Proposition 3

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

Let $f$ be a mapping of $X$ into a topological space $G$, and let $S$ be the ($\mu$-measurable) set of points $t \in T$ such that $g(t) > 0$. In order that $f$ be $\nu$-measurable, it is necessary and sufficient that the restriction of $f \circ \pi$ to $S$ be $\mu$-measurable.

Suppose first that $f$ is $\nu$-measurable. By hypothesis, the set $\mathscr{K}$ of compact subsets $K$ of $S$, such that the restriction of $\pi$ to $K$ is continuous, is $\mu$-dense in $S$ (Ch. IV, §5, No. 10, Prop. 15). To show that the restriction of $f \circ \pi$ to $S$ is $\mu$-measurable, it therefore suffices to prove that for every $K \in \mathscr{K}$, the set of compact subsets $H$ of $K$, such that the restriction of $f \circ \pi$ to $H$ is continuous, is $\mu$-dense in $K$ (Ch. IV, §5, No. 8, Prop. 13). But by hypothesis, there exists a partition of the compact set $\pi(K)$ formed by a $\nu$-negligible set $N$ and a sequence of compact sets $(C_n)$ such that the restriction of $f$ to each $C_n$ is continuous. Under these conditions, $K \cap \overline{\pi}(N)$ and the sets $K \cap \overline{\pi}(C_n)$ form a partition of $K$; but $K \cap \overline{\pi}(N)$ is $\mu$-negligible by virtue of the Cor. of Th. 1 of No. 2, the sets $K \cap \overline{\pi}(C_n)$ are compact, and the restriction of $f \circ \pi$ to each of the latter sets is continuous, which proves that the restriction of $f \circ \pi$ to $S$ is $\mu$-measurable.

Conversely, suppose this to be the case; to show that $f$ is $\nu$-measurable, it suffices to prove that the set $\mathscr{L}$ of compact subsets $L$ of $X$, such that the restriction of $f$ to $L$ is continuous, is $\nu$-dense (Ch. IV, §5, No. 10, Prop. 15). Let $N$ be a subset of $X$ such that $N \cap L$ is $\nu$-negligible for every $L \in \mathscr{L}$, and let us show that $N$ is locally $\nu$-negligible. For this, we must show that $\overline{\pi}(N) \cap S$ is locally $\mu$-negligible (Cor. of Th. 1 of No. 2). Now, the set $\mathscr{H}$ of compact subsets $H$ of $S$, such that the restrictions to $H$ of $\pi$ and $f \circ \pi$ are continuous, is by hypothesis $\mu$-dense in $S$ (Ch. IV, §5, No. 10, Prop. 15). It therefore suffices to prove that $\overline{\pi}(N) \cap H$ is $\mu$-negligible for every $H \in \mathscr{H}$. Now, $\pi(H)$ is compact and may be identified with the quotient space of $H$ by the equivalence relation $\pi(t) = \pi(t')$, $\pi$ being identified with the canonical mapping of $H$ onto this quotient space (GT, I, §5, No. 2, Prop. 3). Since the restriction of $f \circ \pi$ to $H$ is continuous, the restriction of $f$ to $\pi(H)$ is therefore continuous, in other words $\pi(H) \in \mathscr{L}$, consequently $N \cap \pi(H)$ is $\nu$-negligible. By the Cor. of Th. 1 of No. 2, $\overline{\pi}(N \cap \pi(H)) \cap S$ is locally $\mu$-negligible; the same is therefore true of the set

but since $H$ is compact, $H \cap \overline{\pi}(N)$ is $\mu$-negligible, which completes the proof.