Proposition 6

For every $\alpha \in A$, let $\mu_\alpha$ be a positive measure on $G_\alpha$. Suppose that $\varphi_{\beta\alpha}(\mu_\alpha) = \mu_\beta$ for $\alpha \geq \beta$. Then, there exists one and only one positive measure $\mu$ on $G$ such that $\varphi_\alpha(\mu) = \mu_\alpha$ for all $\alpha \in A$.

Uniqueness follows immediately from Lemma 2 b). Let us prove the existence of $\mu$. Let $V$ be the vector space of functions belonging to $\mathscr{K}(G)$ and constant on the cosets of some $K_\alpha$ ($\alpha$ may depend on the function). It follows from Lemma 2 a) that $V$ satisfies the condition (P) of Ch. III, §1, No. 7, Prop. 9: for, let $K$ be a compact set in $G$ and choose $f \in \mathscr{K}_+(G)$ with $f(x) > 0$ for all $x \in K$; let $a > 0$ be the smallest value of $f$ on $K$; by Lemma 2 a), there exists a function $g \in V \cap \mathscr{K}_+(G)$ such that $|f - g| \leq a/2$, therefore $g(x) > 0$ for all $x \in K$, and condition (P) is verified. Let $f \in V$. There exists an $\alpha \in A$ such that $f$ is constant on the cosets of $K_\alpha$. By passage to the quotient, $f$ defines a function $f_\alpha \in \mathscr{K}(G_\alpha)$. The number $\mu(f) = \mu_\alpha(f_\alpha)$ does not depend on the choice of $\alpha$: for, let $\beta$ be any index such that $f$ is constant on the cosets of $K_\beta$; let $\gamma \in A$ be such that $\gamma \geq \alpha, \gamma \geq \beta$; then $f$ defines functions $f_\beta \in \mathscr{K}(G_\beta)$, $f_\gamma \in \mathscr{K}(G_\gamma)$ such that $f = f_\beta \circ \varphi_\beta = f_\gamma \circ \varphi_\gamma$; then $f_\alpha \circ \varphi_{\alpha\gamma} = f_\gamma$, therefore $\mu_\gamma(f_\gamma) = (\varphi_{\alpha\gamma}(\mu_\gamma))(f_\alpha) = \mu_\alpha(f_\alpha)$, and similarly $\mu_\gamma(f_\gamma) = \mu_\beta(f_\beta)$, whence our assertion. This established, it is clear that $\mu$ is a linear form on $V$ and that $\mu(f) \geq 0$ for $f \geq 0$. By Prop. 9 of Ch. III, §1, No. 7, $\mu$ may be extended to a positive measure on $G$, which we again denote by $\mu$. One has $\varphi_\alpha(\mu) = \mu_\alpha$ for all $\alpha \in A$ by the very construction of $\mu$.