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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 $C$. 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.
show/hide this revision's text 1

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 $C$. 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.