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http://en.wikipedia.org/wiki/Steinhaus_theorem

Hi, I was trying to do it the different, please tell me if it is OK.

Since $m(A)>0$, there is a closed $K$ in $A$, with $m(K)>0$. Then $m(K) = m(\text{boundary of }K) + m(\mathrm{int}(K)) = m(\mathrm{int}(K))>0$. Since there is an open subset of $A$ with positive measure. Since in $\mathbb{R}$, every open set countable union of open intervals then we've $(a,b)$ in $A$.

Thus $(-b,-a) \in -A \implies (-(b-a),b-a) \in A-A$.

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2 Answers 2

Your mistake is $m(\text{boundary of }K)+m(\text{int}(K))=m(\text{int}(K))$. There are compact sets (e.g. "fat Cantor sets") with positive measure but empty interior.

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It is false that a subset of $\mathbb{R}$ of positive measure must contain an open subset of positive measure, or that the measure of a closed set equals the measure of its interior.

A counterexample is given by a fat Cantor set, which is closed, has empty interior, but positive Lebesgue measure. So your assertion that $m(K) = m(\mathrm{int}(K))$ is unwarranted. So your argument does not hold.

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