Exit time of Brownian motion with random variables.

Question

Let $(\Omega,\mathcal{F},\mathbb{P})$ $B$ a Brownian motion, $U$ a non-negative random variable and $V$ a random variable, which $V\overset{d}{=}-U$. I am trying to understand why the next random variable is a stopping time \begin{equation} \tau=\inf\{t\geq0|B_t\notin(V,U)\} \end{equation} for some filtration $\mathbb{F}=\{\mathcal{F}_t|t\geq0\}$. My guess is \begin{equation} [\tau\leq t]=[B_t\leq V]\cup[B_t\geq U] \end{equation} then if $\mathbb{F}$ is a filtration for which $B$ is Brownian motion and $U$ is $\mathcal{F}_0$ measurable, then $\tau$ is a stopping time. But I am not very sure if I am right, any help or advice is welcome. Thanks!.