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A volume-preserving diffeomorphism preserves its induced Borel measure
Statement
Assume countable choice. Let be a Hausdorff second-countable smooth -manifold and let be a positive smooth density, with Borel coefficients as in Pointwise Borel nonnegative densities. Let be the intrinsic chart-gluing Borel measure of Intrinsic density measure and its chart restriction. If is a diffeomorphism satisfying , then for every Borel set . The total measure may be infinite.
Facts & Assumptions
Given: Countable choice, a positive smooth density on a Hausdorff second-countable smooth manifold, and a diffeomorphism preserving that density.
Adopt as stated in The Axiom of Countable Choice ().
Intrinsic density measure and its chart restriction: under , the positive density induces an intrinsic Borel measure, and in every chart the measure of a Borel subset is the Lebesgue integral of the chart coefficient of .
A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions: under , a diffeomorphism between open Euclidean sets transforms nonnegative Lebesgue integrals by its absolute Jacobian.
Proof
Given: [A1], [F1], and [F2].
Choose a countable chart cover , and replace it by the disjoint Borel partition . Choose a countable chart cover for the target and form . These are Borel partitions of . For fixed , the coordinate map is a diffeomorphism from the open set onto the open set . The pieces are Borel subsets of its coordinate domain.
By [F1] and countable additivity, is the sum over of the chart integrals of on . The equality says in these charts that . Applying [F2] to the indicator of times the nonnegative coefficient makes each chart integral equal the integral of on . The latter sets, as varies for a fixed , partition because is bijective; summing over therefore gives . All sums have nonnegative terms, so the argument also holds when either total is infinite.
Remarks
The coordinate comparison used in step 2.1 is the commuting square below. The proof uses its coordinate-map Jacobian through [F2]; the diagram is only an aid to locating that map.
S_{ij}\subseteq U_i \arrow[r,"\phi"] \arrow[d,"x_i"'] &
\phi(S_{ij})\subseteq V_j \arrow[d,"y_j"] \\
x_i(S_{ij}) \arrow[r,"y_j\circ\phi\circ x_i^{-1}"'] &
y_j(\phi(S_{ij}))