I am studying a proof of the existence of Haar measure on locally compact groups.


In this proof (at the top of page 7) when proving finite additivity of a content (which is like a measure but is only defined over compact sets) we separate the disjoint compact sets $K_1$ and $K_2$ by an open set $O^{-1}$ such that $K_1O^{-1}$ and $K_2O^{-1}$ are still disjoint. I am unsure why we are allowed to do this.

There is a general theorem where if $K \subseteq U$ where $K$ compact and $U$ open then there is an open set $V$ such that $KV \subseteq U$. So if the space is Hausdorff the compact sets are closed and this can be applied. But I don't think we are assuming Hausdorff. Regularity might be useful as well, but it is not quite the same. And unless we can generate some closed sets it cannot be applied.

Can two compact sets always be separated by an open set in a locally compact topological group? I see why they can is the group is Hausdorff. But I also believe the Haar measure exists on a non Hausdorff group.

Can someone either explain to me why the sets can be separated, or direct me to another proof of why Haar measure is additive in a non Hausdorff Group? And then I can patch it in.

Maybe you take the Kolmogorov Quotient of your group which is Hausdorff, prove Haar measure on it, and then use this to define a measure on the non Kolmogorovised group?

  • $\begingroup$ Take an indiscrete group. You can't separate anything with open sets there. I guess Hausdorffness is assumed. $\endgroup$ Oct 29 '13 at 19:28
  • $\begingroup$ This is something that has bothered me as well. It seems like a lot of the time hausdorff is covertly assumed when talking about locally compact topological groups. $\endgroup$ Oct 30 '13 at 2:17

As far as I can see, the text you are referring to defines a Haar measure as a special kind of Radon measure, which is usually only defined for Hausdorff spaces.

In any case, there are simple examples of locally compact non-Hausdorff groups where Haar measure doesn't make sense. To be precise: there are locally compact topological groups where the only locally finite, translation-invariant measure in which compact sets are measurable is the zero measure.

The simplest example I can think of is the $(\mathbb{Z}, +)$ with the indiscrete topology. Since the smallest neighbourhood of $0$ is $\mathbb{Z}$, the measure has to be finite in order to be locally finite. On the other hand, any singleton is compact, and by translation-invariance each has the same measure $m$. But then $\sum_{n \in \mathbb{Z}} m$ is finite, therefore $m = 0$.

Addendum: If you only require the open sets to be measurable, in stead of compact sets, you could use the Haar measure of the Kolmogorov quotient, but there would not necessarily be a way to make any proper subset of an indiscrete subgroup measurable.

  • $\begingroup$ For the moment at least I only require the Borel sets -- which are formed from the open sets by countable intersections, unions, and complements -- to be measurable. I have been trying to prove you can push forward the measure to the KQ and have reduced it to showing the Borel sets contain none, or all elements of each indistinguishibility class. But this has proved tricky since the definition of a sigma algebra is slightly unclear. $\endgroup$
    – Daron
    Oct 30 '13 at 13:15
  • $\begingroup$ This should not be a problem. The collection of arbitrary unions of equivalence classes is a $\sigma$-algebra that clearly contains all open sets. Therefore it also contains all Borel sets, but it does not contain a non-empty proper subset of any equivalence class. $\endgroup$ Oct 30 '13 at 15:09
  • $\begingroup$ I have it now. You have to show the Borel sets satisfying the "contains none or all of each equivalence class" is a sigma algebra, which implies it is the entier Borel algebra by mutual inclusion. $\endgroup$
    – Daron
    Oct 31 '13 at 8:59

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