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Are there sets A and B such that any $x \in \mathbb{R}$ can be decomposed as $x = a+b$, $a \in A$ and $b \in B$, where the Lebesgue measure of $A$ and of $B$ is null.

There is an indication that this should follow from the fact that $C_q = \{z \in[0,1]; z_i\in\{0,2,\cdots,q-1\}\}$ has null measure for every $q \in \mathbb{N}$, where $z_i$ are the numbers in the q-adic expansion of $z$, that is $$z=\sum_{i=1}^{\infty}\frac{z_i}{q^i}$$

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  • $\begingroup$ This was asked recently on this site. Anyway, let $A=\bigcup_{n\in\mathbb Z}(C+2n) $ and $B=C $, where $C $ is the Cantor set, and use that $C+C=[0,2] $. $\endgroup$
    – Andrés E. Caicedo
    Aug 31, 2017 at 15:09
  • $\begingroup$ Where was it asked recently? Can you please link it? $\endgroup$
    – Francisco Maion
    Aug 31, 2017 at 16:15
  • $\begingroup$ I finally found the question I had in mind. It is not the same, but one of the answers addresses your question. See here. $\endgroup$
    – Andrés E. Caicedo
    Aug 31, 2017 at 16:43

2 Answers 2

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Yes. The following example was given here by Davide Giraudo.

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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  • $\begingroup$ I do not understand the definition of sets $A$ and $B$. Can you give an example? $\endgroup$
    – Idonknow
    Aug 31, 2017 at 13:15
  • $\begingroup$ Can you give an indication of the proof of this fact? $\endgroup$
    – Francisco Maion
    Aug 31, 2017 at 13:38
  • $\begingroup$ For $n>0$ let $D(2n)$ be the set of finite binary sequences of length $2n.$ For $z\in \mathbb Z$ and $d\in D(n)$ let $R(z,d)$ be the set of $x\in [z,z+1)$ that have $d$ as their first $2n$ binary digits to the right of the decimal point. For distinct $d,d'\in D(2n)$ the sets $R(z,d'),R(z,d')$ are disjoint and each is a translate of the other so each has measure $4^{-n}.$ CONTINUED IN NEXT COMMENT. $\endgroup$
    – DanielWainfleet
    Aug 31, 2017 at 17:27
  • $\begingroup$ CONTINUED. Let $E(2n)$ be the set of members of $D(2n)$ with $0$ in all even places. $E(2n)$ has $2^n$ members. Now $A\cap [z,z+1))\subset \cup_{d\in E(2n)}R(z,d)$ and the measure $m(\cup_{d\in E(2n)}R(z,d))=2^n4^{-n}=2^{-n}.$ So $m(A\cap [z,z+1)\leq 2^{-n}$ for all $n\in \mathbb N.$ $\endgroup$
    – DanielWainfleet
    Aug 31, 2017 at 17:35
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    $\begingroup$ Intuitively the probability that a sequence of $2n$ coin-tosses will come up heads on every even-numbered toss is $2^{-n}$ so the probability that any $x\in [z,z+1) $ belongs to A should be less than $2^{-n}$ for all $n.$..... I am taking $A$ as those reals that have $0$ in every even binary digit to the right of the decimal point. $\endgroup$
    – DanielWainfleet
    Aug 31, 2017 at 17:40
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Consider the classical cantor set $C$, then $$C-C = \{x-y\mid x,y\in C\} = [-1,1]$$

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  • $\begingroup$ Sure, this is very easy to see. But I want it to form the Real Line, not the unitary interval. $\endgroup$
    – Francisco Maion
    Aug 31, 2017 at 13:39
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    $\begingroup$ @FranciscoMaion You can extend $C$ to all of $\mathbb{R}$ by making it periodic with period 1. Then after extension, $C-C = \mathbb{R}$. $\endgroup$
    – pisco
    Aug 31, 2017 at 15:53

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