Let {$ \xi _t(\omega), t\in[0,\infty)$} be a random process and $ \xi _t(\omega)\in \{\mathfrak F_t\}$ (some filtration). If $ \xi _t(\omega) $ is $ \mathfrak F_t $ measurable then $\int_0^t\xi _s(\omega)L(ds)$ not necessarily $ \mathfrak F_t $ measurable. I am looking for a process, that will demonstrate this statement. Thanks in advance for any tips.
2026-03-25 14:25:32.1774448732
An example of the fact that from measurability of a random process does not follow measurability of its integral
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