I want to show the following property:
Let $(X_t)_{t\geq 0}$ be a uniformly integrable martingale, then there exists a random variable $X_\infty \in L^1(\Omega)$ with $X_\infty = \lim_{t \to \infty} X_t$ a.s. and such that $X_t = E[X_\infty \mid \mathscr{F}_t]$.
Here is my proof:
The first statement follows from the fact that all $X_t$ are integrable together with the martingale convergence theorem.
By dominated convergence,
\begin{align*}
\lim_{t \to \infty} \int_\Omega |X_t – X_\infty| dP = \int_\Omega\lim_{t \to \infty} |X_t – X_\infty| dP =0,
\end{align*}
so $X_t \to X_\infty$ in $L^1$ as well. It remains to show the conditional expectation property. But since for any $A_s \in \mathscr{F}_s$ we have
\begin{align*}
\int_{A_s} X_t dP = \int_{A_s} X_s dP,
\end{align*}
the property follows from $\int_{A_s} X_t dP \stackrel{t \to \infty}{\longrightarrow} \int_{A_s} X_\infty dP$.
I didn't use the uniform integrability…But I can't see my mistake. Can somebody enlighten me?
Best Answer
The existence of $X_{\infty}=\lim_{t \rightarrow \infty} X_t$ a.s. is because of the martingale convergence theorem (since a uniform martingale is a martingale bounded in $L^1$).
For a proof of $X_t \rightarrow X_{\infty}$ in $L^1$, you can check my answer here: if $M$ is a UI - martingale then $M_t \rightarrow M_{\infty}$ in $L^1$.
Once we have proved that $X_s \rightarrow X_{\infty}$ in $L^1$ we can prove the last claim. Let $0 \leq s \leq t$ and let $A \in \mathscr{F}_s.$ Then \begin{align*} E[(X_\infty - X_s)1_A] &= E[(X_\infty-X_t+X_t - X_s)1_A] \\ &=E[(X_\infty-X_t)1_A]+E[(X_t-X_s)1_A]\\ &= E[(X_\infty-X_t)1_A] \\ &\leq E[|X_\infty-X_t|] \rightarrow 0 \end{align*} as $t \rightarrow \infty.$