Q:

"Geometric" Decomposition of Wiener Space

Let $C_0([0,1];\mathbb{R}^d)$ be the classical Wiener space (of continuous paths with initial value $0$) and let $\nu$ be the Wiener measure on this space.  Does there exist a countable family $\left\{U_n\right\}_{n \in \mathbb{N}}$ of open subsets of $C_0([0,1];\mathbb{R}^d)$ such that

$\nu(U_n)= \frac1{2^n}$
$U_n \cap U_m=\emptyset$ if $n\neq m$
$\nu\left( C_0([0,1];\mathbb{R}^d) - \cup_{n \in \mathbb{N}} U_n\right) =0$

A:

For natural $n$, let 
$$U_n:=\{x=(x_1,\dots,x_d)\in C_0([0,1];\mathbb{R}^d)\colon \Phi(x_1(1))\in\delta_n\},
$$
where $\Phi$ is the standard normal pdf and $\delta_n:=(1-1/2^{n-1},1-1/2^n)$. Then the family $(U_n)$ has all the desired properties.

Indeed, 
(i) for each natural $n$, the set $U_n$ is open, because $\delta_n$ is open; 
(ii) the $U_n$'s are disjoint, because the $\delta_n$'s are disjoint; 
(iii) 
$$\nu(U_n)=P(\Phi(B(1))\in\delta_n)=1/2^n,
$$
where $B$ is the standard Brownian motion -- because $\Phi(B(1))$ is uniformly distributed on $(0,1)$; 
(iv) the property $\nu(C_0([0,1];\mathbb{R}^d)-\cup_{n\in\mathbb{N}} U_n)=0$ follows immediately from (ii) and (iii). 

