How the Lebesgue change-of-variables formula relates to the published formula for Jordan content

Two determinant formulas are now in force, for two different set functions, and this remark says how they meet.

The published one is about Jordan content. A linear endomorphism of ${\mathbb{R}}^n$ sends bounded Jordan sets to bounded Jordan sets and scales their content by the absolute determinant states that a linear endomorphism $T$ of ${\mathbb{R}}^n$ with standard matrix $A$ sends every bounded Jordan set $E$ to a bounded Jordan set with $\mathrm{cont}(T(E))=\mathrm{\mid }\det A\mathrm{\mid }\mathrm{cont}(E)$, and that a singular linear image has content zero (Jordan inner and outer content and Jordan measurable bounded sets in ${\mathbb{R}}^m$).

This page's is about Lebesgue measure. Assuming the Axiom of Countable Choice, A linear map $T$ of ${\mathbb{R}}^n$ sends Lebesgue measurable sets to Lebesgue measurable sets, with ${\lambda }_n(T[E])=\mathrm{\mid }\det T\mathrm{\mid }{\lambda }_n(E)$ when $T$ is invertible and $T[E]$ Lebesgue null when it is not states the same identity with $\mathrm{cont}$ replaced by ${\lambda }_n$, for every Lebesgue measurable $E$, bounded or not, and with the singular case stated as nullity of $T[E]$ rather than as a product.

Where the two agree, and why that is not an accident. On a bounded Jordan set the two set functions take the same value, by Lebesgue outer measure is at most Jordan outer content, and a bounded Jordan measurable set is Lebesgue measurable with Lebesgue measure equal to its Jordan content, so on that class the two formulas are the same equation read twice. Neither implies the other: the published formula says nothing about a Lebesgue measurable set that is not Jordan measurable, and this page's formula says nothing about Jordan measurability of an image, which the published one asserts.

What the extension costs, and where it is spent. Passing from bounded Jordan sets to arbitrary Lebesgue measurable sets is not a matter of taking limits: the Lebesgue proof runs through the uniqueness of a normalised translation-invariant Borel measure, the factorisation of an invertible matrix into elementary matrices, and the fact that a Lipschitz image of a null set is null. The last of these is what carries the argument across the gap between Borel sets and the larger Lebesgue class, and it is why the change of variables holds on all of $\mathcal{L}({\mathbb{R}}^n)$ and not merely on the Borel sets.