MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question

closed as off topic by Bill Johnson, Andrés E. Caicedo, Qiaochu Yuan, Benoît Kloeckner, Ryan Budney Nov 30 '12 at 19:51

Questions on MathOverflow are expected to relate to research level mathematics within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here.If this question can be reworded to fit the rules in the help center, please edit the question.

This is a version of Steinhaus's 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): – 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 $x\in H$.

share|cite|improve this answer
+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

Not the answer you're looking for? Browse other questions tagged or ask your own question.