Example 5.47.1802396DC16057

Example 5.47.18.02396DC16057 For every $n\ge 1$ there is a plurifine open subset of $\mathbb C^ n$ that is not a Borel set.

Let $K\subset [0,1/10]\subset \mathbb C$ be the compact polar set and $\mu $ the atom-free probability measure carried by $K$ of proposition 4.1.12, and put $F(x)=\mu (K\cap (-\infty ,x])$ for $x\in \mathbb R$. The function $F$ is nondecreasing, $F=0$ on $(-\infty ,0)$ and $F=1$ on $[1/10,\infty )$. It is continuous: it is right continuous because $\mu $ is finite, and its jump at $x$ is $\mu (K\cap \{ x\} )=0$, because $\mu $ has no atoms. For $y\in (0,1)$ the set $\{ x\in \mathbb R:F(x)\ge y\} $ is closed, nonempty and bounded below, and we let $g(y)$ be its smallest element. Then $F(g(y))\ge y$, and $F(x)<y$ for $x<g(y)$, so $F(g(y))=y$ because $F$ is continuous at $g(y)$. Hence $g$ is injective; it is also nondecreasing, so $g$ is strictly increasing and in particular Borel measurable. Moreover $g(y)\in K$. Otherwise, since $K$ is closed, there is $\delta >0$ with $[g(y)-\delta ,g(y)]\cap K=\varnothing $; then $F(g(y)-\delta )=F(g(y))=y$, which contradicts the minimality of $g(y)$.

Let $N\subset (0,1)$ be a set that is not Lebesgue measurable, and put $A=g(N)\subset K$. Since $g$ is injective, $N=g^{-1}(A)$; so if $A$ were a Borel subset of $\mathbb C$, then $A=A\cap \mathbb R$ would be a Borel subset of $\mathbb R$, and $N$ would be a Borel set. Hence $A$ is not a Borel subset of $\mathbb C$. As a subset of the polar set $K$, the set $A$ is polar, hence finely closed in $\mathbb C=\mathbb R^2$ by proposition 4.2.6. Since the fine and the plurifine topologies of $\mathbb C$ coincide, as shown at the beginning of this section, $W_1=\mathbb C\setminus A$ is a plurifine open subset of $\mathbb C$ which is not Borel. For $n\ge 2$, the set $W=W_1\times \mathbb C^{n-1}$ is the preimage of $W_1$ under the holomorphic projection $z\mapsto z_1$, so it is plurifine open by proposition 5.47.15(3). It is not Borel, because $W_1$ is its preimage under the continuous map $\zeta \mapsto (\zeta ,0,\ldots ,0)$.