Context: Let $G$ be a l.c. topological group and consider $\mu$ a (left) Haar measure. Denote by $\Sigma$ the $\mu-$completion of the Borel $\sigma-$algbebra. My question is as follows: let $\varphi:G_1\to G_2$ denote a group morphism such that $\varphi$ is $\Sigma-\mathcal{B}(G_2)$ measurable ($\mathcal B$ denotes the Borel algebra). It is known that this hypothesis imply $\varphi$ continuous.
Question: I was wondering if one could prove directly that $\varphi$ is $\mathcal{B}(G_1)-\mathcal{B}(G_2)$ measurable.
A tentative approach: It is easy to prove that $\varphi$ has a locally a.e. equal modification $\tilde \varphi$ which is Borel-measurable; if we were to prove that $\tilde \varphi$ can be choosen to be a morphism then the claim follows, since $\{g:\varphi(g)=\tilde\varphi(g)\}$ is a subgroup with locally null complement, so it must necessarily be the whole $G_1$.
If it helps, one can assume $G$ to be secound countable (although I would be also very interested in the general result).