Existence of the Brownian Motion using the Kolmogorov extension theorem Kolmogorov extension theorem:
Let $T$ denote some interval (thought of as "time"), and let $n \in \mathbb{N}.$ For each $k \in \mathbb{N}$ and finite sequence of times $t_{1}, \dots, t_{k} \in T$, let $\nu_{t_{1} \dots t_{k}}$ be a probability measure on $(\mathbb{R}^{n})^{k}.$ Suppose that these measures satisfy two consistency conditions:


*

*for all permutations $\pi$ of $\{ 1, \dots, k \}$ and measurable sets $F_{i} \subseteq \mathbb{R}^{n}$,
$$\nu_{t_{\pi (1)} \dots t_{\pi (k)}} \left( F_{\pi (1)} \times \dots \times F_{ \pi(k)} \right) = \nu_{t_{1} \dots t_{k}} \left( F_{1} \times \dots \times F_{k} \right);$$

*for all measurable sets $F_{i} \subseteq \mathbb{R}^{n},m \in \mathbb{N}$
$$\nu_{t_{1} \dots t_{k}} \left( F_{1} \times \dots \times F_{k} \right) = \nu_{t_{1} \dots t_{k} t_{k + 1}, \dots , t_{k+m}} \left( F_{1} \times \dots \times F_{k} \times \mathbb{R}^{n} \times \dots \times \mathbb{R}^{n} \right).$$
Then there exists a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and a stochastic process $X : T \times \Omega \to \mathbb{R}^{n}$ such that
$$   \nu_{t_{1} \dots t_{k}} \left( F_{1} \times \dots \times F_{k} \right) = \mathbb{P} \left( X_{t_{1}} \in F_{1}, \dots, X_{t_{k}} \in F_{k} \right)
$$
for all $t_{i} \in T, k \in \mathbb{N}$ and measurable sets $F_{i} \subseteq \mathbb{R}^{n},$ i.e. $X$ has $\nu_{t_{1} \dots t_{k}}$ as its finite-dimensional distributions relative to times $t_{1} \dots t_{k}.$
Brownian motion:
The Brownian motion $B_t$ is characterised by four facts:


*

*$B_0=0$

*$B_t$ is almost surely continuous 

*$B_t$ has independent increments

*$B_t-B_s\sim \mathcal{N}(0,t-s)$ (for $0 \leq s \le t$)


$\mathcal{N}(\mu,\sigma^2)$ denotes the normal distribution with expected value $\mu$ and variance $\sigma^2.$ The condition that it has independent increments means that if $0 \leq s_1 \leq t_1 \leq s_2 \leq t_2$ then $B_{t_1}-B_{s_1}$ and $B_{t_2}-B_{s_2}$ are independent random variables.
MY QUESTION: How to proof the existence of the Brownian Motion using the Kolmogorov extension theorem?
 A: Basically, the construction is divided into three steps:
Step 1: Constructing a consistent set of finite dimensional distributions.
Consider any starting point $x\in\mathbb{R}$ and a set of times $0<t_1<t_2<\cdots<t_n<T$. Define a measure on finite dimensional space $\mathbb{R}^n$ as
$$
\nu_{t_1,\cdots,t_n}(F_1,\cdots,F_n)
\triangleq\int_{F_1}dx_1\cdots\int_{F_n}dx_n
\prod_{i=1}^np_{t_i-t_{i-1}}~(x_i,x_{i-1})~~~~~~(1)
$$
where each $F_i$ is a measurable set in $\mathbb{R}$, $x_0=x$ and the transition probability $p$ is Gaussian, i.e.   
$$
p_t(x,y)\triangleq (2\pi t)^{-1/2}e^{-(y-x)^2/2t}
$$
Obviously, $p$ is a valid transition probability. Also, it is not hard to verify that for any starting point $x$ such construction satisfies the two consistent assumptions in Kolmogorov extension theorem. Thus, we have constructed a consistent set of finite dimensional distributions.
Step 2: Applying Kolmogorov extension theorem to construct a probability measure on the space of functions with rational domains. 
Consider the space 
$$\Omega_q=\{\textrm{functions}~~\omega:~\mathbb{Q}\rightarrow\mathbb{R}\},$$ 
where $\mathbb{Q}$ is the set of rationals. Let $\mathcal{F}_q$ be the $\sigma$- algebra generated by all the finite dimensional measurable sets. Then, Kolmogorov extension theorem tells us that there exists a probability measure $\nu_x$ on $(\Omega_q,~\mathcal{F}_q)$ such that 
$$
\nu_x\{\omega:~\omega(0)=x\}=1,
$$
and
$$
\nu_x\{w:~\omega(t_i)\in F_i,~i=1,2,\cdots,n\}=\nu_{t_1,\cdots,t_n}(F_1,\cdots,F_n).
$$
Furthermore, we have the following theorem due to Kolmogorov again:
Theorem: The probability measure $\nu_x$ assigns probability 1 to sample paths $\omega:~\mathbb{Q}\rightarrow\mathbb{R}$ that are uniformly continuous on $\mathbb{Q}\cap [0,T)$.
The proof of above theorem is too tedious and I skipped here for brevity. You might want to check chapter 8.1 of Durrett's book: Probability: Theory and Examples for the proof of this theorem. According to the theorem, we get probability one continuous sample paths on the space $(\Omega_q,~\mathcal{F}_q)$ and the rest three properties are easily verified.
Step 3: Translating probability measure to space of continuous sample paths.
Let $C$ be the space of continuous mappings from $[0,T)$ to $\mathbb{R}$ and $\mathcal{C}$ be the $\sigma$-algebra generated by the coordinate maps $t\rightarrow w(t)$. Here, we apply the following fact: Given $\mathbb{Q}\cap [0,T)$ is a dense subset of $[0,T)$ and $\omega$ is a uniform continuous mapping from $\mathbb{Q}\cap [0,T)$ to $\mathbb{R}$, then $\omega$ has a unique uniformly continuous extension on $[0,T)$. We denote such extension as mapping $\phi$. Furthermore, the mapping $\phi$ is measurable. (I know there are a few jumps, but, anyway, you get the big picture:)). Finally, let
$$P_x\triangleq \nu_x\cdot\phi^{-1}$$
be the probability measure on $(C,\mathcal{C})$, and we finish the construction. All four property checks out.
Remark: Intuitively, it is not clear why we have to go through this step 2, whereas it is possible to directly construct a probability measure on the space 
$$\{\textrm{functions}~~\omega:~[0,T)\rightarrow\mathbb{R}\}$$
from the finite dimensional distribution by Kolmogorov extension theorem. Well, it turns out if we do this, then, only property 1, 3, 4 of Brownian motion checks out and almost sure continuity can never be verified. See chapter 8.1 of Durrett's book: Probability: Theory and Examples for more detailed discussion:)
