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 $A$?

  • $\begingroup$ Unlike topological neighborhoods, $A\subseteq B$ is not required in order for $B$ to be a density neighborhood of $A$, right? $\endgroup$ – bof Sep 12 at 11:43
  • $\begingroup$ @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. $\endgroup$ – Taras Banakh Sep 12 at 12:46
  • $\begingroup$ 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? $\endgroup$ – bof Sep 13 at 0:58
  • $\begingroup$ @bof Very good question. If $A$ is $\sigma$-compact (or maybe even Menger), then the answer is affirmative. $\endgroup$ – Taras Banakh Sep 13 at 6:12

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