I think I found a way to prove a result but I have doubts about it. I may have made a mistake. ($X$ is a geometric Brownian motion and I do not want to use the strong Markov property)
I have proved that for any $t \in [0,\infty)$ $$ \mathbb E\bigg[\int_t^\infty G(X_u)~\bigg|~F_t\bigg]= U_t \tag 1 $$ for a certain function $G$ such that it is well defined and a certain process $U$.
Can I conclude that for any finite stopping time $\tau$: $$ \mathbb E\bigg[\int_\tau^\infty G(X_u)~\big|~F_\tau] = U_\tau \text{ ?} $$ I think it is true since for any $\omega \in \Omega$ and (I am not sure about this part because it looks too obvious) we get:
\begin{align} \mathbb E\bigg[\int_\tau^\infty G(X_u)~\bigg|~F_\tau\bigg](\omega)&= \mathbb E\bigg[\int_{\tau(\omega)}^\infty G(X_u)~\bigg|~F_{\tau(\omega)}\bigg](\omega) \tag 2 \\ &= U_{\tau(\omega)}(\omega) \end{align} since $\tau$ is finite and by $(1)$.
But is $(2)$ true and do you know how to prove it? I read on one forum that it is true and it is called "local property of conditional expectation" but I did not find it anywhere else.
Thank you in advance for your help.
Edit: Does anyone know? It is important because I proved a result using this property and my proof will be wrong if this property is not correct.