Proposition 1

Let $\pi$ be a $\mu$-measurable mapping of $T$ into a locally compact space $X$. The following two properties are equivalent:

a) for every function $f \in \mathscr{K}(X)$, $f \circ \pi$ is essentially $\mu$-*integrable*.

b) for every compact set $K \subset X$, $\overline{\pi}^{-1}(K)$ is essentially $\mu$-*integrable*.

We have just observed that a) implies that the pair $(\pi, 1)$ is $\mu$-adapted. Consequently (§ 4, No. 4, Th. 2), for every compact set $K \subset X$, the function $\varphi_K \circ \pi = \varphi_A$, where $A = \overline{\pi}^{-1}(K)$, is essentially $\mu$-integrable, in other words, a) implies b).

Conversely, suppose that $\overline{\pi}^{-1}(K)$ is essentially $\mu$-integrable for every compact subset $K$ of $X$, and let us show that a) is verified. Indeed, let $S$ be the support of $f$; since $S$ is compact, setting $A = \overline{\pi}^{-1}(S)$ we have, by hypothesis,

Since $f \circ \pi$ is $\mu$-measurable (Ch. IV, § 5, No. 3, Th. 1), we see that $f \circ \pi$ is essentially $\mu$-integrable (§ 1, No. 3, Prop. 9).

Property b) is obviously equivalent to the following (which is therefore also equivalent to a)):

c) For every point $x$ of $X$, there exists a neighborhood $V$ of $x$ such that $\overline{\pi}^{-1}(V)$ is essentially $\mu$-*integrable*.