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Lebesgue measure on $R^{n}$ is invariant under every orthogonal linear map

Statement

Let $n \geq 1$ , assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), and let $T$ be an orthogonal operator on $R^{n}$ with the Euclidean inner product (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ). Then $T [E]$ is Lebesgue measurable for every Lebesgue measurable $E$ and

$λ_{n} (T [E]) = λ_{n} (E) .$

Both the orientation-preserving operators, of determinant $1$ , and those of determinant $- 1$ are covered, since only the absolute value of the determinant enters; nothing is asserted here about which matrices occur in either class.

Facts & Assumptions

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

[L1]

Assuming countable choice, an invertible linear $T$ with matrix $A$ sends Lebesgue measurable sets to Lebesgue measurable sets with $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E)$ (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).

[F1]

An invertible linear isometry from a real finite-dimensional inner product space to itself is an orthogonal operator (Linear isometries, and orthogonal or unitary operators on finite-dimensional inner product spaces).

[F2]

Every orthogonal or unitary operator $T$ satisfies $∣ \det T ∣ = 1$ ; over $R$ , this says $\det T \in {- 1, 1}$ (Orthogonal and unitary operators form groups, and their determinants have modulus one, For $n \geq 1$ , the determinant over a commutative ring by the Leibniz formula, and $∣ \det A ∣$ for a real matrix).

[F3]

For every linear $L : R^{m} \to R^{n}$ there is a unique matrix $A$ such that $(L h)_{i} = \sum_{j < m} a_{i j} h_{j}$ (Every Euclidean linear map has a unique matrix and satisfies $∥ L h ∥_{2} \leq K ∥ h ∥_{2}$ for some $K \geq 0$ ).

Proof

technique · direct

1.1F1F2F3

An orthogonal operator is by definition an invertible linear map of $R^{n}$ to itself, and its matrix $A$ satisfies $∣ \det A ∣ = 1$ , so in particular $\det A \neq 0$ .

2.1step 1.1L1F2∎

The linear change of variables therefore applies in its invertible clause and gives $λ_{n} (T [E]) = ∣ \det A ∣ λ_{n} (E) = λ_{n} (E)$ for every Lebesgue measurable $E$ , with $T [E]$ measurable.

Depends on

Used by

Polar integration may discard the cut locus Corollary
The Solovay model has no Banach–Tarski decomposition Corollary
An orientation-reversing homeomorphism need not be quasiconformal Counterexample
Matching C¹ pieces across a hyperplane have no jump derivative Example
The volume contradiction for an alleged Banach–Tarski decomposition Example
All sets of reals in Solovay L(R) have LM, BP, and PSP Theorem

Dependency tree · two levels

71 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 2.31 (standard reference, not scraped)
E. A. Carlen, Notes on Lebesgue Measure on $\mathbb{R}^n$ and $S^{n-1}$ (Rutgers Math 501), Theorem 3.2 (standard reference, not scraped)

Lebesgue measure on RnRn is invariant under every orthogonal linear map