# Haar measure on $\mathbb{R}\backslash\{0\}$

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 $$f \mapsto \int f\omega$$?

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

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 $$dy = a^{-1}\;dx$$, $$\int f(x)\frac{dx}{x} = \int f(ay) \frac{a dy}{ay} = \int f(ay)\frac{dy}{y} .$$
• 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 $f$. Mar 15, 2021 at 21:04