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$(a)$ Let $E\subset [0,1]$ be Lebesgue measurable. Suppose there exists a fixed $\epsilon > 0$ such that $m(E\cap (a,b))\geq \epsilon|a-b|$ for any interval $(a,b)\subset [0,1]$. Show that $m(E) = 1$

$(b)$ Give an example of a Lebesgue measurable set $E\subset [0,1]$ with $0<m(E)<1$ and $m(E\cap(a,b))>0$ for all nontrivial intervals $(a,b)\subset [0,1]$. Explain why this example does not contradict part $(a)$

My thoughts:

(a) if $m(E)<1$, then $\exists U$ open s.t. $U$ is the countable union of intervals, $E\subset U$, and $m(U)\leq (1-m(E))/2$. Then $U^c$ will be disjoint with $E$ and will contain some interval whose intersection with $E$ will have positive measure.

(b) I'm at a lost on how to construct $E$.

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  • See the answer to this question. – bof May 21 '18 at 04:10
  • I think that the proof for (a) you present is missing. It is possible that U is a countable union of intervals and is not the whole [0,1], and still [0,1]\U does not contain any interval. For instance, U may be a union of intervals around the rationals and of measure, say, 1/2. – N A-R Jan 01 '23 at 06:58
  • For part a, your proof is indeed not correct, for a proof that works see this question https://math.stackexchange.com/questions/4590931/prove-that-there-do-not-exist-a-lebesgue-measurable-set-with-the-following-prope/4613350#4613350

    The proof there is for $\varepsilon =1/2$ but the idea generalizes

     – Connor Lockhart Jan 07 '23 at 21:39

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Check page $74$ of Fremlin’s “Measure Theory - Volume $1$”, part $(b)$ of $134J$ (“more examples”). Here’s a link to the pdf.