Measurability of subspace of set of all functions

Question

Set $X=\mathbb{R}^n$ and let $X^{I}$, the space of maps from the (bounded or unbounded) interval $I$ to $X$, be endowed with the locally convex topology of pointwise convergence.

Is it true that the subspace $C(I, X)$ of continuous maps is not in the Borel algebra of $X^I$?

It is clear from the proof of the above continuity statement that for each (dense) countable subset $D \subset I$, the space of maps that are continuous from $D$ to $X$ (or even Hölder-continuous for exponents $< 1/2$) is measurable and has measure $1$. However, the set of dense subsets is not countable, so we cannot take the intersection.

How to prove that the set of continuous functions are not measurable?

/Edit: Sorry about this. I meant to ask a somewhat different question but changed it while writing; then forgot to change the title.

Maybe some background: With Kolmogorovs Extension theorem, you get a unique measure on the above function space that is defined by its finite-dimensional distributions. This is the Wiener measure. Then there is the continuity theorem that says, you get a version of the associated stochastic process with continuous paths. From the measure perspective, this fits only together for me if the set of continuous paths is not measurable, as you could otherwise just restrict your measure to this space.

Answer 1

For $(x_j)_{j\in\Bbb N}\subset I$ and $B\in\mathcal B((\Bbb R^n)^{\Bbb N})$, define $S((x_j)_{j\in\Bbb N},B):=\{f\in X^I,(f(x_j))_{j\in\Bbb N}\in B\}$. The collection of sets of this form contains the $\sigma$-algebra generated by the semi-norms $f\mapsto |f(x)|$, $x\in I$, which is the same as the $\sigma$-algebra generated by the evaluation maps $f\mapsto f(x)$.

We now show that the collection $\mathcal C=\{S(\mathbf x,B), \mathbf x\in\Bbb R^{\infty},B\in \Bbb R^{\infty}\}$ is a $\sigma$-algebra. The main difficulty is to show that this collection is stable for countable unions. Let $\tau\colon\Bbb N^2\to\Bbb N$ be a bijection, and $(S(\mathbf x_{k,\cdot},B_k))_{k\in\Bbb N}$ elements of $\mathcal C$. Then take $y_k:=x_{\tau^{-1}(k)}$ and $B\in\mathcal B((\Bbb R^n)^{\Bbb N}$ such that $B_k=\pi_{\{k\}\times\Bbb N}^{-1}(B)$, where $\pi_I((x_n)_{n\in\Bbb N})=(x_n,n\in I)$.

The subset of continuous functions cannot be expressed in this form, because we don't control the behavior of a function of $S((x_j)_{j\in\Bbb N},B)$ outside an uncountable set.