TITLE: Doob's Optional stopping time Theorem
QUESTION [1 upvotes]: There is a version of Doob's Optional stopping time theorem, which is stated as:
Let T be a stopping time. Let $X$ be a martingale. Then $X_T$ is integrable and $$E[X_T]=E[X_0]$$ if X is bounded and T is a.s. finite.
Question: Why do we need to require $T$ is a.s. finite? As $X_n$ is bounded, it will converge against some $X_{\infty}$. And we know that $X_{T\land n}$ is a martingale and use dominated convergence theorem and the fact $lim_{n\rightarrow \infty} X_{T\land n}=X_T$, one can proof the claim. Which argument would fail to work if $T$ can be infinite?

REPLY [0 votes]: The subset  $A=\bigcup_{n\geq 1} \{T= n\}$ should have full measure in the probability space. The sequence $X_{T\wedge n}$ converges pointwise on $A$ only. You will never attain the subset on which $T=+\infty$.
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