Does there exists a $\mathbb{Q}$ vector space automorphism of $\mathbb{R}$ which is not Lebesgue measurable?

Question

I know the existence of non-Lebesgue-measurable sets can be proved using certain representatives of $\mathbb{R}/\mathbb{Q}$. So my naive guess is that the answer to the title question is Yes. Not sure how to prove the existence.

All help is appreciated.

score 3 · Accepted Answer · answered Jun 26 '18 at 15:24

Every Lebesgue-measurable additive function $f \colon \mathbb{R} \to \mathbb{R}$ is contiuous, so the Lebesgue-measurable $\mathbb{Q}$-automorphisms of $\mathbb{R}$ are precisely the $\mathbb{R}$-automorphisms of $\mathbb{R}$, i.e. the $x \mapsto cx$ for $c\in \mathbb{R}\setminus \{0\}$.

Since there are $2^{\mathfrak{c}} > \mathfrak{c}$ $\mathbb{Q}$-automorphisms of $\mathbb{R}$ (every permutation of a Hamel basis induces a $\mathbb{Q}$-automorphism, and there are $2^{\mathfrak{c}}$ of those; on the other hand there are only $2^{\mathfrak{c}}$ maps $\mathbb{R}\to \mathbb{R}$), most - in the cardinality sense - $\mathbb{Q}$-automorphisms of $\mathbb{R}$ are not Lebesgue-measurable.

score 1 · Answer 2 · answered Jun 26 '18 at 13:36

1

The identity function from $\mathbb R$ to itself is a $\mathbb Q$-vector space automorphism of $\mathbb R$. And it is Lebesgue measurable.

answered Jun 26 '18 at 13:36

José Carlos Santos

427,504

1

Actually, you can also modify it arbitrarily on any countable subset of a basis, all in all affecting only countably many points. – Berci Jun 26 '18 at 13:49
Apologies for the vague question. – Chertopkhanov on Malek Adel Jun 26 '18 at 13:52
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@Berci No, every Lebesgue measurable additive function is continuous. So $x \mapsto cx$ for $c \in \mathbb{R}\setminus {0}$ are the only Lebesgue-measurable $\mathbb{Q}$-automorphisms of $\mathbb{R}$. – Daniel Fischer Jun 26 '18 at 13:53
Hmm.. I was not entirely sure. However, where is exactly the sloppiness in my reasoning? – Berci Jun 26 '18 at 13:55
If someone puts Daniel's comment as an answer, I will accept it. – Chertopkhanov on Malek Adel Jun 26 '18 at 13:59
@JosephKabila That someone should be Daniel himself, of course. – José Carlos Santos Jun 26 '18 at 14:00
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@Berci Changing the value of a basis element $\beta$ affects all points having a nonzero coefficient of $\beta$, and there are $2^{\aleph_0}$ of them (for example all $\beta + \gamma$ where $\gamma$ runs over the other basis elements). – Daniel Fischer Jun 26 '18 at 14:08
Ahh, indeed. Thanks for explanation. – Berci Jun 26 '18 at 14:12

Does there exists a $\mathbb{Q}$ vector space automorphism of $\mathbb{R}$ which is not Lebesgue measurable?

2 Answers2