I have stumbled while solving the following problem. It seems simple, therefore your hints would be much valuable.
Let $C$ denote the set of all continuos functions $x.$ from $t\in[0,\infty)$ to $\mathbb{R}$. The value of $x.$ evaluated at point $t$ is denoted by $x_t$.
Let $\mathfrak{U}(C)$ denotes minimal $\sigma$-algebra, that contains cylindrical sets, that is sets of the form $$ \{x.\,:\, x_t\in B\}, $$ where $B\in \mathfrak{B}(\mathbb{R})$ is a Borel set.
Let $\Pi$ be a a family of subsets of $\mathbb{R}$ such that $\sigma(\Pi)=\mathfrak{B}(\mathbb{R})$.
Consider the family of subsets of $C$ of the form $$ \{x.\,:\, x_t\in B\}, $$ where $B\in \Pi$. Prove that $\sigma$-algebra, generated by this family of subsets, is precisely $\mathfrak{U}(C)$.
$\mathfrak{U}(C)$ is by definition the minimal $\sigma$-Algebra such that the functions $p_t: x\mapsto x_t$ are $\mathfrak{U}(C)$-$\mathscr{B}(\mathbb{R})$-measurable for $t \in [0, \infty)$. Now, it can easily be shown that the system of sets \begin{align*} \mathscr{A}:=\left\{ A\in \mathscr{B}(\mathbb{R}): p_t^{-1}(A) \in \sigma(\left\{ p_t^{-1}(B):B\in \prod \right\}) \right\} \end{align*} is a $\sigma$-Algebra which contains $\prod$. Therefore $\sigma(\prod) = \mathscr{B}(\mathbb{R}) \subset \mathscr{A}$, i.e $\mathscr{A} = \mathscr{B}(\mathbb{R})$. From this it follows that any $p_t$ is $\sigma(\left\{ p_t^{-1}(B):B\in \prod \right\})$-$\mathscr{B}(\mathbb{R})$-measurable, i.e. $\mathfrak{U}(C)\subset \sigma(\left\{ p_t^{-1}(B):B\in \prod \right\})$ since $\mathfrak{U}(C)$ is the minimal $\sigma$-Algebra which satisfies this.
BS