Corollary 3

— Suppose that $H$ is unimodular and that there exists on $G/H$ a nonzero bounded positive measure $\nu$ relatively invariant under $G$. Then $\nu$ is invariant, and $G$ is unimodular.

Let $\chi$ be the multiplier of $\nu$. For every $s \in G$, $\nu$ and $\gamma(s)\nu$ have the same finite total mass (§ 1, No. 1, formula (6)); since $\gamma(s)\nu = \chi(s)^{-1}\nu$, we have $\chi(s) = 1$. Thus $\nu$ is invariant. By Cor. 2, $\Delta_G(s) = 1$ for all $s \in H$. Let $G'$ be the set of $t \in G$ such that $\Delta_G(t) = 1$. This is a closed normal subgroup of $G$ containing $H$. Let $\pi$ be the canonical mapping of $G/H$ onto $G/G'$. Then $\pi(\nu)$ is a nonzero, bounded positive measure invariant under $G$. Therefore the left Haar measure of the group $G/G'$ is bounded, so that $G/G'$ is compact (§ 1, No. 2, Prop. 2). Consequently the image of $G$ under $\Delta_G$ is a compact subgroup of $\mathbf{R}_+^*$; this subgroup is reduced to $\{1\}$, thus $\Delta_G = 1$ on all of $G$.