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We wish to find a Lebesgue measurable subset of $[0,1]$ that is in dense in $[0,1]$ with measure exactly $\epsilon$, where $\epsilon \in (0,1)$. My idea is to let $I=(0,\epsilon)$ and let $I'=\mathbb{Q} \cap (\epsilon, 1)$. Then set $A = I \cup I'$. Is this correct? If so, is there another such set?

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    $\begingroup$ Why not choose $[0,\epsilon) \cup \mathbb{Q}$. Or any set $(a, a+\epsilon) \cup \mathbb{Q}$ as long as $0 \le a$ and $a+\epsilon \le 1$. Or any other chopped up interval. The denseness comes from the rationals which add nothing to the measure. The measure comes from the interval. $\endgroup$ – copper.hat Aug 1 '14 at 3:11
  • $\begingroup$ @copper.hat Because $[0,\epsilon]\cup\mathbb Q$ is not a subset of $[0,1]$. $\endgroup$ – bof Aug 1 '14 at 3:14
  • $\begingroup$ Yes, but I was adding in the rationals in the unit interval. $\endgroup$ – copper.hat Aug 1 '14 at 3:15
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    $\begingroup$ @bof: Yes, you are right, but I was getting tired of mathjax :-). $\endgroup$ – copper.hat Aug 1 '14 at 3:16
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    $\begingroup$ Choose $((a, a+\epsilon) \cup \mathbb{Q}) \cap [0,1]$. $\endgroup$ – copper.hat Aug 1 '14 at 3:16
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Yes, that works fine. For an open dense set, put intervals of width $2^{-k + 100} \epsilon$ around the $k$-th rational number (according to your favorite enumeration of $\mathbb{Q} \cap (0, 1)$), adjusted appropriately if the interval doesn't lie in $[0,1]$. Then lengthen the interval around $1/2$ as needed to make up the remaining measure.

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