Proposition 5

Let g be a $\mu$-integrable positive numerical function and let $\Gamma^0(g)$ be the set of measures $\alpha \in \Gamma^0$ such that $\int^* g(xs)\, d\alpha(s) \geq 1$ for all $x \in G$. Then the mapping $\alpha \mapsto \|\mu_\alpha\|$ of $\Gamma^0(g)$ into $\overline{\mathbf{R}}$ is vaguely continuous.

For every measure $\alpha \in \Gamma^0(G)$, recall (Ch. VII, §2, No. 3, Prop. 5) that the function

is defined $\mu_\alpha$-almost everywhere on $Q_\alpha$, is $\mu_\alpha$-integrable, and

In view of Prop. 4, it suffices to prove that, in $\Gamma^0(g)$, $\alpha \mapsto \|\mu_\alpha\|$ is upper semi-continuous. Fix a measure $\alpha \in \Gamma^0(g)$, and let K be a compact subset of G. There exists on $Q_\alpha$ a continuous function with compact support, taking its values in $[0, 1]$, equal to 1 on the compact set $\pi_\alpha(K)$; since the mapping $f \mapsto f_\alpha$ of $\mathscr{H}_+(G)$ into $\mathscr{H}_+(Q_\alpha)$ is surjective (Ch. VII, §2, No. 1, Prop. 2), one sees that there exists a function $f \in \mathscr{H}_+(G)$ such that

Since $\beta \mapsto f * \beta$ is a continuous mapping of $\mathscr{M}_+(G)$, equipped with the vague topology, into $\mathscr{C}(G)$ equipped with the topology of compact convergence (§ 4, No. 2, Remark 1), one sees that for every $\varepsilon > 0$, the set $U_\varepsilon$ of $\beta \in \Gamma^0(G)$ such that

is an open neighborhood of $\alpha$ in $\Gamma^0(g)$; for every $\beta \in U_\varepsilon$, we then have, by virtue of the formula (2), Given a number $\varepsilon > 0$, let us choose a function $h \in \mathscr{K}_+(G)$ such that $\int_G |g(x) - h(x)|\, d\mu(x) \leq \varepsilon$, and let us take $K = \mathrm{Supp}(h)$ in the foregoing. For every $\beta \in \Gamma^0(g)$, by hypothesis $g_\beta(\dot{x}) \geq 1$ almost everywhere (for $\mu_\beta$) in $Q_\beta$, therefore by virtue of (4); since $h$ is zero outside $K$, and $a fortiori$ outside $KH_\beta$, it follows that combining this result with (5), one sees that when $\beta \in U_\varepsilon$, which completes the proof.