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I am studying a theorem about $L^p$ convergence and some equivalent conditions. What I understood is that given an $\mathcal{F}_n$-martingale $X_n$, if $X_n$ converges to $X_{\infty}$ in $L^p$, then $E[X_{\infty}\mid \mathcal{F}_n]=X_n$ for all $n\geq 0$. On the other hand, if we know that $X_n=E[X\mid \mathcal{F}_n]$ for some $X \in L^p$, then $X_n$ converges to some $X_{\infty}$ where $X_{\infty}=E[X\mid \mathcal{F}_{\infty}]$. Is my understanding correct? Am I overlooking something obvious?

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This result appears in Lemma 4.6.6 and Theorem 4.6.8 here (pages 246-247). Specifically, if a martingale $X_n\to X$ in $L^1$, then $X_n=\mathsf{E}[X\mid \mathcal{F}_n]$ a.s. On the other hand, if $X_n=\mathsf{E}[X\mid \mathcal{F}_n]$ a.s. for some integrable r.v. $X$, then $X_n\to\mathsf{E}[X\mid \mathcal{F}_{\infty}]$ a.s. and in $L^1$, where $\mathcal{F}_{\infty}=\sigma(\cup_{n\ge 1}\mathcal{F}_n)$.

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