Proposition 4

— Let T, T', T'' be three locally compact spaces, $\mu$ a positive measure on T, $\pi$ a $\mu$-measurable mapping of T into T', $\pi'$ a mapping of T' into T'', and $\pi'' = \pi' \circ \pi$.

a) Suppose that $\pi$ is $\mu$-proper and let $\mu' = \pi(\mu)$. For $\pi'$ to be $\mu'$-proper, it is necessary and sufficient that $\pi''$ be $\mu$-proper, in which case $\pi''(\mu) = \pi'(\pi(\mu))$ ('transitivity of the image of a measure').

b) Suppose that $\pi'$ is continuous, and that $\pi''$ is $\mu$-proper; then $\pi$ is $\mu$-proper, $\pi'$ is $\pi(\mu)$-proper and $\pi''(\mu) = \pi'(\pi(\mu))$.

Under the hypotheses of a), for $\pi''$ to be $\mu$-measurable it is necessary and sufficient that $\pi'$ be $\mu'$-measurable (No. 2, Prop. 3). On the other hand if $K$ is a compact subset of $T''$, then $\pi''(K) = \pi^{-1}(\pi'(K))$; for $\pi''(K)$ to be essentially $\mu$-integrable, it is necessary and sufficient that $\pi'(K)$ be essentially $\mu'$-integrable, by the Cor. of Th. 1. Finally, if $\pi''$ is $\mu$-proper then, setting $\mu'' = \pi''(\mu)$, we have, for every function $f \in \mathscr{H}(T'')$,

by Th. 1 of No. 2, which completes the proof of a).

Under the hypotheses of b), let $K'$ be a compact subset of $T'$. Then $K'' = \pi'(K')$ is compact, therefore $\pi''(K'')$ is essentially $\mu$-integrable, therefore $\pi^{-1}(K') \subset \pi''(K'')$ is essentially $\mu$-integrable (Ch. IV, §5, No. 5, Prop. 7), so that $\pi$ is $\mu$-proper. The proof is then concluded by applying part a) of the statement.