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?

Brownian motionif its FDD are as required. I think you set out to prove "ForeveryBrownian motion process, (almost) all paths are continuous" but you discovered that that is false. Instead, as the two answers below show, you get the theorem "Thereexistsa Brownian motion process whose (almost) all paths are continuous." $\endgroup$ – user95282 Mar 12 '20 at 18:11