Lemma 1

Let $X$ be a locally compact space countable at infinity, $G$ a group acting continuously and properly on $X$, $\mu$ a measure on $X$ invariant under $G$, $G'$ a subgroup of $G$, $U$ and $U'$ two open subsets of $X$ with finite non-zero measure. Assume that the $sU$ for $s \in G$ (resp. the $s'U'$ for $s' \in G'$) are pairwise disjoint and that their union is of negligible complement. Then $G'$ is of finite index in $G$ and $(G : G') = \mu(U')/\mu(U)$.

Let $(s_\lambda)_{\lambda \in \Lambda}$ be a family of representatives of the right cosets of $G'$ in $G$. Let $U_1$ be the union of the $s_\lambda U$. Then the $s'U_1$, for $s' \in G'$, are pairwise disjoint and have union $M = \bigcup_{s \in G} sU$. Let $M' = \bigcup_{s' \in G'} s'U'$. The union of $U'$ (resp. $U_1$) and a suitable subset of $X - M'$ (resp. $X - M$) is a fundamental domain, evidently $\mu$-measurable, for $G'$. By Integration, Chap. VII, § 2, no. 10, Cor. of Th. 4, $\mu(U') = \mu(U_1)$. This proves that Card $\Lambda = (G : G')$ is finite, and that $\mu(U') = (\text{Card } \Lambda)\mu(U)$.