Can sets of cardinality $\aleph_1$ have nonzero measure? $\aleph_1$ 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.
 A: 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 $2^{\omega_2}$.
A: The following result is due to Martin and Solovay, and can be found in Jech's Set Theory (3rd Millennium edition) as theorem 26.39.

If Martin's Axiom holds, then the union of fewer than $2^{\aleph_0}$ null sets is null, and the union of fewer than $2^{\aleph_0}$ meager sets is meager.

It follows, if so, that under $\sf ZFC+MA+\lnot CH$ the union of $\aleph_1$ singletons has measure zero, so every set of size $\aleph_1$ has measure zero.
It suffices, if so, to show that $\sf ZFC+MA+\lnot CH$ is a consistent theory (relative to the consistency of $\sf ZFC$, of course). This is a result of Solovay and Tennenbaum, which appears in the same book as Theorem 16.13.

Assume $\sf GCH$ and let $\kappa$ be a regular cardinal greater than $\aleph_1$. There is a c.c.c. notion of forcing $P$ such that the generic extension $V[G]$ by $P$ satisfies Martin's Axiom and $2^{\aleph_0}=\kappa$.

By taking, for example, $V=L$ as the ground model and $\kappa=\aleph_2$, the result is a model where the continuum is $\aleph_2$ and the union of $\aleph_1$ null sets is a null set.
A: Of course assuming CH they can.  But if $\aleph_1 < 2^{\aleph_0}$ then there is no measurable set of cardinality $\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no.    We can prove the following in ZFC without requiring additional axioms.
Proposition (ZFC).  Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$.
This is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\aleph_0}$).  For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.
Corollary (ZFC).  Every Lebesgue measurable subset of $\mathbb{R}$ with cardinality less than $2^{\aleph_0}$ has Lebesgue measure zero.
Proof.  Every Lebesgue measurable set $E$ can be written $E = B \cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$.  But if $|E| < 2^{\aleph_0}$ then $|B| < 2^{\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.
