Proposition 10

With the above notations, suppose that $\pi$ is $\nu$-measurable. Let $g$ be the function $t \mapsto \lambda_t^*(1)$ on $T$. For $\pi$ to be $\nu$-proper, it is necessary and sufficient that $g$ be locally $\mu$-integrable, in which case

We begin by arguing on the assumption that $g$ is finite locally $\mu$-almost everywhere; we will rid ourselves of this auxiliary hypothesis at the end of the proof. Since $\pi$ is by hypothesis $\nu$-measurable, to say that $\pi$ is $\nu$-proper is equivalent to saying that $\nu^*(f \circ \pi) < +\infty$ for every function $f \in \mathscr{K}_+(T)$; $g$ being finite locally almost everywhere, we are under the conditions for applying assertion c) of Prop. 5 of §3, No. 2. Therefore

from the fact that $\lambda_t$ is concentrated on $\overline{\pi}^{-1}(t)$. We know that $g$ is $\mu$-measurable, since $\Lambda$ is $\mu$-adequate (§ 3, No. 1, Def. 1). To say that the first member is finite for every $f \in \mathscr{K}_+(T)$ is therefore equivalent to saying that $g$ is locally $\mu$-integrable (§ 5, Prop. 1), and in this case (14) follows at once from the above relations.

It therefore only remains to eliminate the auxiliary hypothesis. If $g$ is locally $\mu$-integrable, then $g$ is finite locally $\mu$-almost everywhere, and the hypothesis is indeed satisfied. Let us assume that $\pi$ is $\nu$-proper, and let us show that $g$ is finite locally almost everywhere. Let $\mathfrak{K}$ be the $\mu$-dense set of compact sets $K$ such that $\Lambda|K$ is vaguely continuous; since $g$ is measurable, we are reduced to showing that every compact set $K \in \mathfrak{K}$ such that $g|K = +\infty$ is $\mu$-negligible. Now, let $\mathscr{H}$ be the set of functions $h \in \mathscr{K}_+(X)$ such that $h \leq 1$; set $g_h(t) = \lambda_t(h)$, denote by $\Lambda_h$ the $\mu$-adequate mapping $t \mapsto h \cdot \lambda_t$, by $\nu_h$ the integral of $\Lambda_h$, and by $f$ an element of $\mathscr{K}_+(T)$ such that $f \geq \varphi_K$. Applying formula (14) to $\Lambda_h$, which does satisfy the auxiliary hypothesis, we obtain:

But the functions $fg_h|K$ form an increasing directed set of continuous functions on $K$, whose upper envelope has the value $+\infty$; by Dini’s theorem (GT, X, §4, No. 1, Th. 1), one can choose $h$ so that $fg_h|K$ is greater than or equal to an arbitrary positive number $n$, and it follows that $\int (f \circ \pi)\, d\nu \geq n\, \mu(K)$. Since the first member is finite because $\pi$ is $\nu$-proper, it then follows that $\mu(K) = 0$.