[Math] What exactly is a Haar measure

fourier analysisfunctional-analysismeasure-theory

I've come across at least 3 definitions, for example:

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Taken from here where $\Gamma$ is a topological group. Apparently, this definition doesn't require the Haar measure to be finite on compact sets.

Or from Wikipedia:
"… In this article, the $\sigma$-algebra generated by all compact subsets of $G$ is called the Borel algebra…."
Then $\mu$ defined on this sigma algebra is a Haar measure if it's outer and inner regular, finite on compact sets and translation invariant.

So I gather the important property of a Haar measure is that it's translation invariant.

Question 1: What I don't gather is, what do I get if I define it on the Borel sigma algebra as opposed to defining it on the sigma algebra generated by compact sets (as they do on Wikipedia)?

Question 2: Can I put additional assumptions on $G$ so that I can drop the requirement that $\mu$ has to be finite on compact sets?

Question 3: As you can guess from my questions I'm poking around in the dark trying to find out how to define a Haar measure suitably. Here suitably means, I want to use it to define an inner product so I can have Fourier series. Are there several ways to do this which lead to different spaces? By this I mean, if I define it on the Borel sigma algebra, can I do Fourier series for a different set of functions than when I have a measure on the sigma algebra generated by compact sets? Or what about dropping regularity? Or dropping finiteness on compact sets?

Thanks.

Best Answer

Summarizing some comments, and continuing: the main point is that Haar measure is translation-invariant (and for non-abelian groups, in general, left-invariant and right-invariant are not identical, but the discrepancy is intelligible).

Unless you have intentions to do something exotic (say, on not-locally-compact, or not-Hausdorff "groups", which I can't recommend), you'll be happier later to have a regular measure, so, yes, the measure of a set is the inf of the measures of the opens containing it, and is the sup of the measures of the compacts contained in it, and, yes, the measure of a compact is finite. Probably you will also want completeness, especially when taking products, so subsets of measure-zero sets have measure zero.

Probably you'll want your groups to be countably-based, too, to avoid some measure-theoretic pathologies.

Then, for abelian topological groups (meaning locally compact, Hausdorff, probably countably-based), the basics of "Fourier series/transforms" work pretty well, as in Pontryagin and Weil. The non-abelian but compact case also turns out very well.

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