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It is well-known that $[0,1]$ is not a nontrivial disjoint union of closed intervals -- e.g.: https://math.stackexchange.com/questions/1195179/the-interval-0-1-is-not-the-disjoint-countable-union-of-closed-intervals

Using fat Cantor sets, I can construct disjoint unions of closed intervals in $[0,1]$ with arbitrarily large Lebesgue measure. Question: Can full Lebesgue measure be achieved?

Edit: Here is my precise question: Does there exist an infinite collection of disjoint, nonempty, closed intervals $I_n=[a_n,b_n]$ such that $\mu([0,1]\setminus\bigcup_{n\ge1}I_n)=0$, where $\mu$ is the Lebesgue measure?

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  • $\begingroup$ Thanks, I clarified. $\endgroup$
    – Aryeh Kontorovich
    Commented Jan 26, 2022 at 12:01

1 Answer 1

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Yes, sure: If you have a finite union of closed intervals, the complement in [0,1] is a finite union of open (in [0,1]) intervals. Hence, you find in this complement a finite union of closed intervals such that, if you add these intervals to your previous collection, the measure of the complement is only half of the measure it was before. Proceeding by induction and taking the union over all intervals you have chosen, you obtain a countable union of disjoint closed intervals of full measure.

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    $\begingroup$ An explicit description of one such set is the set of all reals in $[0,1]$ whose base-$4$ expansion contains a $1$ or $2$ at some position (allowing also $0{.}x_1\dots x_n2333\dots$ for dyadic rationals). $\endgroup$
    – Emil Jeřábek
    Commented Jan 26, 2022 at 12:48

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