Corollary

— *Let X and Y be two topological spaces, $\pi$ a $\mu$-proper mapping of T into X, and $\pi'$ a $\pi(\mu)$-proper mapping of X into Y. The mapping $\pi'' = \pi' \circ \pi$ is then $\mu$-proper, and $\pi''(\mu) = \pi'(\pi(\mu))$ ('transitivity of images of measures').

For, $\pi''$ is $\mu$-measurable (Prop. 5). Set $\mu' = \pi(\mu)$; the image encumbrance $\pi'({\mu'}^\bullet) = \pi'(\pi(\mu^\bullet))$ is obviously equal to $\pi''(\mu^\bullet)$. Since it is locally bounded, $\pi''$ is $\mu$-proper. The measures $\pi''(\mu)$ and $\pi'(\mu')$ then have the same essential upper integral, hence are equal.