TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26

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A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not

Statement

Let $n \geq 1$ , assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )) and let $T : R^{n} \to R^{n}$ be linear with matrix $A$ (Linear map between vector spaces over the same field, Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

Invertible case. If $\det A \neq 0$ , then $T [E]$ is Lebesgue measurable for every Lebesgue measurable $E$ and $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E),$ both sides possibly $+ \infty$ ; the product is defined in $\overline{R}$ because $∣ \det A ∣ > 0$ .
Singular case. If $\det A = 0$ , then $T [E]$ is Lebesgue measurable with $λ_{n} (T [E]) = 0$ for every $E \subseteq R^{n}$ .

The singular clause is stated as nullity and not as a product. When $\det A = 0$ and $λ_{n} (E) = + \infty$ the expression $∣ \det A ∣ λ_{n} (E)$ is $0 \cdot (+ \infty)$ , which The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined leaves undefined; writing the conclusion as $λ_{n} (T [E]) = 0$ says the same thing wherever the product is defined and remains a statement where it is not.

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, a linear map $T$ of $R^{n}$ with matrix $A$ , and a set $E \subseteq R^{n}$ .

[L1]

An invertible linear map carries Borel sets to Borel sets, and there is a strictly positive real $c (T) = λ_{n} (T [(0, 1]^{n}])$ with $λ_{n} (T [E]) = c (T) λ_{n} (E)$ for every Borel $E$ , with $c (S \circ T) = c (S) c (T)$ and $c (i d) = 1$ (An invertible linear map of $R^{n}$ scales the Lebesgue measure of every Borel set by a positive constant depending only on the map).

[L2]

$λ_{n} (D_{p} (c) [(0, 1]^{n}]) = ∣ c ∣ = ∣ \det D_{p} (c) ∣$ and $λ_{n} (E_{p q} [(0, 1]^{n}]) = 1 = ∣ \det E_{p q} ∣$ (A coordinate scaling and a coordinate transposition send the unit cube to a set of measure equal to the absolute value of the determinant).

[L3]

For $n \geq 2$ a shear satisfies $λ_{n} (T_{i j} (t) [(0, 1]^{n}]) = 1 = ∣ \det T_{i j} (t) ∣$ (A shear sends the unit cube to a set of Lebesgue measure one).

[L4]

Every proper linear subspace $W ⊊ R^{n}$ is Lebesgue measurable with $λ_{n} (W) = 0$ (Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null).

[L5]

A Lipschitz self-map of $R^{n}$ carries a set of Lebesgue outer measure zero to a Lebesgue measurable set of measure zero (A Lipschitz self-map of $R^{n}$ carries Lebesgue null sets to Lebesgue null sets, Lipschitz map, $α$ -Hölder map for rational $0 < α \leq 1$ , and contraction).

[L6]

$E$ is Lebesgue measurable if and only if $E = H \cup W$ for an $F_{σ}$ set $H$ and a set $W$ with $λ_{n}^{*} (W) = 0$ (Assuming countable choice, four equivalent descriptions of a Lebesgue measurable subset of $R^{n}$ , condition 4; $G_{δ}$ and $F_{σ}$ subsets of a topological space, agreeing with the real-line notion).

[L7]

Assuming countable choice, $λ_{n}$ is a complete measure on $L (R^{n})$ (Assuming countable choice, $L (R^{n})$ is a sigma-algebra containing every elementary set and $λ_{n}$ is a complete measure extending elementary volume) and every Borel set is Lebesgue measurable (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable, Measures on sigma-algebras).

[F1]

Every invertible matrix $A \in M_{n} (R)$ is a finite product of elementary matrices, the identity being the empty product (Every invertible finite square real matrix is a finite product of elementary matrices), and every elementary matrix is invertible (Every elementary matrix is invertible, with inverse given by the reverse elementary operation, Elementary matrices obtained by applying one elementary row operation to an identity matrix).

[F2]

For $n \geq 1$ and $A, B \in M_{n} (R)$ over a commutative ring, $\det (A B) = \det (A) \det (B)$ (For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ , For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix, Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes), and a triangular matrix has determinant the product of its diagonal entries (The determinant of a triangular matrix is the product of its diagonal entries).

[F3]

For every $n \geq 1$ and every real matrix $A \in M_{n} (R)$ , $A$ is invertible if and only if $\det (A) \neq 0$ (A finite square real matrix is invertible if and only if its determinant is nonzero).

[F4]

For every linear $L : R^{m} \to R^{n}$ there is a unique matrix $A$ with $(L h)_{i} = \sum_{j < m} a_{i j} h_{j}$ , and there is $K \geq 0$ with $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for every $h$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ , Each $∥ \cdot ∥_{p}$ is a norm on $R^{n}$ , and the induced metrics are exactly $d_{1}$ , $d_{2}$ and $d_{\infty}$ of the published metric-spaces page, The $p$ -norms $∥ x ∥_{p}$ for rational $p \geq 1$ , and $∥ x ∥_{\infty}$ , $R^{n}$ as the set of functions $n \to R$ , and $d_{1}$ , $d_{2}$ , $d_{\infty}$ are metrics on it).

[F5]

For a linear map $T : V \to W$ , $im T : = {T (v) : v \in V}$ (Kernel and image of a linear map), and it is a linear subspace (The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial).

[F6]

Every product with one factor $0$ and the other $\pm \infty$ is left undefined in $\overline{R}$ (The extended real line $\overline{R} = R \cup {- \infty, + \infty}$ , its order, and the arithmetic that is left undefined).

[F7]

Let $S \subseteq N$ ; if $0 \in S$ and $σ (n) \in S$ whenever $n \in S$ , then $S = N$ (The principle of mathematical induction).

Proof

technique · direct

1.1

Suppose $\det A = 0$ . Then $im T$ is a proper linear subspace of $R^{n}$ : were $T$ surjective, each standard vector $e_{i}$ would be $T (v_{i})$ for some $v_{i}$ , finitely many instantiations, and the matrix $B$ with $b_{j i} : = (v_{i})_{j}$ would satisfy $(A B)_{k i} = \sum_{j} a_{k j} (v_{i})_{j} = (T v_{i})_{k} = (e_{i})_{k}$ , so $A B = I_{n}$ and $\det A \det B = \det I_{n} = 1$ , contradicting $\det A = 0$ .

F2F3F4F5

1.2

$T$ is Lipschitz, since $∥ T x - T y ∥_{2} = ∥ T (x - y) ∥_{2} \leq K ∥ x - y ∥_{2}$ for a real $K \geq 0$ .

1.3

Every elementary matrix $M$ of $M_{n} (R)$ satisfies $c (M) = ∣ \det M ∣$ : at $n = 1$ the only elementary matrices are the scalings $D_{0} (c)$ , and for $n \geq 2$ the three types are the scalings, the transpositions and the shears, whose unit-cube images have the measures $∣ c ∣$ , $1$ and $1$ , matching $∣ \det ∣$ in each case.

L1L2L3F1

2.1

Suppose $\det A \neq 0$ , so $A$ is invertible and factors as a finite product $M_{1} \dots M_{r}$ of elementary matrices, each invertible. Multiplicativity of $c$ and of the determinant then give $c (T) = \prod_{s} c (M_{s}) = \prod_{s} ∣ \det M_{s} ∣ = ∣ \det A ∣$ by induction on $r$ , the empty product giving $c (i d) = 1 = ∣ \det I_{n} ∣$ .

step 1.1step 1.3L1F1F2F3F7

2.2

For $\det A = 0$ , step 1.1 makes $im T$ a proper linear subspace, hence Lebesgue null; every $T [E]$ is a subset of it, so completeness makes $T [E]$ Lebesgue measurable with $λ_{n} (T [E]) = 0$ , which is claim 2; stating it as a product would require the undefined $0 \cdot (+ \infty)$ when $λ_{n} (E) = + \infty$ .

step 1.1L4L7F6

3.1

For $\det A \neq 0$ and $E$ Lebesgue measurable, write $E = H \cup W$ with $H$ an $F_{σ}$ set, hence Borel, and $λ_{n}^{*} (W) = 0$ ; then $T [E] = T [H] \cup T [W]$ , where $T [H]$ is Borel and $T [W]$ is Lebesgue measurable of measure $0$ by step 1.2 and the Lipschitz lemma, so $T [E]$ is measurable. Since $H \subseteq E$ and $E ∖ H \subseteq W$ , and $T [H] \subseteq T [E]$ with $T [E] ∖ T [H] \subseteq T [W]$ , both pairs differ by null sets, so $λ_{n} (E) = λ_{n} (H)$ and $λ_{n} (T [E]) = λ_{n} (T [H]) = c (T) λ_{n} (H) = ∣ \det A ∣ λ_{n} (E)$ , which is claim 1; claim 2 is step 2.2.

step 1.2step 2.1step 2.2L1L5L6L7∎

Depends on

Used by

Dependency tree · two levels

156 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

John K. Hunter, Measure Theory (UC Davis lecture notes), Theorem 2.33 (standard reference, not scraped)
T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.2.21 (standard reference, not scraped)
E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Theorem 3.1 (standard reference, not scraped)

A linear map TT of RnRn sends Lebesgue measurable sets to Lebesgue measurable sets, with λn(T[E])=∣det⁡T∣ λn(E)λn​(T[E])=∣detT∣λn​(E) when TT is invertible and T[E]T[E] Lebesgue null when it is not