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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, Suvrit, Andreas Blass, Mark Meckes Sep 7 '12 at 1:31

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1 Answer 1

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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