TheoremStatement: AI-adaptedProof: AI-adaptedprecheck 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

1.1F1F6

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 $K^{'} : = K + 1 > 0$ .

2.1step 1.1F1F5F6

$T$ 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 $z \in T [U]$ .

3.1step 2.1F4

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.

3.2step 2.1L2L3L4F1F5F6

$T [(0, 1]^{n}]$ 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.

4.1step 3.1L1L2F2F3

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.

5.1step 3.2step 4.1L1L3F3∎

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 $c (i d) = λ_{n} ((0, 1]^{n}) = 1$ .

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

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

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