I got a problem that seems to be quite standard and easy, but I have lots of problems with it. I do already know that $T_a:=\inf\{t\geq 0: B_t=a\}$ is a stopping time for any $a\in\mathbb{R}$ fixed, $B_t$ being a standard BM.
But what about for instance $T_{B^{o}_1}$, where $B_t^{o}$ is an BM independent from $B_t$? Clearly $\{T_{B_1}\leq t, B^{o}_1=x\}=\{T_{-x}\leq t\}\in\mathcal{F}_t$. But the union over all $x$ is uncountable and I'm not sure if it could work if I only look at rational $x$.
Any suggestions? Thanks!
Assuming that $B_t^o$ is adapted to $F_t$ and $t > 1$ then the classical $\mathbb{Q}$ trick works because of the continuity of brownian paths: \begin{align*} \lbrace T_{B_1^o} \leq t \rbrace &= \lbrace \max_{s \in [0,t]}{B_s \geq B_1^o} \rbrace \cap \lbrace B_1^o \geq 0 \rbrace \cup \lbrace \min_{s \in [0,t]} B_s \leq B_1^o \rbrace \cap \lbrace B_1^o \leq 0 \rbrace \\&= \cup_{q \in \mathbb{Q}^+} \lbrace{\max_{s \in [O,t]} B_s \geq q\rbrace} \cap \lbrace q \geq B^o_1 \geq 0 \rbrace \cup \cup_{q \in \mathbb{Q}^-} \lbrace{\min_{s \in [O,t]} B_s \leq q\rbrace} \cap \lbrace q \leq B^o_1 \leq 0 \rbrace \end{align*} The "max" and "min" sets are in $F_t$ because $B$ is $F_t$-adapted. $B_1^o$ is adapted also so the sets $\lbrace q \geq B_1^o \geq 0 \rbrace$ and $\lbrace q \leq B_1^o \leq 0 \rbrace$ are in $F_1 \subseteq F_t$ because $t > 1$.
In hindsight $\lbrace \max_{s \in [O,t]} B_s \geq B_1^o \rbrace$ is $F_t$ measurable since the max and $B_1^o$ are. This property is shown by the $\mathbb{Q}$-trick early in a course of measure theory.