Suppose $B$ is a Brownian motion and $S$ is a stopping time (i.e. $\{S < t\} \in \mathcal{F}_t$, which is equivalent to $\{S \leq t\} \in \mathcal{F}_t$). The information at stopping time is given by $\mathcal{F_S} = \{A: A \cap \{S \leq t\} \in \mathcal{F}_t \text{ for all } t \geq 0\}$.
In Theorem 7.3.8 of Durrett's Probability: Theory and Examples 5ed, he proved $B_S \in \mathcal{F}_S$. He considered $S_n = ([2^nS]+1)/2^n$ (i.e. $S_n = (m+1)2^{-n} \text{ if } m2^{-n}\leq S < (m+1)2^{-n}$). He has shown before:
Theorem 7.3.5: $S_n$ is a stopping time.
Theorem 7.3.6: If $S \leq T$ are stopping times, then $\mathcal{F}_S \subseteq \mathcal{F}_T$.
Theorem 7.3.7: If $T_n \downarrow T$ are stopping times, then $\mathcal{F_T} = \cap\mathcal{F}_{T_n}$.
In the main proof of $B_S \in \mathcal{F}_S$ he showed if $A$ is a Borel set then $\{B(S_n)\in A\} = \cup_{m=1}^{\infty}\{S_n = m2^{-n}, B(m2^{-n}) \in A\} \in F_{S_n}$.
"Now let $n \rightarrow \infty$ and use Theorem 7.3.7". My question is how does $B_S \in \mathcal{F}_S$ follow from the last sentence?
We want to show $B(S) \in \mathcal{F}_S = \cap\mathcal{F}_{S_n}$. Equivalently, we want to show for each n, $B(S) \in \mathcal{F}_{S_n}$. But how to relate $B(S)$ with $B(S_n)$? I know the Brownian paths are continuous and $S_n \downarrow S$ for all $\omega$, so $B(S_n) \rightarrow B(S)$ for all $\omega$. If $A$ is open, then $\{B(S) \in A\} = \cup_{n=1}^{\infty}\cap_{m=n}^{\infty}\{B(S_m) \in A\}$. And for all $m \geq n, \{B(S_m) \in A\} \in \mathcal{F}_{S_m} \subseteq F_{S_n}$, so $\cap_{m=n}^{\infty}\{B(S_m) \in A\} \in \mathcal{F}_{S_n}$. But this only leads to $\{B(S) \in A\} \in \cup \mathcal{F}_{S_n} = \mathcal{F}_{S_1}$.
Best Answer
$(S_n)$ decreases to $S$. Let $n \geq m$. Then $\{B(S_n) \in A\} \in F_{S_n} \subseteq F_{S_m}$. Since $B(S_n )\to B(s)$ it follows that $\{B(S) \in A\} \in F_{S_m}$. [Use the fact that point-wise limits of measurable functions are measurable. Do not try to do this for fixed set $A$]. This is true for each $m$. By Theorem 7.3.7 we get $\{B(S) \in A\} \in \cap_m F_{S_m}=F_{S}$.