Is the measure of the sum equal to the sum of the measures?

Another example: $$A=\mathbb{Z},\quad B=[0,1],\quad A+B=\mathbb{R}. $$

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 $A+B=[0,4]$)

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 $(0,2)$.

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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 $$ \forall n\in\mathbb{Z},\qquad n\in \left(C+C+C+C+C\right).$$