Show that if $\{ T_n \}_n$ is a sequence of optional times and $T = \inf_n T_n$, then $T$ is also an optional time and $\mathscr{F}_{T+} = \cap_n \mathscr{F}_{T_n+}$.
The first part is easy. We can see that $T$ is an optional time with the following relationship. For any $t \ge 0$:
\begin{align*} \{ T < t \} = \left\{ \left( \inf\limits_{n \ge 1} T_n \right) < t \right\} &= \bigcup_{n=1}^\infty \{ T_n < t \} \\ \end{align*}
Since all $T_n$ are optional, then all sets in the above are $\mathscr{F}_t$ measurable and therefore $T$ is optional.
For the second part, demonstrating that $\mathscr{F}_{T+} = \cap_n \mathscr{F}_{T_n+}$, I'm having trouble.
Consider any set $A$ for any $t \ge 0$:
\begin{align*} \{ T < t \} = \left\{ \left( \inf\limits_{n \ge 1} T_n \right) < t \right\} &= \bigcup_{n=1}^\infty \{ T_n < t \} \\ A \cap \left\{ T < t \right\} &= A \cap \bigcup_{n=1}^\infty \{ T_n < t \} \\ A \cap \left\{ T < t \right\} &= \bigcup_{n=1}^\infty \left( A \cap \{ T_n < t \} \right) \\ A \cap \left\{ T < t \right\} \in \mathscr{F}_{t} &\leftrightarrow \bigcup_{n=1}^\infty \left( A \cap \{ T_n < t \} \right) \in \mathscr{F}_{t} \\ A \in \mathscr{F}_{T+} &\leftrightarrow A \in \cup_n \mathscr{F}_{T_n+} \\ \end{align*}
I seem to have demonstrated that $\mathscr{F}_{T+} = \cup_n \mathscr{F}_{T_n+}$. I was supposed to demonstrate that $\mathscr{F}_{T+} = \cap_n \mathscr{F}_{T_n+}$. I'm pretty sure that I did something wrong but I can't see it.
Note $T \leq T_n$ for all $n$. Therefore, we get $\mathcal{F}_{T+} \subseteq \mathcal{F}_{T_n+}$ for all $n$.
Hence, $\mathcal{F}_{T+} \subseteq \cap_n \ \mathcal{F}_{T_n+}$.
If $A \in \mathcal{F}_{T_n+} $ for all $n$, then $A\cap \{T_n > t\} \in \mathcal{F}_{t+}$. (Check the definition in the book)
So $A\cap \{T > t\} = \cap_m \cap_n \left[A \cap \{T_n > t+1/m \}\right] \in \mathcal{F}_{t+}$ (from right continuity of filtration $\mathcal{F}_{t+}$)
This implies $A \in \mathcal{F}_{T+}$