Proposition 1

If the pair $(\pi, g)$ is $\mu$-adapted, then the mapping $\Lambda : t \mapsto \lambda_t = g(t)\varepsilon_{\pi(t)}$ of T into $\mathscr{M}_+(X)$ is scalarly essentially $\mu$-integrable, vaguely $\mu$-measurable and $\mu$-adequate. Conversely, if $\Lambda$ is scalarly essentially $\mu$-integrable and vaguely $\mu$-measurable, then the function $g$ is $\mu$-measurable and the restriction of $\pi$ to the set $S$ of $t \in T$ such that $g(t) \neq 0$ is $\mu$-measurable.

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: