Proposition 2

*Let T be a locally compact space countable at infinity, $\mu$ a positive measure on T.

1) For every $\mu$-measurable equivalence relation R in T, there exist a locally compact space B and a $\mu$-measurable mapping p of T into B such that R is equivalent to the relation $p(x) = p(y)$.

2) If, moreover, T admits a countable base, one can suppose that B admits a countable base.*

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.