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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

1.1

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 $\det A \neq 0$ .

F1F2F3

2.1

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.

step 1.1L1F2∎

Depends on

Used by

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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