# Relationship between martingale and conditional expectation

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?

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 $$\mathcal{F}_{\infty}=\sigma(\cup_{n\ge 1}\mathcal{F}_n)$$.