If E is a null set in $(\mathbb{R};M_l; \mu_l)$,($M_l$ is a sigma algebra and $\mu_l$ is Lebesgue measure) prove that $E^c$ is dense in $\mathbb{R}$.


For every open interval $I$ in $\mathbb{R}$; $\mu_l(I) > 0$ (property of Lebesgue measure). If $\mu_l(I) = 0$, then by the monotonicity of $\mu_l(I)$, $E$ cannot contain any open interval as a subset. This implies that $E^c \cap I = \emptyset$ for any open interval $I$ in $\mathbb{R}$ . Thus $E^c$ is dense in $\mathbb{R}$

How can $E^c \cap I = \emptyset$ since $E^c$ is the compliment of $E$ and $E$ does not contain any interval?

$E^c$ must contain an interval, right?

  • $\begingroup$ $E^c \cap I =\emptyset$ is not true for every open interval $I$. You can check with $E=\mathbb Q$. $\endgroup$ – ntt Apr 9 '17 at 18:02
  • $\begingroup$ The solution is not mine. If what you say is true which I agree, the solution is wrong, right? $\endgroup$ – Pedro Gomes Apr 9 '17 at 18:07
  • $\begingroup$ First follow-up exercise: Prove that $E^c$ has uncountably many elements in every nonempty open set. Second follow-up exercise: Prove that $E^c$ has $c$ (cardinality of the reals) many elements in every nonempty open set. The second exercise probably requires results you might not yet have available, such as the fact that every uncountable closed set has cardinality $c$ and $F_{\sigma}$ approximations of measurable sets, but I believe the first requires only a little more work than what you're trying to prove. $\endgroup$ – Dave L. Renfro Apr 10 '17 at 21:29

It should be $E^c\cap I\neq\emptyset$, and THAT means that $E^c$ is dense. So in short, the solution contains the right idea, but also a typo.

  • $\begingroup$ I think you are right! Thank for confirming there is a typo! $\endgroup$ – Pedro Gomes Apr 10 '17 at 20:33
  • $\begingroup$ @PedroGomes Glad to help! $\endgroup$ – haemi Apr 10 '17 at 22:14

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