Is $A \cap \{\tau > t\} \in \mathcal{F}_{t}$?

Question

Assume $A \in \mathcal{F}_{\tau}$, where $\tau$ is a stopping time. By definition of the $\sigma$-algebra of the stopping time, we know that $$A \cap \{\tau \leq t \} \in \mathcal{F}_t.$$ Question: Is $A \cap \{\tau > t \} \in \mathcal{F}_t$? If it is, how to prove it?

Answer 1

No, this need not be the case. Let $s < t$ and consider the constant stopping time $\tau = t$. If $A \in \mathcal F_t$ and $A \not \in \mathcal F_s$, we have $A \cap \{\tau > s\} = A \not \in \mathcal F_s$.

For a less trivial example of why this should not be the case intuitively, consider a Brownian motion $B_t$ and $\tau := \inf\{t \ge 0 : |B_t| = 1\}$. Then the event $A = \{B_\tau = 1\}$ is in $\mathcal F_\tau$, but $A \cap \{\tau > 0\} = A$ is not in $\mathcal F_0$.