Lemma 4

Let $\lambda$ be a measure $\neq 0$ on $G/H$, and $\mu$ a left Haar measure on $G$. The following properties are equivalent: a) $\lambda$ is quasi-invariant under $G$; b) for a subset $A$ of $G/H$ to be locally $\lambda$-negligible, it is necessary and sufficient that $\overline{\pi}^{-1}(A)$ be locally $\mu$-negligible; c) the measure $\lambda^\#$ is equivalent to $\mu$.

Suppose this to be the case and let $\lambda^\# = \rho \cdot \mu$, where $\rho$ is a locally $\mu$-integrable function that is everywhere nonzero. Then, for every $s \in G$, the density $\theta_s$ of $\gamma_{G/H}(s)\lambda$ with respect to $\lambda$ is such that

locally $\mu$-almost everywhere on $G$.

c) $\Rightarrow$ b): This follows at once from Prop. 6 a).

b) $\Rightarrow$ a): If property b) holds, then the set of locally $\lambda$-negligible subsets of $G/H$ is invariant under $G$, thus $\lambda$ is quasi-invariant under $G$.

a) $\Rightarrow$ c): Assume $\lambda$ is quasi-invariant under $G$; for every $s \in G$, $\lambda$ and $\gamma_{G/H}(s)\lambda$ are equivalent, therefore $\lambda^\#$ and $\gamma_G(s) \cdot \lambda^\# = (\gamma_{G/H}(s) \cdot \lambda)^\#$ are equivalent (Cor. 1 of Prop. 6); since $\lambda^\# \neq 0$, $\lambda^\#$ is equivalent to $\mu$ (§ 1, No. 9, Prop. 11).

Moreover, for every $s \in G$,

whence (17).

The equivalence of a) and b) implies first the uniqueness result already announced, and even a more precise result: