A full-dimensional flow cannot represent a line-supported law exactly

Given: Countable Choice, the interval probability measure $\nu$, the continuous embedding $i(t)=(t,0)$, and its image measure $\mu$.

The Axiom of Countable Choice, written ${\mathrm{A}\mathrm{C}}_{\omega }$, asserts that every sequence of nonempty sets has a choice function. The Axiom of Countable Choice (${\mathrm{A}\mathrm{C}}_{\omega }$)

A nonnegative measurable density $f$ defines a measure by $A\mapsto {\int }_{A}fd\lambda$. The measure with density $f$ relative to $\mu$

A flow from a Lebesgue density through a global $C^1$ diffeomorphism has density $p_U\circ T^{-1}\mathrm{\mid }\det D(T^{-1})\mathrm{\mid }$. A diffeomorphic flow has the inverse-Jacobian density

A continuous map has Borel preimages of Borel sets. A continuous map has Borel preimages of Borel sets

Every closed box $[a,b]$ in ${\mathbb{R}}^n$ is Borel and Lebesgue measurable; its measure is ${\prod }_{i<n}(b_i-a_i)$, so $[0,1]$ has one-dimensional Lebesgue measure $1$. A box in ${\mathbb{R}}^n$ with parameters $a_i\le b_i$ is Lebesgue measurable of measure ${\prod }_{i<n}(b_i-a_i)$, whichever of its faces are included

A rectangle with a degenerate side is Lebesgue null under Countable Choice. A box with a degenerate side is Lebesgue null, and so is every coordinate hyperplane in ${\mathbb{R}}^n$

A nonnegative measurable density integrates to zero over a measurable null set. A nonnegative integral over a null set vanishes

A normalizing flow is a finite composition of global $C^1$ diffeomorphisms between open equal-dimensional Euclidean domains; its output measure is obtained by taking preimages under the flow. A normalizing flow and its base density