I'm trying to construct Brownian motion using the Kolmogorov extension theorem.
I am happy with the construction of a process with the required FDDs as (the canonical process associated with) a random function in $D([0, \infty), R)$ - (the set of all functions from $R_+$ to $R$, not just cadlag functions). I am also happy with the fact that the set of continuous functions is not measurable within the $\sigma$-algebra generated by 'cylindrical sets'.
So my understanding is that it does not make sense to talk about the probability that such a process is continuous?
But on the other hand it seems we can naively apply the Kolmogorov continuity theorem the constructed process to construct a (continuous) Brownian motion.
So what is going on here? When I have constructed a process with the required FDDs can I naively apply the Kolmogorov continuity theorem the complete the construction of Brownian motion? If not, why not? What goes wrong?
Edit: By naively apply Kolmogorov continuity theorem, I mean the following:
Kolmogorov continuity theorem: (from Le Gall)
Let $X = (X_t)_{t \in I}$ be a random process indexed by a bounded interval $I$ of $R$, and taking values in a complete metric space $(E, d)$. Assume that there exist three reals $q, \epsilon, C > 0$ such that, for every $s, t \in I$,
$E[d(X_s,X_t)^q] \leq C|t - s|^{1 + \epsilon}$ :
Then, there is a modification $\tilde{X}$ of $X$ whose sample paths are Hölder continuous with exponent $\alpha \in (0, \frac{\epsilon}{q})$: This means that, for every $\omega \in \Omega$ and every $\alpha \in (0, \frac{\epsilon}{q})$ there exists a finite constant $C_\alpha(\omega)$ such that, for every $s, t \in I$,
$d(\tilde{X}_s(\omega), \tilde{X}_t(\omega) \leq C_\alpha(\omega)|t-s|^{1+ \alpha}$
In particular, $\tilde{X}$ is a modification of $X$ with continuous sample paths (by the preceding observations such a modification is unique up to indistinguishability).
[end of theorem]
So once we have the random process with the FDDs of Brownian motion taking values in R (a complete metric space, we can just apply the distribution properties of Brownian motion to satisfy the requirements of the theorem and produce a continuous modification (which has the same FDDs since it is a modification).
So where does the above argument go wrong?