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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)?

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    $\begingroup$ Maybe I'm missing something obvious, but why do you restrict attention to abelian groups? If I'm not completely mistaken Cartan's existence proof of Haar measure, as presented e.g. in Hewitt-Ross, Abstract Harmonic Analysis I, Theorem 15.5, p.185ff, is entirely constructive and it applies to any locally compact (Hausdorff) group. $\endgroup$
    – Theo Buehler
    Commented Jul 17, 2011 at 13:00
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    $\begingroup$ Solovay's model will have all subsets measurable as long as the group is (locally compact) metrizable. For "big" groups (non-metrizable) the Haar measure is naturally defined on the Baire sets (the least sigma-algebra so that all continuous real-valued functions are measurable), and extension even to the Borel sets may not be unique. $\endgroup$
    – Gerald Edgar
    Commented Jul 17, 2011 at 13:26
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    $\begingroup$ @Theo: You have a good point there, my original question was about Polish groups as well... One generalization at a time :-) $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 13:32
  • $\begingroup$ @Gerald: Thanks, can you put that in the form of an answer? $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 13:46
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    $\begingroup$ You do need a little bit of choice to construct the Haar measure. If the real line happens to be a countable union of countable sets (which is consistent with ZF), then there is no sigma-additive measure that vanishes on singletons. I think that ZF plus DC is the usual axiom system if you want to do measure theory. $\endgroup$
    – Goldstern
    Commented Jul 17, 2011 at 22:57

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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.

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  • $\begingroup$ @Juris: Thanks a lot of the answer, is there anything to say on the non-metrizable case? $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 20:55
  • $\begingroup$ As Gerald Edgar points out, the result for nonmetrizable groups follows from the metric case because the measurability of a set only depends on its intersection with compact sets. $\endgroup$
    – Juris Steprans
    Commented Jul 17, 2011 at 21:39
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Solovay's model will have all subsets measurable as long as the group is (locally compact) metrizable. For "big" groups (non-metrizable) the Haar measure is naturally defined on the Baire sets (the least sigma-algebra so that all continuous real-valued functions are measurable), and extension even to the Borel sets may not be unique.

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  • $\begingroup$ @Gerald: Thanks, do you have any references for that? Could you somewhat elaborate on the relationship between these facts and the construction of Solovay's model if there is any (as Juris remarks below, for Polish groups it follows without the need to consider the internal works of Solovay's model, however if there is any can you please give some general picture of the relation). Thanks! $\endgroup$
    – Asaf Karagila ♦
    Commented Jul 17, 2011 at 18:55
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    $\begingroup$ Iin Solovay's model, all subsets of a Polish space are universally measurable. So you could quote Solovay for this, or you could use Solovay only for Lebesgue measure and then deduce the general case as Juris does. If your metrizable locally compact group is not separable, then for Haar measure a set is measurable if and only if its intersection with each compact set is measurable, and such a compact set is Polish. So we get the conclusion in that non-Polish case as well. $\endgroup$
    – Gerald Edgar
    Commented Jul 17, 2011 at 19:38
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    $\begingroup$ Is there a well-known example where the extension to Borel sets is not unique? $\endgroup$
    – François G. Dorais
    Commented Jul 17, 2011 at 21:14
  • $\begingroup$ I just asked this question here - mathoverflow.net/questions/70761/… $\endgroup$
    – François G. Dorais
    Commented Jul 19, 2011 at 18:22

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