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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.

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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.

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  • $\begingroup$ Dear Jose, How to verify its Regular and locally finite? $\endgroup$ – Vkay Apr 22 at 13:26
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    $\begingroup$ $\mu$ 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). $\endgroup$ – Jose Avilez Apr 22 at 14:02

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