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How can I Construct a measurable set A contained in C = [ 0, 1] such that for an arbitrary open interval B contained in C, $m(A\cap B)> 0$ and $m(CA\cap B)> 0$, CA is the complement respect to C.

My ideas give some of the measures zero, or can I construct with borel sets? Thanks.

J. Natch
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  • Is that funny symbol y in your condition meant to be AND (or $\wedge$)? – Mark Fischler Aug 15 '17 at 22:33
  • yes the word is and. – J. Natch Aug 15 '17 at 22:35
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    https://math.stackexchange.com/a/961770/169852 –  Aug 15 '17 at 22:43
  • but $F_\sigma$ does not apply. – J. Natch Aug 15 '17 at 23:45
  • if $F_\sigma \subset [0,1]$ and $F_\sigma \neq [0,1]$, $F_\sigma \cap B= \varnothing$ for some B. – J. Natch Aug 15 '17 at 23:50
  • @J.Natch If by $F_\sigma$ you mean the set $M$ constructed in the answer linked above, then what you have written is false. If $B$ is any open interval then it contains one of the $I_k$'s (an interval with rational endpoints) and therefore $B$ contains $M_k$. Then, since $M_K \subseteq M$, we have $M_k = M_k \cap B \subseteq M \cap B$, whence $0 < m(M_k) \leq m(M \cap B)$. –  Aug 16 '17 at 01:48
  • soorry i read wrong, is necessary that $M_k$ be nowhere dense? – J. Natch Aug 16 '17 at 05:43
  • I dont see nowhere density as required condition in my problem. – J. Natch Aug 16 '17 at 18:22

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