Proposition 2

— Let $\Lambda : t \mapsto \lambda_t$ be a scalarly essentially $\mu$-integrable mapping of $T$ into $\mathscr{M}_+(X)$, and let $\nu = \int \lambda_t \, d\mu(t)$.

a) If $\Lambda$ is vaguely continuous, then the mapping $t \mapsto \lambda_t^\bullet(f)$ is lower semi-continuous for every lower semi-continuous function $f \geqslant 0$ defined on $X$, $\Lambda$ is $\mu$-adequate, and we have the relation

b) If $\Lambda$ is vaguely $\mu$-measurable, then $\Lambda$ is $\mu$-adequate.

c) If the topology of $X$ admits a countable base, then $\Lambda$ is vaguely $\mu$-measurable (hence also $\mu$-adequate).

Let $f$ be a lower semi-continuous function $\geqslant 0$ defined on $X$. Let $F$ be the set, directed for the relation $\leqslant$, of functions $g \in \mathscr{K}(X)$ such that $0 \leqslant g \leqslant f$. For $g \in F$, denote by $h_g$ the function defined on $T$ by $h_g(t) = \lambda_t(g)$. Similarly, set

(§1, No. 1, Prop. 4). Let us make the following hypothesis, weaker than that of a): assume only that the restriction of $\Lambda$ to $S$ is vaguely continuous, where $S$ is a closed subset of $T$ that contains the support of $\mu$. For $g \in F$, denote by $\overline{h}_g$ the numerical function that coincides with $h_g$ on $S$ and has the value $+\infty$ on $C_S$. Set $\overline{h}_f = \sup_{g \in F} \overline{h}_g$; then $\overline{h}_f = h_f$ on $S$. For every $g \in F$, the function $\overline{h}_g$ is lower semi-continuous; $\overline{h}_f$ is therefore lower semi-continuous and, since the family $(\overline{h}_g)_{g \in F}$ is directed,

(Ch. IV, §1, No. 1, Th. 1 and §2, No. 3, Prop. 6). Since $h_f = \overline{h}_f$ on $S$, hence almost everywhere, this may also be written $\mu^*(h_f) = \nu^*(f)$, an equality identical to (5). Similarly, $f$ and $\overline{h}_f$ being lower semi-continuous, the preceding relations yield the equality $\mu^\bullet(\overline{h}_f) = \nu^\bullet(f)$ (§1, No. 1, Prop. 4); since $\overline{h}_f = h_f$ on $S$, it follows that $\mu^\bullet(h_f) = \nu^\bullet(f)$ (§1, No. 1, Prop. 1), an equality identical to (4). The mapping $\Lambda$ is therefore $\mu$-pre-adequate; but one could have replaced everywhere in this argument $\mu$ by $\mu' \leqslant \mu$, and $\nu$ by $\nu' = \int \lambda_t \, d\mu'(t)$, because $\Lambda$ is also scalarly essentially $\mu'$-integrable and $S$ contains the support of $\mu'$. It follows from this that $\Lambda$ is $\mu$-adequate.

Suppose $\Lambda$ is vaguely continuous; we may take $S = T$; then $h_f = \overline{h}_f$ is lower semi-continuous, which completes the proof of part a) of the statement.

Assume that $\Lambda$ is vaguely $\mu$-measurable and let us prove b). The set $\mathscr{K}$ of compact subsets K of T such that the restriction of $\Lambda$ to K is continuous being $\mu$-dense (Ch. IV, §5, No. 10, Prop. 15), there exists a summable family $(\mu_\alpha)_{\alpha \in A}$ of measures on T such that $\mu = \sum_{\alpha \in A} \mu_\alpha$ and the support of each of the measures $\mu_\alpha$ belongs to $\mathscr{K}$ (§2, No. 3, Prop. 4). For every $\alpha \in A$, the mapping $\Lambda$ is scalarly essentially $\mu_\alpha$-integrable, and we set $\nu_\alpha = \int \lambda_t d\mu_\alpha(t)$; the family $(\nu_\alpha)$ is summable, and its sum is equal to $\nu$ (Cor. of Prop. 1). If $f$ is a positive lower semi-continuous function defined on X, the first part of the proof, applied to the measures $\mu_\alpha$ and the closed sets $S_\alpha$, shows that:

$1^\circ$ $h_f$ is $\mu_\alpha$-measurable for every $\alpha \in A$, hence is $\mu$-measurable (§2, No. 2, Prop. 2), and

$2^\circ$ $\int^* f(x) d\nu_\alpha(x) = \int^* d\mu_\alpha(t) \int^* f(x) d\lambda_t(x)$.

The formula (4) follows on summing over $\alpha$ (§2, No. 2, Prop. 1). Applying the preceding argument to an arbitrary measure $\mu' \leq \mu$ (which is legitimate, since $\Lambda$ is scalarly essentially $\mu'$-integrable and vaguely $\mu'$-measurable, cf. §2, No. 2, Prop. 2), we conclude that $\Lambda$ is $\mu'$-pre-adequate, and b) is proved.

Finally, assuming that the topology of X admits a countable base, let us show that every scalarly essentially $\mu$-integrable mapping $\Lambda : t \mapsto \lambda_t$ of T into $\mathscr{M}_+(X)$ is vaguely $\mu$-measurable. This will result from the following lemma: