Background:
Let $\{B(t) : t\ge 0\}$ be a Brownian motion on probability space $(\Omega,\mathcal{F}, \mathbb{P})$. We can interpret $B$ as a random variable $\Omega\rightarrow \prod_{[0,\infty)}\mathbb{R}$ that is $\mathcal{F}-\bigotimes_{[0,\infty)}\text{Borel}(\mathbb{R})$ measurable.
We require that $t\mapsto B(t)$ be almost surely continuous. The book that I am reading defines this to mean that the subset $\{\omega\in \Omega : t\mapsto B(t,\omega)\text{ is continuous}\} = \{\omega : B(\cdot,\omega)\in C([0,\infty),\mathbb{R})\}$ contains an almost sure subset $A\in \mathcal{F}$.
I would like to say that $B^{-1}(C([0,\infty)))$ has measure $1$. However, since $C([0,\infty))$ is not $\bigotimes_{[0,\infty)}\text{Borel}(\mathbb{R})$ measurable, the $B^{-1}(C([0,\infty)))$ may not be measureable.
Question:
Is there a larger and natural $\sigma$-algebra on $\prod_{[0,\infty)}\mathbb{R}$ that makes $C([0,\infty))$ a measurable subset?
(Optional) Is there a reason why we define $B$ to be $\mathcal{F}-\bigotimes_{[0,\infty)}\text{Borel}(\mathbb{R})$ measurable and not measurable from $\mathcal{F}$ to this larger $\sigma$ algebra?