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Let $\tau$ be a stopping time and fix $t \in \mathbb{R}_+$. Let $A \in \mathcal{F}_t$ satsify $A \subseteq \{\tau \leq t\}$.

Now, I want to prove that $A \in \mathcal{F}_\tau$.

A stopping time is a random variable $\tau: \Omega \to [0,\infty]$ such that $\{\omega: \tau(\omega) \leq t\} \in \mathcal{F}_t$ for all $0 \leq t < \infty$, where $\mathcal{F}_\tau = \{A \in \mathcal{F} : A \cap \{\tau \leq t\} \in \mathcal{F}_t \text{for all $0 \leq t < \infty$} \}.$

Does the prove follows from the fact that since, $A \subseteq \{\tau \leq t\} \in \mathcal{F}_t$, $A \cap \{\tau \leq t\}\in \mathcal{F}_t$. So, $A \in \mathcal{F}_\tau$?

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    $\begingroup$ Your proof is correct only if your first statement holds for every $t>0$ and not for only one fixed $t$, as your statement is ambiguous in this regard I can't tell if it is correct or not. Best regards $\endgroup$
    – TheBridge
    Commented Mar 8, 2016 at 16:41

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Your hope in unfounded, in general. Here's an example. Let $(\mathcal F_s)_{s\ge 0}$ be the natural filtration of a Brownian motion $B=(B_s)_{s\ge 0}$ starting at $B_0=0$. Let $\tau=\inf\{s:B_s=1\}$. Now fix $t>0$ and define $A=\{B_t\le 0, \tau\le t\}$. Then $A\subset\{\tau\le t\}$, $A\in\mathcal F_t$, but $A\notin\mathcal F_{\tau}$.

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  • $\begingroup$ Hmmpz, true that. The question is, which statements are founded and result in the prove? $\endgroup$
    – iJup
    Commented Mar 8, 2016 at 17:16

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