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Jan 28, 2023 at 21:08 history edited Christophe Leuridan CC BY-SA 4.0
Correction
Jan 28, 2023 at 21:05 comment added Christophe Leuridan Yes you are right. I read the assumptions too quickly and I will modify my answer.
Jan 28, 2023 at 13:56 comment added Anton Sorokovsky The following is not necessary true "For every $t$, the map $(s,\omega) \mapsto f(s,\omega)$ is measurable on $[0,t] \times \Omega$. " The process is not progressively measurable unless it somehow follow from the separability
Jan 27, 2023 at 19:55 history answered Christophe Leuridan CC BY-SA 4.0