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I do not see how $\frac{dx}{|x|}$ can be a Haar measure on $\mathbb{R}\backslash\{0\}$ given that a Haar measure $\mu$ is supposed to have the property that $\int f\mu=\int L_yf \mu$ for all $y\in G$ where $L_{y}f(x)=f(y^{-1}x)$ and $f$ is a positive, compactly supported function on G. Take $f(x)=x^{2}$, for example (ignoring the fact that this is not compactly supported). Then $\int \frac{(y^{-1}x)^{2}}{y^{-1}x}dx=\int \frac{x}{y}dx\neq\int xdx=\int \frac{x^{2}}{x}dx$ for any $y\neq 1$. Heuristically, this would seem to suggest that this definition doesn't work. I suppose $dx$ should be thought of as a $1$-form so that $d(y^{-1}x)=y^{-1}dx$, but this is not the meaning of symbols like $\frac{dx}{x}$ in measure theory, so what is the right way to describe this situation?

EDIT: More generally, let $G$ be an open subset of $\mathbb{R}^{n}$ with left multiplication given by $xy=A(x)y+b(x)$ where $A$ is an invertible linear transformation. When it is said that "$|$det$A(x)|^{-1}dx$ is a Haar measure on $G$", should this be interpreted as "$|$det$A(x)|^{-1}dx$ is a left-invariant $n$-form" with the understanding that integration is with respect to differential forms as opposed to the integration developed in measure theory?

My confusion is that in measure theory when you say "$d\mu=f(x)d\lambda$ is a measure" this means $\int g(x) d\mu=\int g(x)f(x) d\lambda$. This is purely a conceptual question: how do you think of these "Haar measures" in a measure theoretic way? In other words, given a left-invariant differential $n$-form $\omega$ on an $n$-dimensional Lie group $G$, how can you describe the Haar measure arising from the positive linear functional $f \mapsto \int f\omega$?

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  • $\begingroup$ Should Haar measure be $dx/x$ on $x>0$ and $-dx/x$ on $x<0$? $\endgroup$
    – GEdgar
    Mar 14, 2021 at 23:01

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I think this is the right way to view it. Start with $$\int f(x)\frac{dx}{x}$$ For some $a \ne 0$, change variables $y = a^{-1}x$, so that $dy = a^{-1}\;dx$, $$ \int f(x)\frac{dx}{x} = \int f(ay) \frac{a dy}{ay} = \int f(ay)\frac{dy}{y} . $$

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  • $\begingroup$ Yes, the left shift $L_y$ acts on $f$ only, $$\int L_yf(x)d\mu(x)=\int f(y^{-1}x)d\mu(x)$$ and then $\mu$ is a Haar measure provided $$\int f(y^{-1}x)d\mu(x)=\int f(x)d\mu(x)$$ for all $f$. $\endgroup$ Mar 15, 2021 at 21:04

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