[Math] Progressively measurable vs adapted

Question

I often see in stochastic calculus books the terms 'adapted process' and 'progressively measurable process'. I know there is a small difference between them (every progressively measurable process is adapted but the converse is not necessarily true) but I can't get it from the mathematical definitions.

The definition of progressively measurable process is the following:
A stochastic process $X$ defined on a filtered probability space $(\Omega ,{\mathcal F},{({{\mathcal F}_t})_{t \ge 0}},P)$ is progressively measurable with respect to ${({{\mathcal F}_t})_{t \ge 0}}$, if the function $X(s,\omega):[0,t]\times \Omega \rightarrow \mathbb{R}$ is $\cal{B}([0,t]) \times \cal{F}_t$ measurable for every $t\ge 0$.

The definition of adapted process:
A stochastic process $X$ on $(\Omega ,{{\mathcal F}},{({{\mathcal F}_t})_{t \ge 0}},P)$ is adapted to the filtration ${({{\mathcal F}_t})_{t \ge 0}}$ (or ${\mathcal F}_t$-adapted) if $X(t)\in {\mathcal F}_t$ for each $t \ge 0$.

Can someone explain me the difference in simple words? I think I understand the definition of the adapted process quite well but I'm probably confused of the role of the Borel sigma algebra in the definition of the progressively measurable process.

Answer 1

@Conrado Augusto: You may find an example of a measurable, adapted but not progressively measurable process in these lecture notes of Michael Scheutzow (Example 1.38):

Now we provide an example of an adapted and measurable process which is not progressive.

Example 1.38. Let $(\Omega, \mathcal F ) = ([0, 1], \mathcal L)$, where $\mathcal L$ denotes the $\sigma$-algebra of Lebesgue sets in [0, 1] which – by definition – is the completion of $\mathcal B[0,1]$ with respect to Lebesgue measure. Let $\mathcal L_0$ be the $\sigma$-algebra on $[0,1] $ containing all sets in $\mathcal L$ which have Lebesgue measure $0$ or $1$. Define $\mathcal F_t :=\mathcal L_0$ for all $t\ge0$. Define the set $A\subset[0,\infty)\times\Omega$ by $A=\{(x,x)\mid x\in[0,1/2]\}$. Then $A \in \mathcal B[0,\infty )\otimes \mathcal F_t$ but for each $t > 0$, $A\cap ([0,t]\times\Omega ) \not\in \mathcal B[0,t]\otimes \mathcal F_t$ (otherwise the projection of the intersection onto $\Omega$ would be in $\mathcal F_t$ by Theorem 1.36 which is however not the case). Therefore the indicator of $A$ is measurable and (check!) adapted but not progressive.