In the proof of the optional sampling theorem they define for a stopping time $\tau$ the sigma algebra $\mathcal{G}=\sigma(\cup_n \mathcal{F}_{\tau\wedge n})$.
Then they use the fact that for the event $A\in F_\tau$ it holds that $A\cap \{\tau<\infty\}\in \mathcal{G}$.
I tried to show this with the definition of the sigma algebra at the stopping time: $\mathcal{F}_\tau=\{A\in \mathcal{F}_\infty : A \cap \{\tau\leq t\}\in \mathcal{F}_t \text{for all} t \in T\}$, but without succes.
Can anyone help me?
Thanks
Since $$A\cap\{\tau<\infty\}=\cup_{n}(A\cap\{\tau\leq n\}),$$ you just have to prove that, for all $n$, $A\cap\{\tau\leq n\}$ belongs to $\mathcal{F}_{\min(\tau,n)}$. So, let us fix some $t$, and observe that $$A\cap\{\tau\leq n\}\cap\{\min(\tau,n)\leq t\}=A\cap\{\tau\leq \min(n,t)\}\in \mathcal{F}_{\min(n,t)},$$ since $A$ belongs to $\mathcal{F}_{\tau}$. Hence, the result follows by inclusion of $\sigma$-algebra.