Proof of Linear Change of Variables for Lebesgue Measure on ℝⁿ

Throughout, $\mathbf{u}=(1,\dots,1)\in\mathbb{R}^{n}$ and $\mathcal{M}^{+}=\mathcal{M}^{+}(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$; $\det A\neq0$ (A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (i)); $\lambda_{n}(R(0,\mathbf{u}))=1$ (Lebesgue Measure on $\mathbb{R}^n$ (i) and Existence and Uniqueness of Lebesgue Measure on ℝⁿ (iii)); and $B+a\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(B+a)=\lambda_{n}(B)$ for $B\in\mathcal{B}(\mathbb{R}^{n})$ and $a\in\mathbb{R}^{n}$ (Translation and Reflection Invariance of Lebesgue Measure on ℝⁿ (i)). Matrix-vector products obey the usual laws, among them $M(x+y)=Mx+My$, $M(sx)=s(Mx)$ for $s\in\mathbb{R}$, $(MN)x=M(Nx)$ and $I_{n}x=x$ (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (v)), and hence $M(-x)=-Mx$, taking $s=-1$; and a product $MN$ of invertible matrices is invertible with $(MN)^{-1}=N^{-1}M^{-1}$ (Rules for Matrix Inverses (iv)).

Step 1: affine bijections. Let $M\in\mathcal{M}_{n}(\mathbb{R})$ be invertible, $t\in\mathbb{R}^{n}$, and $S(x)=Mx+t$. Then $S$ is measurable: it is the sum of the function $x\mapsto Mx$ and the constant $t$ (Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (ii) and Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (i)). It is a bijection with inverse $S'(y)=M^{-1}y-M^{-1}t$, by Basic Properties of Functions: Equality, Composition, Identity, Inverse and Restriction (vii), since

$$M^{-1}(Mx+t)-M^{-1}t=x,\qquad M\bigl(M^{-1}y-M^{-1}t\bigr)+t=y .$$

As $S'$ has the same form, with the invertible matrix $M^{-1}$ (Rules for Matrix Inverses (iii)) and the vector $-M^{-1}t$, it is measurable too, and for $B\in\mathcal{B}(\mathbb{R}^{n})$

$$S[B]=(S')^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n}),\qquad S^{-1}(B)=S'[B],$$

by Images and Preimages under a Bijection and Its Inverse (i) applied to $S$ and to $S'$. In particular $T$ is measurable, which is the first assertion of (ii), and $T[B]\in\mathcal{B}(\mathbb{R}^{n})$ for every Borel set $B$.

Step 2: the factor of a matrix. Let $M\in\mathcal{M}_{n}(\mathbb{R})$ be invertible, and define $\mu_{M}(B)=\lambda_{n}(M[B])$ for $B\in\mathcal{B}(\mathbb{R}^{n})$. By Step 1 with $t=0$, $M[B]=L^{-1}(B)$ for the measurable function $L(y)=M^{-1}y$, so $\mu_{M}=\lambda_{n}\circ L^{-1}$ is a measure (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i)). It is translation invariant: $M[B+a]=M[B]+Ma$, so

$$\mu_{M}(B+a)=\lambda_{n}(M[B]+Ma)=\lambda_{n}(M[B])=\mu_{M}(B).$$

It is finite on the open unit cube: for $x\in R(0,\mathbf{u})$ we have $\lVert x\rVert^{2}\le n$, hence $\lVert Mx\rVert^{2}\le sn$ with $s=\sum_{i,j}M_{ij}^{2}$ (Algebraic Rules for Matrix Products, Matrix-Vector Products and Transposes (ix)) and so $\lVert Mx\rVert\le1+sn$; thus $M[R(0,\mathbf{u})]$ is bounded and has finite measure (Lebesgue Measure on ℝⁿ: σ-Finiteness, Bounded Sets, Open Sets and Coordinate Hyperplanes (ii)). By Translation-Invariant Measures on ℝⁿ Are Multiples of Lebesgue Measure (i), with the real number $c(M)=\lambda_{n}(M[R(0,\mathbf{u})])\ge0$,

$$\lambda_{n}(M[B])=c(M)\,\lambda_{n}(B)\qquad(B\in\mathcal{B}(\mathbb{R}^{n})).\tag{1}$$

Step 3: the factor is multiplicative. Let $M,N\in\mathcal{M}_{n}(\mathbb{R})$ be invertible. Then $MN$ is invertible, with inverse $N^{-1}M^{-1}$, and $(MN)[B]=M[N[B]]$ because $(MN)x=M(Nx)$. Applying (1) three times,

$$c(MN)\,\lambda_{n}(B)=\lambda_{n}\bigl(M[N[B]]\bigr)=c(M)\,\lambda_{n}(N[B])=c(M)\,c(N)\,\lambda_{n}(B),$$

and $B=R(0,\mathbf{u})$ gives

$$c(MN)=c(M)\,c(N).\tag{2}$$

Step 4: diagonal and orthogonal factors. Let $d\in\mathbb{R}^{n}$ have positive coordinates, let $d'\in\mathbb{R}^{n}$ have coordinates $d'_{i}=1/d_{i}$, and let $D=\operatorname{diag}(d)$. Since $\operatorname{diag}(d)\operatorname{diag}(d')=\operatorname{diag}(\mathbf{u})=I_{n}$ (Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices (ii)), $D$ is invertible with inverse $\operatorname{diag}(d')$ (Rules for Matrix Inverses (ii)). By Scaling Lebesgue Measure on ℝⁿ by a Diagonal Matrix (i), $\lambda_{n}(D[B])=|d_{1}\cdots d_{n}|\,\lambda_{n}(B)=d_{1}\cdots d_{n}\,\lambda_{n}(B)$, the absolute value being redundant because every $d_{i}>0$; with $B=R(0,\mathbf{u})$ this gives $c(D)=d_{1}\cdots d_{n}$.

Let $U\in\mathrm{O}(n)$. Then $U$ is invertible, $U^{\top}\in\mathrm{O}(n)$ and $UU^{\top}=I_{n}$, and $\lVert Ux\rVert=\lVert x\rVert$ for every $x$ (Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion (i) and Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion (iii)). Let $K=\{x\in\mathbb{R}^{n}:\lVert x\rVert\le1\}$, a Borel set because $x\mapsto\lVert x\rVert$ is measurable (Sums, Products, Transposes, Dot Products and Norms of Measurable Vector- and Matrix-Valued Maps (iii)). The isometry property gives $U[K]\subseteq K$, and every $y\in K$ is $U(U^{\top}y)$ with $\lVert U^{\top}y\rVert=\lVert y\rVert\le1$, so

$$U[K]=K .$$

The set $K$ is bounded, so $\lambda_{n}(K)<\infty$ (Lebesgue Measure on ℝⁿ: σ-Finiteness, Bounded Sets, Open Sets and Coordinate Hyperplanes (ii)); and it contains the open box $R(-n^{-1}\mathbf{u},n^{-1}\mathbf{u})$, on which $\lVert x\rVert^{2}<n\cdot n^{-2}\le1$, so $\lambda_{n}(K)\ge(2/n)^{n}>0$. Now (1) gives $\lambda_{n}(K)=\lambda_{n}(U[K])=c(U)\,\lambda_{n}(K)$, and dividing by $\lambda_{n}(K)$,

$$c(U)=1 .$$

Step 5: the factor of $A$. By Every Invertible Real Matrix Is an Orthogonal Matrix Times a Positive Diagonal Matrix Times an Orthogonal Matrix (i), there are $U,V\in\mathrm{O}(n)$ and $d\in\mathbb{R}^{n}$ with positive coordinates such that $A=UDV^{\top}$ with $D=\operatorname{diag}(d)$, and $|\det A|=d_{1}\cdots d_{n}$ by Every Invertible Real Matrix Is an Orthogonal Matrix Times a Positive Diagonal Matrix Times an Orthogonal Matrix (ii). Here $V^{\top}\in\mathrm{O}(n)$ (Rules for Orthogonal Matrices: Inverse, Products, Isometry, Columns, Determinant, Orthonormal Extension and Expansion (i)), $D$ is invertible by Step 4, and so $UD$ is invertible. Applying (2) to $UD$ and $V^{\top}$, then to $U$ and $D$, and using Step 4,

$$c(A)=c(U)\,c(D)\,c(V^{\top})=d_{1}\cdots d_{n}=|\det A| .$$

Step 6: images of Borel sets. Let $B\in\mathcal{B}(\mathbb{R}^{n})$. Then $T[B]=A[B]+b$, so translation invariance and (1) give

$$\lambda_{n}(T[B])=\lambda_{n}(A[B])=|\det A|\,\lambda_{n}(B),$$

which, with Step 1, is (i).

Step 7: the image measure. By Step 1, $T^{-1}(B)=T'[B]$ for the function $T'(y)=A^{-1}y-A^{-1}b$. The matrix $A^{-1}$ is invertible (Rules for Matrix Inverses (iii)) with $\det(A^{-1})=1/\det A$ (A Square Real Matrix Is Invertible if and only if Its Determinant Is Nonzero (ii)). Steps 1 to 6 hold for every invertible matrix and every vector, in particular for $A^{-1}$ and $-A^{-1}b$, so Step 6 applied to $T'$ gives

$$(\lambda_{n}\circ T^{-1})(B)=\lambda_{n}(T^{-1}(B))=|\det A^{-1}|\,\lambda_{n}(B)=|\det A|^{-1}\lambda_{n}(B)$$

(Image Measure (i)), which is (ii).

Step 8: nonnegative functions. Let $f\in\mathcal{M}^{+}$. Then $f\circ T\in\mathcal{M}^{+}$ as a composition of measurable functions (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (ii)). Put $\gamma=|\det A|^{-1}$, a positive real number. By (ii), $\lambda_{n}\circ T^{-1}$ and the constant multiple $\gamma\lambda_{n}$ of $\lambda_{n}$, a measure by Integrals of Finite Sums of Nonnegative Functions, and Constant Multiples of a Measure (ii), agree at every Borel set, so

$$\lambda_{n}\circ T^{-1}=\gamma\lambda_{n}.$$

Therefore (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (ii) and Integrals of Finite Sums of Nonnegative Functions, and Constant Multiples of a Measure (iii))

$$\int f\circ T\,d\lambda_{n}=\int f\,d(\lambda_{n}\circ T^{-1})=\int f\,d(\gamma\lambda_{n})=\gamma\int f\,d\lambda_{n},$$

which is (iii).

Step 9: integrable functions. Let $f:\mathbb{R}^{n}\to\mathbb{R}$ be measurable. By the change of variables for real functions (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (iii)), $f\circ T$ is measurable, and $f\circ T\in\mathcal{L}^{1}(\lambda_{n})$ if and only if $f\in\mathcal{L}^{1}(\lambda_{n}\circ T^{-1})=\mathcal{L}^{1}(\gamma\lambda_{n})$, in which case $\int f\circ T\,d\lambda_{n}=\int f\,d(\gamma\lambda_{n})$. As $\gamma>0$, Integrals of Finite Sums of Nonnegative Functions, and Constant Multiples of a Measure (iv) gives $f\in\mathcal{L}^{1}(\gamma\lambda_{n})$ if and only if $f\in\mathcal{L}^{1}(\lambda_{n})$, and then $\int f\,d(\gamma\lambda_{n})=\gamma\int f\,d\lambda_{n}$. Together, $f\circ T\in\mathcal{L}^{1}(\lambda_{n})$ if and only if $f\in\mathcal{L}^{1}(\lambda_{n})$, and in that case

$$\int f\circ T\,d\lambda_{n}=\int f\,d(\gamma\lambda_{n})=\gamma\int f\,d\lambda_{n},$$

which is (iv).