Let $C[0,\infty)$ denote the space of continuous functions $f : [0,\infty) \rightarrow \mathbb{R}$. Let $\mathcal{C} [0,\infty)$ denote the $\sigma$-algebra of subsets of $C[0, \infty)$ generated by sets of the form $$ \lbrace f ∈ C[0,\infty) : f(t) ∈ A\rbrace, $$ with $t\in [0,\infty)$ and $A \in \mathcal{B}(\mathbb{R})$. Find a consistent family of probability measures $\mu_{t}$ indexed by finite sequences $t = (t_1,..., t_n)$ with $0\leq t_1 < ... < t_n$ with the property that: there is no measure $\mu$ on $C[0,\infty)$ satisfying $$ \mu(\lbrace f : (f(t_1), ... , f(t_n)) \in A\rbrace) = \mu_{(t_1,...,t_n)}(A) $$ for all $n\in\mathbb{N}, t_1 \leq ...\leq t_n$ and $A\in \mathcal{B}(\mathbb{R}^n)$
My approach was thinking about properties of $C[0,\infty)$, because if this space would have been changed to just all functions from $[0,\infty)$ to $\mathbb{R}$, then by Kolmogorov extension theorem there is no consistent family with the property above. Then it seems logical to think about sets like $\lbrace f\in C[0,\infty): f(q_i)=1 \mbox{ iff } q_i\neq 0, \mbox{ otherwise } 0\rbrace$, where $q_i$ is a sequence of all rationals and $r_i$ a sequence of real numbers. This set is empty, but maybe this is a limit of a decreasing sequence whose $\mu$ measure is not $0$ (for contradiction).
However, I cannot find such a family of measures. The one time it leads to non-consitency, the other time the limit set has $\mu$-measure $0$. Is there anyone who can find a good measure family? Thanks in advance.