# A finely open set, not open up to polar set?

I already asked this on M.SE, but get no answers.

Is there a (simple) example of a finely open set (i.e. w.r.t. the fine topology in potential theory) $O$ in $\mathbb R^n$, $n \ge 2$, which is not open up to a polar set (i.e. zero capacity), i.e., there does not exist a polar set $M$, such that the symmetric difference of $O$ and $M$ is open?

I found some examples of finely open sets (e.g. constructed using the Lebesgue spine), but all were "almost" open (in the above sense).

I use the following definition of the ($H_0^1$)-capacity: \begin{equation*} \operatorname{cap}(A) = \inf\big( \|\nabla v\|_{L^2(\Omega)}^2 : v \in H_0^1(\Omega) \text{ and } v \ge 1 \text{ on a neighbourhood of } A\big). \end{equation*} Here, $\Omega$ may be $\mathbb R^n$ or a bounded, open set.

I like the following abstract construction, which relies on a connection between quasi-continuity and the fine topology and also works in the setting of $$p\ne 2$$. It is debatable whether the resulting example is “simple”. The standard reference for the required techniques is Adams, Hedberg: Function spaces and potential theory, 1995.
Let $$(B_k)$$ be a sequence of open balls contained in the open unit ball $$D\subset\mathbb{R}^n$$ such that $$A:=\bigcup_{k} B_k$$ is dense in $$\overline{D}$$, but such that $$\operatorname{cap}(A)<\operatorname{cap}(D)$$. This can be arranged by centering the balls in a countable dense subset of $$D$$ and letting the radii decrease sufficiently rapidly. [As we are in the Hilbert space setting $$p=2$$, it is here where we need $$n\ge2$$.] Let $$K:=\overline{D}\setminus A$$. Note that $$K$$ is compact and has no interior points, but is necessarily still large in measure.
By Theorem 11.3.2 (in Adams, Hedberg), there exists a nontrivial nonnegative Sobolev function $$u$$ supported in $$K$$. In fact, let $$w\in H^1(\mathbb{R}^n)$$ be the capacitary extremal of $$A$$. Then $$0\le \tilde{w}\le 1$$ and $$\tilde{w}=1$$ quasi-everywhere on $$A$$, where $$\tilde{w}$$ is the quasi-continuous representative of $$w$$. As $$\operatorname{cap}(A)<\operatorname{cap}(D)$$, we cannot have $$w=1$$ a.e. on $$D$$. Let $$\varphi\in C^\infty_c(\mathbb{R}^n)$$ be such that $$\varphi(x)>0$$ for all $$x\in D$$ and $$\varphi(x)=0$$ for all $$x\notin D$$. Set $$u := \varphi(1-w)$$. Then $$u\in H^1(\mathbb{R}^n)$$ has the desired properties.
It follows that $$U := \{x\in\mathbb{R}^n: \tilde{u}(x)>0\}$$ is quasi-open and (possibly after removing a polar set) $$U\subset K$$. Let $$O$$ be the fine interior of $$U$$. Then $$O\subset U$$ and $$U\setminus O$$ is polar; see Section 6.4 and Proposition 6.4.12 (in Adams, Hedberg). As $$u$$ is nontrivial, both $$U$$ and $$O$$ have positive measure.
Consequently $$O$$ is a nontrivial finely open set that has no Euclidean interior (as it is contained in $$K$$). Let $$V$$ be an open set. If $$V\cap O$$ is nonempty, then $$V\cap A$$ has positive measure and clearly $$V\cap A\subset V\setminus O$$. Otherwise [i.e. if $$V\cap O=\emptyset$$] the set $$O\setminus V=O$$ has positive measure. It follows that the symmetric difference of $$O$$ and $$V$$ cannot be polar. So $$O$$ is not equivalent (up to a polar symmetric difference) to any open set.