[proofplan] The proof establishes the equivalence of three characterisations of $L^p$-bounded martingales by a cyclic argument (i) $\Rightarrow$ (ii) $\Rightarrow$ (iii) $\Rightarrow$ (i). The main work is in (i) $\Rightarrow$ (ii): the $L^p$ bound implies $L^1$ boundedness via conditional Jensen, which yields almost sure convergence; Fatou's lemma promotes the [limit](/page/Limit) to $L^p$; Doob's $L^p$ inequality provides a dominating [function](/page/Function) in $L^p$, enabling the [Dominated Convergence Theorem](/theorems/4) to upgrade pointwise convergence to $L^p$ convergence. The implication (ii) $\Rightarrow$ (iii) identifies the limit as the generating random variable via the $L^p$ contraction property of conditional expectation, and (iii) $\Rightarrow$ (i) is an immediate application of the conditional [Jensen inequality](/theorems/515). [/proofplan] [step:Deduce $L^1$ boundedness from the $L^p$ bound and obtain almost sure convergence] **(i) $\Rightarrow$ (ii), first part.** Assume $\sup_n \|X_n\|_p < \infty$. Since $p > 1$, the function $\varphi: \mathbb{R} \to [0,\infty)$, $t \mapsto |t|$ is convex, and the [Conditional Jensen Inequality](/theorems/1149) (part (iv), applied with $\varphi(t) = |t|$ and $\mathcal{G} = \{\varnothing, \Omega\}$) gives $\mathbb{E}[|X_n|] \leq \mathbb{E}[|X_n|^p]^{1/p}$ by the $L^p$ contraction stated in the final line of that theorem. In particular, \begin{align*} \sup_n \mathbb{E}[|X_n|] \leq \sup_n \|X_n\|_p < \infty, \end{align*} so $(X_n)_{n \geq 0}$ is bounded in $L^1$. Since a martingale is in particular a supermartingale, the hypotheses of the [Almost Sure Martingale Convergence Theorem](/theorems/1157) are satisfied: $(X_n)$ is a supermartingale bounded in $L^1$. We conclude that there exists a random variable $X_\infty \in L^1(\mathcal{F}_\infty, \mathbb{P})$, where $\mathcal{F}_\infty = \sigma(\mathcal{F}_n : n \geq 0)$, such that $X_n \to X_\infty$ a.s. [guided] The first task is to reduce the $L^p$ hypothesis to the $L^1$ setting, where the almost sure convergence theorem applies. We need $\sup_n \mathbb{E}[|X_n|] < \infty$. The connection between $L^p$ and $L^1$ norms comes from [Jensen's inequality](/theorems/9) (or, equivalently, from Hölder's inequality with exponents $p$ and $p/(p-1)$ applied to $|X_n|$ and the constant function $1$ on the probability space). Since $\mathbb{P}(\Omega) = 1$, Jensen's inequality applied to the convex function $t \mapsto |t|^p$ gives \begin{align*} \mathbb{E}[|X_n|] \leq \mathbb{E}[|X_n|^p]^{1/p} = \|X_n\|_p. \end{align*} This is the $L^p$ contraction property stated in the final line of the [Conditional Convergence Theorems](/theorems/1149) (taking $\mathcal{G} = \{\varnothing, \Omega\}$). Therefore \begin{align*} \sup_n \mathbb{E}[|X_n|] \leq \sup_n \|X_n\|_p < \infty. \end{align*} Why does this suffice? The [Almost Sure Martingale Convergence Theorem](/theorems/1157) requires a supermartingale bounded in $L^1$. Every martingale is a supermartingale (the supermartingale inequality $\mathbb{E}[X_{n+1} \mid \mathcal{F}_n] \leq X_n$ holds with equality), and we have just established $L^1$ boundedness. So the hypotheses are satisfied, and we obtain $X_n \to X_\infty$ a.s. for some $X_\infty \in L^1(\mathcal{F}_\infty, \mathbb{P})$. [/guided] [/step] [step:Verify $X_\infty \in L^p$ via Fatou's lemma] **(i) $\Rightarrow$ (ii), second part.** Since $|X_n|^p \geq 0$ and $|X_n|^p \to |X_\infty|^p$ a.s., the [Fatou Lemma](/theorems/510) applied to the [sequence](/page/Sequence) $(|X_n|^p)_{n \geq 0}$ on the probability space $(\Omega, \mathcal{F}, \mathbb{P})$ gives \begin{align*} \int_\Omega |X_\infty|^p \, d\mathbb{P} \leq \liminf_{n \to \infty} \int_\Omega |X_n|^p \, d\mathbb{P} = \liminf_{n \to \infty} \|X_n\|_p^p \leq \sup_n \|X_n\|_p^p < \infty. \end{align*} Therefore $X_\infty \in L^p(\Omega, \mathcal{F}, \mathbb{P})$. [guided] We know $X_n \to X_\infty$ a.s., but the Almost Sure Convergence Theorem only guarantees $X_\infty \in L^1$. We need the stronger conclusion $X_\infty \in L^p$. The natural tool is the [Fatou Lemma](/theorems/510): for non-negative [measurable functions](/page/Measurable%20Functions), the [integral](/page/Integral) of the $\liminf$ is bounded by the $\liminf$ of the integrals. We apply this to $f_n := |X_n|^p$, which is non-negative and converges a.s. to $|X_\infty|^p$ (since $t \mapsto |t|^p$ is continuous and $X_n \to X_\infty$ a.s.): \begin{align*} \|X_\infty\|_p^p = \int_\Omega |X_\infty|^p \, d\mathbb{P} \leq \liminf_{n \to \infty} \int_\Omega |X_n|^p \, d\mathbb{P} = \liminf_{n \to \infty} \|X_n\|_p^p \leq \sup_n \|X_n\|_p^p < \infty. \end{align*} This establishes $X_\infty \in L^p$. Note that Fatou's lemma is optimal here: we cannot directly apply the Dominated Convergence Theorem because we do not yet have a dominating function (that comes next). [/guided] [/step] [step:Construct an $L^p$ dominating function via Doob's $L^p$ inequality] **(i) $\Rightarrow$ (ii), third part.** Define the maximal function \begin{align*} X_\infty^*: \Omega &\to [0, \infty] \\ \omega &\mapsto \sup_{n \geq 0} |X_n(\omega)|. \end{align*} Since $(|X_n|)_{n \geq 0}$ is a non-negative submartingale (by the conditional Jensen inequality in the [Conditional Convergence Theorems](/theorems/1149) applied with $\varphi(t) = |t|$), [Doob's $L^p$ Inequality](/theorems/1159) applies. The hypotheses of that theorem require $p > 1$ (given) and that $(|X_n|)$ is a non-negative submartingale. With $X_n^* = \sup_{k \leq n} |X_k|$, the inequality gives \begin{align*} \|X_n^*\|_p \leq \frac{p}{p-1} \|X_n\|_p \leq \frac{p}{p-1} \sup_n \|X_n\|_p < \infty \end{align*} for every $n \geq 0$. Since $X_n^* \uparrow X_\infty^*$ pointwise as $n \to \infty$, the [Monotone Convergence Theorem](/theorems/509) yields \begin{align*} \|X_\infty^*\|_p = \lim_{n \to \infty} \|X_n^*\|_p \leq \frac{p}{p-1} \sup_n \|X_n\|_p < \infty. \end{align*} In particular, $X_\infty^* \in L^p(\Omega, \mathcal{F}, \mathbb{P})$. [guided] The goal of this step is to find a single $L^p$ function that dominates the entire sequence $(X_n)$, so that the Dominated Convergence Theorem can be applied in the next step. The natural candidate is the maximal function $X_\infty^* = \sup_{n \geq 0} |X_n|$. To show $X_\infty^* \in L^p$, we use [Doob's $L^p$ Inequality](/theorems/1159). That theorem requires: 1. $p > 1$ — which is a standing hypothesis. 2. $(|X_n|)_{n \geq 0}$ is a non-negative submartingale — which holds because the [Conditional Jensen Inequality](/theorems/1149) (part (iv)) applied with the convex function $\varphi(t) = |t|$ gives $\mathbb{E}[|X_{n+1}| \mid \mathcal{F}_n] \geq |\mathbb{E}[X_{n+1} \mid \mathcal{F}_n]| = |X_n|$ a.s. Doob's inequality for the finite-horizon maximal function $X_n^* = \sup_{k \leq n} |X_k|$ gives \begin{align*} \|X_n^*\|_p \leq \frac{p}{p-1} \|X_n\|_p. \end{align*} The constant $p/(p-1)$ is sharp (it is the conjugate exponent to $p$), and it diverges as $p \to 1^+$ — which is why this argument fails for $p = 1$. To pass from the finite-horizon $X_n^*$ to the infinite-horizon $X_\infty^*$, note that $(X_n^*)^p \uparrow (X_\infty^*)^p$ pointwise. The Monotone Convergence Theorem gives \begin{align*} \|X_\infty^*\|_p^p = \lim_{n \to \infty} \|X_n^*\|_p^p \leq \left(\frac{p}{p-1}\right)^p \sup_n \|X_n\|_p^p < \infty, \end{align*} establishing $X_\infty^* \in L^p$. This is the key estimate that separates the $p > 1$ theory from the $p = 1$ case: for $p = 1$, $L^1$-bounded martingales converge a.s. but need not converge in $L^1$ (consider a martingale that converges a.s. to zero while maintaining unit $L^1$ norm). [/guided] [/step] [step:Upgrade to $L^p$ convergence via the Dominated Convergence Theorem] **(i) $\Rightarrow$ (ii), conclusion.** We have $X_n \to X_\infty$ a.s. and therefore $|X_n - X_\infty|^p \to 0$ a.s. The pointwise bound \begin{align*} |X_n - X_\infty| \leq |X_n| + |X_\infty| \leq X_\infty^* + |X_\infty| \leq 2X_\infty^* \end{align*} holds a.s. for all $n$ (since $|X_\infty| = \lim_n |X_n| \leq \sup_n |X_n| = X_\infty^*$ a.s.), so \begin{align*} |X_n - X_\infty|^p \leq (2X_\infty^*)^p = 2^p (X_\infty^*)^p \in L^1(\Omega, \mathcal{F}, \mathbb{P}). \end{align*} By the [Dominated Convergence Theorem](/theorems/4) applied to the sequence $(|X_n - X_\infty|^p)_{n \geq 0}$ with dominating function $2^p (X_\infty^*)^p$, \begin{align*} \lim_{n \to \infty} \|X_n - X_\infty\|_p^p = \lim_{n \to \infty} \int_\Omega |X_n - X_\infty|^p \, d\mathbb{P} = \int_\Omega \lim_{n \to \infty} |X_n - X_\infty|^p \, d\mathbb{P} = 0. \end{align*} This completes the proof of (i) $\Rightarrow$ (ii): $X_n \to X_\infty$ both a.s. and in $L^p$. [guided] We now have all the ingredients for the Dominated Convergence Theorem: 1. **Pointwise convergence:** $|X_n - X_\infty|^p \to 0$ a.s., since $X_n \to X_\infty$ a.s. and $t \mapsto |t|^p$ is continuous. 2. **Dominating function:** We need a single integrable function that bounds $|X_n - X_\infty|^p$ for all $n$. By the triangle inequality, $|X_n - X_\infty| \leq |X_n| + |X_\infty|$. Since $|X_n| \leq X_\infty^*$ by definition and $|X_\infty| \leq X_\infty^*$ a.s. (as $|X_\infty| = \lim_n |X_n| \leq \sup_n |X_n| = X_\infty^*$), we obtain $|X_n - X_\infty| \leq 2X_\infty^*$ a.s. for all $n$. Raising to the $p$-th power: \begin{align*} |X_n - X_\infty|^p \leq 2^p (X_\infty^*)^p. \end{align*} The function $2^p (X_\infty^*)^p$ is integrable because $X_\infty^* \in L^p$ (established in the previous step). The [Dominated Convergence Theorem](/theorems/4) now applies on the [measure space](/page/Measure%20Space) $(\Omega, \mathcal{F}, \mathbb{P})$: the sequence $(|X_n - X_\infty|^p)$ converges to zero a.s. and is dominated by the integrable function $2^p (X_\infty^*)^p$, so \begin{align*} \|X_n - X_\infty\|_p^p = \int_\Omega |X_n - X_\infty|^p \, d\mathbb{P} \to 0 \quad \text{as } n \to \infty. \end{align*} This establishes $L^p$ convergence. The entire implication (i) $\Rightarrow$ (ii) is now complete. [/guided] [/step] [step:Identify the limit as a conditional expectation via the $L^p$ contraction] **(ii) $\Rightarrow$ (iii).** Set $Z := X_\infty \in L^p(\Omega, \mathcal{F}, \mathbb{P})$. We must show $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. for every $n \geq 0$. Fix $n \geq 0$. For every $m \geq n$, the martingale property and the [Tower Property](/theorems/1150) give \begin{align*} X_n = \mathbb{E}[X_m \mid \mathcal{F}_n] \quad \text{a.s.} \end{align*} We estimate the distance between $X_n$ and $\mathbb{E}[X_\infty \mid \mathcal{F}_n]$. By linearity of conditional expectation ([Basic Properties](/theorems/1148), part (v)), \begin{align*} X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n] = \mathbb{E}[X_m \mid \mathcal{F}_n] - \mathbb{E}[X_\infty \mid \mathcal{F}_n] = \mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n] \quad \text{a.s.} \end{align*} The $L^p$ contraction of conditional expectation (the final line of [Theorem 1149](/theorems/1149): $\|\mathbb{E}[Y \mid \mathcal{G}]\|_p \leq \|Y\|_p$ for $1 \leq p < \infty$, applied with $Y = X_m - X_\infty$ and $\mathcal{G} = \mathcal{F}_n$) gives \begin{align*} \|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_p = \|\mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n]\|_p \leq \|X_m - X_\infty\|_p. \end{align*} The left-hand side is independent of $m$, and the right-hand side tends to zero as $m \to \infty$ by the $L^p$ convergence established in (ii). Therefore $\|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_p = 0$, which gives $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. Setting $Z = X_\infty$ completes the proof of (ii) $\Rightarrow$ (iii). [guided] We claim that the limit $X_\infty$ itself serves as the generating random variable $Z$. The task is to verify the identity $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ for each fixed $n$. The key idea is to use the martingale property to express $X_n$ as $\mathbb{E}[X_m \mid \mathcal{F}_n]$ for any $m \geq n$, then send $m \to \infty$. Specifically, for $m \geq n$, the martingale property gives $\mathbb{E}[X_m \mid \mathcal{F}_n] = X_n$ a.s. (this follows by iterating the defining identity $\mathbb{E}[X_{k+1} \mid \mathcal{F}_k] = X_k$ and applying the [Tower Property](/theorems/1150)). Now compare $X_n$ with $\mathbb{E}[X_\infty \mid \mathcal{F}_n]$: \begin{align*} X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n] &= \mathbb{E}[X_m \mid \mathcal{F}_n] - \mathbb{E}[X_\infty \mid \mathcal{F}_n] \\ &= \mathbb{E}[X_m - X_\infty \mid \mathcal{F}_n], \end{align*} where the second equality uses linearity of conditional expectation ([Basic Properties](/theorems/1148), part (v)). Why can we conclude? The $L^p$ contraction property of conditional expectation states that $\|\mathbb{E}[Y \mid \mathcal{G}]\|_p \leq \|Y\|_p$ for any $Y \in L^p$ — this is the final line of [Theorem 1149](/theorems/1149), a consequence of the conditional Jensen inequality applied with $\varphi(t) = |t|^p$. Applying it with $Y = X_m - X_\infty$ and $\mathcal{G} = \mathcal{F}_n$: \begin{align*} \|X_n - \mathbb{E}[X_\infty \mid \mathcal{F}_n]\|_p \leq \|X_m - X_\infty\|_p \to 0 \quad \text{as } m \to \infty. \end{align*} The left-hand side does not depend on $m$, so it must equal zero. An $L^p$ function with zero norm vanishes a.s., giving $X_n = \mathbb{E}[X_\infty \mid \mathcal{F}_n]$ a.s. [/guided] [/step] [step:Close the cycle via conditional Jensen] **(iii) $\Rightarrow$ (i).** Assume $X_n = \mathbb{E}[Z \mid \mathcal{F}_n]$ a.s. for some $Z \in L^p(\Omega, \mathcal{F}, \mathbb{P})$. The $L^p$ contraction of conditional expectation (the final line of [Theorem 1149](/theorems/1149): $\|\mathbb{E}[Y \mid \mathcal{G}]\|_p \leq \|Y\|_p$, applied with $Y = Z$ and $\mathcal{G} = \mathcal{F}_n$) gives \begin{align*} \|X_n\|_p = \|\mathbb{E}[Z \mid \mathcal{F}_n]\|_p \leq \|Z\|_p \end{align*} for every $n \geq 0$. Therefore $\sup_n \|X_n\|_p \leq \|Z\|_p < \infty$, which is (i). This completes the cyclic proof of the equivalence (i) $\Leftrightarrow$ (ii) $\Leftrightarrow$ (iii). [/step]