TITLE: Consistency Conditions of the Kolmogorov Extension Theorem
QUESTION [8 upvotes]: Kolmogorov's extension theorem allows for the construction of a variety of measures on infinite-dimensional spaces, and its conditions are supposedly "trivially satisfied by any stochastic process". However, I'm a little confused about the extent of generality of these conditions. 
Following wikipedia's notation, let $\nu_{t_1 \dots t_k}$ be a probability measure on $\mathbb R^k$. The first consistency condition requires that for every $k \in \mathbb N$ and for all permutations $\pi$ of $\{1 \dots k\}$ and measurable sets $F_i \in \mathbb R$, 
$$
\nu_{t_{\pi(1)}\dots t_{\pi(k)}}(F_{\pi(1)} \times \dots \times F_{\pi(k)}) = \nu_{t_1\dots t_k}(F_1 \times \dots \times F_k)
$$ 
Doesn't this condition imply that all finite-dimensional marginals of the process are exchangeable? If so, then this condition cannot be said to hold for any random process. Am I missing something? 
Thank you very much!

REPLY [13 votes]: No, Kolmogorov's consistency condition doesn't imply that the random variables are exchangeable. If we define
$$\nu_{t_1,\ldots,t_k}(F_1 \times \ldots \times F_k) := \mathbb{P}(X_{t_1} \in F_1,\ldots, X_{t_k} \in F_k)$$
then the consistency condition reads
$$\mathbb{P}(X_{t_{\pi(1)}} \in F_{\pi(1)},\ldots, X_{t_{\pi(k)}} \in F_{\pi(k)}) = \mathbb{P}(X_{t_1} \in F_1,\ldots, X_{t_k} \in F_k). \tag{1}$$
In contrast, exchangeability means
$$\mathbb{P}(X_{t_{\pi(1)}} \in F_{\color{red}{1}},\ldots, X_{t_{\pi(k)}} \in F_{\color{red}{k}}) = \mathbb{P}(X_{t_1} \in F_1,\ldots, X_{t_k} \in F_k). \tag{2} $$
So, in $(1)$ we apply the permutation both to the times $(t_j)_{j=1,\ldots,n}$ and the sets $(F_j)_{j=1,\ldots,n}$ whereas in $(2)$ we only apply the permutation to $(t_j)_{j=1,\ldots,n}$.