In the stochastic processes course I am taking, the following result was stated without proof.
Claim. $\quad$ Suppose that $B_t$ is a standard brownian motion with respect to the filtration $(\mathcal{F}_t)_{t\geq 0}$, and let $$ A_n = \left\{\limsup_{t \searrow 0 } \frac{B_t}{\sqrt{t}} > n \right\}. $$ Then $A_n \in \mathcal{F}_0^+$, where $\mathcal{F}_0^+ = \bigcap_{t > 0} \mathcal{F}_t$.
I can see that since $B_t$ and $\frac{1}{\sqrt{t}}$ are measurable with respect to $\mathcal{F}_t$ for all $t$. I also know that $\limsup: \mathbb{R}^\mathbb{N} \to \mathbb{R}$ is a measurable function, but I don't see how to prove it for $\limsup: \mathbb{R}^\mathbb{R} \to \mathbb{R}$. In fact I don't really even know how $\limsup$ is defined on an uncountable index set. Is anyone able to help me with a proof of this statement?