The Haar measure problem

Abstract

An old problem asks whether every compact group has a Haar-nonmeasurable subgroup. A series of earlier results reduce the problem to infinite metrizable profinite groups. We provide a positive answer, assuming a weak, potentially provable, consequence of the Continuum Hypothesis. We also establish the dual, Baire category analogue of this result.2010 Mathematics Subject Classification. 28C10, 28A05, 22C05, 03E17.

On a cardinal invariant related to the Haar measure problem

“…In this work we refute the conjecture above, thus demonstrating that the strategy of [6] does not suffice for a general solution to the Haar Measure Problem.…”

“…It is consistent with ZFC that there exists an infinite metrizable profinite group G * such that: non(N ) > fm(G * ). Notice that in the aforementioned work from [1], the exibithed models of ZFC witnessing that the Haar Measure Problem has consistently a positive answer do not satisfy CH, while, despite the failure of the main conjecture in [6] proved in this paper, the work of [6] shows the remarkable result that in all the models of ZFC satisfying CH the Haar Measure Problem has a positive answer.…”

“…We prove here that it is consistent with ZFC that there is a metrizable profinite group G * such that non(N ) > fm(G * ), thus demonstrating that the strategy of [6] does not suffice for a general solution to the Haar Measure Problem.…”

“…Very recently, progress has been made toward a solution to the Haar measure problem for infinite metrizable profinite groups. In fact, in [6] the authors introduced a certain cardinal invariant of the continuum fm(G), depending on a metrizable profinite group G, and proved (see Sec. 2 for definitions):…”

“…

In [6], given a metrizable profinite group G, a cardinal invariant of the continuum fm(G) was introduced, and a positive solution to the Haar Measure Problem for G was given under the assumption that non(N ) fm(G).We prove here that it is consistent with ZFC that there is a metrizable profinite group G * such that non(N ) > fm(G * ), thus demonstrating that the strategy of [6] does not suffice for a general solution to the Haar Measure Problem.

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