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


These notes written by Julien Melleray help us to solve the problem. I just state the results which will help in our case.
As a consequence:
Back to the problem. Of course, $\Bbb R$ with the addition is a Polish group. Let $H$ a subgroup of $\Bbb R$ which is nonmeagre 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 subgroup. It's wellknow 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 nonempty interior, say $(r,r)$. Let $x\in \Bbb R$, and $x'\in H$ such that $xx'\lt r$. Then $xx'\in H$ and $x\in H$. 

