Do sets with positive Lebesgue measure have same cardinality as R? 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? 
 A: I found the answer in the paper "Measure and cardinality" by Briggs and Schaffter. 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.
A: I am not sure if this answer can be helpful or not, but since it is a very elemantary approach to the problem, it might be useful. Assume we are given a subset of R such as A with cardinality smaller than R. Then you can show that the cardinality of A and A+A for any infinite set A is the same. Hence the cardinality of A+A is the same as the cardinality of A which is smaller than R. But you can prove that for any set A of positive measure, A+A has at lease one open interval as a subset. This will be a contradiction with the fact that cardinality of A+A is smaller than R.
A: The answer to the question is that it is independent of ZFC, if one is speaking of outer measure.
The right context for the question and its answer is the very active research area known as Cardinal Characteristics of the Continuum. The point of this subject is to investigate exactly how the dichotomy between countable and continuum plays out in situations when CH fails. For example, in this area, researchers define a number of cardinal invariants:
The bounding number b is the size of the smallest unbounded family of functions from ω to ω. There is no function that bounds every member of the family.
The dominating number d is the size of the smallest dominating family of functions ω to ω. Every function is dominated by a member of the family.
The additivity number for measure is the smallest number of measure zero sets whose union is not measure zero.
The covering number for measure is the smallest number of measure zero sets whose union is all of R.
The uniformity number for measure is the size of the smallest non-measure zero set.
The cofinality number for measure is the smallest size of a family of measure zero sets, such that every measure zero set is contained in one of them. 
Remarkably, none of these numbers is provably equal to any other. In addition, there are models of set theory separating each of them both from ω1 and from the continuum. In the case of the uniformity number, this is the answer that Jonas Meyer has pointed to above. 
One can define similar numbers using the ideal of meager sets in place of the ideal of measure zero sets, and the relationships between all these cardinal characteristics are precisely expressed by Cichon's diagram. In particular, no two of them are provably equal, and there are models of set theory exhibiting wide varieties of possible relationships.
There are dozens of other cardinal characteristics, whose relationships are the focus of intense study by set theorists working in this area. The main tool for separating these cardinal characteristics is the method of forcing and especially iterated forcing.
A: I'm interpreting the question as: Measurable, with positive measure, not as "having positive outer measure" (for which the answer is independent of the basic axioms of set theory, as pointed out by Joel).  
The answer is yes. By elementary properties of Lebesgue measure (regularity), for any $\epsilon>0$, any set $C$ of positive measure contains compact subsets $C_\epsilon$ of measure within $\epsilon$ of the measure of $C$ (interpret this as "arbitrarily large" if $C$ has infinite measure). Any set of positive measure is obviously uncountable. It is straightforward to see that a compact uncountable set of the reals contains a perfect set, and that perfect sets have the same size as the reals. Therefore, $C$ must also have the size of the reals. (I guess the last step uses the Schroeder-Bernstein theorem.)
(On a side note, Cantor proved the result that closed uncountable subsets of ${\mathbb R}$ have the size of the reals. This extends to larger collections of sets, e.g., to all uncountable Borel sets. The first approach to the continuum hypothesis was to try to keep on extending this result.)
To see that perfect sets have the size of the reals: Check that any perfect set has a "copy" of Cantor's set; this is standard; baby Rudin essentially shows how in an exercise in Chapter 1 or 2. Cantor's set, by construction, obviously has size $2^{|{\mathbb N}|}$. Check that the reals also have this size, e.g., by noticing that ${\mathbb R}$ and $(0,1)$ have the same size, and identifying reals in $(0,1)$ with their infinite binary expansion. I suppose this may also use Schroeder-Bernstein depending on how one fleshes this outline out.
A: A set of positive measure contains a closed subset of positive measure. It is known that closed subset of reals may have cardinality either continuum or at most countable
(http://en.wikipedia.org/wiki/Perfect_set_property)
