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A diffeomorphic flow has the inverse-Jacobian density

Statement

Assume the Axiom of Countable Choice. Let d≥1, let U,V⊆Rd be nonempty open sets, and let T:U→V be a bijection such that T and T−1 are both C1. Let pU:U→[0,∞] be Lebesgue measurable with ∫UpU dλd=1, and define μU(A)=∫ApU dλd for Lebesgue-measurable A⊆U and μX(B)=μU(T−1(B)) for Borel B⊆V. Then

q(x)=pU(T−1(x)) ∣det⁡D(T−1)(x)∣

is Lebesgue measurable and satisfies μX(B)=∫Bq dλd for every Borel B⊆V. Thus q is a version of the density of the transformed measure.

Let K≥1, let U0,…,UK be nonempty open subsets of Rd, and let Tj:Uj−1→Uj be a C1 diffeomorphism for j=1,…,K. For T=TK∘⋯∘T1, u∈U0, and z0=u, zj=Tj(zj−1), the log-Jacobian terms add:

log⁡∣det⁡DT(u)∣=∑j=1Klog⁡∣det⁡DTj(zj−1)∣.

Equivalently, for zK=x∈UK and zj−1=Tj−1(zj),

log⁡∣det⁡DT−1(x)∣=∑j=1Klog⁡∣det⁡DTj−1(zj)∣.

Statement references

Facts & Assumptions

Given: Countable Choice, a flow T:U→V, and a Lebesgue probability density pU as in the Statement.

[A1]

The Axiom of Countable Choice, written ACω, asserts that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice (ACω)

[F1]

A normalizing flow is a finite composition of bijections between open equal-dimensional Euclidean domains whose maps and inverses are C1; a base density pU defines μU(A)=∫ApU dλd, and the transformed measure is μX(B)=μU(T−1(B)). A normalizing flow and its base density

[F2]

For a nonnegative measurable function f, the measure with density f relative to λd assigns A the value ∫Af dλd. The measure with density f relative to μ

[F3]

If S:U0→V0 is a C1 diffeomorphism and f:V0→[0,∞] is Lebesgue measurable, then under Countable Choice ∫V0f(y) dλd(y)=∫U0f(S(x))∣det⁡DS(x)∣ dλd(x). A C^1 diffeomorphism satisfies the change-of-variables formula for nonnegative Lebesgue measurable functions

[F4]

Under Countable Choice, Borel sets are Lebesgue measurable, and a C1 diffeomorphism maps every Lebesgue-measurable subset of its domain to a Lebesgue-measurable subset of its range. Assuming countable choice, every Borel subset of Rn is Lebesgue measurable, A C^1 diffeomorphism maps Lebesgue measurable sets to Lebesgue measurable sets

[F5]

For totally differentiable maps, D(g∘f)(a)=Dg(f(a))∘Df(a). The chain rule for total derivatives: D(g∘f)(a)=Dg(f(a))∘Df(a)

[F6]

For same-sized square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B). For same-sized finite square matrices over a commutative ring, det⁡(AB)=det⁡(A)det⁡(B)

[F7]

The Jacobian matrix is the matrix of the total derivative in standard bases, and its Jacobian determinant is the determinant of that matrix. The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case, The Jacobian determinant of a square-dimensional C1 map is the determinant of its Jacobian matrix

[F9]

Products of extended-real measurable functions are measurable whenever defined, including the nonnegative density and determinant factors here. Arithmetic and lattice operations preserve measurability whenever they are defined

Proof

technique · direct
1.1A1F1F4F7F8F9

For every a∈R, the superlevel set Ea={u∈U:pU(u)>a} is Lebesgue measurable, and {x∈V:pU(T−1(x))>a}=T(Ea) is Lebesgue measurable by [F4]. Thus pU∘T−1 is measurable. The entries of D(T−1) are continuous by the C1 hypothesis, and [F8] makes its determinant continuous. Thus [F9] shows that q is nonnegative measurable.

1.2givenF1F5F7

For z0=u and zj=Tj(zj−1), repeated use of [F5] gives DT(u)=DTK(zK−1)⋯DT1(z0). For x=zK, applying the chain rule to T−1=T1−1∘⋯∘TK−1 gives D(T−1)(x)=DT1−1(z1)⋯DTK−1(zK).

2.1A1F1F2F3F4F9step 1.1

Fix a Borel set B⊆V and put E=T−1(B). By [F4], E is Lebesgue measurable. The function h(u)=pU(u)1E(u) is nonnegative measurable by [F9]. Apply [F3] to S=T−1:V→U and h to obtain μX(B)=∫EpU(u) dλd(u)=∫VpU(T−1(x))1B(x)∣det⁡D(T−1)(x)∣ dλd(x)=∫Bq(x) dλd(x). Therefore q is a density version of μX.

3.1F5F6F8step 1.2algebra∎

By [F6], determinants of each derivative composition are the products of the stage determinants. Each factor is nonzero: differentiating Tj−1∘Tj=id⁡ by [F5] gives a product equal to the identity matrix, whose determinant is 1 by [F8]. Taking absolute values and then real logarithms yields both displayed additive formulas.

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