Proof of The Partial Derivatives of a Lipschitz Function on $\mathbb{R}^n$ Exist Almost Everywhere

We use the notation of the statement.

Claim 1. Being Lipschitz, $f$ is continuous from $(\mathbb{R}^{n},d_{E})$ to $(\mathbb{R},d_{\mathbb{R}})$ by A Lipschitz Map is Uniformly Continuous and Continuous, where $d_{\mathbb{R}}(s,t)=|s-t|$ is the absolute value metric. Hence Borel Structure of the Set Where a Partial Derivative Exists applies to $f$ and $i$, and by Borel Structure of the Set Where a Partial Derivative Exists (iii) the set $N=\mathbb{R}^{n}\setminus E_{i}$ belongs to $\mathcal{B}(\mathbb{R}^{n})$.

Fix $y\in\mathbb{R}^{n}$ and define $g:\mathbb{R}\to\mathbb{R}$ by $g(t)=f(y+te_{i})$. For $t,t'\in\mathbb{R}$ we have $(y+te_{i})-(y+t'e_{i})=(t-t')e_{i}$, so by claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and $\lVert e_{i}\rVert=1$, which holds by Orthonormal Families, Standard Basis Vectors, and Plane Rotations of Euclidean Space,

$$|g(t)-g(t')|\le L\,\lVert(t-t')e_{i}\rVert=L\,|t-t'| ,$$

so $g$ is Lipschitz with constant $L$ on every subset of $\mathbb{R}$.

Let $m\in\mathbb{N}$ and let $I_{m}$ be the open interval with endpoints $-m$ and $m$, a nonempty open interval. By A Lipschitz Function on an Open Interval is Differentiable Almost Everywhere (i), applied to the restriction of $g$ to $I_{m}$, the set $I_{m}\setminus D_{m}$ is null, where $D_{m}$ is the set of those $t\in I_{m}$ at which that restriction has a derivative.

We show that $\partial_{i}f(y+te_{i})$ exists for every $t\in D_{m}$. Let $t\in D_{m}$ and let $L_{0}$ be the derivative there. Let $\varepsilon>0$. By Derivative of a Real Function on an Interval there is $\delta_{1}>0$ such that every $h\in\mathbb{R}$ with $0<|h|<\delta_{1}$ and $t+h\in I_{m}$ satisfies $|(g(t+h)-g(t))/h-L_{0}|<\varepsilon$. Since $I_{m}$ is open and $t\in I_{m}$ there is $\delta_{2}>0$ with $t+h\in I_{m}$ whenever $|h|<\delta_{2}$; let $\delta$ be the smaller of $\delta_{1}$ and $\delta_{2}$. Writing $w=y+te_{i}$ we have $w+he_{i}=y+(t+h)e_{i}$, so $(g(t+h)-g(t))/h=(f(w+he_{i})-f(w))/h$, and this quotient differs from $L_{0}$ by less than $\varepsilon$ for every real $h$ with $0<|h|<\delta$. By Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$, applied with the open set $\mathbb{R}^{n}$, the partial derivative of $f$ with respect to the $i$th variable exists at $w$ with value $L_{0}$; that is, $y+te_{i}\in E_{i}$.

Consequently $\{t\in\mathbb{R}:y+te_{i}\in N\}\cap I_{m}\subseteq I_{m}\setminus D_{m}$, which is null, so this intersection is null by claim 5 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$ used in dimension $1$. By The Archimedean Property of the Real Numbers every $t\in\mathbb{R}$ satisfies $|t|<m$ for some $m\in\mathbb{N}$, so $\mathbb{R}=\bigcup_{m\in\mathbb{N}}I_{m}$ and therefore

$$\{t\in\mathbb{R}:y+te_{i}\in N\}=\bigcup_{m\in\mathbb{N}}\Bigl(\{t\in\mathbb{R}:y+te_{i}\in N\}\cap I_{m}\Bigr),$$

a countable union of null sets, which is null by claim 5 of Elementary Properties of Lebesgue Outer Measure on $\mathbb{R}^n$.

This holds for every $y\in\mathbb{R}^{n}$, so A Borel Set Whose Lines in One Coordinate Direction Are Null Is Null (i) gives $\lambda_{n}(N)=0$, and $N$ is null.

Claim 2. Let $y\in E_{i}$ and write $L_{0}=\partial_{i}f(y)$. Suppose $L<|L_{0}|$ and put $\varepsilon=|L_{0}|-L$, a positive real number. By Partial Derivative of a Real-Valued Function on an Open Subset of $\mathbb{R}^n$ there is $\delta>0$ such that every real $s$ with $0<|s|<\delta$ satisfies

$$\Bigl|\frac{f(y+se_{i})-f(y)}{s}-L_{0}\Bigr|<\varepsilon .$$

Fix such an $s$. By claim 7 of Properties of the Absolute Value in an Ordered Field the left-hand side is at least $|L_{0}|-|(f(y+se_{i})-f(y))/s|$, so

$$\Bigl|\frac{f(y+se_{i})-f(y)}{s}\Bigr|>|L_{0}|-\varepsilon=L .$$

On the other hand the Lipschitz hypothesis, claim 5 of Properties of the Euclidean Norm on $\mathbb{R}^n$: Square, Distance, Vanishing, Coordinate Bounds, Homogeneity, Triangle Inequalities and Expansion of a Square and $\lVert e_{i}\rVert=1$ give $|f(y+se_{i})-f(y)|\le L\lVert se_{i}\rVert=L|s|$, so by claim 4 of Properties of the Absolute Value in an Ordered Field the same quotient has absolute value at most $L$. This contradiction shows $|\partial_{i}f(y)|\le L$.