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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.

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  • $\begingroup$ I think I got the point but still I'm not 100% sure. As continuous functions are Borel-measurable could a process that is adapted but not progressively measurable be e.g. $X(t) = {{\bf{1}}_{\{ \sin (t) \le \cos (t)\} }}$ ? I also assume that continuous stochastic processes involving Wiener process $W(t)$ are always progressively measurable (and by definition also adapted). Am I correct? $\endgroup$ – Fragile Jul 20 '14 at 19:14
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    $\begingroup$ If the process has continuous sample paths (actually just left continuous or right continuous is enough), then adapted and progressively measurable are the same thing. See Proposition 4.8 in Revuz & Yor. $\endgroup$ – Martin Hairer Jul 20 '14 at 19:26
  • $\begingroup$ OK, thanks! I mostly deal with Cadlag & Browninan processes in finance and so all of them should be progressively measurable. $\endgroup$ – Fragile Jul 20 '14 at 19:40
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    $\begingroup$ @MartinHairer do you know an example of adapted measurable process which is not progressively measurable? The hypothesis of adapted excludes the example of Stephan Sturm Since adapted means that $X: [0, \infty) \times \Omega \to E$ is $\mathcal{B}[0,\infty] \otimes \mathcal{F}_\infty$ measurable $\endgroup$ – Conrado Costa Aug 25 '15 at 1:25
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@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.

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  • $\begingroup$ Please concider addign at least a good part of the answer from the link to your answer so your answer here won't become invalid in case the url to your destination changes. $\endgroup$ – yukashima huksay Jun 22 '18 at 18:07
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Theorem T46 in Meyer's book "Probability and Potentials" says: For any $(X_t)$ a measurable real-valued process adapted to the family of $\sigma$-fields $(\mathcal{F}_t)$, there exists a modification of the process $(X_t)$ progressively measurable with respect to the same family $(\mathcal{F}_t)$.

In the above example, you can easily see this is true. Change $A$ to a modification $\bar{A}=\{(x,0),x\in[0,\frac{1}{2}]\}$.

There are important relations between them: (Continuous+adapted)$\rightarrow$(mean-square continuous+adapted)$\rightarrow$predictable$\rightarrow$optional$\rightarrow$progressive$\rightarrow$adapted.

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    $\begingroup$ The theorem in Meyer is true, but Meyer's proof is incorrect. $\endgroup$ – iglick Apr 16 '18 at 6:09

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