Does there exist a measurable subset $A \subset \mathbb{R}$, such that $\mu(A)$ is finite, but $\mu(\{a+b|a,b\in A\}) = \infty$?

Question

Does there exist a measurable subset $A \subset \mathbb{R}$, such that $\mu(A)$ is finite, but $\mu(\{a+b|a,b\in A\}) = \infty$? Here $\mu$ stands for Lebesgue measure.

If such subset exists, it can not be bounded: Suppose it is. Then there exists such $a \in \mathbb{R}$ such that $A \subset [-a;a]$. That results in $\{a+b|a,b\in A\} \subset [-2a;2a]$, from which follows that $\mu(\{a+b|a,b\in A\}) \leq 4a$ is finite.

However, I do not know, how to solve the problem in general.

Any help will be appreciated.

@bof why not an answer ? – Henno Brandsma Sep 20 '18 at 13:13 — Henno Brandsma, Sep 20 '18 at 13:13

bof · Answer 1 · 2018-09-21T08:44:53.437

9

$A=[0,1]\cup\mathbb Z$ is a closed set, $\mu(A)=1$, and $\{a+b:a,b\in A\}=\mathbb R.$

edited Sep 21 '18 at 08:44

answered Sep 20 '18 at 21:39

bof

82,298

score 8 · Accepted Answer · answered Sep 20 '18 at 21:45

8

Interesting enough you can even find such a set of measure $0$.

Indeed, if $C$ is the tertiary Cantor set then $C+C=\{ a+b : a,b \in C\} =[0,2]$ see this question

Then $$A= \bigcup_{n \in \mathbb Z} 2n+C $$ has measure $0$ but $A+A=\mathbb R$.

answered Sep 20 '18 at 21:45

N. S.

134,609

2

Very cute, I like it! Who needs Banach–Tarski when you get these weird paradoxical results free of choice? :) – Asaf Karagila Sep 20 '18 at 22:03
+1, this is such a cool result :-) – Sep 20 '18 at 22:06

Does there exist a measurable subset $A \subset \mathbb{R}$, such that $\mu(A)$ is finite, but $\mu(\{a+b|a,b\in A\}) = \infty$?

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