TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-26

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An invertible linear map of $R^{n}$ scales the Lebesgue measure of every Borel set by a positive constant depending only on the map

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 an invertible linear map (Linear map between vector spaces over the same field). Then:

$T [E]$ is a Borel set for every Borel set $E$ , and $T$ carries open sets to open sets;
there is a strictly positive real $c (T)$ , namely $c (T) = λ_{n} (T [(0, 1]^{n}])$ , with $λ_{n} (T [E]) = c (T) λ_{n} (E) for every E \in B (R^{n});$
$c (S \circ T) = c (S) c (T)$ for invertible linear $S$ and $T$ , and $c (i d) = 1$ .

Facts & Assumptions

Given: A natural number $n \geq 1$ , the Axiom of Countable Choice, and an invertible linear map $T$ of $R^{n}$ .

[L1]

Assuming countable choice, a measure $μ$ on $B (R^{n})$ with $μ (E + h) = μ (E)$ for every Borel $E$ and every $h$ , and with $μ ((0, 1]^{n}) = 1$ , equals $λ_{n}$ on $B (R^{n})$ ; in particular the theorem notes that the restriction of $λ_{n}$ to $B (R^{n})$ satisfies these hypotheses (A translation-invariant measure on the Borel sets of $R^{n}$ giving the unit cube measure one is the restriction of Lebesgue measure).

[L2]

Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable (Assuming countable choice, every Borel subset of $R^{n}$ is Lebesgue measurable), and $λ_{n}$ is a 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).

[L3]

Every set $R$ with $R^{\circ} \subseteq R \subseteq \overline{R}$ is Lebesgue measurable with $λ_{n} (R) = \prod_{i < n} (b_{i} - a_{i})$ (A box in $R^{n}$ with parameters $a_{i} \leq b_{i}$ is Lebesgue measurable of measure $\prod_{i < n} (b_{i} - a_{i})$ , whichever of its faces are included).

[L4]

Every bounded subset of $R^{n}$ has finite outer measure (Lebesgue measure is sigma-finite, and every metrically bounded subset of $R^{n}$ has finite outer measure).

[F1]

For every linear $L : R^{m} \to R^{n}$ there is a unique matrix $A$ such that $(L h)_{i} = \sum_{j < m} a_{i j} h_{j}$ , 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 $K \geq 0$ , Linear map between vector spaces over the same field).

[F2]

The translate of $E \subseteq R^{n}$ by $a$ is $E + a : = {x + a : x \in E}$ (Translation of a subset of $R^{n}$ ).

[F3]

A measure on $(X, A)$ is a function $μ : A \to [0, + \infty]$ with $μ (\emptyset) = 0$ that is countably additive on pairwise disjoint sequences (Measures on sigma-algebras), and a scalar multiple $c μ$ is again a measure (Nonnegative scalar multiples and countable weighted sums of measures are measures, Nonnegative scalar multiples and countable weighted sums of measures).

[F4]

The Borel sigma-algebra is the sigma-algebra generated by the open sets (The Borel sigma-algebra of a topological space), $σ_{X} (E)$ is the smallest sigma-algebra containing $E$ (Nonempty intersections of sigma-algebras are sigma-algebras, so the generated sigma-algebra exists and is minimal), and a sigma-algebra is closed under complements and countable unions (Sigma-algebras).

[F5]

A subset $U$ is open in $(X, d)$ when every $x \in U$ has a ball $B (x, r) \subseteq U$ (The metric topology: a set is open when every one of its points has a ball around it inside the set; closed means open complement, Open ball, closed ball and sphere in a metric space), and $A$ is bounded when it is empty or lies in some ball (Bounded subset, diameter, distance from a point to a set, and distance between two sets in a metric space).

[F6]

For every $x \in R^{n}$ , $∥ x ∥_{\infty} \leq ∥ x ∥_{2} \leq ∥ x ∥_{1} \leq n ∥ x ∥_{\infty}$ , and $∥ x - y ∥_{2} = d_{2} (x, y)$ , $∥ x - y ∥_{\infty} = d_{\infty} (x, y)$ (The finite and reverse triangle inequalities for a norm; and for $n \geq 1$ every norm $N$ on $R^{n}$ satisfies $N (x) \leq C ∥ x ∥_{1}$ and is Lipschitz, hence continuous, for $d_{2}$ , claim 3; 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, claim 3; 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).

Proof

technique · direct

1.1

The inverse $T^{- 1}$ is linear, so there are reals $K \geq 0$ and $K_{T} \geq 0$ with $∥ T^{- 1} y ∥_{2} \leq K ∥ y ∥_{2}$ and $∥ T x ∥_{2} \leq K_{T} ∥ x ∥_{2}$ for all $x, y$ ; put $K^{'} : = K + 1 > 0$ .

F1F6

2.1

$T$ carries open sets to open sets: if $U$ is open, $y = T x \in T [U]$ and $B (x, r) \subseteq U$ , then $d_{2} (y, z) < r / K^{'}$ gives $d_{2} (x, T^{- 1} z) = ∥ T^{- 1} (y - z) ∥_{2} \leq K d_{2} (y, z) < r$ , so $T^{- 1} z \in U$ and $z \in T [U]$ .

step 1.1F1F5F6

3.1

The family of $E \subseteq R^{n}$ with $T [E]$ Borel is a sigma-algebra, because $T$ is a bijection and so $T [\cdot]$ commutes with complements and with countable unions, and it contains every open set by step 2.1; minimality of $B (R^{n})$ over the open sets gives claim 1.

step 2.1F4

3.2

$T [(0, 1]^{n}]$ is bounded, being contained in the ball about the origin of radius $K_{T} \sqrt{n} + 1$ , so it has finite measure; and it contains $T [V]$ for the nonempty open box $V : = {x : 0 < x_{i} < 1 (i < n)}$ , which is open and nonempty by step 2.1, hence contains a ball $B (y, r)$ and with it the open box ${x : ∣ x_{i} - y_{i} ∣ < r / n}$ , whose measure $(2 r / n)^{n}$ is a strictly positive real. So $c (T) : = λ_{n} (T [(0, 1]^{n}])$ is a strictly positive real.

step 2.1L2L3L4F1F5F6

4.1

The assignment $ν (E) : = λ_{n} (T [E])$ is well defined on $B (R^{n})$ by claim 1, and it is a measure: $ν (\emptyset) = 0$ , and $T$ being injective carries a pairwise disjoint sequence to a pairwise disjoint sequence with $T [⋃_{k} E_{k}] = ⋃_{k} T [E_{k}]$ , so countable additivity of $λ_{n}$ transfers. It is translation invariant, since $T [E + h] = T [E] + T (h)$ by linearity and $λ_{n}$ is translation invariant.

step 3.1L1L2F2F3

5.1

By step 3.2 the scalar multiple $c (T)^{- 1} ν$ is a measure on $B (R^{n})$ , it is translation invariant, and it gives the unit cube the value $1$ , so the uniqueness theorem identifies it with $λ_{n}$ on the Borel sets; that is claim 2. Claim 3 follows by evaluating at the unit cube: $c (S \circ T) = λ_{n} (S [T [(0, 1]^{n}]]) = c (S) λ_{n} (T [(0, 1]^{n}]) = c (S) c (T)$ , and the identity map gives $c (i d) = λ_{n} ((0, 1]^{n}) = 1$ .

step 3.2step 4.1L1L3F3∎

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Sources

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)
T. Tao, An Introduction to Measure Theory (GSM 126), Exercise 1.2.21 (standard reference, not scraped)

An invertible linear map of RnRn scales the Lebesgue measure of every Borel set by a positive constant depending only on the map