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It is known that all locally compact groups, and therefore compact groups, have a left-invariant Haar measure which is unique up to scalar constant, also a right-invariant one. Is there a strictly wider class of groups that has such a measure? What about weakening of these measures? What about hypergroups?

Thank you. HTTT.

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For general locally compact hypergroups, the existence of a Haar measure is still an open problem as far as I know. It has been answered affirmatively for Abelian, compact, or discrete hypergroups and those arising as coset spaces from locally compact groups.

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  • $\begingroup$ I don't understand "it has been answered affirmatively..." $\endgroup$ – Marc Palm Nov 20 '13 at 6:59
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    $\begingroup$ Every Abelian, compact, or discrete hypergroup has a translation invariant measure. $\endgroup$ – Ben Willson Nov 20 '13 at 7:00
  • $\begingroup$ +1 thank you. It is still early in the morning. $\endgroup$ – Marc Palm Nov 20 '13 at 7:01
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    $\begingroup$ Every locally compact hypergroup does have a subinvariant measure. (Measure of x*f is less than or equal to measure of f) The lack of inverses means subinvariant measures are not necessarily invariant. $\endgroup$ – Ben Willson Nov 20 '13 at 8:03
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For hyperfinite groups - ultraproducts $G$ of sequences of finite groups $G_n$ - one has the Loeb probability measure, which is analogous to Haar measure in that it is a bi-invariant probability measure. The catch though is that the Loeb measure is not measurable with respect to a Borel sigma algebra (indeed, there is no natural topology to place on a hyperfinite group), but instead on the Loeb sigma algebra generated by applying the Caratheodory construction to the Boolean algebra of internal sets (the ultraproduct of subsets $E_n$ of $G_n$, which have measure equal to the (standard part of the) ultralimit of $|E_n|/|G_n|$).

One way to think of this is that while a hyperfinite group $G$ is not a topological space, it is a sigma-topological space - it has a collection of "open" sets (countable union of internal sets) which obey a weakened version of the topology axioms in which one has closure only under countable unions rather than arbitrary unions. Loeb measure is then the analogue of Haar measure with respect to this sigma-topology. I discuss this a bit in my paper with Bergelson.

One can also generalise from the hyperfinite case to ultraproducts of compact groups, or even locally compact groups if one normalises things properly.

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    $\begingroup$ Hi Terry. This confused me for a while, owing to the classical theorem of Weil (expounded in English in the final chapter of Halmos) which asserts roughly that every group $G$ possessing a nontrivial left-invariant measure $\mu$ is a full-measure subgroup of some locally compact group $\hat{G}$ whose Haar measure restricted to $G$ is $\mu$. The catch is that the group operation is assumed to measurable in the strong sense that the sheer $(x,y)\mapsto (x,xy)$ is a measurable map $G\times G\to G\times G$, and I guess this fails for the Loeb $\sigma$-algebra. $\endgroup$ – Sean Eberhard Jan 23 '14 at 13:31
  • $\begingroup$ The shear is measurable from $G \times G$ to itself if one uses the Loeb sigma algebra ${\mathcal L}_{G \times G}$ on the product group $G \times G$, but not if one uses the coarser space $(G \times G, {\mathcal L}_G \times {\mathcal L}_G)$ that is the usual product of the factor groups $(G,{\mathcal L}_G)$. The distinction between ${\mathcal L}_{G \times G}$ and ${\mathcal L}_G \times {\mathcal L}_G$ is closely tied to the graph and algebraic regularity lemmas, see terrytao.wordpress.com/2013/12/07/… $\endgroup$ – Terry Tao Jan 29 '14 at 6:25
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For a Hausdorff topological group, it is equivalent to admit a (quasi-)leftinvariant Radon measure and being locally compact.

Reference: http://www.ams.org/journals/proc/1972-035-02/S0002-9939-1972-0318777-5/S0002-9939-1972-0318777-5.pdf

I don't know about Hypergroups.

It seems that quotient of topological groups give Hypergroups: Note that $G/H$ admits in general only a $G$-invariant measure if the modular functions of $G$ and $H$ coincide on $H$. I am not sure if that helps in your context.

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