Proof of Lipschitz Images and Lebesgue Outer Measure in $\mathbb{R}^n$

Claim 1. Let $A\subseteq E$. If $\lambda_{n}^{\ast}(A)=\infty$ there is nothing to prove, so assume $a=\lambda_{n}^{\ast}(A)$ is real, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$.

By outer regularity of the outer measure there is an open $U\subseteq\mathbb{R}^{n}$ with $A\subseteq U$ and $\lambda_{n}(U)\le a+\varepsilon$. By the dyadic decomposition of an open set there is a sequence $(P_{m})_{m\in\mathbb{N}}$ of pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$, each empty or a dyadic cell contained in $U$, with $\bigcup_{m}P_{m}=U$ and $\lambda_{n}(U)=\sum_{m}\lambda_{n}(P_{m})$.

Fix $m$ with $A\cap P_{m}\ne\varnothing$; then $P_{m}$ is a dyadic cell, say of generation $k$, so $\lambda_{n}(P_{m})=(2^{-k})^{n}$ and any two of its points are at distance at most $\sigma_{n}2^{-k}$, by the grid claim. Choose $x^{m}\in A\cap P_{m}$. For $x\in A\cap P_{m}$ we have $\lVert x-x^{m}\rVert\le\sigma_{n}2^{-k}$ and hence $\lVert T(x)-T(x^{m})\rVert\le L\sigma_{n}2^{-k}$; since $|y_{l}|\le\lVert y\rVert$ for every $l$ by Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$, this gives

The set $D_{m}$ is a Borel rectangle, and by claim 1 of Finite Products of Lebesgue Measure and Coordinate Integration on $\mathbb{R}^l$ together with claim 4 of Existence of Lebesgue Measure on the Real Line,

For $m$ with $A\cap P_{m}=\varnothing$ put $D_{m}=\varnothing$, so that the same inequality $\lambda_{n}(D_{m})\le(2\sigma_{n}L)^{n}\lambda_{n}(P_{m})$ holds trivially.

Since $A\subseteq U=\bigcup_{m}P_{m}$ we have $T(A)=\bigcup_{m}T(A\cap P_{m})\subseteq\bigcup_{m}D_{m}$, so by monotonicity, countable subadditivity and agreement on Borel sets of $\lambda_{n}^{\ast}$,

Write $c=(2\sigma_{n}L)^{n}$, a positive real. Since $\lambda_{n}(U)\le a+\varepsilon<\infty$, every $\lambda_{n}(P_{m})$ is real and, by the finite additivity and monotonicity of $\lambda_{n}$ (claims 1 and 2 of Basic Properties of a Measure), every finite partial sum satisfies $\sum_{m=1}^{p}\lambda_{n}(P_{m})\le\lambda_{n}(U)$. Hence every partial sum of $(\lambda_{n}(D_{m}))_{m\in\mathbb{N}}$ is real and

By the convention for sums of sequences in $[0,\infty]$ fixed in Measure, Measure Space, and Probability Measure, the sum of a sequence whose terms are real with partial sums bounded above is the least upper bound of those partial sums. Therefore $\lambda_{n}^{\ast}(T(A))\le c\,(a+\varepsilon)$.

This holds for every $\varepsilon>0$. If we had $\lambda_{n}^{\ast}(T(A))>(2\sigma_{n}L)^{n}a$, then by The Archimedean Property of the Real Numbers there would be $m\in\mathbb{N}$ with $(2\sigma_{n}L)^{n}/m<\lambda_{n}^{\ast}(T(A))-(2\sigma_{n}L)^{n}a$, and taking $\varepsilon=1/m$ would contradict the displayed bound. Hence $\lambda_{n}^{\ast}(T(A))\le(2\sigma_{n}L)^{n}a$, as asserted.

Claim 2. If $A\subseteq E$ is $\lambda_{n}$-null then $\lambda_{n}^{\ast}(A)=0$ by the null-set claim, so claim 1 gives $\lambda_{n}^{\ast}(T(A))\le(2\sigma_{n}L)^{n}\cdot 0=0$, and the same claim shows $T(A)$ is $\lambda_{n}$-null.

Claim 3. Let $\mathcal{G}$ be the set of pairs $(k,j)\in\mathbb{N}\times\mathbb{Z}^{n}$ such that $Q_{k,j}\subseteq U$ and the restriction of $T$ to $Q_{k,j}$ is Lipschitz for the Euclidean metrics with some constant, where $Q_{k,j}$ denotes the dyadic cell of generation $k$ and index $j$.

The cells indexed by $\mathcal{G}$ cover $U$. Indeed, let $x\in U$. Since $T$ is locally Lipschitz there are $r,L'\in\mathbb{R}$ with $0<r$, $0\le L'$, $\bar{B}(x,r)\subseteq U$, where $\bar{B}(x,r)$ is the closed ball of $(\mathbb{R}^{n},d_{E})$ with centre $x$ and radius $r$, and $\lVert T(y)-T(y')\rVert\le L'\lVert y-y'\rVert$ for $y,y'\in\bar{B}(x,r)$. By the small-cell claim there is $k\in\mathbb{N}$ with $\sigma_{n}2^{-k}\le r$, and if $Q_{k,j}$ denotes the cell of generation $k$ containing $x$, every $y\in Q_{k,j}$ satisfies $\lVert y-x\rVert\le r$; so $Q_{k,j}\subseteq\bar{B}(x,r)\subseteq U$ and $T$ is Lipschitz with constant $L'$ on $Q_{k,j}$. Thus $(k,j)\in\mathcal{G}$ and $x\in Q_{k,j}$.

Now let $A\subseteq U$ be $\lambda_{n}$-null. By claim 1 of the dyadic lemma the set $\mathbb{N}\times\mathbb{Z}^{n}$ is countable and infinite; let $((k_{m},j_{m}))_{m\in\mathbb{N}}$ enumerate it bijectively and put

Each $A\cap Q_{k_{m},j_{m}}$ is a subset of a $\lambda_{n}$-null set, hence $\lambda_{n}$-null; and when $(k_{m},j_{m})\in\mathcal{G}$ the restriction of $T$ to $Q_{k_{m},j_{m}}$ is Lipschitz with a constant which we may take to be positive (enlarging it if necessary), so claim 2 applied with $E=Q_{k_{m},j_{m}}$ shows $Y_{m}$ is $\lambda_{n}$-null. Since the cells indexed by $\mathcal{G}$ cover $U\supseteq A$, we have $T(A)=\bigcup_{m\in\mathbb{N}}Y_{m}$, which is $\lambda_{n}$-null by the null-set claim.