Proposition 8

a) Let $\mu$ and $\mu'$ be two positive measures on $G$ such that, for every $\alpha$ and for every coset $C$ of $K_\alpha$, $\mu(C) = \mu'(C)$. Then $\mu = \mu'$.

b) Fix an $\alpha_0 \in A$. For every $\alpha \geq \alpha_0$ let $n_\alpha$ be the number of elements of the finite group $K_{\alpha_0}/K_\alpha$. There exists one and only one positive measure $\mu$ on $G$ such that, for every $\alpha \geq \alpha_0$, each coset of $K_\alpha$ has measure $n_\alpha^{-1}$. Moreover, $\mu$ is a Haar measure on $G$, such that $\mu(K_{\alpha_0}) = 1$.

Let $\mu$ and $\mu'$ be two positive measures on $G$ satisfying the condition a). The points of the discrete group $G_\alpha$ then have the same measure for $\varphi_\alpha(\mu)$ and $\varphi_\alpha(\mu')$, whence $\varphi_\alpha(\mu) = \varphi_\alpha(\mu')$, and this for all $\alpha$. Therefore $\mu = \mu'$ (Lemma 2 b)).

Let us prove b). For every $\alpha \geq \alpha_0$, let $\mu_\alpha$ be the Haar measure of the discrete group $G_\alpha$ such that every point has measure $n_\alpha^{-1}$. Let $\alpha, \beta$ be such that $\alpha \geq \beta \geq \alpha_0$. Then $K_\beta/K_\alpha$ has $n_\alpha/n_\beta$ elements. Therefore $\varphi_{\beta\alpha}(\mu_\alpha)$ is the measure on $G_\beta$ such that each point has measure $n_\alpha^{-1} \cdot \frac{n_\alpha}{n_\beta} = n_\beta^{-1}$; in other words, $\varphi_{\beta\alpha}(\mu_\alpha) = \mu_\beta$. It then suffices to apply Props. 6 and 7.