Show $L$ is not a stopping time

Question

Let $L = \sup\{ n : n \le 10; A_n \in B \}$, $B \in \mathcal B$, $\sup\{\emptyset \}=0$. $(A_n)_{n \ge1}$ is a process adapted by a natural filtration $\{\mathcal F_n\}.$ Show that $L$ is NOT a stopping time (unless A is 'freaky').

Answer 1

As I said in the comments I think you need some more conditions on $B$ for your statement to be completely true. I'm not sure what the weakest conditions on $B$ can be.

If L is a stopping time then $\{L \le 9\} \in \mathcal{F}_9$, this in turn indicates that $\{A_{10} \in B^c\} \in \mathcal{F}_{9}$. In general this could be a contradiction, but for some specific $B$ it is not.

Or perhaps to make my criticism more concrete suppose that $A_n$ is your fortune at the $n$th step of a game where you toss a coin every round and double your money on heads and lose it all on tails. Also suppose you start with $A_0 = 1$. If $B$ were the set $(-\infty, 2^{10})$ then above you get a contradiction; but if it is the set $(-\infty, 2^{10} + 1)$ then no matter what $A_{10} \in B$. Even at time $n=0$ you will know this, so in fact $\{A_{10} \in B^c\} \in \mathcal{F}_{n}$ for all $n\ge 0$.