Let $(X_t)$ a progressively measurable process, i.e. $[0,t]\times \Omega \ni (t,\omega )\mapsto X_t(\omega )\in (\mathbb R, \mathcal B(\mathbb R))$ is $\mathcal B([0,t])\otimes \mathcal F$ measurable. Prove that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$.
Q1) What does it mean that $X_T$ is $\mathcal F_T$ measurable on $\{T<\infty \}$ ? Does it mean that for all Borel set $B$, we have $(X_T\boldsymbol 1_{\{T<\infty \}})^{-1}(B)\in \mathcal F_T$ ?
Suppose my definition is correct. Then that's enough to prove
$\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\in \mathcal F_T$ for all $x$. ($*$)
Q2) Any idea on how to prove ($*$) ?
I don't see how I can prove that
$\{X_T\boldsymbol 1_{\{T<\infty \}} \} \in \mathcal F_\infty $ and $\{X_T\boldsymbol 1_{\{T<\infty \}}\leq x\}\cap \{T\leq t\}\in \mathcal F_t$ for all $t$.
Proof: provided that $T$ takes a finite value, for every positive value $t$ it will take, we will be able to decide whether $X_t$ belongs to a Borel set $B$, whatever set we choose.
Formaly, we need to prove for all Borel set $B$ and $t\ge0$ that $$ \left\{\omega\in\Omega,\;X_{T(\omega)}(\omega)\in B\right\}\cap\left\{\omega\in\Omega,\;T(\omega)\le t\right\} $$ is $\mathcal F_t$-measurable. But \begin{alignat*}{2} \left\{X_{T}\in B\right\}\cap\left\{T\le t\right\} & = \left\{\omega\in\Omega:\; T(\omega) = s, s \le t, X_s(\omega) \in B \right\} \end{alignat*} which is $\mathcal F_t$-mesurable.