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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1Perhaps it would be helpfull putting the definition of Minkowski sum. – matgaio Jun 04 '12 at 20:28
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1@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. – Jun 04 '12 at 20:32
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1@matgaio Added ! – Lierre Jun 04 '12 at 20:36
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@Marvis — Well, it's not clear to me neither ;) – Lierre Jun 04 '12 at 20:37
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4For 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 07 '12 at 16:20
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@DaveL.Renfro Thanks for this ! – Lierre Jun 07 '12 at 18:36
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@Lierre if $A, B \subseteq \mathbb{R}$ are measurable, then $A + B = m(A x B)$, where $m \colon \mathbb{R}^{2} \to \mathbb{R}$ is the continuous function $m(x,y) = x+y$. In particular, as the continuous image of the measurable $A \times B$, the set $A+B$ is 'analytic'. It is a basic fact in descriptive set theory, that analytic subsets of Polish spaces are universally measurable. – Thomas Feb 10 '23 at 09:12
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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.
Davide Giraudo
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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 (additive inverses) of elements of the Cantor ternary set, then $A+B=[-1,1]$.
Cameron Buie
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1@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 04 '12 at 20:42
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Sorry, I didn't understand what you mean by "set of opposites of the Cantor ternary" ! – Fardad Pouran Nov 29 '15 at 09:58
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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?
hmakholm left over Monica
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