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?
[Math] Dense subset of $[0,1]$ with Lebesgue measure $\epsilon$
lebesgue-measuremeasure-theory
Best Answer
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.