If $T$ is a stopping time, prove that $X_T$ is $\mathcal F_T$ measurable. 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 $t$.  
 A: Hint
Let $B$ a Borel set.
$$\{X_{T}\in B\}\cap \{T\leq t\}=\{X_{T\wedge t}\in B\}\cap \{T\leq t\}.$$
Therefore, if you can prove that the stopped process is progressively measurable, the claim follow.
A: 
As mentionned Surb, it isn't a complete proof as it is for discrete and not continuous time. I leave it for sake of learning purpose.

After some research by myself, I was finally able to solve that question :
Write :
$$
\begin{align}
\small (X^T_n)^{-1}(B)&=\bigcup_{k\geq 0}(X_{k\wedge n}^{-1}(B)\cap \{\omega\mid T(\omega)=k\})
\\&=\bigcup_{k=0}^{n-1}(X_k^{-1}(B)\cap\{\omega\mid T(\omega)=k\})
\cup (X_n^{-1}(B)\cap \{\omega\mid T(\omega)\leq n\}^c)
\\& \in\mathcal F_n.
\end{align}
$$
Notice that the first equality is true iff $ T < \infty $. If you need more detail let me know.
A: 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.
