Kolmogorov extension alone does not give a continuous version

Take $I=[0,1]$. For each finite $F\subseteq I$, let ${\mu }_F$ be the uniform probability on $\{0,1{\}}^F$. Its mass is $2^{-\mathrm{\mid }F\mathrm{\mid }}$ at every point; for $F=\mathrm{\varnothing }$ this is the unique probability on the singleton $\{0,1{\}}^{\mathrm{\varnothing }}$. Marginalizing from $G$ to $F\subseteq G$ sums over $2^{\mathrm{\mid }G\setminus F\mathrm{\mid }}$ extensions and gives $2^{\mathrm{\mid }G\setminus F\mathrm{\mid }}{2}^{-\mathrm{\mid }G\mathrm{\mid }}=2^{-\mathrm{\mid }F\mathrm{\mid }}$, so the family is consistent. By [F1], under AC it has a probability extension on the cylinder sigma-algebra of $\mathrm{\Omega }=\{0,1{\}}^{[0,1]}$, and $X_t(\omega )=\omega (t)$ is a measurable coordinate process with these finite laws. For a finite $J\subseteq I$ and sets $A_j\subseteq \{0,1\}$, uniform counting gives $P\text{\hspace{0.17em}⁣}({\bigcap }_{j\in J}\{X_j\in A_j\})=\frac{{\prod }_{j\in J}\mathrm{\mid }{A}_j\mathrm{\mid }}{2^{\mathrm{\mid }J\mathrm{\mid }}}={\prod }_{j\in J}P(X_j\in A_j)\mathrm{.}$ This also gives $1$ for $J=\mathrm{\varnothing }$, by the empty-product convention. Since every subset of $\{0,1\}$ is measurable, [F1] makes the whole coordinate family independent, and each coordinate is a fair bit.

For $n\in \mathbb{N}$, put $q_n=1\mathrm{/}(n+1)$. These are distinct points of $(0,1]$. Given $\epsilon >0$, [F6] gives a positive integer $N$ with $1\mathrm{/}N<\epsilon$; whenever $n+1\ge N$, positivity and order reversal under inversion give $0<q_n\le 1\mathrm{/}N<\epsilon$. Hence $q_n\to 0$.

Put $A_n=\{X_{q_n}=1\}$. By the independence and fair laws in step 1.1, the events $(A_n)$ are pairwise independent and $P(A_n)=1\mathrm{/}2$. For each pair, [F2] also makes their complements $A_n^c=\{X_{q_n}=0\}$ independent, and $P(A_n^c)=1\mathrm{/}2$. Both probability series diverge because their first $m$ terms sum to $m\mathrm{/}2$. Applying [F2] twice gives probability-one events $E_1=\{A_n\text{ occurs infinitely often}\},E_0=\{A_n^c\text{ occurs infinitely often}\}\mathrm{.}$ Thus on $E_0\cap E_1$ the bit sequence $(X_{q_n})$ has infinitely many zeros and infinitely many ones.

Suppose for contradiction that $Y=(Y_t{)}_{t\in [0,1]}$ is a continuous modification of $X$: for some measurable event $C$ with $P(C)=1$, every path $t\mapsto Y_t(\omega )$ with $\omega \in C$ is continuous on $[0,1]$. By [F4], each $H_n=\{Y_{q_n}=X_{q_n}\}$ is measurable and has probability one. The complement of $H={\bigcap }_n{H}_n$ is the countable union of the null events $H_n^c$, so [F3] gives $P(H)=1$. A finite union bound likewise gives $P(C\cap H\cap E_0\cap E_1)=1$, so this intersection is nonempty. Fix $\omega$ in it.

The path $f(t)=Y_t(\omega )$ is continuous at the endpoint $0$. Since $q_n\to 0$, the choice-free forward implication in [F5] gives $Y_{q_n}(\omega )\to Y_0(\omega )$. But $\omega \in H\cap E_0\cap E_1$, so $Y_{q_n}(\omega )=X_{q_n}(\omega )$ for every $n$, with both values $0$ and $1$ occurring infinitely often. This sequence cannot converge: if it converged to $a$, its tail would eventually lie within $1\mathrm{/}3$ of $a$; a tail containing both $0$ and $1$ would then give $1\le \mathrm{\mid }a\mathrm{\mid }+\mathrm{\mid }1-a\mathrm{\mid }<2\mathrm{/}3$, a contradiction. Therefore no such continuous modification $Y$ exists.

The witness has consistent finite laws and a genuine cylinder-space Kolmogorov extension, yet lacks a continuous modification, which refutes the statement. The empty finite support was checked in step 1.1; $t=0$ is the continuity endpoint in step 4.1; $t=1=q_0$ and the values $0,1$ occur in the construction; repeated coordinates are handled by the coordinate process rather than treated as independent copies. There is no biconditional. AC is used exactly through the arbitrary-index extension invoked in step 1.1, while Borel--Cantelli, the fixed sequence, and the countable intersection add no choice.