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Indeed some results of this Acta paper were announced in the Comptes Rendus note C. R. Acad. Sci., Paris, Sér. I 293, 677-680 (1981) (and possibly in the Maurey-Schwartz seminar as well).
Thanks for the clarification, but ... may I suggest to use precise wording in a question? For the record, in stochastic analysis a process $(\xi_s)_s$ is called measurable if $(s,\omega)\mapsto \xi_s(\omega)$ is measurable for the product $\sigma$-algebra. Incidentally, $\xi_s(\omega)$ is a number, not a process. Returning to what you wanted to ask, it seems to me now that the question is: If $s\mapsto K(s,\omega)$ is measurable (and integrable) for each $\omega$ and $\omega\mapsto K(s,\omega)$ is measurable for each $s$, need $\omega\mapsto \int_0^t K(s,\omega)\,ds$ be measurable?
I think your question doesn't really have to do with filtrations and stochastic processes. You are given a mapping $K: (s,\omega)\mapsto \xi_s(\omega)$ on $[0,t]\times \Omega$, and you seem to be assuming ("measurable random process") that $K$ is measurable for the product $\sigma$-algebra of $[0,t]\times \Omega$. But then it is embodied in the proof of Fubini's theorem that $\omega\mapsto \int_0^t K(s.\omega)\,ds$ is measurable. Maybe I'm missing something?
@TarasBanakh: There are infinite compact $K$ for which $C(K)$ is a dual space: these are precisely the hyperstonean $K$, e.g., $\beta\mathbb{N}$. (On the other hand there are non-dual $C(K)$ for which the unit ball is the norm-closed convex hull of its extreme points, e.g. $\alpha\mathbb{N}$. These are precisely the totally disconected $K$.)
