Proposition 4

Let $\pi$ be a $\mu$-proper mapping of T into a topological space X. There exists one and only one measure $\nu$ on X such that $\nu^\bullet$ is equal to the image encumbrance $\pi(\mu^\bullet)$ (§1, No. 1), in other words, such that $\nu^\bullet(g) = \mu^\bullet(g \circ \pi)$ for all $g \in \mathscr{F}_+(X)$.

Uniqueness is obvious (§1, No. 2, Cor. of Prop. 2). To establish existence, we shall first treat the case that $\mu$ is carried by a compact set K, such that the restriction of $\pi$ to K is continuous. Then $L = \pi(K)$ is compact; let $\pi'$ be the continuous mapping of K into L induced by $\pi$, and let $\nu'$ be the image measure $\pi'(\mu_K)$ on L, $\nu$ the measure on X defined by $\nu'$ (§1, No. 3, Example 2). For every $g \in \mathscr{F}_+(X)$,

(we have used successively formula (3) of §1, No. 3; Prop. 2 of Ch. V, §6, No. 2; the definition of $\mu_K^\bullet$; and the fact that $\mu$ is carried by K). In other words, $\nu^\bullet = \pi(\mu^\bullet)$.

Let us now pass to the general case; by Props. 10 and 9 of §1, No. 8, $\mu$ is the sum of a summable family $(\mu_\alpha)_{\alpha \in A}$ of measures with compact support, such that the restriction of $\pi$ to the support $K_\alpha$ of $\mu_\alpha$ is continuous for every $\alpha \in A$. The special case treated above permits associating to each measure $\mu_\alpha$ on T a measure $\nu_\alpha$ on X such that $\nu_\alpha^\bullet = \pi(\mu_\alpha^\bullet)$. Then, for $g \in \mathscr{F}_+(X)$,

The encumbrance $\pi(\mu^\bullet)$ is locally bounded, since $\pi$ is $\mu$-proper; the family $(\nu_\alpha)_{\alpha \in A}$ is therefore summable (§1, No. 7, Prop. 7), and its sum $\nu$ satisfies the statement.