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Let $A\subset \mathbb{R}$ be of positive Lebesgue measure, i.e. $\mu(A)>0$. Is it then true that $\mu(\mathbb{R}\setminus (A+\mathbb{Q})) = 0$?

I am quite sure that if $\mu(A)>0$, then $A-A$ contains a rational, as well as an irrational number, but I'm not sure if this actually helps.

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About your second paragraph: $A-A$ contains an open interval. – user127096 Apr 4 '14 at 23:19

Suppose to the contrary that the set $ \mathbb R\setminus (A+\mathbb Q)$ has positive measure. By the Lebesgue density theorem, it has a point of density, call it $x$. Also pick a point of density of $A$, call it $a$. For sufficiently small $r>0$, we have $$ \mu((x-2r,x+2r)\cap (A+\mathbb Q))<r $$ and $$ \mu((a-r,a+r)\cap A)> r $$ Let $q$ be a rational number such that $|a+q-x|<r$. Then $$((a-r,a+r)\cap A)+q \,\subseteq \,(x-2r,x+2r)\cap (A+q) $$ but the set on the left has greater measure, a contradiction.

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