question on Kolmogorov's extension theorem I'm reading Bernt Oksendal's "Stochasticc Differential Equations" and got confused at the Kolmogorov's extension theorem.
In chapter 2, page 11 (sixth edition) it says:
Theorem 2.1.5 (Kolmogorov's extension theorem)
For all $t_1, \cdots , t_k \in T$, $k \in \mathbb{N}$ let $\nu_{t_1,\cdots,t_k}$ be probability measures on $\mathbb{R}^{nk}$ s.t.
$$\nu_{t_{\sigma(1)},\cdots,t_{\sigma(k)}} (F_1 \times \cdots \times F_k) = \nu_{t_1,\cdots,t_k} (F_{\sigma^{-1}(1)} \times \cdots \times F_{\sigma^{-1}(k)})  \tag{K1}$$
for all permutations $\sigma$ on $\{1,2, \cdots,k\}$ and
$$\nu_{t_1,\cdots,t_k}(F_1\times \cdots \times F_k) = \nu_{t_1,\cdots,t_k, t_{k+1}, \cdots, t_{k+m}}(F_1\times \cdots \times F_k\times \mathbb{R}^n \times \cdots \times\mathbb{R}^n )\tag{K2}$$
for all $m \in \mathbb{N}$, where (of course) the set on the right hand side has a total of $k + m$ factors.
Then there exists a probability space $(\Omega,\mathscr{F},P)$ and a stochastic process $\{X_t\}$ on $\Omega$, $X_t:\Omega \rightarrow \mathbb{R}^n$, s.t.
 , 
$$\nu_{t_1,\cdots,t_k} (F_1 \times \cdots \times F_k) = P[X_{t_1} \in F_1, \cdots , X_{t_k} \in F_k] $$
for all $t_i \in T$, $k\in \mathbb{N}$ and all Borel sets $F_i$. 
I don't understand here why (K1) is necessary?
(K2) is clearer to me: it simply says, given $k$ observation, $t_1$ to $t_k$ and $F_1$ to $F_k$, calculate the chance  $\nu_{t_1,\cdots,t_k} (F_1 \times \cdots \times F_k)$, then it is equal to a broader observation of $n+m$. This is to ensure the consistency.
But what does (K1) mean?
If I choose $k=2$ and the permutation $\sigma$ that $\sigma(1)=2, \sigma(2)=1$, (K1) becomes
$$\nu_{t_2, t_1} (F_1 \times   F_2) = \nu_{t_1, t_2} (F_2 \times  F_1)  \tag{K1.k=2}$$
But since $\nu_{t_i}(F_i) =P(X_{t_i} \in F_i)$, isn't this (K1.k=2) obvious? 
Since
$$\nu_{t_1}(F_2) =P(X_{t_1} \in F_2)$$
$$\nu_{t_2}(F_1) =P(X_{t_2} \in F_1)$$
so,
$$\nu_{t_2, t_1} (F_1 \times   F_2) \\ =  P(X_{t_2} \in F_1)*  P(X_{t_1} \in F_2) \\ =P(X_{t_1} \in F_2) * P(X_{t_2} \in F_1) \\ = \nu_{t_1, t_2} (F_2 \times  F_1)  \tag{K1.k=2}$$
This is a trivial conclusion? Or, did I mistake here -- that  (K1.k=2) means the measure $\nu_1, \nu_2$ are independent, and this is not obvious?
 A: Actually, for $k=2$, (K1) does not say what is in the question but that, for every measurable $F_1$ and $F_2$, one has $\nu_{t_2, t_1} (F_2 \times   F_1) = \nu_{t_1, t_2} (F_1 \times  F_2)$. This is a nonempty condition on $\nu_{t_2, t_1}$ and $\nu_{t_1, t_2}$.
More generally, (K1) says that each $\nu_{t_1,\cdots,t_k}$ determines completely $k!-1$ other probability measures, namely, the measures $\nu_{t_{\sigma(1)},\cdots,t_{\sigma(k)}}$ for every $\sigma$ in $\mathfrak S_k\setminus\{\mathrm{Id}\}$.
Edit: The revised version assumes that random variables $(X_t)$ exist with marginals $\nu_{t_1,\cdots,t_k}$. This is taking the conclusion as a hypothesis: to ensure the existence of such random variables is the whole point of conditions (K1) and (K2) hence to notice that if $(X_t)$ exist then (K1) holds is moot. 
Additionally, the post seems to assume that $(X_t)$ is independent, hence that $\nu_{t_1,\cdots,t_k}$ is always the product of the probability measures $\nu_{t_i}$. The setting of Kolmogorov extension theorem is much wider than the independent case (in fact the independent case does not require this theorem) hence all the computations after "But what does (K1) mean?" are moot as well.
