Proposition 6

— Let T and X be two locally compact spaces, $(\lambda_\alpha)_{\alpha \in A}$ a family of positive measures on T, directed for the relation $\leq$, admitting in $\mathscr{M}(T)$ a supremum $\mu$. In order that a mapping $\pi$ of T into X be $\mu$-proper, it is necessary and sufficient that it be $\lambda_\alpha$-proper for all $\alpha \in A$, and that the family $(\pi(\lambda_\alpha))_{\alpha \in A}$ be bounded above in $\mathscr{M}(X)$. In this case,

For $\pi$ to be $\mu$-measurable, it is necessary and sufficient that $\pi$ be $\lambda_\alpha$-measurable for all $\alpha \in A$ (§ 1, No. 4, Cor. 2 of Prop. 11). Suppose that this condition is satisfied; to say that $\pi$ is $\mu$-proper is then equivalent to saying that, for every function $f \in \mathscr{K}_+(X)$,

Now,

(§ 1, No. 4, Prop. 11); the first member is thus finite for every $f \in \mathscr{K}_+(X)$ if and only if the family $(\pi(\lambda_\alpha))$ admits a supremum $\theta$ in $\mathscr{M}(X)$, in which case $\int (f \circ \pi)\, d\mu = \int f\, d\theta$, a relation equivalent to (5).