Are measurable automorphism of a locally compact group topological automorphisms? 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). 
 A: You should look up "automatic continuity."  Here is a paper by J.W. Lewin that may be of interest:
http://www.jstor.org/pss/2044356
A: Some partial results: By Hewitt+Ross, Theorem 22.18 (http://www.ams.org/mathscinet-getitem?mr=551496) a Borel measurable homomorphism between two locally compact groups is continuous if the codomain is separable or $\sigma$-compact.  (I think this result goes back to Banach).
A nice paper (see edit below!), which proves some similar results, is:
MR1473172 (98i:22003)
Neeb, Karl-Hermann(D-ERL-MI)
On a theorem of S. Banach. (English summary)
J. Lie Theory 7 (1997), no. 2, 293–300.
http://www.ams.org/mathscinet-getitem?mr=1473172
I'm afraid that I don't know the limits of these sort of results (i.e. a counter-example in the non-separable case, say), or if being an automorphisms gives anything more.
Edit: As Julien points out in a comment, this paper of Neeb is a little suspect, so I withdraw my recommendation.  André Henriques shows, in a short argument, that given a bijective group homomorphism which is measurable for the completed Haar measure, the homomorphism must be continuous.
I was a bit worried about the difference between "measurable" in the sense of "inverse image of open set is Borel or Haar measurable" and this stronger sense.  But I think uniqueness of Haar measures rescues us.  Indeed, if $\tau:G\rightarrow G$ is a continuous automorphism of $G$, then the map $A\mapsto |\tau(A)|$ will be a left invariant measure; as $\tau$ is a homeomorphism, this measure also assigns finite measure to compacts, and non-zero measure to open sets.  Thus it will be proportional to the Haar measure.  As $\tau$ preserves Borel sets, it follows that it will preserve all the Haar measurable sets, and so will be measurable in this strong sense.  Note that the example of Robin Chapman shows that this isn't necessarily so for a merely injective, continuous homomorphism.
A: 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.
A: Here is a result by Adam Kleppner (Measurable homomorphisms of locally compact groups, 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 $U\subset H$.
