TITLE: Construction of Brownian Motion
QUESTION [7 upvotes]: In Wiener's construction of Brownian Motion, it is assumed that there exists a probability space $(\Omega,\mathcal F,\mathbb P)$ and random variables $X_n:\Omega\rightarrow\mathbb R$ for $n\in\mathbb N,$ which are independent and normally distributed. Does anyone know what this probability space actually is and why it exists? Is it unique? Can Brownian Motion be constructed on any measurable space? What role plays then the probability measure $\mathbb P$?

REPLY [3 votes]: For simplicity consider the case of all the variables independent and following $\mathcal{N}(0,1)$. You can consider the space $(\Omega, \mathcal{F}, P) = (\mathbb{R}^\mathbb{N}, \mathcal{B}(\mathbb{R}^\mathbb{N}), \mu)$, where $\mu$ is defined by the finite dimensional distributions 
$$\mu_{i_1, \ldots, i_k}(A) = \int_A \frac{1}{\sqrt{2\pi}^k} e^{\frac{-\sum_{j=0}^k x_j^2}{2}} dx_1 \ldots dx_j \quad A \in \mathbb{R}^k$$
That is, the finite dimensional laws are multivariate normal random variables with independent components each following $\mathcal{N}(0,1)$.
The existence of the measure $\mu$ is a consequence of  the  Kolmogorov's consistency theorem which assures that given a set of consistent finite dimensional laws the measure $\mu$ will exist in the infinite product space and it will be consistent with the specified finite dimensional laws.
Explicitly being consistent is that on the product space the measure of a cilinder
$$ C = A_{i_1} \times A_{i_2} \cdots \times A_{i_k} \times \mathbb{R}^{\mathbb{N} - \{i_1 \, \ldots, i_k\}}$$
will be given by
$$ \mu(C) = \mu_{i_1, \ldots, i_k}(A_{i_1} \times \cdots \times A_{i_k})$$
In plain english, every finite dimensional component will have the specified measure, in this case independent normal elements since
$$ P(X_{i_1} \in A_{i_1}, \ldots, X_{i_k} \in A_{i_k}) = \mu(C)$$
You can even use this theorem directly to construct the Brownian Motion in $(\mathbb{R}^{[0, T]}, \mathcal{B}(\mathbb{R}^{[0, T]}))$ by specifying the finite dimensional distributions as normal with mean zero and the covariance structure of the BM. 
But in my opinion is one of the unfunnier ways to construct it ;).