Karatzas and Shreve Problem 1.3.16. Proving nonnegative right continuous supermartingale convergence

Question

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?

Answer 1

Hint: There is no need to prove uniform integrability. Just apply Fatou's lemma (for conditional expectations).

Answer 2

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.