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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 $n\neq m$
  • $\nu\left( C_0([0,1];\mathbb{R}^d) - \cup_{n \in \mathbb{N}} U_n\right) =0$
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  • $\begingroup$ Since $\nu$ is a probability measure then $K_n\triangleq \nu^{-1}[\frac1{2}^n,\frac1{2}^n-\frac1{2}^{n+1}]$ is $\nu$-measureable and of $\nu$ measure $\frac1{2^n}$. Since $\nu$ is an *outer-*regular Borel measure then there are open subsets $U_n^k$ satisfying - $\nu(U_n^k)\leq \nu(U_n^{k+1})\mapsto \nu(K_n)$ - $K_n\subseteq U_n^k$. Some more work probably shows the result... $\endgroup$
    – ABIM
    Commented Oct 30, 2019 at 11:21

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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).

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  • $\begingroup$ Thank you Iosif, this is very helpful. $\endgroup$
    – ABIM
    Commented Oct 30, 2019 at 14:43

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