Haar measure is a measure on locally compact abelian groups which is invariants to translations. For example, the Lebesgue measure on the reals is such measure.
It can be shown without the use of the axiom of choice that the Haar measure exists and it is unique up to a scalar, that is if we want the measure of the unit interval (for example) to be $1$ then it is really unique.
While the measure is defined on Borel subsets, but we can complete the measure in a unique way by adding all the subsets of measure zero sets (and in the case of the real numbers we once again have the Lebesgue algebra)
As with the Lebesgue measure, when the axiom of choice is present there are cases in which non-measurable sets can be constructed. In the Solovay model, however, we have that all subsets of reals are measurables.
Are there any similar results about Haar measures of general LCA groups? Is there a model in which all Haar measures (perhaps under some limitations on the groups) are "full measures" (in the sense that every subset is measurable)?
Best Answer
If $(X,\mathcal{S})$ is standard Borel space and $\mu$ a continuous measure on $(X,\mathcal{S})$ then there is a Borel isomorphism $F:X\to [0,1]$ that sends $\mu$ to Lebesgue measure on $[0,1]$. (See Kechris 17.41.) Since the isomorphism preserves measure this shows that any measurable subset of $[0,1]$ has measurable preimage under $F$. In other words, if all sets of reals are measurable then so are all subsets of $X$. So one does not have to look at the specifics of Solovay's model, nor at Haar measure for particular groups, as long as one restricts attention to Polish measure spaces.
It should be added that these arguments only depend on DC which does hold in Solovay's model. Without DC I expect very strange behaviour is possible.