Consider a locally compact group $G$, considered as a measurable space with the completed Borelstructure wrt. the Haarmeasure. Consider a map $f:G \to G$, which is measurable and has an inverse, which is then also measurable. Is $f$ an homeomorphism?


What if $G$ is abelian? If not, what are necessary conditions on $G$, such that this is the case.


For $\mathbb{R}$, it seems to be true: see <http://math.uchicago.edu/~henderson/additive.pdf> ([Wayback Machine](http://web.archive.org/web/20180131033819/http://math.uchicago.edu/~henderson/additive.pdf)). 



Are measurable automorphism of a locally compact group topological automorphisms?
Here is a result by Adam Kleppner ([Measurable homomorphisms of locally compact groups](https://doi.org/10.1090/S0002-9939-1989-0948154-8), Proc. Amer. Math. Soc., vol. 106, no. 2, 1989, 391-395): any measurable homomorphism between locally compact groups is continuous. Actually what he really needs, for a homomorphism $\alpha:G\rightarrow H$, is that $\alpha^{-1}(U)$ is measurable in $G$ for every open subset $U\subset H$.


