Proposition 3

— Let f be a numerical function $\geqslant 0$ defined on X. a) The following inequalities hold:

b) If $\Lambda$ is vaguely continuous, then

c) If $\lambda_t^\bullet(1) < +\infty$ locally $\mu$-almost everywhere, then

Let g be a lower semi-continuous function on X such that $f \leqslant g$. For every $t \in T$,

therefore, by (4) and Prop. 4 of §1,

The first of the inequalities (6) then follows from the definition of $\int^* f(x)\, d\nu(x)$ (Ch. IV, §1, No. 3, Def. 3), and the second follows immediately from it. The inequality (7) is proved in an analogous way if $\Lambda$ is vaguely continuous, using (5) instead of (4).

Let us pass to the proof of (8). The mapping $t \mapsto \lambda_t^*(1)$ is measurable, and finite locally $\mu$-almost everywhere. The set $\mathscr{K}$ of compact subsets of $T$ such the restriction of $t \mapsto \lambda_t^*(1)$ to $K$ is finite and continuous is therefore $\mu$-dense, and Prop. 4 of §2, No. 3 implies the existence of a summable family $(\mu_\alpha)_{\alpha \in A}$ of positive measures, with supports belonging to $\mathscr{K}$, such that $\mu = \sum_{\alpha \in A} \mu_\alpha$. The mapping $\Lambda$ is $\mu_\alpha$-adequate for every $\alpha \in A$; set $\nu_\alpha = \int \lambda_t d\mu_\alpha(t)$. Prop. 1 shows that $\nu = \sum_{\alpha \in A} \nu_\alpha$, and the relation (4), applied to the measure $\mu_\alpha$ and the function 1, shows that $\nu_\alpha$ is a bounded measure (because $\lambda_t^*(1)$ is bounded on $\operatorname{Supp}(\mu_\alpha)$). Let us then write the formula (6) for the measure $\mu_\alpha$, replacing the symbol $\int^*$ in the first member by $\int^\bullet$, which is legitimate by Prop. 7 of §1; then

(the last equality due to the fact that $\lambda_t$ is bounded locally almost everywhere, and Prop. 7 of § 1). The inequality in (8) is then obtained by summing on $\alpha$ (§2, No. 2, Prop. 1).

If no hypothesis analogous to that of c) is made, the inequality (8) may fail (Exer. 2).