# Does $F-F=[0,1] \mod 1$ imply $\mu F = 1$ for $F$ with positive Lebesgue measure?

This question has come up while playing around with the Steinhaus theorem:

Let $F-F$ denote the algebraic difference $\{f-g \mod 1 | f,g \in F\}$. Suppose that $F\subset[0,1]$ with $\mu F>0$ , where $\mu$ is the usual Lebesgue measure. If we know that $F-F = [0,1]$, may we conclude that $\mu F = 1$?

So you are looking for a converse to Steinhaus? No, not true, try $F=$ Cantor set. – Lubin Nov 12 '12 at 5:12
Here is a simple counter-example: $F=[0,\frac{1}{2}]$.