# Do there exist any subsets of $\mathbb{R}$ with positive measure but not being continuum? [duplicate]

As title goes, without assuming continuum hypothesis, is there a subset of $\mathbb{R}$ with positive measure but not being continuum? That is, $$\textrm{Does there }\exists A\subseteq \mathbb{R}, \quad \textrm{such that }\quad \mu(A)>0, \quad \aleph_0<|A|<\aleph_1$$ I believe that maybe in some system without assuming continuum hypothesis, there exists.

And by the regularity of Lebesgue measure, it suffices to deal with compact set.

## marked as duplicate by Asaf Karagila♦ cardinals StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); Feb 28 '18 at 6:06

• Note that there is an important subtlety here: the assumption of measurability. There are, provably in ZFC, sets of reals of cardinality $\aleph_1$ which are not measurable; since null sets are measurable, this means such sets have positive outer measure, and if CH fails they have cardinality $<2^{\aleph_0}$. So if we drop the implicit measurability assumption, we get independence. – Noah Schweber Feb 28 '18 at 3:49
There's a standard exercise in measure theory that says that if $A$ has positive measure, then $A-A$ contains an interval. It follows from this that $A$ must have the same cardinality as the reals.
I find a solution due to This question. Due to the regularity of Lebesgue measure, it suffices to deal with compact set. And a standard result is a closed subset of $\mathbb{R}$ is either at most countable or continuum.