TITLE: Are measurable automorphism of a locally compact group topological automorphisms?
QUESTION [10 upvotes]: 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).

REPLY [24 votes]: The basic fact about locally compact groups is that you can recover the topology from the underlying measure space:
This is because, for any measurable subset $X\subset G$ of positive measure, the set
$$
X^{-1}X:=\{x^{-1}y\mid x\in X, y\in X\}
$$
is a neighborhood of the neutral element.
Letting $X$ vary along all measurable subsets of positive measure you get a basis of neighborhoods of $e\in G$. By translating by group elements, you get a basis of neighborhoods of any element $g\in G$. And so you recover the topology on $G$.
Corollary:
Since the topology is entirely encoded in the measurable structure, an automorphism that respects the measurable structure, will also respect the topology, i.e., be continuous.