Let $M \subset \mathbb{R}$ be a nonempty set of Lebesgue measure zero. Does it follow that $M$ is totally disconnected in the sense that for any $x<y$, with $x,y\in M,$ there exists $z\notin M$ such that $x<z<y$?

I think the answer to the questions is yes, since otherwise one could argue by contradiction and say that then there is an interval contained in $M$ and hence it cannot have measure zero.

Is the reasoning above sound? Also just as side question, connectedness of nonempty sets does not imply positive measure in $\mathbb{R}^n$, since a line in $\mathbb{R}^2$ is connected and has measure zero, right?

Thank you for your time and appreciate any feedback.

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    $\begingroup$ Hint: Can a set of measure zero contain, as a subset, an interval that contains more than one point? Follow-up Hint: The measure of an interval containing more than one point is {negative, zero, positive} (pick one), and measure is a monotone set-function. $\endgroup$ – Dave L. Renfro Feb 16 at 9:11
  • $\begingroup$ @DaveL.Renfro I think it cannot, since that would contradict the set having measure zero. Please correct me if I am wrong. $\endgroup$ – Gaby Alfonso Feb 16 at 9:13
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    $\begingroup$ You are correct. This is one of those questions that I imagine initially seems more difficult than it is, possibly because all the Cantor set weirdness makes you extra cautious and the notion "totally disconnected" is possibly something you deal with a lot more in topology than in an analysis course. $\endgroup$ – Dave L. Renfro Feb 16 at 9:16

You are right: since, in $\mathbb R$, the non-empty connected sets are the intervals and since the non-degenerated intervals have positive measure, a set with measure $0$ cannot contain a non-degenerated interval and therefore it is totally disconnected.

You are are right too about $\mathbb{R}^n$, when $n>1$.


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