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All examples of a dense and co-dense set I have seen are either of full Lebesgue measure or of measure zero. For instance, in restriction to the unit interval $\Bbb I=[0\,\pmb,\,1]$, we could have respectively $\Bbb I\cap\Bbb Q$ or $\Bbb I\setminus\Bbb Q$. What I am looking for is a dense and co-dense subset $A\subset\Bbb I$ such that $$\operatorname{m}(A)=\operatorname{m}(\Bbb I\setminus A)=\tfrac12.$$ I have attempted this task sequentially by, ever more finely, nibbling holes out of subintervals of $\Bbb I$ and partially back-filling the previously created holes. It's easy to approach half measure at each step, but I can't see how to to get convergence.

John Bentin
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  • Even more, there exists a set such that its intersection with any interval has positive measure and intersection of its complement with any interval also has positive measure. I bet someone already asked this question on this site – Aleksei Kulikov Dec 20 '21 at 05:11
  • @AlekseiKulikov : Indeed, the question and answers here are relevant. – John Bentin Dec 20 '21 at 09:30

2 Answers2

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You can take $A=\left[0,\frac12\right]\cup\left(\left[\frac12,1\right]\cap\Bbb Q\right)$.

José Carlos Santos
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  • +1 for being a perfectly good answer to the question as it was formulated originally, which unfortunately was mis-worded. I missed the co-dense condition. – John Bentin Dec 19 '21 at 18:42
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Let $C$ be a fat Cantor set with measure $1/2$. Set $A = C \cup \mathbb{Q} \cap [0,1]$ and you're done.

Jose Avilez
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    Thank you! (+1) This answer is so simple that I feel stupid for asking the question. It also caught the intention of my question, which wasn't clear but is now made explicit in the edit. I'll wait a bit before accepting it, because I'd like to see if there are other kinds of set with this property. – John Bentin Dec 19 '21 at 16:29