Let $X\subset \mathbb{R}$ Lebesgue measurable, $|X|<2^{\aleph_0}$, is it true that $X$ is null?

Of course I am not assuming the Continuum Hypothesis.

EDIT: It might be helpful to know that all Borel measurable sets have cardinality either $\aleph_0$ or $2^{\aleph_0}$. Then a measurable set of cardinality strictly between those two must be Lebesgue but not Borel measurable.

  • $\begingroup$ Have you seen that Lebesgue measure is countably additive? $\endgroup$ – TomGrubb May 22 '15 at 22:31

Yes. This is a trivial consequence of a theorem by Steinhaus:

Suppose that $X$ has a positive measure, then $X-X=\{x-y\mid x,y\in X\}$ contains an interval around $0$.

It is not hard to prove that if $X$ is infinite, then $X$ and $X-X$ are equipotent (there is a surjection from $X^2$ onto $X-X$, and there is an obvious injection from $X$ into $X-X$). Therefore if $X-X$ contains an interval, it has size continuum, and so must $X$.

And so it follows that if $|X|<2^{\aleph_0}$ and $X$ is measurable, then it has to have measure zero.

  • $\begingroup$ @AsafKaragila Is the theorem still true if you replace "$X$ is measurable with positive measure" with "$X$ has positive outer measure"? I'm just wondering if a set with cardinality less than the continuum is necessarily null. $\endgroup$ – JLA May 24 '15 at 21:37
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    $\begingroup$ @JLA: Good question. The answer is not necessarily. If $\sf CH$ holds, then the answer is of course positive. In fact if $\sf CH$ fails but Martin's Axiom holds the answer is still positive. Namely, every set of size $<2^{\aleph_0}$ is null. But it is also consistent that there are sets of cardinality $\aleph_1<2^{\aleph_0}$ which are not measurable. And if they are not measurable, their outer measure cannot be $0$. $\endgroup$ – Asaf Karagila May 24 '15 at 21:44

Asaf's answer is quite fine, but perhaps too specific to Lebesgue measure.

More basically, you can use just the regularity of Lebesgue measure. This tells you that if your measurable set $X$ is not null, then it contains a closed set $C$ which is not null either.

This $C$ is certainly uncountable, and so (as for any uncountable Polish space) one can embed $\{ 0,1\}^{\mathbb N}$ into $C$ using a "Cantor-like" construction. So $C$ (and hence $X$) has cardinality at least $2^{\aleph_0}$.

Not sure, however, that this is "easier" than Asaf's answer...

  • $\begingroup$ It's probably not an easier theorem. It also requires heavier tools from descriptive set theory (e.g. every uncountable closed set has a copy of the Cantor set). But it's a good alternative! $\endgroup$ – Asaf Karagila May 24 '15 at 21:25
  • $\begingroup$ @Asaf I think I agree with You : it's not easier (especially because to perform the Cantor-like construction, one first needs to show that the uncountable closed set $C$ contains a closed subset with no isolated points; and also the regularity is not easy!). Anyway, I would never have thought to use Steinhaus' Theorem here! $\endgroup$ – Etienne May 24 '15 at 21:31

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