Let $A$ and $B$ be two subsets of $\Bbb R$ of measure zero. Is it true that the Minkowski sum $A+B = \{ a + b \mid a \in A, b \in B \}$ has measure zero as well? I think so but I can't prove it. The usual trick with the convolution $\mathbf 1_A \star \mathbf 1_B$ does not seem to lead to something interesting.
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$\begingroup$If $A$ is the set of real numbers such that in their proper binary expansion, the even terms are $0$, and $B$ the same with odd numbers, then $A$ and $B$ have measure $0$ but their sum is the whole real line.
$\endgroup$ 1 $\begingroup$Assuming I understand Minkowski sum correctly, this is not the case. For example, if $A$ is the Cantor ternary set and $B$ the set of opposites of the Cantor ternary set, then $A+B=[-1,1]$.
$\endgroup$ 5 $\begingroup$The following i think works too: Let $A=\cup_{n\in \mathbb{Z}^+\cup\{0\}}\left(n+\frac{1}{2}C\right)$ and $B=\cup_{m\in\mathbb{Z}^-}\left(m+\frac{1}{2}C\right)$, where $C$ is the ternary cantor set. Then, since $\frac{1}{2}C+\frac{1}{2}C=[0,1]$ (which is not hard to prove), it would follow that $A+B=\mathbb{R}$, with $\mu(A)=\mu(B)=0$.
Can someone comment on this solution?
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