A selection principle in measure theory

Question

A Borel subset $B$ of the unit interval $\mathbb I=(0,1)$ is defined to be a density neighborhood of a set $A\subseteq\mathbb I$ if for every $a\in A$ we have $$\lim_{\varepsilon\to0}\frac{\lambda(B\cap[a-\varepsilon,a+\varepsilon])}{2\varepsilon}=1$$where $\lambda$ denotes the Lebesgue measure on $\mathbb I$.

Problem. Let $A\subseteq\mathbb I$ be a set of Lebesgue measure zero and $(B_n)_{n\in\omega}$ be a sequence of Borel density neighborhoods of $A$. Is there a sequence of compact sets $(K_n)_{n\in\omega}$ such that $K_n\subseteq B_n$ for all $n\in\omega$ and the set $K=\bigcup_{n\in\omega}K_n$ is a density neighborhood of $A$?

Answer 1

Professor Wladyslaw Wilczynski kindly informed me that the answer to this problem is negative.

Take any Lebesgue null dense $G_\delta$-set $A$ in the real line $\mathbb R$. Choose a countable dense subset $\{x_n\}_{n\in\omega}$ in $\mathbb R\setminus A$. Since the density topology $\tau_d$ on the real line is Tychonoff, for every $n\in\omega$ there exist disjoint $\tau_d$-open Borel sets $D_n,E_n\subseteq\mathbb R$ such that $x_n\in D_n$ and $A\subseteq E_n\subseteq\mathbb R\setminus\{x_k\}_{k\in\omega}$.

For every $n\in\omega$ consider the Borel set $B_n=\bigcap_{k\le n}E_k$. It is clear that $B_n$ is a density neighborhood of $A$. It can be shown that the sequence $(B_n)_{n\in\omega}$ has the required property: for any sequence of compact sets $K_n\subseteq B_n$, the union $\bigcup_{n\in\omega}K_n$ is not a density neighborhood of $A$.