I have been trying to solve this question but reaching nowhere

Starting from a countable basis of $\mathbb R$ ,I am asked to construct a Borel set such that $0<m(E \cap I)<m(I)$ for every non empty segment I.

And then must $E$ be of infinite measure?

Here $m$ denotes the Lebesgue measure

  • $\begingroup$ such that what? $\endgroup$ – Marios Gretsas Oct 3 '19 at 11:54

Hint: for a basis, take the intervals $(a,b)$ where $a < b$ are rational.
For your set $E$, take a union of "fat Cantor sets", one for each of these intervals.

  • $\begingroup$ I am not allowed to use the Cantor set $\endgroup$ – يمنى الحاج Oct 3 '19 at 12:03
  • $\begingroup$ Why not? Anyway, a "fat Cantor set" is not the Cantor set. $\endgroup$ – Robert Israel Oct 3 '19 at 12:06
  • $\begingroup$ Is it obvious that $m(E\cap I)<m(I)$ for all $I$? @RobertIsrael $\endgroup$ – Kavi Rama Murthy Oct 3 '19 at 12:18
  • 1
    $\begingroup$ @KaviRamaMurthy No, but if you choose the measures of the fat Cantor sets appropriately you can arrange it. $\endgroup$ – Robert Israel Oct 3 '19 at 14:04

I think the following works; I'll let you worry about verifying that.

Say $(r_j)$ is a dense sequence. Choose $a_j>0$ so that $$a_k>\sum_{j=k+1}^\infty a_j,$$for example $a_j=1/3^j$, then define $$I_j=(r_j-a_j,r_j+a_j),$$ $$E_k=I_k\setminus\bigcup_{j=k+1}^\infty I_j$$and $$E=\bigcup E_k.$$


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