Proof of The Lebesgue Measure of a Lipschitz Image of a Compact Subset of $\mathbb{R}^n$

Throughout, $B$, $Q_{m,j}$, $J_{m}$, $E_{m}$ and $\sigma_{n}$ are as in Uniform Grids on a Half-Open Box and Grid Hulls of a Compact Set in $\mathbb{R}^n$, for the parameters $c$ and $s$ chosen in step 2 below. We use the bilinearity and symmetry of the dot product (The Dot Product on $\mathbb{R}^n$ Is Symmetric, Bilinear and Positive Definite) and the properties of the Euclidean norm collected in Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square.

Step 1: $K$ and $T(K)$ are compact Borel sets of finite measure. Since $K$ is nonempty, so is $T(K)$. Being compact, $K$ belongs to $\mathcal{B}(\mathbb{R}^{n})$ and satisfies $\lambda_{n}(K)<\infty$ by claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure; moreover $K$ is bounded and closed by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere. Fix $z\in\mathbb{R}^{n}$ and a real $R>0$ with $d_{E}(z,x)\le R$ for every $x\in K$, and fix $x_{\ast}\in K$.

The set $T(K)$ is bounded: for $x\in K$, the triangle inequality for $d_{E}$ gives $\lVert x-x_{\ast}\rVert=d_{E}(x_{\ast},x)\le d_{E}(x_{\ast},z)+d_{E}(z,x)\le 2R$, whence $d_{E}\bigl(T(x_{\ast}),T(x)\bigr)=\lVert T(x)-T(x_{\ast})\rVert\le 2LR$.

The set $T(K)$ is closed. Let $(w^{(q)})_{q\in\mathbb{N}}$ be a sequence in $T(K)$ converging to $w\in\mathbb{R}^{n}$, and for each $q$ choose $x^{(q)}\in K$ with $T(x^{(q)})=w^{(q)}$. By The Bolzano-Weierstrass Theorem in $\mathbb{R}^n$ there are $x\in\mathbb{R}^{n}$ and a strictly increasing sequence $(q_{t})_{t\in\mathbb{N}}$ in $\mathbb{N}$ with $(x^{(q_{t})})$ converging to $x$; since $K$ is closed, $x\in K$ by Sequential Characterization of Closed Subsets of a Metric Space. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$, and choose $t$ with $\lVert x^{(q_{t})}-x\rVert<\varepsilon/(2L+2)$ and $\lVert w^{(q_{t})}-w\rVert<\varepsilon/2$. Then

$$\lVert w-T(x)\rVert\le\lVert w-w^{(q_{t})}\rVert+\lVert T(x^{(q_{t})})-T(x)\rVert<\frac{\varepsilon}{2}+L\,\frac{\varepsilon}{2L+2}\le\varepsilon .$$

As $\varepsilon$ was an arbitrary positive real number and $0\le\lVert w-T(x)\rVert$, we get $\lVert w-T(x)\rVert=0$, hence $w=T(x)\in T(K)$ by claim 3 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. By Sequential Characterization of Closed Subsets of a Metric Space the set $T(K)$ is closed, and being bounded it is compact by The Heine-Borel Theorem in $\mathbb{R}^n$, and Compactness of Closed Balls and the Unit Sphere; claim 3 of Balls Have Positive Lebesgue Measure and Bounded Sets Have Finite Lebesgue Measure then gives $T(K)\in\mathcal{B}(\mathbb{R}^{n})$ with $\lambda_{n}(T(K))<\infty$.

Step 2: a half-open box containing $K$. Put $R_{1}=\lVert z\rVert+R$. For $x\in K$ the triangle inequality gives $\lVert x\rVert\le\lVert z\rVert+\lVert x-z\rVert\le R_{1}$, hence $|x_{i}|\le\lVert x\rVert\le R_{1}$ for every $i\in[n]$ by claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square. Let $c\in\mathbb{R}^{n}$ have every component equal to $-R_{1}-1$ and put $s=2R_{1}+2$, a positive real number. Then for $x\in K$ and $i\in[n]$,

$$c_{i}=-R_{1}-1<-R_{1}\le x_{i}\le R_{1}<R_{1}+1=c_{i}+s ,$$

so $K\subseteq B$, the half-open box of Uniform Grids on a Half-Open Box and Grid Hulls of a Compact Set in $\mathbb{R}^n$ determined by $c$ and $s$.

Step 3: the covering estimate. Let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Since $K$ is nonempty, compact and contained in $B$, the exhaustion of a compact set by grid hulls provides a mesh index $m\in\mathbb{N}$ (one of the $m_{k}$ there) with

$$\lambda_{n}(E_{m})\le\lambda_{n}(K)+\varepsilon .$$

Write $h=s/m$, $J=J_{m}$ and $E=E_{m}$. By the grid-hull properties, $J$ is nonempty and finite with $\lambda_{n}(E)=|J|\,h^{n}$; put $p=|J|$ and let $g:[p]\to J$ be a bijection, as in Number of Elements of a Set.

For every $u\in[p]$ the set $Q_{m,g(u)}\cap K$ is nonempty by the definition of $J$, so Choice for a Family Indexed by a Finite Set provides points $z^{(u)}\in Q_{m,g(u)}\cap K$, one for each $u\in[p]$. Put $\beta=L\,\sigma_{n}\,h$, a nonnegative real number, and

$$W_{u}=\bigl\{w\in\mathbb{R}^{n}:\bigl|w_{i}-T(z^{(u)})_{i}\bigr|\le\beta\ \text{ for every }i\in[n]\bigr\}\qquad(u\in[p]).$$

Each $W_{u}$ is the product over $i\in[n]$ of the sets $\{t\in\mathbb{R}:T(z^{(u)})_{i}-\beta\le t\le T(z^{(u)})_{i}+\beta\}$, which are intervals with endpoints differing by $2\beta$ and hence Borel subsets of $\mathbb{R}$ of Lebesgue measure $2\beta$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness. By Lebesgue Measure on $\mathbb{R}^n$, $W_{u}\in\mathcal{B}(\mathbb{R}^{n})$ and $\lambda_{n}(W_{u})=(2\beta)^{n}$.

Let $x\in K$. Since $K\subseteq E$ there is $u\in[p]$ with $x\in Q_{m,g(u)}$, and $z^{(u)}$ also lies in that cell, so the diameter bound for a grid cell gives $\lVert x-z^{(u)}\rVert\le\sigma_{n}h$. Both $x$ and $z^{(u)}$ lie in $K$, so

$$\bigl\lVert T(x)-T(z^{(u)})\bigr\rVert\le L\,\lVert x-z^{(u)}\rVert\le L\,\sigma_{n}h=\beta ,$$

and claim 4 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square gives $|T(x)_{i}-T(z^{(u)})_{i}|\le\beta$ for every $i\in[n]$; that is, $T(x)\in W_{u}$. Hence

$$T(K)\subseteq\bigcup_{u=1}^{p}W_{u},$$

and by claims 2 and 4 of Basic Properties of a Measure (the latter applied to the sequence $W_{1},\dots,W_{p},\varnothing,\varnothing,\dots$),

$$\lambda_{n}\bigl(T(K)\bigr)\le\sum_{u=1}^{p}\lambda_{n}(W_{u})=p\,(2\beta)^{n}=p\,\bigl(2\sigma_{n}L\bigr)^{n}h^{n}=\bigl(2\sigma_{n}L\bigr)^{n}\lambda_{n}(E)\le\bigl(2\sigma_{n}L\bigr)^{n}\bigl(\lambda_{n}(K)+\varepsilon\bigr).$$

Step 4: conclusion. Put $a=\lambda_{n}(T(K))$, $b=\lambda_{n}(K)$ and $C=(2\sigma_{n}L)^{n}$; by step 1 these are nonnegative real numbers. Step 3 shows that $a\le C\,(b+\varepsilon)$ for every positive real $\varepsilon$. Suppose $C\,b<a$. If $C=0$ then $a\le 0=C\,b$, contradicting $C\,b<a$; so $0<C$, and taking $\varepsilon=(a-C\,b)/(2C)$, a positive real number, gives

$$a\le C\,b+C\,\varepsilon=C\,b+\frac{a-C\,b}{2}=\frac{a+C\,b}{2}<a,$$

a contradiction. Hence $a\le C\,b$, that is, $\lambda_{n}(T(K))\le(2\sigma_{n}L)^{n}\lambda_{n}(K)$.