I have just started studying Fourier analysis on topological groups from Folland's A course in abstract harmonic analysis and Rudin's Fourier analysis on groups. It seems that Rudin defines the Haar measure as a regular measure on the Borel sets, while Folland defines it as (what he calls) a Radon measure. A regular measure is a (non-negative) measure $\mu$ that satisfies$$\mu(\mathbb{B}) = \sup\{\mu(K)\colon K\subseteq\mathbb{B},K~\text{compact}\} = \inf\{\mu(G)\colon G\supseteq\mathbb{B},G~\text{open}\},\forall\mathbb{B}~\text{Borel},$$ and a Radon measure is a measure $\mu$ which satisfies each of the following:$$\mu(K)<\infty,\forall K~\text{compact},$$ $$\mu(\mathbb{B}) = \inf\{\mu(G)\colon G\supseteq\mathbb{B},G~\text{open}\},\forall\mathbb{B}~\text{Borel},$$ $$\mu(G) = \sup\{\mu(K)\colon K\subseteq G,K~\text{compact}\},\forall G~\text{open}.$$ Obviously, the two are not the same. Furthermore, Wikipedia agrees with Folland, and even gives an example where the (left-invariant) Haar measure is not regular. All three sources seem to work on locally compact Hausdorff groups, the only major difference being that Rudin restricts to Abelian groups. I understand that the Abelian assumption makes the left-invariance and right-invariance distinction redundant, but does it also make regularity and "Radon-ity" equivalent?


Unless Rudin assumes the group is $\sigma$-compact, or something like that, his definition is wrong: the example given in the Wikipedia article you linked to is of an Abelian group.

If $G=\mathbb{T}\times\Bbb{R}_d$, where $\Bbb{T}$ is the circle group with the usual topology and $\Bbb{R}_d$ is the additive group of the reals with the discrete topology, then $\{1\}\times[0,1]$ is a Borel (non-open) subset which is not inner regular with respect to the Haar measure of $G$ (where one uses the usual definition, given by Folland). By uniqueness of Haar measures, we see $G$ doesn't admit a nontrivial left-invariant regular measure.

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  • $\begingroup$ This is actually a very subtle point. It seems to me that the two approach are correct. This gives you two Haar measures that might be different. However, as you say, the Haar measure people usually deal with is the one that is Radon, not the one that is regular. $\endgroup$ – M. Dus Sep 25 '18 at 19:23
  • $\begingroup$ Yeah, these are pretty subtle issues. In fact, by regularity of the Radon Haar measure, you can see that $\mathbb{T}\times\Bbb{R}_d$ has no regular Haar measure, because such a measure would also be Radon, but the Radon Haar measure is not regular. $\endgroup$ – Cronus Sep 26 '18 at 20:45
  • $\begingroup$ I don't understand your comment. I don't understand your counterexample as well in your answer. The Haar measure on $\mathbb{T}\times \mathbb{R}^d$ is the product measure of the Haar measure on the circle and the Lebesgue measure on $\mathbb{R}^d$, whatever definition you take. It is both Radon and regular. $\endgroup$ – M. Dus Sep 27 '18 at 18:10
  • $\begingroup$ Oh, so why didn't you say so earlier? In any case, I didn't write $\Bbb{R}^d$ - I wrote $\Bbb{R}_d$, which is the additive group of the reals with the discrete topology. Since this is not a sigma-finite measure space you don't have a unique product measure, and the (Radon) Haar measure of this group is not inner regular. The set $\{1\}\times [0,1]$ provides the counterexample: it's of infinite measure by outer regularity (since every open set which contains it must be of infinite measure), but every compact set if contains is finite and hence of measure zero, contradicting inner regularity. $\endgroup$ – Cronus Sep 28 '18 at 19:28
  • $\begingroup$ I didn't write out the details - I think you can find a more detailed proof in Wikipedia or Folland. If you don't find it/don't understand it I'll try to write them down myself. $\endgroup$ – Cronus Sep 28 '18 at 19:29

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