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.
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.
This proof is for measurable $ A $ :
It suffice to prove it for some measurable $ B \subset A $ and $ B\subseteq I $ where $I$ is an interval and $ m (B) > 0$:
There some measurable $ S$ with $ m (S) > 0$ and by Lebesgue density theorem there some $ x \in S \subseteq I $ and some $ 0<r<\frac{m(I)}{2},1>\epsilon > 0$ such that $ I_r=(x-r, x+r)$, $ m (S \bigcap I_r) > (1-\epsilon)m (I_r)$
It also obvious that for $ S_q =S + q $, $ I_{r, q}=I_r + q $,$ m(S_q)=m (S) , m(I_{r, q})=m (I_r) $, and $ (S_q\bigcap I_{r, q}) >(1-\epsilon)m (I_r)$ ,where $q \in \mathbb {Q} $
Now we can take subsequence {$ I_{r, q} $}that is pairwise disjoint such that $\bigcup_q I_{r, q} \supseteq I $
Define $ B = S \bigcap I_r $ , $ B_q=B+q $, It follows $ m(\bigcup_q B_{q}) >(1-\epsilon)m (I)$
$ m(I-\bigcup_q B_{q}) = m (I) - m ( I\bigcap\bigcup_q B_{q}) $
$ m(I-\bigcup_q B_{q}) \le \epsilon m (I)+ 2r $
$A_q=A+q$
$ I- \bigcup_q A_q \subseteq I-\bigcup_q B_{q} $
Therefore $ m(I-\bigcup_q A_q) \le \epsilon m (I)+2r $
Letting $r,\epsilon \to 0$ the result follows
