Stochastic Processes: Measurable vs Progressively Measurable

Question

I'm reading Karatzas + Shreve, Brownian Motion and Stochastic Calculus.

The definition for measurability is given in 1.6: The stochastic process $X$ is called measurable if, for every $A \in \mathscr{B}(\mathbb{R}^d)$ the set $\{ (t, \omega) ; X_t(\omega) \in A \}$ belongs to the product $\sigma$-field $\mathscr{B}([0,\infty)) \otimes \mathscr{F}$. In other words, if the following mapping is measurable:

\begin{align*} (t, \omega) \mapsto X_t(\omega): ([0,\infty) \times \Omega, \mathscr{B}([0,\infty)) \otimes \mathscr{F}) \to (\mathbb{R}^d, \mathscr{B}(\mathbb{R}^d)) \\ \end{align*}

The definition for progressively measurable is given on the next page in 1.11: The stochastic process $X$ is called progressively measurable with respect to filtration $\{ \mathscr{F}_t \}$ if for each $t \ge 0$ and $A \in \mathscr{B}(\mathbb{R}^d)$, the set $\{ (s,\omega); 0 \le s \le t, \omega \in \Omega, X_s(\omega) \in A \}$ belongs to the product $\sigma$-field $\mathscr{B}([0,t]) \otimes \mathscr{F}_t$. In other words, if the following mapping is measurable for each $t \ge 0$:

\begin{align*} (s, \omega) \mapsto X_s(\omega): ([0,t] \times \Omega, \mathscr{B}([0,t]) \otimes \mathscr{F}_t) \to (\mathbb{R}^d, \mathscr{B}(\mathbb{R}^d)) \\ \end{align*}

These definitions look awfully similar. Can someone help clarify the difference between these two?

Answer 1

"Progressive" is "adapted to $(\mathscr F_t)$ in a useful way".

Suppose (for example) that $X$ is progressively measurable and $|X_t(\omega)|\le C$ for all $(t,\omega)$. Then by Fubini/Tonelli, the integral $$ Y_t(\omega):=\int_0^t X_s(\omega)\,ds $$ defines an $\mathscr F_t$-measurable random variable, for each $t>0$.

If $X$ is only known to be measurable, then you can conclude that $Y_t$ is $\mathscr F$-measurable for each $t$, but not more.