# Stochastic Processes: Measurable vs Progressively Measurable

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?

"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.