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Let $X_s(\omega)$ be measurable and adapted.

Under what conditions will the process $$ F_{t}(\omega) = \int_0^t X_s(\omega) \, ds $$ also be adapted?

To me it seems that adaptedness and measurability should be enough but at the bottom of page 133 in Karatzas and Shreve they say this is not enough. Why?

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Assuming that $\int_0^t|X_s(\omega)|\,ds<\infty$ for all $t>0$ and all $\omega$, and that the filtration satisfies the usual conditions, the process $F_t:=\int_0^t X_s\,ds$ is well defined and adapted (even predictable, being continuous). This matter is discussed in the paper "Un exemple de processus mesurable adapté non progressif" by G. Letta: https://link.springer.com/chapter/10.1007%2FBFb0084150. Letta provides an example in which $F$ is not adapted when the filtration doesn't satisfy the usual conditions. Another source of detail on these things is Chapter 3 of Introduction to Stochastic Integration by K.L. Chung and R.J. Williams.

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  • $\begingroup$ Thanks for your helpful answer. The reason it took me so long to accept it was I was not sure I had understood it. In particular, I was still not sure why in Karatzas and Shreve they explicitly state (on page 133) that the integral process defined above need not be adapted, even when X is adapted. My understanding now is that in their book at that point they are not assuming that the filtration satisfies the usual conditions, which would be consistent with your answer. $\endgroup$
    – user152718
    Commented Mar 11, 2020 at 14:03
  • $\begingroup$ K&S do assume "throughout this chapter" the usual conditions. And they do argue the point in the discussion "part (c)" starting at the foot of page 133 and continuing on to page 134. $\endgroup$
    – John Dawkins
    Commented Mar 11, 2020 at 15:15
  • $\begingroup$ Yes I had seen that they argue this point there which is what was confusing me. It suggested to me that the adaptedness of the process we have discussed above relied on some specific aspect of the context at that part of the book, rather than being a more general point. Anyway thanks for your help, I think I understand now. $\endgroup$
    – user152718
    Commented Mar 11, 2020 at 15:21

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