Minkowski sum of a positive Lebesgue measure set and $\mathbb{Q}$. Let $A\subset \mathbb{R}$ be of positive Lebesgue measure, i.e. $\mu(A)&gt;0$. Is it then true that $\mu(\mathbb{R}\setminus (A+\mathbb{Q})) = 0$? 

I am quite sure that if $\mu(A)&gt;0$, 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 $r&gt;0$, we have 
$$
\mu((x-2r,x+2r)\cap (A+\mathbb Q))&lt;r
$$ 
and 
$$
\mu((a-r,a+r)\cap A)&gt; r
$$ 
Let $q$ be a rational number such that $|a+q-x|&lt;r$. 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) &gt; 0$:

There some measurable $ S$ with $ m (S) &gt; 0$  and by Lebesgue density theorem there some $ x \in S \subseteq I $ and some $ 0&lt;r&lt;\frac{m(I)}{2},1&gt;\epsilon &gt; 0$ such that $ I_r=(x-r, x+r)$,  $ m (S \bigcap I_r)  &gt; (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})  &gt;(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})  &gt;(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