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We know that semifinite measure space $(X,\mathcal{M},\mu)$ is a measure space that for every measurable set $E\in\mathcal{M}$ with measure $\mu(E)=\infty$, there exist a measurable set $B\subseteq E$ such that $0<\mu(B)<\infty$.

by

Haar measures are localizable

it follows that Harar measures are localizable and since by

Why every localizable measure space is semifinite measure space?

every localizable measure is semifinite, it follows that Haar measures are semifinite.

but I'm doubted that every Haar measure is localizable or not!

can someone tell me is my argument true or not?

Amirhossein Haddadian
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    Are you assuming that the underlying group is locally compact? – Keen-ameteur Mar 21 '23 at 18:53
  • yes, of course. are haar measures also defined for general topological group? – Amirhossein Haddadian Mar 21 '23 at 20:14
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    Hey man, you have been posting questions with pretty much no effort in trying to figure them out. If you're doing math at such an advanced level, you need to understand that hard proofs aren't something people on MSE (or any experienced mathematician) just provide for free. I don't want to come off as that guy, but if you want to keep asking questions on here and avoid people closing your question, you have to show more effort than giving a definition and asking a question. – Accelerator Mar 22 '23 at 01:14
  • What's the point of this question? You are given that $A\Rightarrow B$ and $B\Rightarrow C$, isn't it obvious that $A\Rightarrow C$? – Arctic Char Mar 28 '23 at 19:02
  • @Arctic Char. because, I doubt that every Haar measure is localizable or not! – Amirhossein Haddadian Mar 28 '23 at 19:20
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    Then that's a different question, where you have asked in a different post. – Arctic Char Mar 28 '23 at 19:32

1 Answers1

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Haar measures need not be semifinite.

For example, let $G= \mathbb{R}_E\times \mathbb{R}_D$ where $\mathbb{R}_E$ is $\mathbb{R}$ with the usual topology and $\mathbb{R}_D$ is $\mathbb{R}$ with the discrete topology.

Let $\lambda$ be Lebesgue measure on $\mathbb{R}$ and $\sharp$ counting measure on $\mathbb{R}$. Then consider $\mu:= \lambda \times \sharp$ where $\times$ denotes the Radon product of Radon measures (see e.g. section 7.4 in Folland's book "Real analysis: Modern techniques and their applications"). Then $\mu$ is a Haar measure on $G$.

Let $F:= \{0\}\times \mathbb{R}_D\subseteq G$. Then one can show that $\mu(F)=\infty$ and $\mu(K)= 0$ for all compact subsets $K\subseteq F$ (Folland 7.2 exercise 12). Radon measures are inner regular on sets with finite measure, so if $S$ is a Borel set with $\mu(S) <\infty$ and $S\subseteq F$, then necessarily $\mu(S)= 0$. We conclude that $\mu$ is not semi-finite.

J. De Ro
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  • What is the precise definition of Haar measure you are using? It seems to me that you run into problems with uniqueness if you allow such measures. – MaoWao Apr 19 '23 at 09:51
  • Note that for example Fremlin claims that every locally compact group has a strictly localizable (in particular semi-finite) Haar measure. – MaoWao Apr 19 '23 at 10:09
  • @MaoWao I use Folland's definition. Haar measure is a Borel measure that is finite on compacts, inner regular on opens, outer regular on Borel subsets and invariant in the usual sense. Folland proves uniqueness of the Haar measure in his books. – J. De Ro Apr 20 '23 at 17:26
  • Interesting. I read some more of the details in Fremlin's book. He requires Haar measures to be semi-finite, so in particular, your measure would not be a Haar measure (or even a Radon measure) in his sense. But he also does not work with measures on the Borel $\sigma$-algebra, but with complete measures on a $\sigma$-algebra containing the Borel sets. I think that's what allows him to get Haar measures that are always semi-finite (strictly localizable even). – MaoWao Apr 20 '23 at 19:16
  • @MaoWao Interesting indeed. I didn't know that the term Haar measure could be ambiguous! I think in most contexts it is only defined on the Borel subsets, though. – J. De Ro Apr 20 '23 at 20:36
  • That's probably right. But I think in this case the main difference is what one requires from a Radon measure. Folland wants inner regularity (with respect to compacts?) only for open sets, but also outer regularity. Fremlin wants inner regularity with respect to compacts for all measurable sets, but no outer regularity. So by his definition, your set $F$ would have Haar measure $0$, not $\infty$, and the restriction of $\mu$ to the Borel sets might even be semi-finite (haven't checked it, though). – MaoWao Apr 20 '23 at 20:44
  • @MaoWao I don't quite follow. Are you changing the measure when you say that the set $F$ has measure $0$? – J. De Ro Apr 20 '23 at 20:48
  • Yes, I'm talking about the Haar measure one time in Folland's sense and one time in Fremlin's sense. They are not the same measures. I mean, even if one restricts Fremlin's measure to the Borel sets. Notably, Fremlin's product Radon measures satisfy $\pi\times\nu(A\times B)=\pi(A)\nu(B)$ for all measurable sets $A$, $B$. – MaoWao Apr 20 '23 at 20:49