Does the existence of (left) Haar measure on a locally compact topological group require that the group be Hausdorff?
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asked Apr 28, 2010 at 3:29
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4$\begingroup$ 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. $\endgroup$– François G. DoraisCommented Apr 28, 2010 at 3:39
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3$\begingroup$ Perhaps I'm missing something obvious, but I'm not sure about "pull the measure back". Generally one can push measures forward though. $\endgroup$– Todd TrimbleCommented Dec 29, 2012 at 23:56
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$\begingroup$ The group on 2 elements with the indiscrete topology is locally compact, I think it has a Haar measure for any definition I can think of. Anyway, the question is unclear since "Haar measure" should be defined in a way that the question makes sense. $\endgroup$– YCorCommented Mar 6, 2017 at 4:00
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1$\begingroup$ @ToddTrimble since Borel subsets of a topological group are invariant by the closure of $\{1\}$, the definition of "pull the measure back" is clear. In general, if $f:X\to Y$ is a map and you have a $\sigma$-algebra and measure on $Y$, the pull-back of the $\sigma$-algebra is the set of $f^{-1}(B)$, $B$ in the $\sigma$-algebra of $Y$, and the measure is clearly defined. This construction is quite trivial and uninteresting but precisely is fine for this question. $\endgroup$– YCorCommented Mar 6, 2017 at 4:02
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$\begingroup$ @YCor: You say that "the measure is clearly defined". Could you please make this part explicit for the slower ones of us ("lourds d'esprit")? I believe that the pullback measure should be $(f^* \mu) (B) = \mu (f (B))$, and it is not clear why $f(B)$ should be Borel when $B$ is Borel, and why this should give a measure. Or am I misunderstanding what "pullback measure" is supposed to mean here? $\endgroup$– Alex M.Commented Apr 6, 2018 at 20:52
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1 Answer
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No. Simon Rubinstein-Salzedo's "On the existence and uniqueness of invariant measures on locally-compact groups" presents a proof of existence (and uniqueness up to a multiplicative strictly positive constant) of a left Haar measure given a locally-compact, not necessarily Hausdorff, group.
answered Apr 28, 2010 at 3:43
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$\begingroup$ Do you know what definition of Radon measure he is using in that paper? A Borel measure that is locally finite and inner regular? $\endgroup$ Commented Apr 28, 2010 at 4:01
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5$\begingroup$ @JasonDyer This comment is 7 years late: in Definition 5 of page 2 of that link one finds the notation $\mathcal{O}$ used to denote the collection of open sets, and on the 2nd to last line of page 3 it is thereby said that if $O$ is open and $K$ is a compact subset then $O \cap K^c$ is open. That is tantamount to knowing compact sets are closed. How does one know such closedness if the group is not assumed to be Hausdorff? $\endgroup$– nfdc23Commented Mar 6, 2017 at 1:25
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2$\begingroup$ @nfdc23 Isn't the property used in the paper actually equivalent (rather than just tantamount) to all compact sets being closed? It is asserted to work for all open sets $O$. So if we take $O$ to be the whole group, then the assertion is that $K$ compact implies $K^c$ open, right? If all compact sets are closed then the space is $T_1$. But topological groups are always completely regular. And comp. reg. + $T_1$ implies Hausdorff. $\endgroup$– LindenCommented Jun 30, 2017 at 19:43
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1$\begingroup$ @AlexM.: A $T_0$ topological group is $T_{3\frac{1}{2}}$, i.e. completely regular Hausdorff. $\endgroup$– tomaszCommented Feb 18, 2019 at 16:10
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1$\begingroup$ @tomasz: Indeed. And I also know where my comment was mistaken: an algebraic group is semi-topological, but not necessarily topological. $\endgroup$– Alex M.Commented Feb 18, 2019 at 16:26