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.

notmeasurable if both $\mathbb R$ and $\mathbb R^2$ are given their completed sigma-algebras for Lebesgue measure. – Matthew Daws May 6 '11 at 14:00