This lemma is none other than the Cor. to Th. 1 of GT, X, §3, No. 3. One may observe that it also follows from Lemma 1 and the fact that the topology of uniform convergence on $\mathscr{K}(Y)$ is coarser than the direct limit topology of the topologies of the subspaces $\mathscr{K}(Y, \overline{U}_n)$.

For, Lemma 1, applied to $\mathscr{K}(X)$, shows that the mapping $t \mapsto \lambda_t$ is vaguely $\mu$-measurable ($§ 1$, No. 5, Prop. 13).

a) $\| \lambda_b \| = 1$ for all $b \in p(T)$;

b) $\lambda_b$ is concentrated on the set $\overline{p}^{-1}(b)$ (Ch. V, §5, No. 7, Def. 4) for all $b \in B$; in particular, $\lambda_b = 0$ for $b \notin p(T)$;

c) $\mu = \int \lambda_b d\nu(b)$.

Moreover, if $b \mapsto \lambda'_b$ ($b \in B$) is a second $\nu$-adequate family of positive measures on T having the properties b) and c), then $\lambda'_b = \lambda_b$ almost everywhere in B for the measure $\nu$.

1) Uniqueness. For every function $f \in \mathscr{K}(B)$, $f \circ p$ is $\mu$-integrable since p is $\mu$-proper (Ch. V, §6, No. 2, Th. 1); for every function $g \in \mathscr{K}(T)$, the function $t \mapsto g(t)f(p(t))$ is therefore $\mu$-integrable. It follows (Ch. V, §3, No. 3, Th. 1) that for almost every $b \in B$, the function $t \mapsto g(t)f(p(t))$ is $\lambda_b$-integrable and that

But since $\lambda_b$ is concentrated on $\overline{p}^{-1}(b)$, we have, for every $b \in B$, $f(p(t)) = f(b)$ almost everywhere for $\lambda_b$, therefore the second member of (1) is equal to $\int f(b)\langle g, \lambda_b \rangle d\nu(b)$. The analogous formula for $\lambda'_b$ also holds; consequently $\int f(b)\langle g, \lambda_b \rangle d\nu(b) = \int f(b)\langle g, \lambda'_b \rangle d\nu(b)$ for all $f \in \mathscr{K}(B)$ and $g \in \mathscr{K}(T)$. In other words, the two mappings $b \mapsto \lambda_b$ and $b \mapsto \lambda'_b$ of B into $\mathscr{M}(T)$ are equal scalarly locally almost everywhere for $\nu$, hence equal almost everywhere for $\nu$ (Lemma 1 and §1, No. 1, Remark 2).

2) Provisional definition of the family $b \mapsto \lambda_b$. For every function $f \in \mathscr{L}^1(\nu)$, $f \circ p$ is $\mu$-integrable (Ch. V, §6, No. 2, Th. 1), therefore $(f \circ p) \cdot \mu$ is a bounded measure on T, and

(Ch. IV, §4, No. 7, Prop. 12; Ch. V, §5, No. 3, Th. 1 and §6, No. 2, Th. 1). It follows that $(f \circ p) \cdot \mu$ depends only on the class $\tilde{f}$ of $f$ in $L^1(\nu)$ and that $\tilde{f} \mapsto (f \circ p) \cdot \mu$ is an isometric linear mapping of $L^1(\nu)$ into the Banach space $\mathscr{M}^1(T)$ of bounded measures on T, the strong dual of the Banach space $\mathscr{K}(T)$, which is separable (Lemma 2). By the Dunford–Pettis theorem (§2, No. 5, Cor. 2 of Th. 1) there exists a mapping $b \mapsto \lambda_b$ of B into the unit ball of $\mathscr{M}^1(T)$, scalarly $\nu$-measurable (for the topology $\sigma(\mathscr{M}^1(T), \overline{\mathscr{K}(T)})$) and such that, for every function $f \in \mathscr{L}^1(\nu)$,

which may also be written, for every function $g \in \overline{\mathscr{K}(T)}$

If $f \geq 0$ and $g \geq 0$, the first member of (3) is $\geq 0$, which proves that for every function $g \geq 0$ in $\mathscr{K}(T)$, the measure $(\int g(t) \, d\lambda_b(t)) \cdot \nu$ is $\geq 0$, hence that $\int g(t) \, d\lambda_b(t) \geq 0$ except for $b$ belonging to a $\nu$-negligible set $N(g)$ (Ch. V, §5, No. 3, Cor. 3 of Prop. 3). Now, there exists a dense sequence $(g_n)$ in the space $\mathscr{K}_+(T)$ of functions $\geq 0$ of $\mathscr{K}(T)$ (Lemma 1). The union N of the $N(g_n)$ is $\nu$-negligible and, for $b \notin N$, we have $\int g_n(t) \, d\lambda_b(t) \geq 0$ for all $n$, therefore $\int g(t) \, d\lambda_b(t) \geq 0$ for every function $g \in \mathscr{K}_+(T)$, in other words $\lambda_b \geq 0$.

This being so, we may replace $\lambda_b$ by 0 for every $b \in N$ without altering the validity of (3); we can therefore assume this modification to have been made, so that $\lambda_b \geq 0$ for every $b \in B$.

3) Extensions of the formula (3).

$\alpha)$ For every function $f \in \mathscr{L}^1(\nu)$, it follows from (3) that the mapping $b \mapsto \lambda_b$ of B into $\mathscr{M}(T)$ is scalarly integrable for the measure $|f \cdot \nu|$ and the topology $\sigma(\mathscr{M}(T), \mathscr{K}(T))$, therefore (Lemma 3) the family $b \mapsto \lambda_b$ is $|f \cdot \nu|$-*adequate*. Now let $g$ be a numerical function defined on T, integrable for the measure $|(f \circ p) \cdot \mu|$, that is (Ch. V, §5, No. 3, Th. 1), such that $t \mapsto g(t)f((p(t)))$ is $\mu$-integrable; it then follows from (2), from Th. 1 of Ch. V, §3, No. 3 and from Th. 1 of Ch. V, §5, No. 3 that, for almost every $b \in B$, $g$ is integrable for $\lambda_b$, that the function (defined almost everywhere) $b \mapsto \int g(t)\, d\lambda_b(t)$ is integrable for $|f \cdot \nu|$, and that the formula (3) is again valid.

$\beta)$ For every function $g \in \overline{\mathscr{K}(T)}$, it follows from (3), applied to $f \in \mathscr{K}(B)$, that the mapping $p$ is proper for the measure $|g \cdot \mu|$ (Ch. V, §6, No. 1, Def. 1) and the image under $p$ of the measure $g \cdot \mu$ is the measure with density $b \mapsto \int g(t)\, d\lambda_b(t)$ with respect to $\nu$. If $f$ is then taken to be a function such that $f \circ p$ is integrable for the measure $|g \cdot \mu|$, that is, such that $t \mapsto g(t)f(p(t))$ is $\mu$-integrable (Ch. V, §5, No. 3, Th. 1), the formula (3) is again valid (Ch. V, §6, No. 2, Th. 1).

4) Properties of the family $b \mapsto \lambda_b$. By the property $\beta)$, we can apply formula (3) by taking $f = 1,\ g \in \mathscr{K}(T)$; this proves that $b \mapsto \lambda_b$ is scalarly $\nu$-integrable (for the topology $\sigma(\mathscr{M}(T), \mathscr{K}(T))$), hence is $\nu$-*adequate* (Lemma 3), and that $\mu = \int \lambda_b\, d\nu(b)$.

Now let $\psi$ be any function in $\mathscr{K}(B)$; the conditions of property $\alpha)$ are fulfilled by taking $f \in \mathscr{K}(B)$ and $g = \psi \circ p$, because the function $\psi(p(t))f(p(t))$ is $\mu$-integrable since $f\psi$ belongs to $\mathscr{K}(B)$ and $p$ is $\mu$-proper. Then $\psi \circ p$ is $\lambda_b$-integrable for almost every $b \in B$, and

but the first member is by definition $\int f(b)\psi(b)\, d\nu(b)$. We therefore see that for every function $\psi \in \mathscr{K}(B)$, the measure $\psi \cdot \nu$ and the measure with density $b \mapsto \int \psi(p(t))\, d\lambda_b(t)$ are identical. Consequently (Ch. V, §5, No. 3, Cor. 2 of Prop. 3) there exists a $\nu$-negligible set $N'(\psi)$ such that, for every $b \notin N'(\psi)$, the function $\psi \circ p$ is $\lambda_b$-integrable and $\psi(b) = \int \psi(p(t))\, d\lambda_b(t)$.

Let S be a countable subset of $\mathscr{K}(B)$ having the properties stated in Lemma 1 (with $Y = B$), and let $N'$ be the $\nu$-negligible set that is the union of the $N'(\psi)$ for $\psi \in S$. Every function $\psi \geq 0$ of $\mathscr{K}(B)$ is the uniform limit of a sequence $(\psi_n)$ of elements of S with $\psi_n \leq \psi_0$. Consequently for $b \notin N'$, Lebesgue's theorem shows on the one hand that $\psi \circ p$ is $\lambda_b$-integrable, in other words that $p$ is $\lambda_b$-proper, and on the other hand that $\psi(b) = \int \psi(p(t))\, d\lambda_b(t)$. In other terms, the mappings $b \mapsto \varepsilon_b$ and $b \mapsto p(\lambda_b)$ of B into $\mathscr{M}(B)$ (the latter being defined almost everywhere) are scalarly almost everywhere equal for $\nu$ (and for the topology $\sigma(\mathscr{M}(B), \mathscr{K}(B))$); it follows that these mappings are equal almost everywhere for $\nu$ (Lemma 1 and §1, No. 1, Remark 2). Finally, if $p(\lambda_b) = \varepsilon_b$, the set $B - \{b\}$ is $\varepsilon_b$-negligible, therefore the set $T - \bar{p}^{-1}(B)$ is $\lambda_b$-negligible (Ch. V, §6, No. 2, Cor. 2 of Prop. 2), in other words $\lambda_b$ is concentrated on $\overline{p}^{-1}(b)$; and, on the other hand, $\| \lambda_b \| = \int d\lambda_b = \int d((p(\lambda_b))) = \| \varepsilon_b \| = 1$ (Ch. V, §6, No. 2, Th. 1).

5) Modifications of the family $b \mapsto \lambda_b$. We have thus defined a $\nu$-adequate family $b \mapsto \lambda_b$ of measures $\geqslant 0$ on T, satisfying condition c) of the statement and such that, for almost every $b \in B$, $p$ is $\lambda_b$-proper, and $\lambda_b$ is concentrated on $\overline{p}^{-1}(b)$ and has norm 1. Let $N''$ be the $\nu$-negligible set of points $b \in B$ where one of the last three properties is not verified; we can then modify $\lambda_b$ in the following manner. If $b \in B - p(T)$, take $\lambda_b = 0$; if $b \in p(T) \cap N''$, take $\lambda_b = \varepsilon_{\xi(b)}$, where $\xi(b)$ is any point of $\overline{p}^{-1}(b)$. Since $B - p(T)$ is $\nu$-negligible (Ch. V, §6, No. 2, Cor. 3 of Prop. 2), we have only modified $\lambda_b$ at the points of a negligible set, consequently the family $b \mapsto \lambda_b$ is still $\nu$-adequate and has the property c); moreover, it now satisfies a) and b), which completes the proof.

Every $\nu$-adequate family $b \mapsto \lambda_b$ of positive measures on T, having the properties b) and c) of Th. 1, is said to be a disintegration of the measure $\mu$, relative to the $\mu$-proper mapping $p$.

For, there exists a bounded measure $\mu'$ on T equivalent to $\mu$ (Ch. V, §5, No. 6, Prop. 11); $p$ is then $\mu'$-proper.

Moreover, if $\nu' = r \cdot \nu$ is a second pseudo-image measure of $\mu$ under $p$, and if $b \mapsto \lambda'_b$ is a $\nu'$-adequate family of positive measures on T having the properties b) and c) with respect to $\nu'$, then $\lambda_b = r(b)\lambda'_b$ almost everywhere in B (for $\nu$ or $\nu'$).

There exists a continuous and finite numerical function $f$ defined on T, such that $f(t) > 0$ for every $t \in T$ and such that $\mu'' = f \cdot \mu$ is bounded (Ch. V, §5, No. 6, Prop. 11). Let $\nu'' = p(\mu'')$, which is equivalent to $\nu$, and write $\nu'' = g \cdot \nu$, with $g$ finite and locally $\nu$-integrable; one can suppose, moreover, that $g(b) > 0$ for all $b \in B$ (Ch. V, §5, No. 6, Prop. 10). Th. 1 of No. 1, applied to $\mu''$ and $\nu''$, shows that there exists a $\nu''$-adequate family $b \mapsto \lambda''_b$ ($b \in B$) of positive measures on T, such that:

1) $\| \lambda''_b \| = 1$ for $b \in p(T)$; 2) $\lambda''_b$ is concentrated on $\overline{p^{-1}}(b)$ for every $b \in B$; 3) $\mu'' = \int \lambda''_b \, d\nu''(b)$.

For every $b \in B$, let us define a positive measure $\lambda_b$ on T by the formula $\lambda_b = (1/f) \cdot (g(b)\lambda''_b)$. It is clear that the family $b \mapsto \lambda_b$ has the properties a) and b) of the statement. On the other hand, for every function $h \in \mathscr{K}(T)$, $h/f$ belongs to $\mathscr{K}(T)$, therefore

But since the function $b \mapsto \int (h(t)/f(t)) \, d\lambda''_b(t)$ is $\nu''$-integrable, the function $b \mapsto g(b) \int (h(t)/f(t)) \, d\lambda''_b(t)$ is $\nu$-integrable (Ch. V, §5, No. 3, Th. 1).

By the definition of $\lambda_b$, this function is $b \mapsto \int h(t) d\lambda_b(t)$, whence (loc. cit.) $\int h(t) d\mu(t) = \int d\nu(b) \int h(t) d\lambda_b(t)$, which proves that $\mu = \int \lambda_b d\nu(b)$.

To establish the second part of the statement, we remark that one can suppose that $r(b) > 0$ for all $b \in B$ (Ch. V, §5, No. 6, Prop. 10); set $\lambda'''_b = f \cdot \left( (r(b)/g(b)) \lambda'_b \right)$; one shows, as above, that for every function $h \in \mathscr{K}(T)$, the relation

implies

Therefore Th. 1 of No. 1, applied to $\mu''$ and $\nu''$, implies that for almost every $b \in B$, $\lambda'''_b = \lambda''_b$, whence $\lambda_b = r(b)\lambda'_b$.

Since T is countable at infinity, there exists an increasing sequence $(K_n)_{n \geq 1}$ of compact subsets of T such that T is the union of the $K_n$ and a $\mu$-negligible set N, and such that each of the relations $R_{K_n}$ is Hausdorff. Let $B_n$ be the quotient space $K_n / R_{K_n}$, which is compact, and let $B'_n$ be the compact space that is the topological sum of $B_n$ and a point $a_n$. Let $q_n$ be the canonical mapping of $K_n$ onto $B_n$; we extend $q_n$ to a mapping $p_n$ of T into $B'_n$ in the following manner: if $x \in T$ is equivalent mod R to an element $y \in K_n$, set $p_n(x) = q_n(y)$; in the contrary case, set $p_n(x) = a_n$.

Let $B'$ be the product space $\prod_{n=1}^{\infty} B'_n$, which is compact, and let $p'$ be the mapping $x \mapsto (p_n(x))$ of T into B'. Let us show that $p'$ is $\mu$-measurable: it suffices (Ch. IV, §5, No. 3, Th. 1) to prove that each of the mappings $p_n$ is measurable, and for this it suffices that the restriction of $p_n$ to each $K_m$ be measurable. Now, this is obvious if $m \leq n$; if, on the contrary, $m > n$, let $K_{nm}$ be the saturation of $K_n$ for $R_{K_m}$, which is a compact subset of $K_m$ (GT, I, §9, No. 4, Th. 2); since $p_n$ is constant on $K_m - K_{nm}$, it suffices to prove that the restriction of $p_n$ to $K_{nm}$ is continuous, which is obvious on account of the canonical isomorphism between $K_{nm}/R_{K_{nm}}$ and $K_n/R_{K_n}$ (GT, I, §9, No. 4, Cor. 4 of Th. 2).

Let A be the saturation of $\bigcup_n K_n$ for the relation R, and let $N' = T - A \subset N$. We shall see that the relation $p'(x) = p'(y)$ is equivalent to the relation «$R\{x, y\}$ or $(x, y) \in N' \times N'$ ». For, if $R\{x, y\}$ then $p_n(x) = p_n(y)$ for all $n$, therefore $p'(x) = p'(y)$; and if $x \in N'$, $y \in N'$ then $p_n(x) = p_n(y) = a_n$ for all $n$, therefore $p'(x) = p'(y)$. If on the other hand $x$ and $y$ are in A and are not equivalent mod R, then there exist an integer $n$, an element $x' \in K_n$ (resp. $y' \in K_n$) equivalent mod R to $x$ (resp. $y$) such that $x'$ is not equivalent to $y'$ mod $R_{K_n}$; therefore $p_n(x) \neq p_n(y)$, consequently $p'(x) \neq p'(y)$. Finally, if $x \in N'$ and $y \in A$, then $p_n(y) \in B_n$ for $n$ sufficiently large and $p_n(x) = a_n$ for all $n$, therefore $p'(x) \neq p'(y)$, which establishes our assertion.

Consider then the quotient set $B_0 = N'/R_{N'}$; let $q_0$ be the canonical mapping of $N'$ onto $B_0$, $s_0$ a section of $q_0$. Set $p_0(x) = s_0(q_0(x))$ for $x \in N'$ and extend $p_0$ to T by taking $p_0$ to be constant on A equal to an element of T. Then $p = (p', p_0)$ is a $\mu$-measurable mapping of T into the locally compact space $B = B' \times T$; it is immediate that if $x \in N'$, $y \in N'$, the relation $p_0(x) = p_0(y)$ implies $R\{x, y\}$; thus $p$ meets the requirements. Moreover, if T admits a countable base, then so do each of the quotient spaces $B_n$ (GT, IX, §2, No. 10, Prop. 17), therefore B' admits a countable base, hence so does B.

a) $R$ is $\mu$-measurable.

b) There exists a sequence of mappings $p_n : T \to F_n$ into Hausdorff topological spaces, such that each $p_n$ is $\mu$-measurable and such that the relation $R\{x, y\}$ is equivalent to «for all $n$, $p_n(x) = p_n(y)$».

c) There exists a sequence $(A_n)$ of $\mu$-measurable sets, saturated for R, such that for every $x \in T$ the class of x with respect to R is the intersection of those $A_n$ that contain x.

With notations as in the statement of b), set $p(x) = (p_n(x))$; the property b) means that the mapping $p$ of T into the product space $\prod F_n$ is measurable (Ch. IV, §5, No. 3, Th. 1) and that the relation $R\{x, y\}$ is equivalent to $p(x) = p(y)$; thus b) implies a).

Next let us show that c) implies b). Suppose c) verified; then the characteristic functions $\varphi_{A_n}$ are $\mu$-measurable, and the hypothesis c) means that the relation $R\{x, y\}$ is equivalent to «for every $n$, $\varphi_{A_n}(x) = \varphi_{A_n}(y)$».

Finally, let us show that a) implies c). By Prop. 2, there exist a locally compact space B with a countable base, and a $\mu$-measurable mapping $p$ of T into B, such that the relation $R\{x, y\}$ is equivalent to $p(x) = p(y)$. Let $(U_n)$ be a countable base for the topology of B. The sets $A_n = \overline{p^{-1}(U_n)}$ are $\mu$-measurable (Ch. IV, §5, No. 5, Prop. 7) and saturated for R; and if $x, y$ are points of T such that $p(x) \neq p(y)$, there exists an index $n$ such that $p(x) \in U_n$ and $p(y) \notin U_n$, which means that $x \in A_n$ and $y \notin A_n$.

We may clearly suppose that the measure $\mu$ is bounded and that $\mu(T) \leq 1$ (Ch. V, §5, No. 6, Prop. 11). We are going to define a sequence $(S_n)$ of Borel subsets (GT, IX, §6, No. 3) such that each equivalence class with respect to R intersects the union $S'$ of the $S_n$ in at most one point, that for every $n$ the saturation $T_n$ of the union of the $S_p$ with index $p \leq n$ is $\mu$-measurable, and that $\mu(T - T_n) \leq 1/2^n$. The saturation $T'$ of $S'$ will therefore be $\mu$-measurable and $N = T - T'$ will have measure zero. If $S''$ is any section of N for the relation $R_N$, $S = S' \cup S''$ will meet the requirements, since $S'$, being a Borel set, is $\mu$-measurable (Ch. IV, §5, No. 4, Cor. 3 of Th. 2), and $S''$ has measure zero.

By Prop. 2, $R\{x, y\}$ is equivalent to the relation $p(x) = p(y)$, where $p$ is a $\mu$-measurable mapping of T into a locally compact space F. Suppose the $S_k$ defined for $k \leq n$. Since $T - T_n$ is $\mu$-measurable and of measure $\leq 1/2^n$, there exists a compact subset K of $T - T_n$ such that $\mu(T - (T_n \cup K)) \leq 1/2^{n+1}$ and such that the restriction of $p$ to K is continuous. Since the induced relation $R_K$ is closed and K is metrizable, we know that there exists a Borel subset $S_{n+1}$ of K such that, in K, each point is equivalent (mod R) to one and only one point of $S_{n+1}$ (GT, IX, §6, No. 8, Th.4). Therefore $p(S_{n+1}) = p(K)$, which is a compact set in F; the saturation of $S_{n+1}$ for R is the inverse image $\overline{p^{-1}(p(K))}$, which is therefore μ-measurable (Ch. IV, §5, No. 5, Prop. 7); it is clear that this set contains K, therefore the union T_{n+1} of T_n and $\overline{p}^{-1}(p(K))$ is μ-measurable, saturated for R, and is such that $\mu(T - T_{n+1}) \leq 1/2^{n+1}$, which completes the proof.