Show that $\mathbb{E}(\lim_{n\to\infty} X_n|\mathcal{F}_n)\ge X_n$ for $(X_n)_{n\ge 0}$ a UI submartingale Let $(X_n)_{n\ge 0}$ a uniformly integrable submartingale (or supermartingale, respectively). Show that there exists $X\in L^1$ such that $X_n$ converges to $X$ a.s. and in $L^1$. Furthermore, show that for all $n\ge 0,\ \mathbb{E}(\lim_{n\to\infty} X_n|\mathcal{F}_n)\ge X_n\,$   ($\le\,$ in the supermartingale case).

My attempt:  A UI family is bounded in $L^1$ so by Doob's sub/-supermartingale convergence theorem, there exists $X\in L^1$ s.t. $X_n\to X$ a.s. This convergence also takes place wrt. the $L^1$ norm since a UI sequence which converges a.s. (and hence in probability) also converges in $L^1$.
I'm now trying to show that for all $n\ge 0,\ \mathbb{E}(\lim_{n\to\infty} X_n|\mathcal{F}_n)\le X_n\,$ in the supermartingale case.
My try was to use Doob's decomposition and write $X_n = M_n + I_n$ where $M_n$ is a martingale and $I_n$ is previsible and non-increasing. It follows that for all $m\ge n$, $||X_m-X||=||M_m+I_m-X||\ge||\mathbb{E}(M_m+I_m-X|\mathcal{F}_n)||=||M_n + E(I_m|\mathcal{F}_n)-E(X|\mathcal{F}_n)||$ with $E(I_m|\mathcal{F}_n)\le E(I_n|\mathcal{F}_n)=I_n$. This is where I get stuck. How do I continue?
 A: 
If a submartingale $X=\left(X_{n}\right)_{n \in \mathbb{Z}_{+}}$ is "uniformly integrable", then $X_{n} \xrightarrow[a.e.]{L^{1}}X_{\infty}$ and $\left(X_{n}\right)_{n \in \overline{\mathbb{Z}}_{+}}$ becomes a closed submartingale.

To prove the proposition, we need some facts:

*

*(Criterion of $L^{1}$-convergence) Let $X_{n} \in L^{1}$; then $X_{n} \xrightarrow{L^{1}}  X$ if and only if $X_{n}$ are uniformly integrable and
$X_{n} \rightarrow_\mathbb{P} X$.

*(Vitali's convergence theorem) Suppose that $f_{n} \rightarrow f$ a.e. $(\mathbb{P})$. If the $f_{n}$ are uniformly integrable, then $f$ is integrable and such that
$$
 \int f_{n} d \mathbb{P} \rightarrow \int f d \mathbb{P}.
 $$
By the definition of uniform integrability, $(X_n)_{n\in \mathbb{Z}_+}$ is bounded in $L^1$ and, moreover, $\left(X_{n} 1_{A}\right)_{n \in \mathbb{Z}_{+}}$ is also uniformly integrable for any $A \in \mathscr{F}_{\infty}$. Thus, by Doob's submartingale convergence theorem, we have
$$
X_{n} \stackrel{\text { a.e. }}{\longrightarrow} X_{\infty} \in L^{1}\left(\Omega, \mathscr{F}_{\infty}, \mathbb{P}\right) \Rightarrow X_{n} 1_{A} \stackrel{\text { a.e. }}{\longrightarrow} X_{\infty} 1_{A} \text { and } X_{n} 1_{A} \rightarrow_\mathbb{P} X_{\infty} 1_{A}.
$$
Then by fact 1, we see as $n \rightarrow \infty, X_{n} \rightarrow X_{\infty}$ in $L^{1}\left(\Omega, \mathscr{F}_{\infty}, \mathbb{P}\right)$. Moreover, by fact 2, it follows that as $m \rightarrow \infty$,
$$
\text{E}\left[X_{m} 1_{A}\right] \rightarrow \text{E}\left[X_{\infty} 1_{A}\right] ,\quad \forall A \in \mathscr{F}_{\infty}.
$$
Finally, to show $\left(X_{n}\right)_{n \in \overline{\mathbb{Z}}_{+}}$ is a closed submartingale, we only need to prove $\text{E}\left[X_{\infty} \mid \mathscr{F}_{n}\right] \geq X_{n}$
a.e. for all $n \geq 0$. For this, by $\text{E}\left[X_{m} \mid \mathscr{F}_{n}\right] \geq X_{n}$ a.e. for all $m>n$, it follows that for any $A \in \mathscr{F}_{n}$ we have
$$
\text{E}\left[X_{n} 1_{A}\right] \leq \text{E}\left[X_{m} 1_{A}\right] \rightarrow \text{E}\left[X_{\infty} 1_{A}\right] \text { as } m \rightarrow+\infty. 
$$
This completes the proof.

See Proof of Fact 1,
Proof of Fact 2.
