Proof of The Degenerate-Derivative Values of a Locally Lipschitz Map of $\mathbb{R}^n$ Form a Null Set

Throughout, $Q_{k,j}$ denotes the dyadic cell of generation $k$ and index $j$, and $\lambda_{n}^{\ast}$ denotes Lebesgue outer measure. When $n=1$ the factors $(2R)^{n-1}$ and $L^{n-1}$ occurring below are to be read as $1$. We write $\bar{B}(x,r)$ for the closed ball of $(\mathbb{R}^{n},d_{E})$ with centre $x$ and radius $r$.

Step 1: reduction to a single cell. Let $\mathcal{G}$ be the set of pairs $(k_{0},j_{0})\in\mathbb{N}\times\mathbb{Z}^{n}$ such that $Q_{k_{0},j_{0}}\subseteq U$ and the restriction of $T$ to $Q_{k_{0},j_{0}}$ is Lipschitz for the Euclidean metrics with some positive constant. These cells cover $U$. Indeed, let $x\in U$. Since $T$ is locally Lipschitz there are $r,L'\in\mathbb{R}$ with $0<r$, $0\le L'$, $\bar{B}(x,r)\subseteq U$ and $\lVert T(y)-T(y')\rVert\le L'\lVert y-y'\rVert$ for all $y,y'\in\bar{B}(x,r)$. By the small-cell claim there is $k\in\mathbb{N}$ with $\sigma_{n}2^{-k}\le r$, and the cell $Q_{k,j}$ of generation $k$ containing $x$ then satisfies $\lVert y-x\rVert\le r$ for every $y\in Q_{k,j}$, so $Q_{k,j}\subseteq\bar{B}(x,r)\subseteq U$ and $T$ is Lipschitz on $Q_{k,j}$ with the positive constant $\max\{L',1\}$. Thus $(k,j)\in\mathcal{G}$ and $x\in Q_{k,j}$.

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 $Y_{m}=T(S\cap Q_{k_{m},j_{m}})$ if $(k_{m},j_{m})\in\mathcal{G}$ and $Y_{m}=\varnothing$ otherwise. Then $T(S)=\bigcup_{m}Y_{m}$, so by the null-set claim it suffices to prove that $T(S\cap Q_{0})$ is $\lambda_{n}$-null for every dyadic cell $Q_{0}=Q_{k_{0},j_{0}}$ with $(k_{0},j_{0})\in\mathcal{G}$.

Step 2: a maximal disjoint family of cells. Fix such a $Q_{0}$, of generation $k_{0}$, and let $L\in\mathbb{R}$ with $0<L$ be a Lipschitz constant for $T$ on $Q_{0}$. Fix $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$.

Let $x\in S\cap Q_{0}$. By differentiability of $T$ at $x$ there is $\rho\in\mathbb{R}$ with $0<\rho$ such that

$$\lVert T(y)-T(x)-DT(x)(y-x)\rVert\le\varepsilon\lVert y-x\rVert\qquad\text{for all }y\in U\text{ with }\lVert y-x\rVert\le\rho .$$

By the small-cell claim there is $k\in\mathbb{N}$ with $\sigma_{n}2^{-k}\le\rho$, and the same holds for every larger $k$ because $2^{-k}$ decreases as $k$ increases; hence the set $A_{x}=\{k\in\mathbb{N}:k_{0}\le k\text{ and }\sigma_{n}2^{-k}\le\rho\}$ is nonempty; let $k(x)$ be its least element (The Natural Numbers Are Well Ordered) and let $Q(x)$ be the cell of generation $k(x)$ containing $x$. Since $k_{0}\le k(x)$ and $x\in Q_{0}$, nesting gives $Q(x)\subseteq Q_{0}$; and every $y\in Q(x)$ satisfies $\lVert y-x\rVert\le\sigma_{n}2^{-k(x)}\le\rho$ and lies in $U$, so the displayed estimate applies to it.

Let $\mathcal{F}\subseteq\mathbb{N}\times\mathbb{Z}^{n}$ be the set of indices of the cells $Q(x)$ for $x\in S\cap Q_{0}$, and let $\mathcal{M}$ be the set of $(k,j)\in\mathcal{F}$ for which there is no $(k',j')\in\mathcal{F}$ with $k'<k$ and $Q_{k,j}\subseteq Q_{k',j'}$.

Every cell indexed by $\mathcal{F}$ is contained in a cell indexed by $\mathcal{M}$: given $(k,j)\in\mathcal{F}$, the set of $k'\in\mathbb{N}$ with $k'\le k$ such that the cell of generation $k'$ containing $Q_{k,j}$ (unique by nesting) is indexed by a member of $\mathcal{F}$ is nonempty, since it contains $k$; its least element $k^{\ast}$ gives a pair in $\mathcal{M}$, for a strictly coarser cell in $\mathcal{F}$ containing it would contradict minimality.

Distinct members of $\mathcal{M}$ index disjoint cells: if the cells of $(k,j)$ and $(k',j')$ meet, then by nesting one contains the other, say $Q_{k',j'}\subseteq Q_{k,j}$ with $k\le k'$; if $k<k'$ this contradicts $(k',j')\in\mathcal{M}$, and if $k=k'$ then the cells coincide, so $j=j'$ by claim 1 of the dyadic lemma.

All these cells are contained in $Q_{0}$, so, being pairwise disjoint members of $\mathcal{B}(\mathbb{R}^{n})$, they satisfy

$$\sum_{(k,j)\in\mathcal{M}}\lambda_{n}(Q_{k,j})\le\lambda_{n}(Q_{0})=(2^{-k_{0}})^{n},$$

by countable additivity and monotonicity of $\lambda_{n}$ (the sum being over a countable index set, enumerated as above with zero terms inserted for indices outside $\mathcal{M}$).

Step 3: the image of a single cell lies in a thin slab. Let $(k,j)\in\mathcal{M}$ and choose $x\in S\cap Q_{0}$ with $Q(x)=Q_{k,j}$. Write $A=DT(x)$ and let $\nu\in\mathbb{R}^{n}$ with $\lVert\nu\rVert=1$ satisfy $\nu\cdot(Ah)=0$ for every $h\in\mathbb{R}^{n}$, as provided by the definition of $S$. For $y\in Q_{k,j}$ we have $y,x\in Q_{0}$ and $\lVert y-x\rVert\le\sigma_{n}2^{-k}$, hence

$$\lVert T(y)-T(x)\rVert\le L\,\lVert y-x\rVert\le L\sigma_{n}2^{-k},$$

and, since $\nu\cdot(A(y-x))=0$ and by The Cauchy-Schwarz Inequality on $\mathbb{R}^n$,

$$\bigl|\nu\cdot(T(y)-T(x))\bigr|=\bigl|\nu\cdot\bigl(T(y)-T(x)-A(y-x)\bigr)\bigr|\le\lVert T(y)-T(x)-A(y-x)\rVert\le\varepsilon\sigma_{n}2^{-k}.$$

Therefore $T(Q_{k,j})\subseteq\Sigma_{k,j}$, where

$$\Sigma_{k,j}=\bigl\{w\in\mathbb{R}^{n}:\lVert w-T(x)\rVert\le R,\ |\nu\cdot(w-T(x))|\le\delta\bigr\},\qquad R=L\sigma_{n}2^{-k},\ \delta=\varepsilon\sigma_{n}2^{-k}.$$

By the slab bound,

$$\lambda_{n}(\Sigma_{k,j})\le 2\sigma_{n}\delta(2R)^{n-1}=2^{n}\sigma_{n}^{n+1}L^{n-1}\,\varepsilon\,(2^{-k})^{n}=C\,\varepsilon\,\lambda_{n}(Q_{k,j}),\qquad C=2^{n}\sigma_{n}^{n+1}L^{n-1}.$$

Step 4: summation and conclusion. Every $x\in S\cap Q_{0}$ lies in $Q(x)$, whose index belongs to $\mathcal{F}$ and whose cell is therefore contained in a cell indexed by $\mathcal{M}$; hence $S\cap Q_{0}\subseteq\bigcup_{(k,j)\in\mathcal{M}}Q_{k,j}$ and

$$T(S\cap Q_{0})\subseteq\bigcup_{(k,j)\in\mathcal{M}}T(Q_{k,j})\subseteq\bigcup_{(k,j)\in\mathcal{M}}\Sigma_{k,j}.$$

By monotonicity, countable subadditivity and agreement on Borel sets of $\lambda_{n}^{\ast}$, together with Steps 2 and 3,

$$\lambda_{n}^{\ast}\bigl(T(S\cap Q_{0})\bigr)\le\sum_{(k,j)\in\mathcal{M}}\lambda_{n}(\Sigma_{k,j})\le C\,\varepsilon\sum_{(k,j)\in\mathcal{M}}\lambda_{n}(Q_{k,j})\le C\,\varepsilon\,(2^{-k_{0}})^{n}.$$

The constants $C$ and $k_{0}$ do not depend on $\varepsilon$. If $\lambda_{n}^{\ast}(T(S\cap Q_{0}))$ were positive, then by The Archimedean Property of the Real Numbers there would be $m\in\mathbb{N}$ with $C(2^{-k_{0}})^{n}/m<\lambda_{n}^{\ast}(T(S\cap Q_{0}))$, and taking $\varepsilon=1/m$ would contradict the display. Hence $\lambda_{n}^{\ast}(T(S\cap Q_{0}))=0$, so $T(S\cap Q_{0})$ is $\lambda_{n}$-null by the null-set claim. By Step 1, $T(S)$ is $\lambda_{n}$-null.