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On an exam I took recently, I was challenged with the following problem but couldn't come up with an answer. I couldn't find the answer by searching the forum either so I hope it hasn't been asked before. Any help on how to answer, and if possible how to think when challenged with this kind of problem, is much appreciated:

Problem Create a Lebesgue measurable set $E\subseteq [0,1]$ such that $0<m(E \cap I)<m(I)$ for all intervals $I=(a,b)$, where $0\leq a<b\leq 1$ and m is the Lebesgue measure.

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  • $\begingroup$ Construct a fat Cantor set in $[0,1]$. In each of the "removed intervals", insert fat Cantor sets. In each of the removed intervals of the fat Cantor sets inserted in the second step, insert fat Cantor sets. Continue... $\endgroup$
    – David Mitra
    Commented Jan 17, 2014 at 10:06
  • $\begingroup$ @DavidMitra While taking care that the union of those "fat Cantor sets" does not have measure one. $\endgroup$
    – bof
    Commented Jan 17, 2014 at 10:15
  • $\begingroup$ You can find a detailed discussion of such sets in this Math StackExchange question from 13 August 2011: Construction of a Borel set with positive but not full measure in each interval. (Identical comment I gave here, if this rings a bell for anyone.) $\endgroup$
    – Dave L. Renfro
    Commented Jan 17, 2014 at 15:03
  • $\begingroup$ @DaveL.Renfro, thanks, that clears it up! $\endgroup$
    – KOE
    Commented Jan 18, 2014 at 7:32

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I assume you know how to construct a closed nowhere dense set of positive Lebesgue measure, i.e., a so-called "fat Cantor set". Clearly, a fat Cantor set can be constructed inside any given open interval.

Let $\{I_n:n\lt\omega\}$ be the set of all open intervals $\{q,r\}$ where $q,r$ are rational numbers and $0\le q\lt r\le1$. Recursively construct a sequence of pairwise disjoint fat Cantor sets $A_1,B_1,A_2,B_2,\dots,A_n,B_n,\dots$ with $A_n,B_n\subseteq I_n$. The set $E=\bigcup_{n=1}^\infty A_n$ has the desired properties.

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  • $\begingroup$ Thanks @bof, I think I get the idea, also conveyed by David Mitra's comment. However I'm not perfectly clear on the details. Let me check if I got it right: if $n$ indexes the rational intervals and for every $n$, we construct two disjoint fat cantor sets, minus all points already in previously constructed sets, denoted $A_n,B_n$ inside the corresponding interval and then $\cup_n A_n$ has the desired property? I don't get the notation ${I_n:n<\omega}$, but that's what I understood from the text. I just have to convince myself it works. $\endgroup$
    – KOE
    Commented Jan 17, 2014 at 10:35

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