Proposition 2

— Let $\pi$ be a proper continuous mapping (GT, I, §10, No. 1) of T into X, and let g be a continuous, finite numerical function on T such that $g(t) > 0$ for all $t \in T$. The pair $(\pi, g)$ is then $\mu$-adapted, and if one sets

then, for every numerical function $f \geq 0$ defined on $X$,

It is clear that $\pi$ and $g$ are $\mu$-measurable; moreover, for every function $\psi \in \mathscr{K}(X)$, $\psi \circ \pi$ is continuous with compact support, since $\pi$ is proper; the pair $(\pi, g)$ is therefore $\mu$-adapted and, moreover, the mapping $t \mapsto g(t)\varepsilon_{\pi(t)}$ is vaguely continuous.

Let $h$ be a lower semi-continuous function on $T$ such that

We are going to show that

By the definition of upper integral, this will imply the inequality

which, combined with the inequality (7) of §3, No. 2, will prove (7).

To prove (8), let us define a function $\overline{f}$ on $X$ in the following manner: $\overline{f}(x)$ is the infimum of $h(t)/g(t)$ in the set $\overline{\pi}^{-1}(x)$ (infimum equal to $+\infty$ if $\overline{\pi}^{-1}(x) = \varnothing$). The function $\overline{f}$ has the following properties:

$1^\circ$ $\overline{f}(x) \geq f(x)$ for all $x \in X$ (since $g(t) > 0$ for all $t \in T$).

$2^\circ$ $\overline{f}(\pi(t))g(t) \leq h(t)$ for all $t \in T$.

$3^\circ$ The function $\overline{f}$ is lower semi-continuous by virtue of the Lemma, the function $h/g$ being lower semi-continuous on $T$.

Consequently, in view of Prop. 2a) of §3, No. 1:

which establishes (8), and completes the proof.