Theorem 4

Let T be a Polish locally compact space, $\mu$ a positive measure on T, and R a $\mu$-measurable equivalence relation in T. Let B, B' be two Polish locally compact spaces, p, p' two $\mu$-measurable mappings of T into B, B' respectively, such that R$\{x, y\}$ is equivalent to $p(x) = p(y)$ and to $p'(x) = p'(y)$. Let $\nu, \nu'$ be pseudo-image measures of $\mu$ under p, p' respectively; let $b \mapsto \lambda_b, b' \mapsto \lambda_{b'}$ be disintegrations of $\mu$ relative to $\nu, \nu'$ respectively.

Under these conditions, there exist in B (resp. B') a set N (resp. N') negligible for $\nu$ (resp. $\nu'$) and a bijection f of B - N onto B' - N', having the following properties:

a) The mapping f (defined almost everywhere in B) is $\nu$-measurable and its inverse mapping $f'$ is $\nu'$-measurable; every pseudo-image measure of $\nu$ (resp. $\nu'$) under f (resp. $f'$) is equivalent to $\nu'$ (resp. $\nu$).

b) For every $b \in B - N$, the measure $\lambda'_{f(b)}$ on T is of the form $r(b)\lambda_b$, where $r(b) \neq 0$ and r is locally $\nu$-integrable.

To establish a), we may limit ourselves to the case that $\nu$ and $\nu'$ are bounded measures (Ch. V, §5, No. 6, Prop. 11). Let $N_0 = B - p(T)$, $N'_0 = B' - p'(T)$; we know that $N_0$ (resp. $N'_0$) is negligible for $\nu$ (resp. $\nu'$) (No. 2). There exists a bijection f of B - $N_0$ onto B' - $N'_0$ defined by $f(p(t)) = p'(t)$ for all $t \in T$; let $f'$ be the inverse mapping of $f$, so that $f'(p'(t)) = p(t)$. For every subset M of B, the relation « M is $\nu$-measurable » is equivalent to « $\overline{p}^{-1}(M)$ is $\mu$-measurable », that is, to « $p'(f(M))$ is $\mu$-measurable », thus finally to « $f(M)$ is $\nu'$-measurable » (Ch. V, §6, No. 2, Cor. of Prop. 3). We thus see that $f$ (resp. $f'$) transforms every $\nu$-measurable (resp. $\nu'$-measurable) set into a $\nu'$-measurable (resp. $\nu$-measurable) set; since B and $B'$ are metrizable and have countable bases, it follows that $f$ and $f'$ are measurable (Ch. IV, §5, No. 5, Th. 4). Moreover, if $M \subset B$ is $\nu$-negligible then $\overline{p}^{-1}(M) = p'(f(M))$ is $\mu$-negligible, therefore $f(M)$ is $\nu'$-negligible (Ch. V, §6, No. 2, Cor. 2 of Prop. 2); similarly, $f'$ transforms every $\nu'$-negligible set into a $\nu$-negligible set. Consequently, the image of $\nu$ under $f$ (which is defined since $\nu$ is bounded, which implies that $f$ is $\nu$-proper) is equivalent to $\nu'$, and the image of $\nu'$ under $f'$ is equivalent to $\nu$ (Ch. V, §5, No. 6, Prop. 10). It remains to prove b). By virtue of Th. 2 of No. 3, we can restrict ourselves to the case that $\nu' = f(\nu)$. Since $\mu = \int \lambda_{b'}' d\nu'(b')$, we have, for every function $h \in \mathscr{K}(T)$,

(Ch. V, §3, No. 4, Th. 1 and §6, No. 2, Th. 1); in other words, $\mu = \int \lambda_{f(b)}' d\nu(b)$. But since also $\mu = \int \lambda_b d\nu(b)$ and since, for every $b \in B - N_0$, $\lambda_b$ and $\lambda_{f(b)}'$ are carried by $\overline{p}^{-1}(b)$, Th. 2 of No. 3 implies that $\lambda_b = \lambda_{f(b)}'$ for almost every $b \in B - N_0$, hence for almost every $b \in B$. The conditions of Th. 4 are therefore verified by taking for N the union of $N_0$ and the set of $b \in B$ such that $\lambda_b \neq \lambda_{f(b)}'$.