Uniform Integrability and Supermartingal Process

Question

Suppose $\{X_t, \mathcal{F}_t: 0\leq t<\infty\}$ is a right-continuous and nonnegative supermartingal. I want to show that $\lim_{t\to\infty}X_t(\omega)$ exists a.e. and that $\{X_t, \mathcal{F}_t: 0\leq t\leq\infty\}$ is also a supermartingale (here we are defining $X_\infty(\omega)=\lim_{t\to\infty}X_t(\omega)$ when this value exists).

This is one of the very first problems I'm trying to solve about this subject, so detailed answers are welcome.

My attempt: first of all, I'm allowed to use the following result

Theorem (Submartingale Convergence): Let $\{X_t, \mathcal{F}_t: 0\leq t<\infty\}$ be a right-continuous submartingale and assume $\sup_{t\geq 0} E[X_t^+]<\infty$. Then $X_\infty(\omega):=\lim_{t\to\infty}X_t(\omega)$ exists a.e. and $E[\ |X_\infty|\ ] <\infty$.

My idea was to use the fact that $-X_t$ is submartingale. Of course it's is right-continuous. Also, note that $-X_t$ is negative, so $-X_t^+=\max\{0,-X_t\} = 0$ for all $t\geq 0$, therefore $\sup_{t\geq 0} E[-X_t^+] = 0<\infty$. Now we can use the theorem to define $-X_\infty(\omega)=\lim_{t\to\infty}-X_t(\omega)$ a.e., then we have $X_\infty$ satisfying $E[\ |X_\infty|\ ] < \infty$.

This argument establishes the first part of the problem. I have a text here with another solution, but it uses results that I want to avoid. My first question is about my solution, is it right?

For the second part (show that $\{X_t, \mathcal{F}_t: 0\leq t\leq\infty\}$ is also a supermartingale), the idea in the text is to show that $\{X_t\}_{t\geq 0}$ is uniformly integrable, then

$$E[X_\infty|\mathcal{F}_t] = \lim_{s\to\infty}E[X_s|\mathcal{F}_t]\leq X_t.$$

About that, my initial problem is to understand why uniform integrability implies $$E[X_\infty|\mathcal{F}_t] = \lim_{s\to\infty}E[X_s|\mathcal{F}_t].$$

I didn't try to show uniform integrability yet because I don't even know how to use this to show the process is supermartingale.

Thanks.

Answer 1

The first part is fine.

For the second part, uniform integrability isn't needed. Recall the so-called "conditional monotone convergence theorem":

If $Y_m \ge 0$ and $Y_m \uparrow Y$, then $E[Y_m \mid \mathcal{F}] \to E[Y \mid \mathcal{F}]$ a.s.

Fix $t$, and let $t= t_0 \le t_1 \le t_2 \le \dots$ with $t_n \uparrow \infty$. Let $Y_m = \inf_{n \ge m} X_{t_n}$. Note $Y_m \ge 0$ and $Y_m \uparrow X_\infty$.

Now for any $m$ we have $Y_m \le X_{t_m}$ so $E[Y_m \mid \mathcal{F}_t] \le E[X_{t_m} \mid \mathcal{F}_t] \le X_t$ a.s. As $m \to \infty$ the left side converges a.s. to $E[X_\infty \mid \mathcal{F}_t]$.