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There exists a compact set $K \subset \mathbb{R}$ of positive Lebesgue measure such that $m(K \cap I) < |I|$ for every interval $I$ of positive length

For this question, I am thinking about using a fat Cantor set K . I know there is no interval contained in a fat Cantor set K . But how to proceed to argue that $m(K \cap I) <|I|$ then?

Source: it's an exercise from Tao's book.

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  • $\begingroup$ what is $m$? How about $I$ Your question is missing context. $\endgroup$ – mathemagician Oct 26 '14 at 22:02
  • $\begingroup$ @mathemagician, m denotes the lebesgue measure and I is just an interval. $\endgroup$ – user119758 Oct 26 '14 at 22:05
  • $\begingroup$ There should be something wrong with your problem, $m(K\cap I)$ is always less than $m(I)$, at least weakly. $\endgroup$ – mathemagician Oct 26 '14 at 22:08
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    $\begingroup$ One noteworthy property of a Cantor set is that it's "nowhere dense". What does that mean? Can you prove that, if $S\subseteq I$ and $m(S)=m(I)$, then $S$ is dense in $I$? $\endgroup$ – bof Oct 26 '14 at 22:19
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    $\begingroup$ If you read your question, you will find a reason why $I\setminus K \neq \varnothing$ [unless $I =\varnothing$ of course, but then you cannot have a strict inequality anyway]. $\endgroup$ – Daniel Fischer Oct 26 '14 at 22:48
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Summary of the main points made in comments:

  1. Every nonempty open set has positive length.
  2. For every open interval, $I\setminus K$ is open.
  3. If $I\setminus K$ is empty, then $K$ contains $I$, so it has nonempty interior.
  4. Fat Cantor sets have empty interior.
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