathcal{G} = \mathcal{F}_n$, $n \ge 0$. By the [Conditional Expectations are Uniformly Integrable](/theorems/1161) theorem, the full family $\{\mathbb{E}[Z \mid \mathcal{G}] : \mathcal{G} \subset \mathcal{F}\}$ is uniformly integrable. Since uniform integrability is inherited by subfamilies (the supremum over a smaller index set is at most the supremum over the full index set), $(X_n)_{n \ge 0}$ is uniformly integrable. This completes the cyclic chain (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i), and therefore the three conditions are equivalent. [guided] The representation $X_n = \mathbb{E}[Z \mid \mathcal{F}_n]$ means that each $X_n$ is a conditional expectation of the same integrable random variable $Z$, but with respect to different sub-$\sigma$-algebras $\mathcal{F}_0, \mathcal{F}_1, \mathcal{F}_2, \ldots$ The [Conditional Expectations are Uniformly Integrable](/theorems/1161) theorem states: for any fixed $Z \in L^1(\Omega, \mathcal{F}, \mathbb{P})$, the family obtained by conditioning $Z$ on all possible sub-$\sigma$-algebras of $\mathcal{F}$ is uniformly integrable. Its hypothesis is precisely that $Z \in L^1$, which holds by assumption. Since $\{X_n : n \ge 0\} = \{\mathbb{E}[Z \mid \mathcal{F}_n] : n \ge 0\} \subset \{\mathbb{E}[Z \mid \mathcal{G}] : \mathcal{G} \subset \mathcal{F}\}$, and uniform integrability passes to subfamilies (the defining supremum $\sup_n \mathbb{E}[|X_n| \mathbb{1}_{|X_n| > \alpha}]$ is bounded above by the supremum over all sub-$\sigma$-algebras), the family $(X_n)_{n \ge 0}$ is uniformly integrable. The cyclic implication chain is now complete, and the three conditions are equivalent. [/guided] [/step]