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Let $\rho$ be a $\mathscr{F}_t$ stopping time. Then how do we show that $\mathscr{F}_\rho = \sigma(\cup_i (\mathscr{F}_\rho \cap \mathscr{F}_i))$?

I can show this if $\rho$ is a bounded stopping time, since each $A \in \mathscr{F}_\rho$ is $\cup_i (A \cap \{\rho \le i\})$, and $A \cap \{\rho \le i\} \in \mathscr{F}_\rho \cap \mathscr{F}_i$.

However, if $\{\rho = \infty\} \neq \emptyset$, then I am stuck. I would greatly appreciate any help.

nomadicmathematician
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  • Noting that $\mathcal{F}{\varrho} \cap \mathcal{F}_i = \mathcal{F}{\varrho \wedge i}$, this is a duplicate of this question (the proof is somewhat easier if $\varrho$ is a finite stopping time, see here). – saz Jun 26 '20 at 04:19

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