Suppose that the pair $(\pi, g)$ is $\mu$-adapted; for every function $f \in \mathscr{K}(X)$, the function $t \mapsto \langle f, \lambda_t \rangle = f(\pi(t))g(t)$ is then essentially $\mu$-integrable. Let us show that $t \mapsto \lambda_t$ is vaguely $\mu$-measurable. For, we first note that if $\pi$ and $g$ are continuous, then the mapping $t \mapsto \lambda_t$ is vaguely continuous. In the general case, the set of compact subsets $K$ of $T$ such that the restrictions of $\pi$ and $g$ to $K$ are continuous is $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15); if $K$ is such a set, then the restriction of $t \mapsto \lambda_t$ to $K$ is vaguely continuous, whence the first assertion of the statement. Prop. 2 of §3, No. 1 shows that $\Lambda$ is $\mu$-adequate.

Conversely, suppose that $\Lambda$ is scalarly essentially $\mu$-integrable and vaguely $\mu$-measurable; it is then $\mu$-adequate (§3, No. 1, Prop. 2). Since the function 1 is lower semi-continuous on $X$, the function $t \mapsto \lambda_t(1) = g(t)$ is $\mu$-measurable (§3, Def. 1). The set $S$ is therefore measurable (Ch. IV, §5, No. 5, Prop. 7). The set $\mathfrak{K}$ of compact sets $K \subset S$ such that $g|K$ is continuous and $\Lambda|K$ is vaguely continuous is $\mu$-dense in $S$ (Ch. IV, §5, No. 10, Prop. 15); if $K \in \mathfrak{K}$, then the restriction to $K$ of the mapping $t \mapsto \varepsilon_{\pi(t)} = \frac{1}{g(t)} \lambda_t$ is therefore vaguely continuous, and this implies the continuity of $\pi|K$ (Ch. III, §1, No. 9, Prop. 13). Since $\mathfrak{K}$ is $\mu$-dense in $S$, the restriction of $\pi$ to $S$ is $\mu$-measurable.

We shall make use of the following lemma:

For every (finite) real number $a$, denote by $B_a$ the set of $x \in X$ such that $f(x) \leq a$, and by $A_a$ the set of $t \in T$ such that $g(t) \leq a$; it all comes down to showing that $B_a$ is closed (GT, IV, §6, No. 2, Prop. 1). Now, $A_a$ is closed (same ref.) and the proper mapping $\pi$ is closed (GT, I, §10, No. 1, Prop. 1); we are thus reduced to proving that $\pi(A_a) = B_a$. The obvious relation $f(\pi(t)) \leq g(t)$ for all $t \in T$ implies that $\pi(A_a) \subset B_a$. On the other hand, let $x \in B_a$; the set $\overline{\pi}^{-1}(x)$ is quasi-compact (GT, I, §10, No. 2, Th. 1) and nonempty, therefore there exists a $t \in \overline{\pi}^{-1}(x)$ such that

A) Suppose first that the measure $\mu$ has compact support K and that the restrictions to K of the functions $g$ and $\pi$ are continuous. By formula (4) of $§ 3$, No. 1, $\nu^\bullet(1) = \int_K g(t)\, d\mu(t) < +\infty,$ so that all of the measures that figure in formula (1) are bounded. We may therefore replace $\int^\bullet$ by $\int^*$ in the first member. In view of formula (6) of $§ 3$, No. 2, it all comes down to proving that (2)

where the symbol $\int^\bullet$ in the second member can in turn be replaced by $\int^*$. By the definition of upper integral, it suffices to verify the inequality (3) for every lower semi-continuous function $h$ on T that is $\geq$ the function $t \mapsto f(\pi(t))g(t)$. Now, let $\varepsilon$ be a number $> 0$ and let $u$ be the function $(h + \varepsilon)/g,$ which is lower semi-continuous in K. If $t \in \pi^{-1}(\{x\}) \cap K,$ then $u(t) \geq f(x):$ this is obvious if $g(t) = 0,$ because then $u(t) = +\infty;$ if $g(t) > 0,$ then

whence the asserted inequality. Under these conditions, for every $x \in X$ let $v(x)$ be the infimum of $u(t)$ for $t \in \pi^{-1}(\{x\}) \cap K$. The function $v$ is $\geq f$ by the foregoing, it is lower semi-continuous on $X$ by the Lemma (applied to the restriction of $\pi$ to $K$), and $v(\pi(t))g(t) \leq h(t) + \varepsilon$ for all $t \in K$ (recall that the first member is zero by convention if $g(t) = 0$). Let us then apply to $v$ the formula (4) of §3, No. 1. We obtain:

Since the measure $\mu$ is bounded and $\varepsilon$ is arbitrary, the inequality (3) follows.

B) Let us now pass to the general case. Since the mapping $t \mapsto (\pi(t), g(t))$ of $T$ into $X \times \mathbf{R}_+$ is $\mu$-measurable (Ch. IV, §5, No. 3, Th. 1), the set $\mathscr{K}$ of compact subsets $K$ of $T$ such that the restrictions of $\pi$ and $g$ to $K$ are continuous is $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15). By Prop. 4 of §2, No. 3, $\mu$ is the sum of a summable family $(\mu_\alpha)_{\alpha \in A}$ of measures whose supports are elements of $\mathscr{K}$; the pair $(\pi, g)$ being $\mu_\alpha$-adapted for every $\alpha \in A$, let $\nu_\alpha$ be the measure $\int g(t) \varepsilon_{\pi(t)}\, d\mu_\alpha(t)$. Then by A),

But the $\nu_\alpha$ form a summable family whose sum is equal to $\nu$ (§3, No. 1, Cor. of Prop. 1). Therefore, by Prop. 1 of §2, No. 2,

One has an analogous relation for the second member of (5), and (1) therefore follows from (5) by summing over $\alpha$.

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.

Let $f$ be a mapping of $X$ into a topological space $G$, and let $S$ be the ($\mu$-measurable) set of points $t \in T$ such that $g(t) > 0$. In order that $f$ be $\nu$-measurable, it is necessary and sufficient that the restriction of $f \circ \pi$ to $S$ be $\mu$-measurable.

Suppose first that $f$ is $\nu$-measurable. By hypothesis, the set $\mathscr{K}$ of compact subsets $K$ of $S$, such that the restriction of $\pi$ to $K$ is continuous, is $\mu$-dense in $S$ (Ch. IV, §5, No. 10, Prop. 15). To show that the restriction of $f \circ \pi$ to $S$ is $\mu$-measurable, it therefore suffices to prove that for every $K \in \mathscr{K}$, the set of compact subsets $H$ of $K$, such that the restriction of $f \circ \pi$ to $H$ is continuous, is $\mu$-dense in $K$ (Ch. IV, §5, No. 8, Prop. 13). But by hypothesis, there exists a partition of the compact set $\pi(K)$ formed by a $\nu$-negligible set $N$ and a sequence of compact sets $(C_n)$ such that the restriction of $f$ to each $C_n$ is continuous. Under these conditions, $K \cap \overline{\pi}(N)$ and the sets $K \cap \overline{\pi}(C_n)$ form a partition of $K$; but $K \cap \overline{\pi}(N)$ is $\mu$-negligible by virtue of the Cor. of Th. 1 of No. 2, the sets $K \cap \overline{\pi}(C_n)$ are compact, and the restriction of $f \circ \pi$ to each of the latter sets is continuous, which proves that the restriction of $f \circ \pi$ to $S$ is $\mu$-measurable.

Conversely, suppose this to be the case; to show that $f$ is $\nu$-measurable, it suffices to prove that the set $\mathscr{L}$ of compact subsets $L$ of $X$, such that the restriction of $f$ to $L$ is continuous, is $\nu$-dense (Ch. IV, §5, No. 10, Prop. 15). Let $N$ be a subset of $X$ such that $N \cap L$ is $\nu$-negligible for every $L \in \mathscr{L}$, and let us show that $N$ is locally $\nu$-negligible. For this, we must show that $\overline{\pi}(N) \cap S$ is locally $\mu$-negligible (Cor. of Th. 1 of No. 2). Now, the set $\mathscr{H}$ of compact subsets $H$ of $S$, such that the restrictions to $H$ of $\pi$ and $f \circ \pi$ are continuous, is by hypothesis $\mu$-dense in $S$ (Ch. IV, §5, No. 10, Prop. 15). It therefore suffices to prove that $\overline{\pi}(N) \cap H$ is $\mu$-negligible for every $H \in \mathscr{H}$. Now, $\pi(H)$ is compact and may be identified with the quotient space of $H$ by the equivalence relation $\pi(t) = \pi(t')$, $\pi$ being identified with the canonical mapping of $H$ onto this quotient space (GT, I, §5, No. 2, Prop. 3). Since the restriction of $f \circ \pi$ to $H$ is continuous, the restriction of $f$ to $\pi(H)$ is therefore continuous, in other words $\pi(H) \in \mathscr{L}$, consequently $N \cap \pi(H)$ is $\nu$-negligible. By the Cor. of Th. 1 of No. 2, $\overline{\pi}(N \cap \pi(H)) \cap S$ is locally $\mu$-negligible; the same is therefore true of the set

but since $H$ is compact, $H \cap \overline{\pi}(N)$ is $\mu$-negligible, which completes the proof.

A) We begin by treating the case that the measure $\mu$ has compact support $K$, on which $g$ is bounded. The measures $\mu$ and $\nu$ are then bounded, and one can replace 'essentially integrable' in the statement by 'integrable'. Suppose that $f$ is $\nu$-integrable: the function $f(\pi(t))g(t)$ is then $\mu$-integrable, and the relation (9) is verified, by Th. 1 of §3, No. 3. Conversely, suppose that $f(\pi(t))g(t)$ is $\mu$-integrable: $f$ is then $\nu$-measurable (No. 3, Prop. 3 and Remark), and

(No. 2, Th. 1); $f$ is therefore essentially $\nu$-integrable (§1, No. 3, Prop. 9), hence $\nu$-integrable. Th. 1 of §3, No. 3 then implies (9).

B) Let us pass to the general case. Let $\mathscr{K}$ be the set of compact subsets $K$ of $T$ such that $g|K$ is continuous: $\mathscr{K}$ is $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15), therefore the measure $\mu$ is the sum of a family $(\mu_\alpha)_{\alpha \in A}$ of measures whose supports are elements of $\mathscr{K}$ (§2, No. 3, Prop. 4). The pair $(g, \pi)$ is obviously $\mu_\alpha$-adapted for every $\alpha \in A$, and the measure $\nu$ is the sum of the family of measures $\nu_\alpha = \int \varepsilon_{\pi(t)}g(t) \, d\mu_\alpha(t)$ (§3, No. 1, Cor. of Prop. 1). Since the argument of A) may be applied to the measures $\mu_\alpha, \nu_\alpha$, the first part of the statement then follows from Prop. 3 of §2, No. 2.

For the function $f$ (resp. $t \mapsto f(\pi(t))g(t)$) to be integrable for $\nu$ (resp. for $\mu$), it is necessary and sufficient that it be essentially integrable and that

The second part of the statement therefore follows from the first part and Proposition 2.