# Verifying Haar Measure

I am reading Haar Measure right now.
Definition: For G be locally compact Hausdorff group. A left Haar functional on G is a non trivial positive linear functional on $$C_{c}(G)$$ which is invariant under left translation.
I am reading an example of G=$$\mathbb{R^{*}}$$. The map $$D$$ from $$C_{c}(G)$$ to $$\mathbb{R}$$ defined by $$f \mapsto \int_\mathbb{R} f(x)\frac{dx}{|x|}$$ is left and right Haar functional. It can be verified using classical substitution principle. In last it concluded that D defines a left and right Haar measure on G. I don’t know how this concluded. I have read a theorem that for every Haar functional on G we have a Haar measure in G and conversely. Having some hint or suggestion would really help.

## 1 Answer

Note that for any Borel set $$A \subseteq \mathbb{R}^*$$ we can define $$\mu (A) = \int_A \frac{1}{|x|} dx$$ where $$dx$$ stands for integration against the Lebesgue measure. This measure is Borel regular and locally finite; as advertised, you can argue by change of variables to show that this measure is translation invariant.

You may conclude that this is indeed a Haar measure. Notice that since $$\mathbb{R}^*$$ is an Abelian group this measure is, at once, the left and right Haar measure for this group.

• Dear Jose, How to verify its Regular and locally finite? – Vkay Apr 22 at 13:26
• $\mu$ is locally finite because $\frac{1}{|x|}$ is bounded on compact sets. Since $\mathbb{R}^*$ is second countable, regularity follows (see Folland Real Analysis Thm 7.8). – Jose Avilez Apr 22 at 14:02