Karatzas and Shreve Problem 1.3.16. Proving nonnegative right continuous supermartingale convergence 
This is Problem 3.16 from Chapter 1 of Karatzas and Shreve. 
Let $\{X_t, \mathscr{F}_t: 0 \le t < \infty\}$ be a right-continuous, nonnegative supermartingale; then $X_\infty(\omega) = \lim_{t \to \infty} X_t (\omega)$ exists for $P$-a.e. $\omega \in \Omega$, and $\{X_t , \mathscr{F}_t: 0 \le t \le \infty\}$ is a supermartingale.

I have shown that the limit exists a.e. from the same theorem for submartingales.
However, I am having difficulty showing that 
$$E[X_\infty | \mathscr{F}_t] \le X_t.$$
I know this is the case if $\{X_t\}$ is uniformly integrable since then we have 
$$
E[X_\infty | \mathscr{F}_t]
 = \lim_{s \to \infty} E[X_s | \mathscr{F}_t]
 \le X_t.
$$
So I want to show that $\int_{|X_t| > C} |X_t| dP \to 0$ as $C \to \infty$, but I got stuck here. How can we prove uniform integrability here?
 A: Hint: There is no need to prove uniform integrability. Just apply Fatou's lemma (for conditional expectations).
A: Following the saz's hints.
Remark that  $\left(-X_t  \right)_{0 \leq t < \infty}$  is $\mathscr{F}_t$-supermartingale ( $(\mathscr{F}_t)$ s right continuous ).
Then 
\begin{align}
 \sup_{t \geq 0}{E(-X_t)^+} = 0
\end{align}
hence  there  exists a  $P$-Null  set $A$ such that
\begin{align}
Z_\infty := \lim_{t \to \infty} (-X_t)\mathbb I_{\Omega \backslash A}. \\
\text{(By the sub-Martingale convergence theorem.)}
\end{align} (see for example this post).
But  $Z_\infty$ is  $\mathscr{F}_\infty/\mathcal B_{\mathbb R}$-Measurable,
then
\begin{align}
X_\infty := \lim_{t \to \infty} X_t\mathbb I_{\Omega \backslash A}
\end{align}
And $X_\infty = -Z_\infty$ is also  $X_\infty$ is measurable．
According to Fatou's lemma,
\begin{align}
\forall t \geq 0,   &  \,  \int_A X_\infty\ dP \leq \liminf_{\substack{n \to \infty \\ n > t}} \int_A X_n\ dP \leq \int_A X_t\ dP.
\end{align}
This implies
$\left(X_t  \right)_{0 \leq t < \infty}$  is $\mathscr{F}_t$-supermartingale. 
