# 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 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. – Taras Banakh Sep 12 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 at 0:58
• @bof Very good question. If $A$ is $\sigma$-compact (or maybe even Menger), then the answer is affirmative. – Taras Banakh Sep 13 at 6:12