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Let $G$ be a locally compact Hausdorff group with a left invariant Haar measure $\mu$ and a closed subgroup $H$. It is well-known (and not hard to prove) that $G/H$ possesses an invariant measure if and only if the modular functions of $G$ and $H$ agree on $H$.

I am looking for the exact reason for which the following naive approach to defining such a measure on $G/H$ fails.

Let $f\in C_c(G/H)$ and define

$$\int_{G/H} f\, d\sigma:=\int_G \chi f\circ\pi \, d\mu$$ where $\pi\colon G\to G/H$ is the canonical projection map and $0\le \chi\in C_c(G)$ has the property that $\int_G \chi d\mu=1$ (or perhaps with some similar property.)

Assuming that $G/H$ does have an invariant measure, can it be expressed in the above form?

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What is your proof that such a formula is an invariant measure? – KConrad Feb 23 '14 at 20:33
Keith, I think the problem is that this "definition" doesn't necessarily assign a (finite) number to every $f$. But if it does, then its invariance is clear from the invariance of the Haar measure on $G$. Such a definition exists for groups diffemorphic to $\mathbb{R}^n$ by Fubini. – user47358 Feb 24 '14 at 0:40
For general quotient, you will have to work with quasi-invariant measures. – Marc Palm Feb 24 '14 at 12:51
Surely you'll need some additional conditions on $\chi$ to ensure invariance. As it stands, $\chi$ could concentrate on only "half" of the cosets of $H$. – Andreas Blass Apr 25 '14 at 14:13
you need to use of half forms see G. Heckman and H. Schlichtkrull: Harmonic Analysis and Special Functions on Symmetric Spaces. Academic Press 1994. – baba ab dad May 25 '14 at 14:10

I think the impulse to "lift" a function on $G/H$ to $G$, and define the integral on $G/H$ as the integral on $G$ of that lift, is entirely reasonable... but doesn't quite work out, no matter how one struggles. A prescription for seeing the problem in various (futile) attempts resides in the key point for the successful characterization: letting $\alpha$ be the averaging map $(\alpha f)(x)=\int_H f(hx)\,dh$, with the sided-ness of the measure easily determined... the requirement is exactly as suggested in the question: $$ \int_G f(g)\,dg \;=\; \int_{G/H} (\alpha f)(\dot{g})\,d\dot{g} $$ where the $\dot{g}$ is the coordinate on the quotient. To make this be an unequivocal characterization of the integral/measure on the quotient, one must check that the averaging map is a surjection of $C^o_c(G)$ to $C^o_c(G/H)$, which is indeed so, in general.

Then, to return to the question of making a "section" of the averaging map... if the condition on "modular functions" is met, and if continuity is not necessary, then this is certainly possible. But this might be a less than optimal approach, since we might want to characterize measures/integrals via Riesz' theorem about functionals on $C^o_c$.

NB, as @KConrad suggests, the modular-function condition is necessary, as one fills out the alleged proof that such a formula gives an invariant measure. The popular cases, such as $\mathbb R \to \mathbb R/\mathbb Z$ with section given by multiplying by characteristic function of an interval, so subliminally assume the modular-function condition that we may not have realized.

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I understand that the invariance of the quotient measure depends on the definition of $\chi$. In the case of modular groups, I think that such a $\chi$ exists (by taking a "section" of the averaging map), eg for compact groups $\chi=1$ seems to work. – user47358 Feb 24 '14 at 5:21
@user47349, if you register an account we can merge this unregistered user into it, and then you'll be able to accept the answer. Email and tell them the account numbers you'd like merged. – Scott Morrison Feb 24 '14 at 8:13

For general quotient, you will have to work with quasi-invariant measures. There is a multiplier, which is constant if the measure is invariant. For it to be constant, it is necessary and sufficient that $\Delta_G|_{H} = \Delta_H$.

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