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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 $\mu F = 1$?

Thanks for your input!

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So you are looking for a converse to Steinhaus? No, not true, try $F=$ Cantor set. –  Lubin Nov 12 '12 at 5:12
    
@Lubin, OP asks for example with positive Lebesgue measure. Of course, your example is easily modified. –  Gerry Myerson Nov 12 '12 at 5:27
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1 Answer 1

up vote 1 down vote accepted

Here is a simple counter-example: $F=[0,\frac{1}{2}]$.

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Thanks richard! Should've seen something so straightforward. –  Mathemantissa Nov 12 '12 at 5:28
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