Proposition 8

*Let $T$ and $X$ be two topological spaces, $\pi$ a mapping of $T$ into $X$.

a) Let $\nu$ be a bounded measure on $X$. In order that there exist a measure $\mu$ on $T$ such that $\pi$ is $\mu$-proper and $\pi(\mu) = \nu$, it is necessary and sufficient that there exist, for every number $\varepsilon > 0$, a compact set $K_\varepsilon \subset T$ such that the restriction of $\pi$ to $K_\varepsilon$ is continuous and $\nu^\bullet(X - \pi(K_\varepsilon)) < \varepsilon$.

b) Suppose that $\pi$ is injective; let $\mu$ and $\mu'$ be two measures on $T$, such that $\pi$ is proper for $\mu$ and $\mu'$, and such that $\pi(\mu) = \pi(\mu')$. Then $\mu = \mu'$.

The condition stated in a) is necessary. For, if $\pi$ is $\mu$-proper and $\pi(\mu) = \nu$, the relation $\mu^\bullet(1) = \nu^\bullet(1) < +\infty$ implies that $\mu$ is bounded. Prop. 2 of §1, No. 2, applied to the function 1, implies the existence of a compact subset $K$ of $T$ such that $\mu^\bullet(T - K) < \varepsilon/2$. Since $\pi$ is $\mu$-measurable, there exists a compact set $K_\varepsilon \subset K$ such that the restriction of $\pi$ to $K_\varepsilon$ is continuous, and such that $\mu^\bullet(K - K_\varepsilon) < \varepsilon/2$. Then (No. 3, Prop. 4)

To show that the condition is sufficient, we first treat a special case.