Karatzas and Shreve - Problem 3.3.19

Question

I'm struggling here to solve this problem, but with no success.

I was able to prove a $\implies$ b, but the next implication is troubling me. In the book, they give a solution, but I think there is some details missing in this solution and this details are the big problem. Here is their solution.

First, saying that $X_t$ converges in $L^1$ to $X_\infty$ doesn't imply (as fas as I know) that $X_\infty$ is in $L^1$. If we want to show $\{X_t,\mathcal{F}_t: 0\leq t\leq \infty\}$ is a submartingal, we need $X_\infty$ to be in $L^1$.

Second, what is going on when he makes $t\to\infty$ ? Given any $A\in\mathcal{F}_s$, is he saying $\lim_{t\to\infty}\int_A X_t\ dP = \int_AX_\infty\ dP$ ? If this is the case, I don't know why this is valid.

Thank you!

Answer 1

First: $\|X_\infty\|_1\le\|X_\infty-X_t\|_1+\|X_t\|_1$ and if $X_t$ converges in $L^1$ (to $X_\infty$) then $\sup_t\|X_t\|_1<\infty$.

Second: $\Big|\int_A X_t\,dP-\int_A X_\infty\,dP\Big|\le\int_A|X_t-X_\infty|\,dP\le\|X_t-X_\infty\|_1$.