Theorem 1

Let $(\pi, g)$ be a $\mu$-adapted pair, and let

For every numerical function $f \geq 0$ defined on X, (1)

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 $\alpha$.