Is the measure of the sum equal to the sum of the measures? Let $A,B$ be subsets in $\mathbb{R}$. Is it true that
$$m(A+B)=m(A)+m(B)?$$
Provided that the sum is measurable.
I think it should not be true, but could not find a counterexample. 
 A: Let we consider the ternary representation of some number in $(0,1)$:
$$ x= 0.\overline{1022101}_3 $$
Exploiting $0=\frac{0+0}{2},1=\frac{0+2}{2},2=\frac{2+2}{2}$ digit by digit, we may write $x$ as the average between two numbers $a,b$
$$ x = 0.\overline{1022101}_3 = \frac{0.\overline{0022000}_3+0.\overline{2022202}_3}{2}=\frac{a+b}{2} $$
with the property that all their ternary digit belong to $\{0,2\}$. It follows that if $K$ is a Cantor set in $[0,1]$, $\mu(K)=0$, but $\mu(K+K)\geq 2$, since $K+K$ contains every point of the interval $(0,2)$.

This argument also has a discrete analogue in terms of Sidon sets or additive bases. For instance, if $Q$ is  the set of integer squares and $E=Q+Q$, $E$ has density zero in $\mathbb{N}$, i.e.
$$ \lim_{n\to +\infty}\frac{\left|E\cap[1,n]\right|}{n}=0,$$
but every natural number belongs to $E+E$ by Lagrange's four-squares theorem.
If we take $C$ as the set of integer cubes,
$$ \lim_{n\to +\infty}\frac{\left|C\cap[-n,n]\right|}{2n+1}=0, $$
but
$$ \forall n\in\mathbb{Z},\qquad n\in \left(C+C+C+C+C\right).$$
A: Another example:
$$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}.
$$
A: I preassume that here $A+B:=\{a+b\mid a\in A, b\in B\}$.
Counterexample (for Lebesgue measure):
Take $A=[0,1]\cup[2,3]$ and $B=[0,1]$ (so that $A+B=[0,4]$)
