I have trouble understanding the proof of Proposition 2.18 in Karatzas/Shreve: Brownian Motion and Stochastic Calculus, which states that a $\mathcal{F_t}$-progressively measurable process $X_t$, stopped at a $\mathcal{F_t}$-stopping time $\tau$, remains progressive, i. e. $X_{t\wedge\tau}$ is $\mathcal{F_t}$-progressive.
The main idea is to decompose the map $[0,T]\times\Omega\to\mathbb{R}^d, (t,\omega)\mapsto X_{t\wedge\tau(\omega)}(\omega)$ into $M: (t,\omega)\mapsto (t\wedge\tau(\omega),\omega)$ and $(t,\omega)\mapsto X_t(\omega)$ (for all $T>0$). Then it suffices to show that $M$ is $(\mathcal{B}[0,T]\otimes \mathcal{F}_T)$-measurable, since the other map is measurable because of the assumption of progressive measurability of $X$.
The main idea is clear to me. But I don’t see how $M$ is measurable. It seems intuitively clear to me (thinking of the filtration as information), but I don’t manage to prove it formally.
Best Answer
Recall that the product sigma-algebra $\mathcal{B}[0,T] \otimes \mathcal{F}_T$ is generated by
$$\{[0,t] \times B; t \leq T, B \in \mathcal{F}_T\}.$$
Consequently, it suffices to prove that $M^{-1}([0,t] \times B) \in \mathcal{B}[0,T] \otimes \mathcal{F}_T$ for $t \leq T$, $B \in \mathcal{F}_T$. This follows from
$$\begin{align*} M^{-1}([0,t] \times B) &= \{(s,\omega) \in [0,T] \times \Omega; s \wedge \tau(\omega) \leq t, \omega \in B\} \\ &= \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau \leq t]\} \\ &\quad \cup \{(s,\omega); s \wedge \tau(\omega) \leq t, \omega \in B \cap [\tau>t]\} \\ &= ([0,T] \times (B \cap[\tau \leq t])) \cup ([0,t] \times (B \cap [\tau>t])).\end{align*}$$