# A selection principle in measure theory

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 $$A$$?

• Unlike topological neighborhoods, $A\subseteq B$ is not required in order for $B$ to be a density neighborhood of $A$, right?
– bof
Sep 12, 2020 at 11:43
• @bof You are right: this is not required. I thought how to call such a set. But nothing better than "density neighborhood" could not find. Sep 12, 2020 at 12:46
• What are the trivial cases? I think it's trivial if $(B_n)$ is a constant sequence, am I right? What if $A$ is a countable set, is it true in that case, and is it trivial?
– bof
Sep 13, 2020 at 0:58
• @bof Very good question. If $A$ is $\sigma$-compact (or maybe even Menger), then the answer is affirmative. Sep 13, 2020 at 6:12

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 $$A$$.