Assume that $H$ is a Lebesgue measurable additive subgroup of $\mathbb{R}$. Is $H$ necessarily a Borel subset of $\mathbb{R}$?



Is a Lebesgue measurable subgroup of $\mathbb{R}$ a Borel measurable set?
No. Let $\langle A \rangle$ be a proper Borel subgroup of ${\mathbb R}$ generated by an algebraically independent Borel set $A$ with the cardinality of the continuum (such a set was constructed in


*J. v. Neumann*, MR 1512442 [**Ein System algebraisch unabhängiger zahlen**](http://dx.doi.org/10.1007/BF01459089), *Math. Ann.* **99** (1928), no. 1, 134--141.


, see also the discussion in


*Barthélemy Le Gac*, MR 687640 [**Some properties of Borel subgroups of real numbers**](http://dx.doi.org/10.2307/2043358), *Proc. Amer. Math. Soc.* **87** (1983), no. 4, 677--680.


). By the [Steinhaus lemma](https://en.wikipedia.org/wiki/Steinhaus_theorem), $\langle A \rangle$ has measure zero, thus $\langle A' \rangle$ is a Lebesgue measurable subgroup for every $A' \subset A$. But this gives $2^{\mathfrak c}$ such subgroups, more than the cardinality of the Borel $\sigma$-algebra, so at least one of these subgroups must be non-Borel. 


