Let $G$ be a locally compact group, $H$ be a closed subgroup and $N$ be a normal subgroup of $G$ such that $H\subseteq N$. How can we get $$\int_{G/H} f(xH)d\mu_{G/H}(xH)=\mu_{N/H}(N/H)\int_{G/N}f(xN)d\mu_{G/N}(xN)\ \ ?$$
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closed as too localized by Dan Petersen, Bill Johnson, S. Sra, Andreas Blass, Mark Meckes Sep 7 at 1:31 |
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First, in order for an invariant measure on $G/H$ to exist, you need for the modular functions the equality $$ \Delta_G|_H=\Delta_H. $$ Under these circumstances the left hand side if the equation is defined, provided the function $f$ is $H$-invariant. If in addition $f$ is even $N$-invariant the right hand side is defined as well (the modular condition is automatic, as $N$ is normal). If the measure $\mu_{N/H}(N/H)$ is finite, it then is possible to choose the invariant measures such that the claimed equality holds for every measurable $N$-invariant function which is such that either side of the equality exists. A proof can be found in Deitmar/Echterhoff: Principles of Harmonic Analysis. |
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