Progressive measurability of stopped process Let $(\mathcal{F}_t)_{t\in \mathbb{R}_+}$ be a filtration and let $X$ be a stochastic process progressively measurable with respect to $(\mathcal{F}_t)_{t\in \mathbb{R}_+}$. Let $T$ be a stopping time and define the stopped process $X^T$ by $(X^T)_t=X_{T\wedge t}$. It is straightforward to show (and done so in most books on stochastic calculus) that $X^T$ is progressive with respect to $(\mathcal{F}_t)_{t\in \mathbb{R}_+}$. 


My question is as follows: Is $X^T$ progressive with respect to $(\mathcal{F}_{t\wedge T})_{t\in \mathbb{R}_+}$?  

Revuz and Yor, Continuous Martingales and Brownian Motion, states my question as a proposition left for the reader (Chapter 1, §4, Proposition 4.10), but I have not been able to give a proof.
 A: Not sure if this question is still actual, but I decided to put a proof here.
The process $X=(X_t)_{t\in\mathbb{R}_+}$ takes values in a measurable space $(E,\mathcal{E})$ and is progressively measurable with respect to the filtration $\mathcal{F}=(\mathcal{F}_t)_{t\in\mathbb{R}_+},$ $T:\Omega\to [0,\infty]$ is $\mathcal{F}-$stopping time. We want to show that for any $t\geq 0$ and any $A\in\mathcal{E}$
$$
\{(s,\omega)\in [0,t]\times \Omega: X_{s\wedge T(\omega)}(\omega)\in A\}\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.  \ \ \ (1)
$$
I will use several auxiliary statements.

*

*For any $D\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t},$
$$
D\cap (\mathbb{R}_+\times \{T>t\})\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.
$$
Proof. It is enough to consider the set $D$ of the form $D=B\times C,$ where $B\in\mathcal{B}([0,t]),$ $C\in\mathcal{F}_t.$ Observe that $C\cap \{T>t\}\in \mathcal{F}_{t\wedge T}.$ Indeed,
$$
C\cap \{T>t\}\cap \{t\wedge T\leq s\}=\begin{cases}
C\cap \{T>t\}\in\mathcal{F}_t\subset \mathcal{F}_s, \ \mbox{ if } t\leq s \\
\emptyset \in\mathcal{F}_s, \ \mbox{ if } t>s
\end{cases}.
$$
So,
$$
D\cap (\mathbb{R}_+\times \{T>t\})=B\times (C\cap \{T>t\})\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.
$$


*Let $\Delta_t=\{(s,\omega)\in[0,t]\times \Omega: T(\omega)>s\}.$ Then $\Delta_t\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.$
Proof. Consider $(s,\omega)\in \Delta_t.$ Then either $s=t$ and $t<T(\omega),$ or $s<t$ and then $s<t \wedge T(\omega).$ In the latter case there exists rational $r\in (0,t)$ such that $s\leq r< T(\omega).$ So,
$$
\Delta_t=\left(\{t\}\times \{T>t\}\right)\cup \bigcup_{r\in \mathbb{Q}\cup (0,t)}\left([0,r]\times \{T>r\}\right)\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.
$$
Indeed, in statement 1 we saw that $\{T>r\}\in\mathcal{F}_{r\wedge T}.$


*$\tilde{\Delta}_t=\{(s,\omega)\in [0,t]\times \Omega: T(\omega)\leq s\}\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.$
Proof. $\tilde{\Delta}_t=([0,t]\times \Omega)\setminus \Delta_t,$ where $\Delta_t$ is from previous statement.


*$X_{t\wedge T}$ is $\mathcal{F}_{t\wedge T}-$measurable.

Proof. This is proved e.g. in D.Revuz and M. Yor, Continuous Martingales and Brownian Motion (Chapter 1, §4, Proposition 4.9).
Now we prove (1). Observe that $X_{s\wedge T(\omega)}(\omega)\in A,$ if and only if one of the three following cases takes place:

*

*$T(\omega)\leq s$ and $X_{T(\omega)}(\omega)\in A;$


*$s=t<T(\omega)$ and $X_t(\omega)\in A;$


*$s<t\wedge T(\omega)$ and $X_s(\omega)\in A.$
Correspondingly, we obtain a decomposition:
$$
\{(s,\omega)\in [0,t]\times \Omega: X_{s\wedge T(\omega)}(\omega)\in A\}=D_1\cup D_2\cup D_3,
$$
where
$$
D_1=\{(s,\omega)\in [0,t]\times \Omega: T(\omega)\leq s, X_{T(\omega)}(\omega)\in A\},
$$
$$
D_2=\{(s,\omega)\in [0,t]\times \Omega: s=t<T(\omega), X_{t}(\omega)\in A\},
$$
$$
D_3=\{(s,\omega)\in [0,t]\times \Omega: s<t\wedge T(\omega), X_{s}(\omega)\in A\}.
$$
We will verify that each $D_j\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T},$ $j=1,2,3.$
Proof for $D_1.$ If $T\leq s,$ then $T=t\wedge T.$ Hence,
$$
D_1=\{(s,\omega)\in [0,t]\times \Omega: T(\omega)\leq s, X_{t\wedge T(\omega)}(\omega)\in A\}=
$$
$$
=\tilde{\Delta}_t \cap ([0,t]\times \{X_{t\wedge T}\in A\})\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}
$$
by statemens 3 and 4.
Proof for $D_2$. If $s=t<T,$ then $t=t\wedge T.$ Hence,
$$
D_2=\{(s,\omega)\in [0,t]\times \Omega: s=t, X_{t\wedge T(\omega)}(\omega)\in A, s<T(\omega)\}=
$$
$$
=(\{t\}\times \{X_{t\wedge T}\in A\})\cap \Delta_T\in \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}
$$
by statements 2 and 4.
Proof for $D_3$. If $s<t\wedge T,$ then there exists rational $r\in (0,t)$ such that $s\leq r <T.$ Hence,
$$
D_3=\bigcup_{r\in\mathbb{Q}\cap (0,t)}\left\{(s,\omega)\in[0,t]\times \Omega: s\leq r, X_s(\omega)\in A, r<T(\omega)\right\}=
$$
$$
=\bigcup_{r\in\mathbb{Q}\cap (0,t)}\left(\left\{(s,\omega)\in[0,r]\times \Omega: X_s(\omega)\in A\right\}\cap (\mathbb{R}_+\times \{T>r\})\right)\in  \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}.
$$
Indeed, the set $\{(s,\omega)\in[0,r]\times \Omega: X_s(\omega)\in A\}\in \mathcal{B}([0,r])\otimes \mathcal{F}_r$ by progressive measurability of $X.$ So, by the statement 1,
$$
\{(s,\omega)\in[0,r]\times \Omega: X_s(\omega)\in A\}\cap (\mathbb{R}_+\times \{T>r\})\in \mathcal{B}([0,r])\otimes \mathcal{F}_{r\wedge T}\subset \mathcal{B}([0,t])\otimes \mathcal{F}_{t\wedge T}
$$
P.S. Maybe there exists a more elegant proof, e.g. if we want to prove  $(\mathcal{F}_t)_{t\in\mathbb{R}_+}-$progressive measurability of the stopped process $X^T,$ then it is enough to consider composition $g\circ f,$ where
$$
f:[0,t]\times \Omega\to [0,t]\times \Omega, \ f(s,\omega)=(s\wedge T(\omega),\omega),
$$
$$
g:[0,t]\times \Omega\to E, \ g(s,\omega)=X_s(\omega).
$$
