Given the definitions:
$\mathcal{F}_\tau := \{A ∈ \mathcal{F} : A ∩ \{τ ≤ t\} ∈ \mathcal{F}_t \quad ∀t ≥ 0 \}$
$\mathcal{F}_{\tau+} := \{A ∈ \mathcal{F} : A ∩ \{τ < t\} ∈ \mathcal{F}_t\}$
How do we rigorously prove that $\mathcal{F}_\tau ⊂ \mathcal{F}_{\tau+}$ and $\mathcal{F}_\tau = \mathcal{F}_{\tau+}$ if $(\mathcal{F}_t)$ is right-continuous?
I can do the first part of this question, the inclusion, but I am stuck trying to show equality when the filtration is right-continuous.
For the secnnd part let $E \in \mathcal F_{\tau+}$. Then $E\cap \{\tau <t\} \in \mathcal F_t$ for all $t$. Hence $E\cap \{\tau <t+\frac 1 n\} \in \mathcal F_{t+\frac 1n}$. So $\bigcap_{ n\geq m} E\cap \{\tau <t+\frac 1 n\} \in \mathcal F_{t+\frac 1m}$. But this says $E \cap \{\tau \leq t\} \in \mathcal F_{t+\frac 1m}$. Since this is true for all $m$ we see that $E\cap\{ \tau \leq t\} \in \mathcal F_{t+}=\mathcal F_t$ by hypothesis.