# Borel Group on R [closed]

Last week in class we used the fact that if we have a group within R which is also a Borel Set, then it is either R or meagre. Why is it so? Can you direct me to a proof?

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## closed as off topic by Bill Johnson, Andres Caicedo, Qiaochu Yuan, Benoît Kloeckner, Ryan BudneyNov 30 '12 at 19:51

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This is a version of Steinhaus's theorem: en.wikipedia.org/wiki/Steinhaus_theorem A proof of a natural generalization, Pettis's theorem, is in Kechris's book Classical descriptive set theory (somewhere in chapter 9, I believe): books.google.com/books?id=pE1VMQEACAAJ –  Theo Buehler Nov 29 '12 at 21:13

These notes written by Julien Melleray help us to solve the problem. I just state the results which will help in our case.

Lemma 3.3 (Pettis) Let $G$ a Polish group. For $A\subset G$, define $U(A)$ as the biggest open set $V$ such that $A$ is comeagre $V$. For any subsets $A$ and $B$ of $G$, we have $$U(A)\cdot U(B)\subset A\cdot B.$$

As a consequence:

Theorem 3.4 Let $G$ a Polish group, and $A$ a Baire measurable non-meagre subset of $G$. Then $e$, the neutral element, belongs to the interior of $A\cdot A^{-1}$.

Back to the problem. Of course, $\Bbb R$ with the addition is a Polish group. Let $H$ a subgroup of $\Bbb R$ which is non-meagre and Borel measurable. It's Baire measurable. By the last theorem, $e$ belongs to the interior of $H\cdot H^{-1}=H$ as $H$ is a sub-group.

It's well-know that the subgroups of $\Bbb R$ are either of the form $a\Bbb Z$ (hence meagre) or dense. So we have a subgroup $H$ which is dense and has non-empty interior, say $(-r,r)$. Let $x\in \Bbb R$, and $x'\in H$ such that $|x-x'|\lt r$. Then $x-x'\in H$ and $x\in H$.

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+1 for linking to Julien Melleray's notes, they look very nice at a first glance. I saw that you started with your PhD thesis: Good luck with it! Best wishes (t.b. from math.SE). –  Theo Buehler Nov 29 '12 at 21:31
Please keep in mind that the notes have not been carefully re-read and probably contain mistakes (and I'll be grateful, should you find any, if you let me know!)... Also, in the case of $\mathbb R$, probably the easiest way to answer the question is to use the Lebesgue measure, via the following statement (due to Steinhaus, I believe): if $A \subset \mathbb R$ is measurable of positive measure, then $A−A$ contains an interval centered around $0$ (this is a classical consequence of the regularity of the Lebesgue measure) –  Julien Melleray Nov 29 '12 at 21:41
Thanks! I hope I will have luck with it, which will be the least we could expect of a probabilist! –  Davide Giraudo Nov 29 '12 at 21:43
@Julien I thought about that, and it proves that a subgroup is either $\Bbb R$ or has $0$ measure. Is there a simple way to see it's meagre? –  Davide Giraudo Nov 29 '12 at 21:45
A word of warning: Baire measurable is not the same as measurable with respect to the Baire $\sigma$-algebra (the $\sigma$-algebra generated by the compact $G_\delta$'s). It means the $\sigma$-algebra generated by the open sets and the meagre sets. There are meagre sets that aren't Borel: every set of reals can be written as the disjoint union of a meagre set $A$ and a Lebesgue null set $B$. If you partition a Vitali set this way: $V = A \cup B$ then $A$ can't even be Lebesgue measurable but is Baire measurable. –  Theo Buehler Nov 29 '12 at 21:46