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

Question

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.

Answer 1

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.

Answer 2

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...