Prove $A+A=\mathbb{R}$ in which the measure of the complement of $A$ is zero Let $A\subset \mathbb{R}$ such that $m(\mathbb{R}\smallsetminus A)=0$. Show
   that $A+A=\mathbb{R} $, where
   $$A+A=\{a+b\mid a,b\in A\}$$.
This is a question in the past qualifying exam in my university. I do not know where to approach. I encountered a similar problem  that if $A$ is measurable and $m(A)>0$, then $A-A$ contains an interval, but I used $m(A)$ finite to do this problem. Can you help?
 A: Suppose not. Then there exists $r\in \mathbb{R}$ s.t. $r-a\notin A$ for any $a\in A$. Let $B:=\{r-a: a\in A\}$. By translation invariance of the Lebesgue measure $m(B)>0$. However, $B\subset \mathbb{R}\setminus A$ and so $m(B)=0$.
A: Assume a point $p$ does not exist in $A + A$. 
That means that  $\forall x, (x \in A)\rightarrow((p-x) \not \in A)$. 
Define $A_p = \{p-x| x\in A \}$ . We have $A_{p} \in \bar{A}$ . 
However, $f(x) = p-x$ is a measure preserving transformation, so $A_p$ is of the same measure as $A$. Therefore the complement of $A$ would have measure greater than or equal to $A$. 
If it had measure $0$, then the union of $A$ and its complement would be of measure $0$. This cannot happen.
A: Against the thesis, let $\ q\in\mathbb R\setminus(A\times A).\ $ Let
$\ p:= \frac q2.\ $ Define,
$$ K\ :=\ [p-1;p]\cap\mathbb R\qquad\mbox{and}\qquad
      L\ := [p;p+1] $$
Furthermore, Let $\forall_{x\in\mathbb R}\ f(x):= q-x\ $ and
$$ K'\ :=\ f(K) $$
so that
$$ K'\ \subseteq [p;p+1]\qquad\mbox{and}\qquad K'\cap L=\emptyset. $$
Clearly, all three sets $\ K\ K'\ L\ $ are measurable, and
$\ m(K')+m(L)\ \le\ 1.\ $ But $\ m(K')=m(K),\ $ hence
$$ m(K\cup L)\ =\ m(K)+m(L)\ \le\ 1 $$
while
$$ \mathbb R\cap[p-1;p+1]\ =\ K\cup L $$
Thus,
$$ m((\mathbb R\cap[p-1;p+1])\ \setminus\ (K\cup L))\,\ \ge\,\ 1 $$
This proves that $\ m(\mathbb R\setminus A)\ \ge\ 1,\ $ in a contradiction to the assumption of the theorem.
Great!
