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 19:51
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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 $x\in H$. 

