# Covering null sets by a finite number of intervals

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.

• 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. – bof Apr 26 at 7:53
• @bof Than you very much. This characterization is something that can go into text books. It is simple to state and simple to prove. – Kavi Rama Murthy Apr 26 at 7:56

## 2 Answers

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.

• (+1) Nice answer! I wish I had thought of that. – José Carlos Santos Apr 26 at 7:25

The set $$\mathbb Z$$ has measure $$0$$ and it is nowhere dense. However, the property $$P$$ doesn't hold for $$\mathbb Z$$.

• 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. – Kavi Rama Murthy Apr 26 at 8:28
• No problem. That answer is better than mine. – José Carlos Santos Apr 26 at 8:29