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?