Proof of Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map

Each result cited below is universally quantified over the data in its own statement.

Conventions. For a natural number $l\ge1$ let $\mathcal{B}_{l}$ and $\lambda_{l}$ be the $\sigma$-algebra and measure on $\mathbb{R}^{l}$ built in Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$. By claim 5 of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets, $\mathcal{B}_{l}=\mathcal{B}(\mathbb{R}^{l})$, and by Lebesgue Measure on $\mathbb{R}^n$ this $\lambda_{l}$ is Lebesgue measure on $\mathcal{B}(\mathbb{R}^{l})$; for $l=1$ it is Lebesgue measure $\lambda$ on $\mathcal{B}_{1}=\mathcal{B}(\mathbb{R})$. By claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ each $\lambda_{l}$ is $\sigma$-finite. For $l\ge2$ we use the identification of $\mathbb{R}^{l}$ with $\mathbb{R}^{l-1}\times\mathbb{R}$ fixed there, writing $x=(x',x_{l})$ with $x'=(x_{1},\dots,x_{l-1})$; under it $\mathcal{B}_{l}=\mathcal{B}_{l-1}\otimes\mathcal{B}(\mathbb{R})$ and $\lambda_{l}=\lambda_{l-1}\otimes\lambda$. A Borel $f:\mathbb{R}^{l}\to[0,\infty]$ is one with $\{f>a\}\in\mathcal{B}(\mathbb{R}^{l})$ for every real $a$ (Measure Spaces and the Lebesgue Integral: Standing Notation §measurable).

Claim A (Composition). If $S:\mathbb{R}^{p}\to\mathbb{R}^{m}$ is measurable with respect to $\mathcal{B}(\mathbb{R}^{p})$ and $\mathcal{B}(\mathbb{R}^{m})$ and $h:\mathbb{R}^{m}\to[0,\infty]$ is Borel, then $h\circ S$ is Borel: for real $a$, $\{h\circ S>a\}=S^{-1}(\{h>a\})\in\mathcal{B}(\mathbb{R}^{p})$ by Measurable Function and Real-Valued Measurable Function.

Claim B (Arithmetic in $[0,\infty]$). For positive reals $a,b$ and $I\in[0,\infty]$ we have $a(bI)=(ab)I$ and $a(a^{-1}I)=I$: for finite $I$ this is field arithmetic, and for $I=\infty$ both sides are $\infty$ by the convention $a\cdot\infty=\infty$ for $0<a$ of Measure Spaces and the Lebesgue Integral: Standing Notation §extended (with $0<ab$ and $0<a^{-1}$ by claims 5 and 7 of Elementary Order Arithmetic in an Ordered Field).

Step 1 (Proof of Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map §borel). Let $T$ be a real $d\times d$ matrix. By claim 3 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product there is a real $C\ge0$ with $\lVert Tv\rVert\le C\lVert v\rVert$ for all $v\in\mathbb{R}^{d}$, and $Tx-Ty=T(x-y)$ by claim 1 there, so $\lVert Tx-Ty\rVert\le C\lVert x-y\rVert$. Since $d_{E}(x,y)=\lVert x-y\rVert$ (claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$), $x\mapsto Tx$ is Lipschitz with constant $C$, hence continuous by A Lipschitz Map is Uniformly Continuous. For $i\in[d]$ the component $x\mapsto(Tx)_{i}$ satisfies $|(Tx)_{i}-(Ty)_{i}|\le\lVert Tx-Ty\rVert\le C\lVert x-y\rVert$ by claim 4 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$. On $\mathbb{R}=\mathbb{R}^{1}$ the Euclidean distance is $|s-t|$: for $u\in\mathbb{R}^{1}$, $\lVert u\rVert^{2}=u^{2}=|u|^{2}$ by claim 1 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and claim 1 of Properties of the Absolute Value in an Ordered Field, both $\lVert u\rVert$ and $|u|$ are nonnegative, so $\lVert u\rVert=|u|$ by claim 3 of Monotonicity of Squaring on the Nonnegative Elements of an Ordered Field. Hence each component is Lipschitz, so continuous (A Lipschitz Map is Uniformly Continuous), from $(\mathbb{R}^{d},d_{E})$ to $(\mathbb{R},d_{E})$, and therefore measurable with respect to $\mathcal{B}_{d}$ and $\mathcal{B}(\mathbb{R})$ by claim 3(a) of The Borel Sigma-Algebra of a Euclidean Space as a Product, and Measurability of Projections, Sequentially Continuous Maps, and Open and Closed Sets. By claim 2 there (with $(X,\mathcal{F})=(\mathbb{R}^{d},\mathcal{B}_{d})$), $x\mapsto Tx$ is measurable with respect to $\mathcal{B}_{d}$ and $\mathcal{B}_{d}$, that is, Borel, as $\mathcal{B}_{d}=\mathcal{B}(\mathbb{R}^{d})$.

Step 2 (Positivity of triangular determinants). Let $T$ be a lower or upper triangular real $l\times l$ matrix with $0<T_{kk}$ for all $k\in[l]$. By claim 1 or claim 2 of The Determinant of a Triangular Matrix is the Product of its Diagonal Entries, whose determinant is that of Determinant of a Real Square Matrix by Row Properties of the Determinant, $\det T=\prod_{k=1}^{l}T_{kk}$. By the recursion in claim 1 of Properties of Finite Products and claim 5 of Elementary Order Arithmetic in an Ordered Field, induction on the number of factors gives $0<\det T$. This proves the first assertion of Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map §triangular. If $l\ge2$, let $T'$ be the $(l-1)\times(l-1)$ matrix with $T'_{ij}=T_{ij}$ for $i,j\in[l-1]$; it is triangular of the same kind with the same first $l-1$ diagonal entries, and by The Determinant of a Triangular Matrix is the Product of its Diagonal Entries and claim 1 of Properties of Finite Products (restriction and recursion),

Step 3 (Splitting off the last coordinate). Let $l\ge2$, $T$ and $T'$ as in Step 2, and $x=(x',x_{l})\in\mathbb{R}^{l}$. By Matrix-Vector Product and the recursion in claim 1 of Properties of Finite Sums, $(Tx)_{i}=\sum_{j=1}^{l-1}T_{ij}x_{j}+T_{il}x_{l}$ for every $i\in[l]$, and $\sum_{j=1}^{l-1}T_{ij}x_{j}=(T'x')_{i}$ for $i\le l-1$.

If $T$ is lower triangular, then $T_{il}=0$ for $i<l$, so, with $\beta(x')=\sum_{j=1}^{l-1}T_{lj}x_{j}$,

If $T$ is upper triangular, then $T_{lj}=0$ for $j<l$, so $\sum_{j=1}^{l-1}T_{lj}x_{j}=0$ (claim 3 of Properties of Finite Sums with $\lambda=0$), and, with $b=(T_{1l},\dots,T_{l-1,l})\in\mathbb{R}^{l-1}$ and the sum and scalar multiple of $\mathbb{R}^{l-1}$ taken coordinatewise,

Step 4 (Proof of the integral identity in Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map §triangular). We prove by induction on $l\ge1$ the statement $P(l)$: for every lower or upper triangular real $l\times l$ matrix $T$ with positive diagonal entries and every Borel $f:\mathbb{R}^{l}\to[0,\infty]$, $\det T\int_{\mathbb{R}^{l}}f(Tx)\,d\lambda_{l}(x)=\int_{\mathbb{R}^{l}}f\,d\lambda_{l}$. Here $f\circ T$ is Borel by Claim A and Step 1, so the left integral is defined. Then $P(d)$ is the claim.

Base case $l=1$. Here $T=(t)$ with $t=T_{11}>0$, $Tx=tx$ by Matrix-Vector Product and claim 1 of Properties of Finite Sums, and $\det T=t$ by Step 2 and claim 1 of Properties of Finite Products. By claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$ with $n=1$ and $c=t$, where $|t|^{1}=|t|=t$ by Natural Number Power of an Element of a Field, claim 1 of Properties of Finite Products and Absolute Value in an Ordered Field,

and multiplying by $t$ gives $P(1)$ by Claim B.

Inductive step. Let $l\ge2$ and assume $P(l-1)$. Let $T$ and $f$ be as in $P(l)$, let $T'$ be as in Step 2, and put $H=f\circ T$, Borel on $\mathbb{R}^{l}=\mathbb{R}^{l-1}\times\mathbb{R}$. We apply Tonelli and Fubini Theorems with $(X,\mathcal{F},\mu)=(\mathbb{R}^{l-1},\mathcal{B}_{l-1},\lambda_{l-1})$ and $(Y,\mathcal{G},\nu)=(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, both $\sigma$-finite, to $H$ and to $f$, which are $\mathcal{B}_{l-1}\otimes\mathcal{B}(\mathbb{R})$-measurable.

Lower triangular $T$. By the Tonelli part of Tonelli and Fubini Theorems and Step 3,

Fix $x'$. The section $k(s)=f(T'x',s)$ is $\mathcal{B}(\mathbb{R})$-measurable by the sections part of Tonelli and Fubini Theorems. By claim 2 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ with $n=1$ and $a=\beta(x')$, $k_{1}(s)=k(s+\beta(x'))$ is measurable with $\int_{\mathbb{R}}k_{1}\,d\lambda=\int_{\mathbb{R}}k\,d\lambda$, and by claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$ with $n=1$ and $c=T_{ll}$ (as in the base case), $\int_{\mathbb{R}}k_{1}(T_{ll}t)\,d\lambda(t)=T_{ll}^{-1}\int_{\mathbb{R}}k_{1}\,d\lambda$. As $k_{1}(T_{ll}t)=f(T'x',T_{ll}t+\beta(x'))$, the inner integral equals $T_{ll}^{-1}g(T'x')$, where $g(z')=\int_{\mathbb{R}}f(z',s)\,d\lambda(s)$ defines a $\mathcal{B}_{l-1}$-measurable $g:\mathbb{R}^{l-1}\to[0,\infty]$ by the Tonelli part of Tonelli and Fubini Theorems. The map $g\circ T'$ is Borel (Claim A and Step 1), so claim 1 of Linearity and Monotonicity of the Lebesgue Integral with the constant $T_{ll}^{-1}$ gives $\int_{\mathbb{R}^{l}}H\,d\lambda_{l}=T_{ll}^{-1}\int_{\mathbb{R}^{l-1}}g(T'x')\,d\lambda_{l-1}(x')$. By $P(l-1)$ applied to $T'$ and $g$, and the Tonelli part of Tonelli and Fubini Theorems applied to $f$,

Using $\det T=\det T'\cdot T_{ll}$ (Step 2) and Claim B, $\det T\int H\,d\lambda_{l}=\det T'\cdot T_{ll}\bigl(T_{ll}^{-1}\int g\circ T'\,d\lambda_{l-1}\bigr)=\det T'\int g\circ T'\,d\lambda_{l-1}=\int f\,d\lambda_{l}$.

Upper triangular $T$. By the Tonelli part of Tonelli and Fubini Theorems, in the other order, and Step 3,

Fix $t$. The section $k(z')=f(z',T_{ll}t)$ is $\mathcal{B}_{l-1}$-measurable by the sections part of Tonelli and Fubini Theorems. By claim 2 of Translation and Reflection Invariance of Lebesgue Measure on $\mathbb{R}^n$ with $n=l-1$ and $a=tb$, $k_{2}(z')=k(z'+tb)$ is measurable with $\int k_{2}\,d\lambda_{l-1}=\int k\,d\lambda_{l-1}$. The inner integrand is $k_{2}(T'x')$, and $k_{2}\circ T'$ is Borel (Claim A, Step 1); by $P(l-1)$ applied to $T'$ and $k_{2}$,

where $g:\mathbb{R}\to[0,\infty]$ is $\mathcal{B}(\mathbb{R})$-measurable by the Tonelli part of Tonelli and Fubini Theorems. By Claim B the inner integral equals $(\det T')^{-1}g(T_{ll}t)$. The map $t\mapsto g(T_{ll}t)$ is measurable by claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$, so by claim 1 of Linearity and Monotonicity of the Lebesgue Integral, claim 2 of Scaling of Lebesgue Measure and the Lebesgue Integral on $\mathbb{R}^n$ (with $n=1$, $c=T_{ll}$) and the Tonelli part of Tonelli and Fubini Theorems,

Multiplying by $\det T=\det T'\cdot T_{ll}$ and using Claim B gives $\det T\int H\,d\lambda_{l}=\int f\,d\lambda_{l}$. This proves $P(l)$ and completes the induction.

Step 5 (Proof of Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map §spd). Let $A\in\mathcal{S}(d)$ be positive definite. By Cholesky Factorisation of a Symmetric Positive Definite Real Matrix there is a lower triangular real $d\times d$ matrix $L$ with $0<L_{ii}$ for all $i$ and $A=LL^{\top}$. Put $U=L^{\top}$; by Transpose of a Real Matrix, $U_{ij}=L_{ji}$, which is $0$ when $j<i$, and $U_{ii}=L_{ii}>0$, so $U$ is upper triangular with positive diagonal. By The Determinant is Multiplicative, Step 2 and claim 5 of Elementary Order Arithmetic in an Ordered Field, $\det A=\det L\cdot\det U>0$.

By claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product, $Ax=(LU)x=L(Ux)$. Let $f:\mathbb{R}^{d}\to[0,\infty]$ be Borel; $f\circ L$ is Borel by Claim A and Step 1. By Lebesgue Measure under a Triangular or Symmetric Positive Definite Linear Map §triangular (proved in Step 4) applied to $U$ and $f\circ L$, and then to $L$ and $f$,

so by Claim B, $\det A\int f(Ax)\,d\lambda_{d}(x)=\det L\bigl(\det U\int f(Ax)\,d\lambda_{d}(x)\bigr)=\int f\,d\lambda_{d}$.

By Invertibility of Symmetric Positive Definite Matrices, $A$ is invertible. For $y\in\mathbb{R}^{d}$, claim 2 of Linearity, Compatibility with the Matrix Product, and a Norm Bound for the Matrix-Vector Product gives $A(A^{-1}y)=(AA^{-1})y=I_{d}y=y$ and likewise $A^{-1}(Ay)=y$, where $(I_{d}y)_{i}=\sum_{j}\delta_{ij}y_{j}=y_{i}$ by Identity Matrix, Matrix-Vector Product and claim 7 of Properties of Finite Sums. Hence $x\mapsto Ax$ is a bijection of $\mathbb{R}^{d}$ onto $\mathbb{R}^{d}$ with inverse $y\mapsto A^{-1}y$.

Let $B\in\mathcal{B}(\mathbb{R}^{d})$. Then $y\in A(B)$ if and only if $A^{-1}y\in B$, so $A(B)$ is the preimage of $B$ under $y\mapsto A^{-1}y$, which is Borel by Step 1; thus $A(B)\in\mathcal{B}(\mathbb{R}^{d})$. The indicator $f=\mathbf{1}_{A(B)}$ is Borel (each set $\{f>a\}$ is $\mathbb{R}^{d}$, $A(B)$ or $\emptyset$), and $f(Ax)=\mathbf{1}_{B}(x)$ for all $x$, by injectivity. The identity just proved and The Integral of an Indicator Function is the Measure of the Set give $\det A\,\lambda_{d}(B)=\lambda_{d}(A(B))$.