Equip $\mathbb{R}^\mathbb{R}$ with the cylindrical $\sigma$-algebra $\mathcal{C}$ and view $C(\mathbb{R})$ as a subset of $\mathbb{R}^\mathbb{R}$.
Question: Is $C(\mathbb{R})\in \mathcal{C}$?
The cylindrical $\sigma$-algebra is defined as follows: Define $ \hat x:\mathbb{R}^\mathbb{R}\rightarrow \mathbb{R}$ by $\hat x(f)= f(x)$. Then $\mathcal{C}=\sigma(\hat x : x\in \mathbb{R})$.
Equip $C(\mathbb{R})$ with the topology of uniform convergence on compacts and denote $\mathcal{B}$ the corresponding Borel $\sigma$-algebra. Then the inclusion $\iota:(C(\mathbb{R}),\mathcal{B})\rightarrow ( \mathbb{R}^ \mathbb{R},\mathcal{C})$ is measurable and we even have $\iota^{-1}(\mathcal{C})=\mathcal{B}$. - that much I know. The question above may thus be reformulated as 'is $\iota$ bimeasurable?'
My motivation is the following: I want to know whether there is an easy way to check whether a measure $\nu$ on $( \mathbb{R}^ \mathbb{R},\mathcal{C})$ is given as the push-forward $\nu =\iota_*\mu$ of a Borel-measure $\mu$ on $C(\mathbb{R}).$ If the answer to the question above is positive, then the condition for this is simply that $\nu(\mathbb{R}^\mathbb{R}\backslash C(\mathbb{R})) =0$.
It seems like this issue should be discussed in the context of stochastic processes - but e.g. in Dudley's book I didn't find anything. Namely, if $(X_t:t\in \mathbb{R})$ is a stochastic process (in $ \mathbb{R}$), then 'sample continuous' is sometime defined as the condition $$ \mathbb{P}(\omega: t\mapsto X_t(\omega) \text{ is continuous}) =1. $$ However, if we first view $X$ as a random variable in $( \mathbb{R}^ \mathbb{R},\mathcal{C})$, then the event can be rewritten as $E=\{\omega: X(\omega)\in C(\mathbb{R})\}$ and we would need this to be measurable to make sense of its probability. In general $E$ is only measurable if $C(\mathbb{R})\in \mathcal{C}$. I suppose one could circumvent this by asking for a measurable subset $E'\subset E$ of probability one instead.