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Let us say that a subset $A$ of $\mathbb R$ has property $P$ if every $\epsilon>0$ there is a finite collection of open intervals $(a_1,b_1),(a_2,b_2),\cdots,(a_n,b_n)$ such that $A \subset \cup_i (a_i,b_i)$ and $\sum (b_i-a_i) <\epsilon$. My question: if $A$ is a nowhere dense set with measure $0$ does it have property $P$?

Some basics: every compact set of measure $0$ (obviously) has property $P$. No dense set of measure $0$ can have property $P$. More generally, if $A$ has property $P$ then $A$ is nowhere dense. Proof: if $(\alpha,\beta) \subset \overset {-} {A}$ take $\epsilon <\beta -\alpha$. If $A \subset \cup_i (a_i,b_i)$ and $\sum (b_i-a_i) <\epsilon$ then $(\alpha,\beta) \subset \cup_i [a_i,b_i]$ so $\beta -\alpha <\epsilon$, a contradiction. My question is if every nowhere dense set of measure $0$ has property $P$. My guess that the implication does not hold but I don't have a counterexample.

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  • $\begingroup$ As I noted in an edit to my answer, your property $P$ has a simple characterization: $A$ has property $P$ iff $A$ is bounded and its closure has Lebesgue measure zero. $\endgroup$ – bof Apr 26 at 7:53
  • $\begingroup$ @bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove. $\endgroup$ – Kavi Rama Murthy Apr 26 at 7:56
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For an example of a bounded nowhere dense set of measure zero which does not have your property $P$, let $F$ be a compact nowhere dense set of positive measure (a fat Cantor set), and let $A$ be a countable dense subset of $F$.

In fact, it's easy to see that a set $A\subseteq\mathbb R$ has property $P$ if and only the closure of $A$ is a compact set of measure zero.

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  • $\begingroup$ (+1) Nice answer! I wish I had thought of that. $\endgroup$ – José Carlos Santos Apr 26 at 7:25
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The set $\mathbb Z$ has measure $0$ and it is nowhere dense. However, the property $P$ doesn't hold for $\mathbb Z$.

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  • $\begingroup$ I am sorry to have removed the approval to your answer. bof has given a simmple necessary and sufficient condition and I would like to draw the attention of other members to that answer. $\endgroup$ – Kavi Rama Murthy Apr 26 at 8:28
  • $\begingroup$ No problem. That answer is better than mine. $\endgroup$ – José Carlos Santos Apr 26 at 8:29

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