Prove that collection $\mathcal{F}_{\tau}$ associated with a stopping time $\tau$ is a $\sigma$-algebra

Question

$\mathcal{F}_{\tau}$ is defined as $$ \mathcal{F}_\tau = \{ A\in \mathcal{F}: A \cap \{ \tau \leq t \}\in \mathcal{F}_t, \forall t\in T \}. $$

I've found in one book that proving that it is a $\sigma$-algebra is "obvious". Still, I have troubles so I would be thankful for any comments on my present solution. I started from the definition of a $\sigma$-algebra:

1. $\emptyset \in \mathcal{F}_\tau$. Ok, if I intersect $\emptyset$ with $\{ \tau \leq t \}$ I will get $\emptyset$ which is in $\mathcal{F}_t$ by definition, so $\emptyset \in \mathcal{F}_\tau$.

2. $A \in \mathcal{F}_\tau \implies \mathcal{F} \backslash A \in \mathcal{F}_\tau$.

3. $A_1, A_2, \ldots \in \mathcal{F}_\tau \implies \bigcup A_i \in \mathcal{F}_\tau$.

If $A_1, A_2, \ldots \in \mathcal{F}_\tau$ then $$ \begin{split} A_1 & \cap \{ \tau \leq t \}\in \mathcal{F}_t,\\ A_2 & \cap \{ \tau \leq t \}\in \mathcal{F}_t,\\ & \ldots \\ & \forall t\in T. \end{split} $$

Then $$ [(A_1 \cap \{ \tau \leq t \}) \cup (A_2 \cap \{ \tau \leq t \}) \cup \ldots ] \in \mathcal{F}_t $$ and by the set formula this is same as $$ [ (A_1 \cup A_2 \cup \ldots) \cap \{\tau \leq t \}] \in \mathcal{F}_t, \hspace{0.2cm} \forall t\in T, $$

so $(A_1 \cup A_2 \cup \ldots)\in \mathcal{F}_\tau$. Is my reasoning correct? If so, what is the way to prove 2. ? Thanks in advance for any help.

Answer 1

Yes, the idea is correct. As for 2, it follows from the fact that $(\tau\leq t)\in\mathcal{F}_t$, which is due to the definition of stopping time.

Indeed, suppose that $A\cap (\tau\leq t)\in \mathcal{F}_t$ for all $t$. Then, for each $t$

note that

$(\tau\leq t)\cap A^c=(\tau\leq t)\cap [A\cap (\tau\leq t)]^c$.

Since $(\tau\leq t), [A\cap (\tau\leq t)]^c\in\mathcal{F}_t$, the result follows.