# Sum of sets of measure zero

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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Perhaps it would be helpfull putting the definition of Minkowski sum. – matgaio Jun 4 '12 at 20:28
@Lierre I have a question. Actually, it is not completely clear to me if the set is even measurable in the first place (?). It might be trivial though but it is not clear to me. – user17762 Jun 4 '12 at 20:32
@matgaio Added ! – Lierre Jun 4 '12 at 20:36
@Marvis — Well, it's not clear to me neither ;) – Lierre Jun 4 '12 at 20:37
For those interested in history, Lebesgue seems to be the first person to have proved this result. See p. 285 of Lebesgue, Sur la recherche des fonctions primitives par l'integration, Atti della Accademia Nazionale dei Lincei, Rendiconti, Classe di Scienze Fisiche, Matematiche e Naturali (5) 16 #1 (1907), 283-290. – Dave L. Renfro Jun 7 '12 at 16:20

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 $A+B=[-1,1]$.

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Awesome example ! – Lierre Jun 4 '12 at 20:38
@Lierre: Glad you like it! Even simpler to prove is that $A+A=[0,2]$, from which the result I quoted follows readily. – Cameron Buie Jun 4 '12 at 20:42
Probably this example is even in your textbook... – GEdgar Sep 19 '12 at 20:10
Sorry, I didn't understand what you mean by "set of opposites of the Cantor ternary" ! – Fardad Pouran Nov 29 '15 at 9:58
@Fardad: I mean $B=\{-x:x\in A\}.$ – Cameron Buie Nov 29 '15 at 11:21

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.

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Even simpler, thanks ! – Lierre Jun 4 '12 at 20:42

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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Yes, this is an extension of Cameron Buie's answer. – Ross Millikan Sep 18 '12 at 23:22