Is the Haar measure on Abelian groups regular?

Question

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$$\mu(B) = \sup\{\mu(K)\colon K\subseteq B,K~\text{compact}\} = \inf\{\mu(G)\colon G\supseteq B,G~\text{open}\},\forall B~\;\text{Borel},$$ and a Radon measure is a measure $\mu$ which satisfies each of the following:$$\mu(K)<\infty,\forall K~\text{compact},$$ $$\mu(B) = \inf\{\mu(G)\colon G\supseteq B,G~\text{open}\},\forall B~\text{Borel},$$ $$\mu(G) = \sup\{\mu(K)\colon K\subseteq G,K~\text{compact}\},\forall G~\text{open}.$$ 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?

Answer 1

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.