I have been thinking about which kind of wild non-measurable functions you can define. This led me to the question:


Is it possible to prove in ZFC, that if a (**Edit**: measurabel) set $A\subset \mathbb{R}$ has positive Lebesgue-measure, it has the same cardinality as $\mathbb{R}$? It is obvious if you assume CH, but can you prove it without CH? 



Do sets with positive Lebesgue measure have same cardinality as R?
I found the answer in the paper ["Measure and cardinality" by Briggs and Schaffter](http://www.jstor.org/pss/2320153). In short: not if I interpret positive measure to mean positive outer measure. A proof is given that every *measurable* subset with cardinality less than that of $\mathbb{R}$ has Lebesgue measure zero. However, they then survey results of Solovay that show that there are models of ZFC in which CH fails and every subset of cardinality less than that of $\mathbb{R}$ is measurable, and that there are models of ZFC in which CH fails and there are subsets of cardinality $\aleph\_1$ that are nonmeasurable. So it is undecidable in ZFC.


If it was intended that our sets are assumed to be measurable, then the answer would be yes by the first part above.


**Edit**: In light of the comment by Konrad I added a couple of lines to clarify.


