By definition, $\mathcal{F}_{T-}=\mathcal{F}_0 \vee \sigma(A\cap \{ t<T\}, A \in \mathcal{F}_t, t \in [0,\infty[)$.

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Why is $X_{T-}$ is $\mathcal{F}_{T-}$ measurable?


1 Answer 1


If $Y$ is predictable, then $Y_T$ is ${\cal F}_{T-}$-measurable. The left-limit process $Y$, as defined above, is left-continuous and adapted, hence predictable.

See this source, Lemma 1, for a proof using a monotone class argument.

  • $\begingroup$ Thanks for the answer ;) $\endgroup$ Jun 25, 2021 at 7:34
  • $\begingroup$ @Anoldmaninthesea. I'm glad it helps. $\endgroup$
    – ir7
    Jun 26, 2021 at 18:29

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