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 $n\neq m$
* $\nu\left( C\_0([0,1];\mathbb{R}^d) - \cup\_{n \in \mathbb{N}} U\_n\right) =0$



"Geometric" Decomposition of Wiener Space
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). 


