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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.

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    The first definition does not require finiteness on compacts, since it already is (the full measure is one).2012-07-20
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    Although not explicitly stated, the first definition was intended to apply to compact topological groups. You may notice that although he starts out by defining group, then topological group, everything else in that file is about compact topological groups.2012-07-20
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    Some authors only require Haar measure to be quasi-Radon. See e.g. David Fremlin's "Measure theory, vol. 4" (freely available). I think that's also where I've seen the theorem that such a measure on a locally compact group is already Radon. I've only skimmed the book, though, so can't tell much about the details.2012-07-20
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    As for your first question, you just get a larger algebra. Sometimes. In some spaces, there are very few compact sets. On the other hand, in a compact space, compact sets do generate the usual Borel algebra (because all closed sets are compact), but not so in an arbitrary locally compact space, for example an uncountable discrete space is locally compact, but the only compact sets are finite, and the sigma algebra generated by them is the countable-cocountable algebra.2012-07-20
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    As for the second question, you can define a trivial measure like $\mu(A)=0$ if $A=\emptyset$, else $\mu(A)=\infty$, or a counting measure which would be infinite on infinite sets, and they will both be invariant, but would not make much sense. For an invariant measure that's not like this you need at least something like local finiteness.2012-07-20
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    @tomasz: In a compact *Hausdorff* space, compact sets generate the Borel $\sigma$-algebra.2012-07-20
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    A lot of these definitions have technical details which only really come into play for wild and crazy groups. We could go into lots of abstract detail about what is needed under what conditions. However, it sounds like you have a particular group or class of groups in mind. If you say what they are, then more specific (and probably much simpler) answers could be given.2012-07-20

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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.