Proof of Borel Structure of the Set Where a Partial Derivative Exists

We use the notation of the statement, with $i$ fixed throughout. We first record an elementary continuity property of the difference quotients.

Sub-step (continuity in the increment). Let $y\in\mathbb{R}^{n}$, let $s\in\mathbb{R}$ with $s\neq0$, and let $\varepsilon\in\mathbb{R}$ with $0<\varepsilon$. Then there is $\tau\in\mathbb{R}$ with $0<\tau$ such that every $u\in\mathbb{R}$ with $u\neq0$ and $|u-s|<\tau$ satisfies $|\Delta_{u}(y)-\Delta_{s}(y)|<\varepsilon$.

Write $g(u)=f(y+ue_{i})$ for $u\in\mathbb{R}$ and $a=f(y)$, so that $\Delta_{u}(y)=(g(u)-a)/u$ for $u\neq0$, and put $M=|g(s)-a|$. For $u\neq0$ a direct computation gives

By claim 2 of Elementary Properties of the Euclidean Norm on $\mathbb{R}^n$ and claim 5 of that lemma, $d_{E}(y+ue_{i},y+se_{i})=\lVert(u-s)e_{i}\rVert=|u-s|\,\lVert e_{i}\rVert=|u-s|$, since $\lVert e_{i}\rVert=1$ by Orthonormal Families, Standard Basis Vectors, and Plane Rotations of Euclidean Space. Hence, $f$ being continuous at $y+se_{i}$, there is $\tau_{1}>0$ such that $|u-s|<\tau_{1}$ implies $|g(u)-g(s)|<\varepsilon|s|/4$. Put

a positive real number. Let $u\neq0$ with $|u-s|<\tau$. From $|u-s|<|s|/2$ and claim 7 of Properties of the Absolute Value in an Ordered Field we get $|u|>|s|/2$, so $1/|u|<2/|s|$, and therefore, using claim 5 and claim 4 of Properties of the Absolute Value in an Ordered Field,

This proves the sub-step.

Claim 1. Fix $j,k$ and let $y\in\mathbb{R}^{n}\setminus E_{j,k}$. Then there are $s,s'\in\mathbb{Q}$ with $0<|s|\le1/j$, $0<|s'|\le1/j$ and $|\Delta_{s}(y)-\Delta_{s'}(y)|>1/k$. Put $\theta=|\Delta_{s}(y)-\Delta_{s'}(y)|-1/k$, a positive real number, and put $\eta=\theta|s|\,|s'|/(4|s|+4|s'|)$, a positive real number.

Since $f$ is continuous at each of the three points $y$, $y+se_{i}$ and $y+s'e_{i}$, and since $d_{E}(z+ce_{i},y+ce_{i})=\lVert z-y\rVert$ for every $c\in\mathbb{R}$, there is $\delta>0$ such that every $z\in\mathbb{R}^{n}$ with $\lVert z-y\rVert<\delta$ satisfies

namely the smallest of the three radii supplied by continuity at those points. For such $z$, claim 5 of Properties of the Absolute Value in an Ordered Field gives

and likewise $|\Delta_{s'}(z)-\Delta_{s'}(y)|<2\eta/|s'|$. Since $2\eta/|s|+2\eta/|s'|=\theta/2$, the triangle inequality gives

so $z\notin E_{j,k}$. Thus $B(y,\delta)\subseteq\mathbb{R}^{n}\setminus E_{j,k}$, so $\mathbb{R}^{n}\setminus E_{j,k}$ is open and $E_{j,k}$ is closed. By the Borel $\sigma$-algebra clause of the setting, closed sets are Borel.

Claim 2. Let $y\in E_{j,k}$ and let $s,s'\in\mathbb{R}$ with $0<|s|\le1/j$ and $0<|s'|\le1/j$. Let $\varepsilon>0$.

We first choose a rational $\sigma$ with $0<|\sigma|\le1/j$ and $|\Delta_{\sigma}(y)-\Delta_{s}(y)|<\varepsilon/2$. If $s\in\mathbb{Q}$ take $\sigma=s$. Otherwise $|s|\neq1/j$, because $1/j$ is rational, so $0<|s|<1/j$; let $\tau$ be as in the sub-step for this $s$ and $\varepsilon/2$, and use The Rational Numbers are Dense in the Real Numbers to choose $\sigma\in\mathbb{Q}$ with $|\sigma-s|<\min\{\tau,|s|,1/j-|s|\}$. Then $|\sigma|\ge|s|-|\sigma-s|>0$ and $|\sigma|\le|s|+|\sigma-s|<1/j$ by claim 5 of Properties of the Absolute Value in an Ordered Field, and the sub-step gives $|\Delta_{\sigma}(y)-\Delta_{s}(y)|<\varepsilon/2$. Choose $\sigma'$ from $s'$ in the same way.

Since $y\in E_{j,k}$ and $\sigma,\sigma'$ are rational with $0<|\sigma|\le1/j$ and $0<|\sigma'|\le1/j$, we have $|\Delta_{\sigma}(y)-\Delta_{\sigma'}(y)|\le1/k$, and therefore

As $\varepsilon>0$ was arbitrary, $|\Delta_{s}(y)-\Delta_{s'}(y)|\le1/k$.

Claim 3. Suppose first $y\in E_{i}$ and put $L=\partial_{i}f(y)$; let $k\in\mathbb{N}$. There is $\delta>0$ with $|\Delta_{u}(y)-L|<1/(2k)$ for every real $u$ with $0<|u|<\delta$, and The Archimedean Property of the Real Numbers provides $j\in\mathbb{N}$ with $1/j<\delta$. For $s,s'\in\mathbb{Q}$ with $0<|s|\le1/j$ and $0<|s'|\le1/j$ we then have $0<|s|<\delta$ and $0<|s'|<\delta$, so by claim 5 of Properties of the Absolute Value in an Ordered Field

so $y\in E_{j,k}$. As $k$ was arbitrary, $y\in\bigcap_{k}\bigcup_{j}E_{j,k}$.

Conversely let $y\in\bigcap_{k}\bigcup_{j}E_{j,k}$ and, for each $k\in\mathbb{N}$, choose $j_{k}\in\mathbb{N}$ with $y\in E_{j_{k},k}$. Put $a_{m}=\Delta_{1/m}(y)$ for $m\in\mathbb{N}$. Given $\varepsilon>0$, The Archimedean Property of the Real Numbers provides $k$ with $1/k<\varepsilon$; if $m,m'\in\mathbb{N}$ satisfy $j_{k}\le m$ and $j_{k}\le m'$ then $1/m$ and $1/m'$ are rational numbers with $0<1/m\le1/j_{k}$ and $0<1/m'\le1/j_{k}$, so $|a_{m}-a_{m'}|\le1/k<\varepsilon$. Hence $(a_{m})_{m\in\mathbb{N}}$ is a Cauchy sequence of real numbers and, by Every Cauchy Sequence of Real Numbers Converges, it converges to some $L\in\mathbb{R}$.

We show that $\partial_{i}f(y)$ exists with value $L$. Let $\varepsilon>0$ and use The Archimedean Property of the Real Numbers to choose $k$ with $1/k<\varepsilon$; put $\delta=1/j_{k}$. Let $s\in\mathbb{R}$ with $0<|s|<\delta$. For every $m\in\mathbb{N}$ with $j_{k}\le m$ we have $0<1/m\le1/j_{k}$, so claim 2 applied to $y\in E_{j_{k},k}$ gives $|\Delta_{s}(y)-a_{m}|\le1/k$, that is, by claim 6 of Properties of the Absolute Value in an Ordered Field,

By Arithmetic of Limits of Real Sequences the sequences $(a_{m}-1/k)_{m}$ and $(a_{m}+1/k)_{m}$ converge to $L-1/k$ and $L+1/k$, so Order Properties of Limits of Real Sequences, applied to these sequences and to the constant sequence with value $\Delta_{s}(y)$, gives $L-1/k\le\Delta_{s}(y)\le L+1/k$, that is $|\Delta_{s}(y)-L|\le1/k<\varepsilon$. Since $\varepsilon>0$ was arbitrary, Partial Derivative on a Euclidean Open Set shows that $y\in E_{i}$ with $\partial_{i}f(y)=L$.

This proves $E_{i}=\bigcap_{k\in\mathbb{N}}\bigcup_{j\in\mathbb{N}}E_{j,k}$. Each $E_{j,k}$ is Borel by claim 1; a $\sigma$-algebra is closed under countable unions and under complements, hence also under countable intersections, so $E_{i}\in\mathcal{B}(\mathbb{R}^{n})$ and $\mathbb{R}^{n}\setminus E_{i}\in\mathcal{B}(\mathbb{R}^{n})$.

Claim 4. Let $y\in E_{i}\cap E_{j,k}$, put $L=\partial_{i}f(y)$, and let $s\in\mathbb{R}$ with $0<|s|\le1/j$. Let $\varepsilon>0$ and choose $\delta>0$ with $|\Delta_{u}(y)-L|<\varepsilon$ for every real $u$ with $0<|u|<\delta$. Let $s'$ be half the smaller of $\delta$ and $1/j$, so that $0<s'<\delta$ and $0<s'\le1/j$. By claim 2, $|\Delta_{s}(y)-\Delta_{s'}(y)|\le1/k$, hence

As $\varepsilon>0$ was arbitrary, $|\Delta_{s}(y)-\partial_{i}f(y)|\le1/k$.