Øksendal states in his book "stochastic differential equations" (Defintion 2.2.1 iii), p.13) that
the set $H = \{\ \omega \mid t → B_t (\omega)\ \text{is continuous}\ \}$ is not measurable with respect to the Borel $\sigma$-algebra $\mathcal{B}$ on $(\mathbb{R}^n)^{[0,\infty)}$ (...) ($H$ involves an uncountable number of $t$'s),
where $B_t$ is a brownian motion and we identify $\omega$ with the path of $B_t(\omega)$.
Unfortunately I don't know much about $\mathcal{B}((\mathbb{R}^n)^{[0,\infty)})$ aside its defintion. According to this question a set $A$ is measurable iff there exists $J\subseteq \mathbb{R}$ with $|J|≤\aleph_0$ and $B\in \mathcal{B}(\mathbb{R}^J)$ such that $A=B \times \mathbb{R}^{\mathbb{R}\setminus J}=\{f \in \mathbb{R}^\mathbb{R} \colon \ (f(j) \colon \ t \in J) \in B\}$.
I would appreciate it, if someone could provide me a reference for the statement above.
Edit: If I am not mistaken the product $\sigma$-algebra $\mathcal{B}((\mathbb{R}^n))^{[0,\infty)}$ is a true subset of the Borel-$\sigma$-algebra $\mathcal{B}((\mathbb{R}^n)^{[0,\infty)})$ and thus the statement above should be correct for the product, but not for the Borel algebra. Prior to the quote Øksendal writes (after Definition 2.1.4, p. 10):
$\mathcal{B}$ [the algebra generated by cylindrical sets] is the same as the Borel $\sigma$-algebra on $\tilde{\Omega}$ [$=(\mathbb{R}^n)^T$] if $T = [0,\infty)$ and $\tilde{\Omega}$ is given the product topology
This should be false or am I missing something?
(I'm going to assume that $\Omega=\Bbb R^{[0,\infty)}$.) As Øksendal notes, if $H$ were measurable then there would be a countable set $(t_n)\subset[0,\infty)$ and a Borel set $B\in\mathcal B({\Bbb R}^{\Bbb N})$ such that $H=\{\omega\in\Omega: (\omega(t_1),\omega(t_2),\ldots)\in B\}$. In particular, this would mean that if $\omega$ and $\omega^*$ were two elements of $\Omega$ with $\omega(t_n)=\omega^*(t_n)$ for all $n\in\Bbb N$, and if $\omega\in H$, then so too $\omega^*\in H$. But clearly if we take any element $\omega\in H$ (that is, a continuous path) and define $\omega^*(t):=\omega(t)$ for all $t$ except one point $t^*\in [0,\infty)\setminus\{t_1,t_2,\ldots\}$ and set $\omega^*(t^*)=\omega(t^*)+17$, then $\omega^*$ passes the test (involving $(t_n)$ and $B$) for inclusion in $H$ but $\omega^*$ is not continuous.