Given a filtered probability space $(\Omega, \mathscr{F}, \{\mathscr{F_n}\}, \mathbb{P})$,
let $X = (X_n)_{n \geq 0}$ be a $(\{\mathscr{F_n}\}, \mathbb{P})$-supermartingale and $T$ be a finite $\mathbb{P}$-a.s. stopping time.
When can we say that $$X_T = \lim_{n \to \infty} X_{T \wedge n} \ \text{a.s.} \tag{*}$$?
It seems that $$\lim_{n \to \infty} T \wedge n = T \tag{**}$$ but I don't think that's enough.
I think I can say that
$$X_{T\wedge n} = ( \ \sum_{k=0}^{n-1} 1_{\{T=k\}} X_k \ ) + 1_{T\ge n} X_n$$
If so, I think we have
$$\lim_{n \to \infty} X_{T\wedge n} = \lim_{n \to \infty} ( \ \sum_{k=0}^{n-1} 1_{\{T=k\}} X_k \ ) + \lim_{n \to \infty} 1_{T\ge n} X_n$$
$$ = \lim_{n \to \infty} ( \ \sum_{k=0}^{n-1} 1_{\{T=k\}} X_k \ ) + \lim_{n \to \infty} 1_{T\ge n} \lim_{n \to \infty} X_n$$
$$ = \lim_{n \to \infty} ( \ \sum_{k=0}^{n-1} 1_{\{T=k\}} X_k \ ) + (0) \lim_{n \to \infty} X_n \text{(*)}$$
$$ = \lim_{n \to \infty} ( \ \sum_{k=0}^{n-1} 1_{\{T=k\}} X_k \ )$$
$$ = \sum_{k=0}^{\infty} 1_{\{T=k\}} X_k = X_T$$
(*) I guess I need to assume that $\lim_{n \to \infty} X_n$ exists. Is that it? $X_T = \lim_{n \to \infty} X_{T \wedge n}$ a.s. provided $\lim_{n \to \infty} X_n$ exists a.s.?
Edit: I found this link that seems to suggest $(*)$ and $(**)$:
Fix $\omega \in \Omega$ such that $T(\omega)<\infty$. Then there exists $N \in \mathbb{N}$ such that $T(\omega) \leq N$. In particular,
$$T(\omega) \wedge n = T(\omega) \qquad \text{for all } n \geq N.$$
This implies
$$X_{T \wedge n}(\omega) = X_{T}(\omega) \qquad \text{for all }n \geq N.$$
Hence, obviously,
$$\lim_{n \to \infty} X_{T \wedge n}(\omega) = X_{T}(\omega).$$
Since this holds for almost all $\omega \in \Omega$, this proves
$$\lim_{n \to \infty} X_{T \wedge n} = X_T \quad \text{a.s.}$$
Remark: Note that this argument applies to any stochastic process $(X_n)_{n \in \mathbb{N}}$; we do not need the supermartingale property.