After learning martingale theory for some time, I am confused with martingale with stopping time. Assume $T$ is a stopping time of the filtration space $\left(\Omega,\mathscr{F},\left\{ \mathscr{F}_{n}\right\} _{n\in \mathbb{N}},P\right)$ and $\left\{ X_{n}\right\} _{n\in\mathbb{N}}$ is a (sub-, super-,)martingale, then we define $X_{T}\left(\omega\right)=X_{T\left(\omega\right)}\left(\omega\right)$.
My question is that since $T\in\mathbb{N}\cup\left\{ \infty\right\}$, if $P\left(T=\infty\right)>0$, what is $X_{T}$ if $T=\infty$ ? If $X_{n}$ converges to $X_{\infty}$ a.e., it is trivial. What if $X_{n}$ doesn't converge? I have searched many textbooks, but none of them explains it clearly. Thanks in advance.
$X_T$ is defined only when one of the following two conditions holds:
1) $T$ is a finite stopping time
2) The martingale converges to a random variable $X_{\infty}$ in some sense ( a.s., in mean etc).
In cae 2) is is understood that we are using $X_{\infty}$ to define $X_T$.
You cannot define $X_T$ without either of these assumptions.