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Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?

flag	asked Apr 28 2010 at 3:29 Beren Sanders 1,375●12●19

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No, you can do the usual construction on the Kolmogorov quotient (which is locally compact Hausdorff) and pull the measure back on the original group. – François G. Dorais♦ Apr 28 2010 at 3:39

Perhaps I'm missing something obvious, but I'm not sure about "pull the measure back". Generally one can push measures forward though. – Todd Trimble Dec 29 at 23:56

Jason Dyer · Accepted Answer · 2010-04-28 03:43:07Z UTC

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No. (Theorem 1 is a proof of existence of a left Haar measure given a locally compact group.)

answered Apr 28 2010 at 3:43

Jason Dyer
1,866●11●23

	Do you know what definition of Radon measure he is using in that paper? A Borel measure that is locally finite and inner regular? – Beren Sanders Apr 28 2010 at 4:01
	I don't know of any other definition. (Is there one?) – Jason Dyer Apr 28 2010 at 4:28
	Of course there is another definition. In fact, "Radon Measure" comes originally from Bourbaki, where it is defined as a certain type of linear functional. – Gerald Edgar Apr 28 2010 at 10:25
	Thanks, do you have a link to a recent paper that uses it that way? – Jason Dyer Apr 28 2010 at 13:00
	I've seen various definitions of Radon measure. They seem to all be some combination of: locally finite, finite on compact subsets, inner regular, regular, quasi-regular (outer regular for all Borel sets but only inner regular for open sets), positive on nonempty open sets... If you do a google book search for Radon measure I think the chances are that no more than half of them will define Radon measure as "a Borel measure that is locally finite and inner regular". But I am ignorant and if someone with more knowledge would like to clarify usage of the term "Radon measure" I would be grateful. – Beren Sanders Apr 28 2010 at 20:54

Must a locally compact group be Hausdorff in order to possess a Haar measure?

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