Lemma 4

Let $X$ be a topological space, $\nu$ a measure on $X$, and $f$ a $\nu$-measurable mapping of $X$ into a topological space $F$ (Hausdorff or not). There exists a universally measurable mapping $f'$ of $X$ into $F$, equal to $f$ locally $\nu$-almost everywhere.

The proof is identical to that of Prop. 7 of Ch. V, §3, No. 4, on taking into account Prop. 10 of §1, No. 8.

Let us pass to the proof of Prop. 13. A) Suppose that $X$ is compact and metrizable and that $p$ is continuous and surjective: The space $T$ is then compact and metrizable (GT, IX, §2, No. 10). By Th. 1 of Ch. VI, §3, No. 1, there exists a mapping $H : t \mapsto \eta_t$ of $T$ into $\mathscr{M}_+(X)$, vaguely $\mu$-measurable and scalarly essentially $\mu$-integrable, such that $\nu = \int_T \eta_t \, d\mu(t)$ and such that $\eta_t$ has total mass 1 and is carried by $\overline{p}(t)$ for every $t \in T$. Let $(S_n)_{n \in \mathbf{N}}$ be a crushing of $T$ for $\mu$, such that the restriction of $H$ to each of the sets $S_n$ is continuous (§1, No. 8, Props. 10 and 11); we denote by $\Lambda : t \mapsto \lambda_t$ the mapping of $T$ into $\mathscr{M}_+(X)$ that is equal to $H$ on $S = \bigcup_{n \in \mathbf{N}} S_n$ and to 0 on $T - S$. It is clear that $\nu = \int_T \lambda_t \, d\mu(t)$ and that $\Lambda$ satisfies condition a) of the statement. Let $\theta$ be a measure on $T$; the mapping $\Lambda$ is vaguely $\theta$-measurable and scalarly essentially $\theta$-integrable, hence also $\theta$-adequate (Ch. V, §3, No. 1, Prop. 2 b)). Let $f$ be a universally measurable positive function on $X$; by Prop. 5 of Ch. V, §3, No. 2 applied to $\int \lambda_t \, d\theta(t)$, the mapping $t \mapsto \lambda_t^\bullet(f)$ is $\theta$-measurable, hence universally measurable in view of the arbitrariness of $\theta$. The formula (12) results from Prop. 5 of Ch. V, §3, No. 2. B) Suppose that there exists a compact subset $X'$ of $X$ carrying the measure $\nu$ and such that $p_{X'}$ is continuous: Set $T' = p(X')$, and $p' = p_{X'}$; let us denote by $\nu'$ the measure $\nu_{X'}$, and by $\mu'$ the image measure $p'(\nu')$ on $T'$. Since $p'$ is continuous and surjective and since $X'$ is compact and metrizable, by A) there exists a mapping $\Lambda' : t' \mapsto \lambda'_{t'}$ of $T'$ into $\mathscr{M}_+(X')$ satisfying the following conditions: a') for every $t' \in T'$, the measure $\lambda'_{t'}$ is carried by $X' \cap \overline{p}'(t')$; b') for every universally measurable positive function $f'$ on $X'$, the function $t' \mapsto {\lambda'_{t'}}^\bullet(f')$ is universally measurable on $T'$ and

Let $t \in T$; if $t$ belongs to $T'$, let us denote by $\lambda_t$ the image of $\lambda'_t$ under the canonical injection of $X'$ into $X$, and if $t$ belongs to $T - T'$ we set $\lambda_t = 0$. The reader will verify without difficulty that the mapping $t \mapsto \lambda_t$ satisfies the conditions a) and b) of the statement.

C) Existence in the general case:

The measure $\nu$ on $X$ being moderated, we may choose a covering $(U_m)_{m \in \mathbf{N}}$ of $X$ consisting of $\nu$-integrable open sets. Let in addition $(X_n)_{n \in \mathbf{N}}$ be a $\nu$-crushing of $X$ such that the restriction of $p$ to each set $X_n$ is continuous (§ 1, No. 8, Props. 10 and 11); denote by $\nu_n$ the measure $\varphi_{X_n} \cdot \nu$ on $X$ and by $\mu_n$ its image under $p$. By B) there exists, for every integer $n \in \mathbf{N}$, a mapping $t \mapsto \alpha^n_t$ of $T$ into $\mathscr{M}_+(X)$ satisfying the following conditions:

$a'')$ The measure $\alpha^n_t$ is carried by $\overline{p}(t)$ for all $t \in T$.

$b'')$ If $f$ is a universally measurable positive function on $X$, the positive function $t \mapsto (\alpha^n_t)^*(f)$ on $T$ is universally measurable and

One has $\nu = \sum_{n \in \mathbf{N}} \nu_n$ and $\mu = \sum_{n \in \mathbf{N}} \mu_n$; it follows immediately from Prop. 3 of No. 2 and the above Lemma 4 that there exists a sequence $(g_n)_{n \in \mathbf{N}}$ of universally measurable positive functions on $T$ such that $\mu_n = g_n \cdot \mu$ for all $n \in \mathbf{N}$ and such that $\sum_{n \in \mathbf{N}} g_n = 1$. For every $t \in T$, let us denote by $\beta^n_t$ the measure $g_n(t) \cdot \alpha^n_t$ on $X$ and by $q_t$ the encumbrance $\sum_{n \in \mathbf{N}} (\beta^n_t)^*$ on $X$. Let $f$ be a universally measurable positive function on $X$; using Prop. 2 of No. 2 and summing over $n$ in (13), we obtain

it is moreover clear that the function $t \mapsto q_t(f)$ on $T$ is universally measurable.

For every $m \in \mathbf{N}$, let $E_m$ be the set of $t \in T$ such that $q_t(U_m) = +\infty$; the set $E_m$ is universally measurable because this is true of the mapping $t \mapsto q_t(U_m)$, and $E_m$ is locally $\mu$-negligible by the formula (14) applied to $f = \varphi_{U_m}$, since $\nu^*(U_m)$ is finite. The set $E = \bigcup_{m \in \mathbf{N}} E_m$ is therefore universally measurable and locally $\mu$-negligible. We set $\lambda_t = 0$ for $t \in E$. Moreover, let $t \in T - E$; the encumbrance $q_t$ is locally bounded since the open sets $U_m$ cover $X$ and since $q_t(U_m)$ is finite; by Prop. 7 of § 1, No. 7, there exists a measure $\lambda_t$ on $X$ such that $q_t = \lambda_t^*$ and $\lambda_t = \sum_{n \in \mathbf{N}} \beta^n_t$. It is immediate that the mapping $t \mapsto \lambda_t$ satisfies the conditions a) and b) of the statement.

D) Proof of c): Let $f$ be a universally measurable function on $X$ that is positive and bounded; we are going to show that the universally measurable function $h_f : t \mapsto \lambda_t^\bullet(f)$ on $T$ is a density for the measure $\mu_f = p(f \cdot \nu)$ with respect to $\mu = p(\nu)$. Let $K$ be a compact subset of $T$ and let $A = \overline{p}^{-1}(K)$. For every $t \in T$, the measure $\lambda_t$ is carried by $\overline{p}^{-1}(t)$; if $t$ belongs to $K$ then $\overline{p}^{-1}(t) \subset A$, whence $\lambda_t^\bullet(f \varphi_A) = \lambda_t^\bullet(f)$; on the other hand, if $t$ belongs to $T - K$ then $\overline{p}^{-1}(t) \subset X - A$, whence $\lambda_t^\bullet(f \varphi_A) = 0$. Applying the formula (12) to $f \cdot \varphi_A$,$^{(1)}$ we obtain

which establishes the relation $\mu_f = h_f \cdot \mu$.

Letting $f = 1$, one sees that the function $h_1 : t \mapsto \| \lambda_t \|$ is a density of the measure $\mu_1 = \mu$ with respect to $\mu$, hence is equal to 1 locally $\mu$-almost everywhere in $T$.

E) Uniqueness: Let $t \mapsto \lambda_t^i$ (for $i = 1, 2$) be two mappings of $T$ into $\mathscr{M}_+(X)$ satisfying the conditions a) and b) of the statement. As in C), choose a $\mu$-crushing $(X_n)_{n \in \mathbf{N}}$ of $X$ such that $p_{X_n}$ is continuous for every $n \in \mathbf{N}$, and set $N = X - \bigcup_{n \in \mathbf{N}} X_n$. For every integer $n \in \mathbf{N}$, choose a countable set $D_n$ of positive functions on $X$, zero outside $X_n$, whose restrictions to $X_n$ form a dense set in the normed space $\mathscr{C}(X_n)$ (apply Th. 1 of GT, X, §3, No. 3 to the metrizable compact space $X_n$). We set $D = \bigcup_{n \in \mathbf{N}} D_n$.

Let $f \in D$; by D), the functions $t \mapsto (\lambda_t^1)^\bullet(f)$ and $t \mapsto (\lambda_t^2)^\bullet(f)$ are densities of the measure $\mu_f$ with respect to $\mu$, and so there exists a locally $\mu$-negligible set $E_f$ in $T$ such that $(\lambda_t^1)^\bullet(f) = (\lambda_t^2)^\bullet(f)$ for $t \in T - E_f$. Moreover, by (12), the set $F_i$ of $t \in T$ such that $(\lambda_t^i)^\bullet(N) \neq 0$ is locally $\mu$-negligible for $i = 1, 2$. Since $D$ is countable, the set $G = (\bigcup_{f \in D} E_f) \cup F_1 \cup F_2$ is locally $\mu$-negligible; for $t \in T - G$, we have $(\lambda_t^1)^\bullet(N) = (\lambda_t^2)^\bullet(N) = 0$ and $(\lambda_t^1)_{X_n} = (\lambda_t^2)_{X_n}$, whence $\lambda_t^1 = \lambda_t^2$ by Prop. 9 of §1, No. 8.

Q.E.D.