Lemma

— a) Let $f$ be an element of $\mathscr{K}_+(X)$; the function $t \mapsto \lambda_t(f)$ is positive, upper semi-continuous, with compact support, and its restriction to $\pi(X)$ is continuous.

b) Let $U$ be an element of $\mathscr{U}$, $\nu$ the integral of the scalarly essentially $\mu$-integrable function $t \mapsto \lambda_t$; the image of the measure $\nu|U$ under $\pi|U$ is equal to $\mu|\pi(U)$.

To establish a), one may reduce by means of a partition of unity (Ch. III, §1, No. 2, Lemma 1) to the case that the support $S$ of $f$ is contained in an open set $U \in \mathscr{U}$. Let $g$ be the mapping $t \mapsto \lambda_t(f)$; $\pi|U$ being a homeomorphism, $g|\pi(U)$ belongs to $\mathscr{K}_+(\pi(U))$, consequently ($\pi(U)$ being an open set in $\pi(X)$) the restriction of $g$ to $\pi(X)$ is continuous. Since $g$ is positive and the restriction of $g$ to the compact set $\pi(S)$ is continuous, one sees that $g$ is upper semi-continuous on $T$. It follows that $g$ is $\mu$-integrable.

To establish b), denote by $g$ an element of $\mathscr{K}(\pi(U))$, by $g^\circ$ its extension by 0 to $T$, by $f$ the function $g \circ (\pi|U)$, and by $f^\circ$ the extension by 0 of $f$ to $X$. The assertion b) is equivalent to the equality $\int g^\circ d\mu = \int f^\circ d\nu$. But $f \in \mathscr{K}(U)$, therefore $f^\circ \in \mathscr{K}(X)$, and the second integral is therefore equal to $\int \lambda_t(f^\circ) d\mu(t)$. Finally $\lambda_t(f^\circ) = g^\circ(t)$, which completes the proof.

We now observe that $\pi(X)$ is open in $T$, hence is $\mu$-measurable; the mapping $\Lambda : t \mapsto \lambda_t$ is vaguely $\mu$-measurable, because its restriction to each of the sets $\pi(X)$ and $\mathbf{C}\pi(X)$ is vaguely continuous. Under these conditions, the formula $\overline{\pi}^{-1}(\mu) = \int \lambda_t d\mu(t)$ defines a decomposition into slices of $\overline{\pi}^{-1}(\mu)$ relative to $\pi$, and Prop. 10 yields the following result: