TITLE: Is a Lebesgue measurable subgroup of $\mathbb{R}$ a Borel measurable set?
QUESTION [12 upvotes]: Assume that $H$ is a Lebesgue measurable additive subgroup of $\mathbb{R}$. Is $H$ necessarily a Borel subset of $\mathbb{R}$?

REPLY [15 votes]: 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, 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, Proc. Amer. Math. Soc. 87 (1983), no. 4, 677--680.
).  By the Steinhaus lemma, $\langle A \rangle$ 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.