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To simplify, let's assume $\mu$ is an invariant Haar measure on a commutative locally compact group $G$. Then, this means that $\mu$ is invariant under translation $\mu(U)=\mu(aU)$.

However, I noticed is that, for cases like $G=(\Bbb R, +)$, or $G=\Bbb R^*, \times)$, $\mu$, is also invariant, up to a minus, under inverses. Specifically:

$$\begin{aligned}x\mapsto -x&&x\mapsto x^{-1}\\ dx\mapsto -dx&&\frac {dx} x \mapsto -\frac {dx} x\end{aligned}$$

This is pretty neat when we are integrating over some domain, because after taking the inverse on the integration variable, we can also get a second minus sign by swapping the limits of integration.

Does this generalize to other locally compact groups?

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    $\begingroup$ Abelian groups, yes. Also compact groups, whether abelian or not. $\endgroup$ – GEdgar Feb 10 '14 at 14:36
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    $\begingroup$ A Haar measure is inversion invariant iff the group $G$ is unimodular (this is an easy exercise (see Handbook of Measure Theory: In two volumes, by E. Pap, p. 1112). $\endgroup$ – Watson Nov 14 '17 at 9:23
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It depends of the value of "generalize". See:

http://en.wikipedia.org/wiki/Haar_measure#The_modular_function

https://mathoverflow.net/questions/130241/how-do-these-two-haar-measures-on-sl2-r-compare

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