TITLE: Can sets of cardinality $\aleph_1$ have nonzero measure?
QUESTION [16 upvotes]: $\aleph_1$ is the cardinality of the countable ordinals. It is the least cardinal number greater than $\aleph_0$, and assuming the continuum hypothesis it's equal to $\mathfrak{c}$, the cardinality of the the real numbers.  
My question is, is it possible for all $\aleph_1$ subsets of $\Bbb{R}$ to have Lebesgue measure $0$?  This is, of course, impossible assuming the continuum hypothesis, because then all sets of real numbers would have measure $0$, which is absurd.  But is $\mathsf{ZFC}$ + $\lnot\mathsf{CH}$ + "all subset of $\Bbb{R}$ of cardinality $\aleph_1$ have Lebesgue measure $0$" consistent?  If not, what if we replaced Lebesgue measure with some other measure?
It may be worth noting that, although there may be subsets of $\Bbb{R}$ with cardinality $\aleph_1$, there are no subsets of $\Bbb{R}$ which have the order-type $\omega_1$ (the order-type of the countable ordinals) under the usual ordering on $\Bbb{R}$.
Any help would be greatly appreciated.
Thank you in Advance.

REPLY [4 votes]: A very simple model of "$2^{\omega} > \omega_1$ and every set of reals of size $\omega_1$ is null" is the Cohen's original model for the failure of CH. It is obtained by adding $\omega_2$ Cohen reals - i.e. forcing with $Fn(\omega_2, 2)$. The proof uses the fact that adding a Cohen real makes the set of old reals null. The category analogue of this can be obtained by adding $\omega_2$ random reals - i.e. forcing with the usual measure algebra on $2^{\omega_2}$.