I'm trying to figure this out, and I feel I'm pretty close to why it's the case. I just can't quite get the details to work.
Let $X$ be an adapted process on $(\Omega,\mathcal{F},\mathbb{P})$ and $T$ a finite stopping time. Show that $X_T$ is $\mathcal{F}$-measurable.
As I understand it, the process $X_T = \{X^T_n\}_{n\geq1}$ is defined as $X^T_n(\omega) := X_{T(\omega) \wedge n}(\omega)$. Since $X$ is an adapted process we have that $X_n$ is $\mathcal{F}_n$-measurable for all $n$, and since $\mathcal{F}_n \subset \mathcal{F}$ this intuitively should carry over to $X_T$. The $\mathcal{F}_n's$ are from the filtration of $X$. My problem is with $T(\omega)$ being dependant on $\omega$, and that $T$ isn't necessarily bounded which means you can't bound $T(\omega)\wedge n$.
Any tips on how I should approach this?
EDIT: some clarification