Proof of Scaling Lebesgue Measure on ℝⁿ by a Diagonal Matrix

Proof of (i). Write $\delta=|d_{1}\cdots d_{n}|$. Since no $d_{i}$ vanishes, $d_{1}\cdots d_{n}\neq0$ (Products over Finite Sets: a Single Factor, Vanishing in a Field, and Signs and Monotonicity in an Ordered Ring (ii)), so $\delta$ is a positive real number; and $\delta=\prod_{i=1}^{n}|d_{i}|$: the absolute value is multiplicative, $|st|=|s|\,|t|$ for all real $s,t$ (a routine law of $\mathbb{R}$), so it preserves the product and therefore iterated products (Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (ii)).

Step 1: $D[B]$ as a preimage. By Eigenvalues: the Determinant Criterion, Diagonal Matrices, Orthogonal Similarity and Orthogonality of Eigenvectors of Symmetric Matrices (ii), $(Dx)_{i}=d_{i}x_{i}$ for $x\in\mathbb{R}^{n}$ and $i\in[n]$. Let $T:\mathbb{R}^{n}\to\mathbb{R}^{n}$ be the function with $T(y)_{i}=y_{i}/d_{i}$. Then $T(Dx)=x$ and $D\,T(y)=y$ for all $x,y\in\mathbb{R}^{n}$, since the components agree. Hence, for $B\subseteq\mathbb{R}^{n}$, $y\in D[B]$ if and only if $T(y)\in B$:

$$D[B]=T^{-1}(B).$$

Step 2: boxes. Let $p,q\in\mathbb{R}^{n}$, and let $u,v\in\mathbb{R}^{n}$ have components $u_{i}=d_{i}p_{i}$, $v_{i}=d_{i}q_{i}$ if $d_{i}>0$, and $u_{i}=d_{i}q_{i}$, $v_{i}=d_{i}p_{i}$ if $d_{i}<0$. Multiplying by $d_{i}$, which reverses the inequalities when $d_{i}<0$, shows that $p_{i}<y_{i}/d_{i}<q_{i}$ if and only if $u_{i}<y_{i}<v_{i}$; so

$$T^{-1}\bigl(R(p,q)\bigr)=R(u,v).$$

Open boxes lie in $\mathcal{B}(\mathbb{R}^{n})$ and generate it (Existence and Uniqueness of Lebesgue Measure on ℝⁿ (i)), so $T$ is $\mathcal{B}(\mathbb{R}^{n})/\mathcal{B}(\mathbb{R}^{n})$-measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (i)), and $\lambda_{n}\circ T^{-1}$ is a measure on $(\mathbb{R}^{n},\mathcal{B}(\mathbb{R}^{n}))$ (Image Measures and Measures with a Density: Measures, Integrals and Change of Variables (i)). If $p_{i}<q_{i}$ for every $i\in[n]$, then $u_{i}<v_{i}$ and $v_{i}-u_{i}=|d_{i}|\,(q_{i}-p_{i})$ in both cases, so by Lebesgue Measure on $\mathbb{R}^n$ (i) and Iterated Operations: Recursion, Splitting, Reordering, Termwise Combination and Homomorphisms (v)

$$(\lambda_{n}\circ T^{-1})\bigl(R(p,q)\bigr)=\prod_{i=1}^{n}|d_{i}|\,(q_{i}-p_{i})=\Bigl(\prod_{i=1}^{n}|d_{i}|\Bigr)\prod_{i=1}^{n}(q_{i}-p_{i})=\delta\,\lambda_{n}\bigl(R(p,q)\bigr).$$

Otherwise $q_{i}\le p_{i}$ for some $i$, so $R(p,q)=\emptyset$ (Open Boxes: Emptiness, Intersections and Preimages under Concatenation (i)) and $T^{-1}(R(p,q))=\emptyset$, and both sides are $0$.

Step 3: comparison of measures. The function $\delta\lambda_{n}:B\mapsto\delta\,\lambda_{n}(B)$ on $\mathcal{B}(\mathbb{R}^{n})$ is a measure (Integrals of Finite Sums of Nonnegative Functions, and Constant Multiples of a Measure (ii)). By Step 2, it agrees with $\lambda_{n}\circ T^{-1}$ on every open box, where both are real numbers. Hence $\lambda_{n}\circ T^{-1}=\delta\,\lambda_{n}$ by Existence and Uniqueness of Lebesgue Measure on ℝⁿ (iv). For $B\in\mathcal{B}(\mathbb{R}^{n})$, Step 1 now gives $D[B]=T^{-1}(B)\in\mathcal{B}(\mathbb{R}^{n})$ and

$$\lambda_{n}(D[B])=(\lambda_{n}\circ T^{-1})(B)=|d_{1}\cdots d_{n}|\,\lambda_{n}(B).$$

Proof of (ii). Let $d\in\mathbb{R}^{n}$ have every component equal to $c$, and $D=\operatorname{diag}(d)$. Then $(Dx)_{i}=c\,x_{i}=(cx)_{i}$ for every $i$, so $Dx=cx$ for every $x\in\mathbb{R}^{n}$ and $D[B]=cB$. Moreover $d_{1}\cdots d_{n}=\prod_{i=1}^{n}c=c^{n}$ and $|c^{n}|=|c|^{n}$ (Powers with Exponents in the Natural Numbers with Zero (i), Powers in a Commutative Ring, a Field and an Ordered Field: Exponent Laws, Factorisation, Geometric Sums, Monotonicity and Bernoulli's Inequality (vi)). So (i) gives $cB\in\mathcal{B}(\mathbb{R}^{n})$ and

$$\lambda_{n}(cB)=|c|^{n}\,\lambda_{n}(B).$$