Question

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?

Answer 1

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 $x\in H$.