Theorem 3
Tag 09CE, permanent. Cite it as [Bourbaki, Tag 09CE].
Integration, chapter VI, § 3. Disintegration of measures
— Let T be a locally compact space with a countable base, a positive measure on T, and R a -measurable equivalence relation in T. Then, there exists a -measurable subset S of T that intersects each class with respect to R at one and only one point (a 'measurable section' for R).
We may clearly suppose that the measure is bounded and that (Ch. V, §5, No. 6, Prop. 11). We are going to define a sequence of Borel subsets (GT, IX, §6, No. 3) such that each equivalence class with respect to R intersects the union of the in at most one point, that for every the saturation of the union of the with index is -measurable, and that . The saturation of will therefore be -measurable and will have measure zero. If is any section of N for the relation , will meet the requirements, since , being a Borel set, is -measurable (Ch. IV, §5, No. 4, Cor. 3 of Th. 2), and has measure zero.
By Prop. 2, is equivalent to the relation , where is a -measurable mapping of T into a locally compact space F. Suppose the defined for . Since is -measurable and of measure , there exists a compact subset K of such that and such that the restriction of to K is continuous. Since the induced relation is closed and K is metrizable, we know that there exists a Borel subset of K such that, in K, each point is equivalent (mod R) to one and only one point of (GT, IX, §6, No. 8, Th.4). Therefore , which is a compact set in F; the saturation of for R is the inverse image , 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 is μ-measurable, saturated for R, and is such that , which completes the proof.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.