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I'm not sure how to prove this:

Show that there exists a measurable set $E \subset [0,1]$ such that

$$0 < \frac{m(E \cap (a,b))}{b-a} < 1 \text{ for every interval } (a,b) \subset [0,1],$$

where $m$ denotes the Lebesgue measure on $\mathbb{R}$.

I'm having trouble understanding what direction to take with this exercise. Instead of constructing $E$ explicitly, the idea is to show that the class of sets $A \subset \mathcal{L}([0,1])$ (the Lebesgue measurable sets in $[0,1]$) which satisfy the inequality is a countable intersection of open dense sets, and then conclude that it's non-empty using the Baire Category Theorem. I honestly have no idea where to start on this, so any guidance or proof ideas would be appreciated.

ozz
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  • one possible approach is to mimic the construction of the fat cantor set, and "insert " many such fat cantor sets in specially chosen places within the unit interval (so that the ratio is strictly less than 1,and the reason we want many such cantor sets is so that every open interval contains atleast one copy, so the quotient is strictly positive) . I'm pretty sure Rudin also has a paper where he explicitly describes how to construct such a compact set. – peek-a-boo Apr 03 '21 at 00:49

1 Answers1

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Hint: Let $(a_n,b_n), n=1,2,\dots$ be a basis for the usual topology on $(0,1).$ For each $n,$ let $K_n\subset (a_n,b_n)$ be a compact set with positive measure and no interior. The $K_n$ can be chosen so that they are disjoint. Think about $\cup_{n=1}^\infty K_n.$

zhw.
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  • And the $K_n$ can be chosen so that $m(\bigcup_{n-1}^\infty K_n)=1$. The problem is to show that the $K_n$ can be chosen so that $m(\bigcup_{n=1}^\infty K_n\cap(a,b)\lt b-a$ for every interval $(a,b)$. – bof Apr 03 '21 at 01:22