Proof of A Nondecreasing Function on an Open Interval is Differentiable Almost Everywhere

Notation. For real $p<q$ write $J=[p,q]$ for the closed interval with these endpoints, $|J|=q-p$ and, when $J\subseteq I$, $\Delta J=g(q)-g(p)$; since $g$ is nondecreasing, $\Delta J\ge0$. By claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness we have $\lambda(J)=|J|$, and $J$ is the closed ball with centre $(p+q)/2$ and radius $(q-p)/2$, so families of such intervals are families of closed balls in the sense of The Vitali Covering Theorem in $\mathbb{R}^n$, an interval of length $\ell$ having radius $\ell/2$.

Step S: increments add up inside an interval. Let $J\subseteq I$ be a closed interval and let $K_{1},\dots,K_{M}\subseteq J$ be pairwise disjoint closed intervals. Then $\sum_{j=1}^{M}\Delta K_{j}\le\Delta J$. Indeed, relabelling the finitely many intervals we may assume their left endpoints increase, writing $K_{j}=[p_{j},q_{j}]$ with $p_{1}\le\dots\le p_{M}$; disjointness then forces $q_{j}<p_{j+1}$ for $j<M$, and $g$ nondecreasing gives $g(q_{j})\le g(p_{j+1})$. Writing $J=[p,q]$ we have $p\le p_{1}$ and $q_{M}\le q$, so

$$\sum_{j=1}^{M}\bigl(g(q_{j})-g(p_{j})\bigr)\le\sum_{j=1}^{M}\bigl(g(q_{j})-g(p_{j})\bigr)+\sum_{j=1}^{M-1}\bigl(g(p_{j+1})-g(q_{j})\bigr)=g(q_{M})-g(p_{1})\le g(q)-g(p),$$

all the discarded terms being nonnegative.

Step 0: reduction to a compact subinterval. For every $x\in I$ there are rationals $a<b$ with $x\in(a,b)$ and $[a,b]\subseteq I$: since $I$ is open there is a real $\rho>0$ with $\{t:|t-x|<\rho\}\subseteq I$, and The Rational Numbers are Dense in the Real Numbers supplies rationals $a$ and $b$ with $x-\rho<a<x<b<x+\rho$. The set of pairs of rationals is countable by The Integers and the Rational Numbers are Countable and Products and Powers of Countable Sets, so $I$ is the union of a countable family of open intervals $(a,b)$ with $a,b$ rational and $[a,b]\subseteq I$. Since a countable union of null sets is null by the null-set claim, it suffices to prove that $(a,b)\setminus D$ is null for each such pair.

Fix such $a<b$, put $J_{0}=[a,b]\subseteq I$ and $T=\Delta J_{0}=g(b)-g(a)\ge0$. All intervals below are contained in $(a,b)$, and every subset of $(a,b)$ has outer measure at most $b-a<\infty$ by monotonicity and agreement on Borel sets.

Step 1: the points of unbounded difference quotients. For $M\in\mathbb{N}$ let $G_{M}$ be the set of $x\in(a,b)$ such that for every real $\eta>0$ there is a closed interval $J$ with $x\in J\subseteq(a,b)$, $0<|J|<\eta$ and $\Delta J>M|J|$. We claim $\lambda^{\ast}(G_{M})\le T/M$.

The pairs carrying these intervals finely cover $G_{M}$ in the sense of the fine covering condition: given $x\in G_{M}$ and $\eta>0$, apply the definition with $2\eta$ in place of $\eta$ to obtain such a $J$ containing $x$ of radius $|J|/2<\eta$. Let $\varepsilon>0$. By the Vitali covering theorem there are finitely many pairwise disjoint intervals $J_{1},\dots,J_{N}$ of this family with $\lambda^{\ast}(G_{M}\setminus\bigcup_{i}J_{i})\le\varepsilon$. Using subadditivity and monotonicity of $\lambda^{\ast}$, then $M|J_{i}|<\Delta J_{i}$, and finally Step S with $J=J_{0}$,

$$\lambda^{\ast}(G_{M})\le\varepsilon+\sum_{i=1}^{N}|J_{i}|\le\varepsilon+\frac{1}{M}\sum_{i=1}^{N}\Delta J_{i}\le\varepsilon+\frac{T}{M}.$$

This holds for every real $\varepsilon>0$; were $\lambda^{\ast}(G_{M})>T/M$, taking $\varepsilon=\bigl(\lambda^{\ast}(G_{M})-T/M\bigr)/2$ would contradict it. Hence $\lambda^{\ast}(G_{M})\le T/M$, as claimed. Consequently $P=\bigcap_{M\in\mathbb{N}}G_{M}$ satisfies $\lambda^{\ast}(P)\le T/M$ for every $M\in\mathbb{N}$, by monotonicity; if $\lambda^{\ast}(P)$ were positive then The Archimedean Property of the Real Numbers would supply $M\in\mathbb{N}$ with $T/M<\lambda^{\ast}(P)$, a contradiction. So $\lambda^{\ast}(P)=0$ and $P$ is null by the null-set claim.

Step 2: the points of oscillating difference quotients. Let $u,v$ be rationals with $0\le u<v$ and let $E_{u,v}$ be the set of $x\in(a,b)$ such that for every real $\eta>0$ there are closed intervals $J,K$ with $x\in J\subseteq(a,b)$, $x\in K\subseteq(a,b)$, $0<|J|<\eta$, $0<|K|<\eta$,

$$\Delta J<u\,|J|\qquad\text{and}\qquad \Delta K>v\,|K| .$$

We claim $\lambda^{\ast}(E_{u,v})=0$. Put $s=\lambda^{\ast}(E_{u,v})$, a real number with $0\le s\le b-a$, and suppose $0<s$. Let $\varepsilon$ be a real number with $0<\varepsilon<s/2$.

By outer regularity of the outer measure there is an open $O'$ with $E_{u,v}\subseteq O'$ and $\lambda(O')\le s+\varepsilon$; put $O=O'\cap(a,b)$, again open, still containing $E_{u,v}$, with $\lambda(O)\le s+\varepsilon$.

First application. Let $\mathcal{F}_{1}$ consist of the pairs carrying the closed intervals $J$ with $J\subseteq O$ and $\Delta J<u|J|$. This family finely covers $E_{u,v}$: for $x\in E_{u,v}$ and $\eta>0$, choose a real $\rho>0$ with $\{t:|t-x|<\rho\}\subseteq O$ and apply the definition of $E_{u,v}$ with $\min\{2\eta,\rho\}$ in place of $\eta$; the resulting $J$ contains $x$, has $|J|<\rho$ and therefore $J\subseteq O$, and has radius $<\eta$. By the Vitali covering theorem there are pairwise disjoint $J_{1},\dots,J_{N}\in\mathcal{F}_{1}$ with $\lambda^{\ast}(E_{u,v}\setminus\bigcup_{i}J_{i})\le\varepsilon$. Being pairwise disjoint Borel subsets of $O$, they satisfy $\sum_{i}|J_{i}|\le\lambda(O)\le s+\varepsilon$, whence

$$\sum_{i=1}^{N}\Delta J_{i}<u\sum_{i=1}^{N}|J_{i}|\le u(s+\varepsilon). \tag{1}$$

Second application. Let $A=E_{u,v}\cap\bigcup_{i=1}^{N}(p_{i},q_{i})$, where $J_{i}=[p_{i},q_{i}]$. The finite set $F$ of the endpoints $p_{i},q_{i}$ is null, being a finite union of singletons, each of measure $0$ by claim 4 of Lebesgue Measure on the Real Line: Existence, Lengths of Intervals and Uniqueness applied to an interval with equal endpoints, and a countable union of null sets being null by the null-set claim; moreover $E_{u,v}\subseteq(E_{u,v}\setminus\bigcup_{i}J_{i})\cup A\cup F$, so subadditivity gives $s\le\varepsilon+\lambda^{\ast}(A)$, that is $\lambda^{\ast}(A)\ge s-\varepsilon$.

Let $\mathcal{F}_{2}$ consist of the pairs carrying the closed intervals $K$ with $K\subseteq(p_{i},q_{i})$ for some $i$ and $\Delta K>v|K|$. This family finely covers $A$: for $x\in A$ pick $i$ with $x\in(p_{i},q_{i})$ and a real $\rho>0$ with $\{t:|t-x|<\rho\}\subseteq(p_{i},q_{i})$, then apply the definition of $E_{u,v}$ with $\min\{2\eta,\rho\}$ in place of $\eta$. By the Vitali covering theorem there are pairwise disjoint $K_{1},\dots,K_{M}\in\mathcal{F}_{2}$ with $\lambda^{\ast}(A\setminus\bigcup_{j}K_{j})\le\varepsilon$, so subadditivity gives

$$\sum_{j=1}^{M}|K_{j}|\ge\lambda^{\ast}(A)-\varepsilon\ge s-2\varepsilon>0 .$$

Each $K_{j}$ lies in exactly one $J_{i}$, the $J_{i}$ being pairwise disjoint; grouping the $K_{j}$ accordingly and applying Step S inside each $J_{i}$ gives $\sum_{j}\Delta K_{j}\le\sum_{i}\Delta J_{i}$. Combining this with $\Delta K_{j}>v|K_{j}|$ and with $(1)$,

$$v\,(s-2\varepsilon)\le v\sum_{j=1}^{M}|K_{j}|<\sum_{j=1}^{M}\Delta K_{j}\le\sum_{i=1}^{N}\Delta J_{i}<u\,(s+\varepsilon).$$

This holds for every real $\varepsilon$ with $0<\varepsilon<s/2$. If $vs>us$ then, $v$ and $u$ being fixed, the Archimedean property supplies such an $\varepsilon$ with $(2v+u)\varepsilon<vs-us$, contradicting the display; hence $vs\le us$, so $(v-u)s\le0$. As $u<v$ and $s>0$ this is impossible, and therefore $s=0$.

Step 3: assembling. Let $x\in(a,b)$ belong to none of the sets $\bigcap_{M}G_{M}$ and $E_{u,v}$ (over rationals $0\le u<v$). We show $x\in D$.

For $h\ne0$ with $x+h\in(a,b)$ let $J^{h}$ be the closed interval with endpoints $x$ and $x+h$ and put $q(h)=\bigl(g(x+h)-g(x)\bigr)/h$. Then $|J^{h}|=|h|$ and $\Delta J^{h}=|g(x+h)-g(x)|$, because $g$ is nondecreasing; as $g(x+h)-g(x)$ and $h$ have the same sign, $q(h)=\Delta J^{h}/|J^{h}|\ge0$.

Since $x\notin\bigcap_{M}G_{M}$ there is $M\in\mathbb{N}$ with $x\notin G_{M}$, so there is a real $\eta_{0}>0$ such that every closed interval $J$ with $x\in J\subseteq(a,b)$ and $0<|J|<\eta_{0}$ has $\Delta J\le M|J|$. Applying this to $J^{h}$ gives $0\le q(h)\le M$ whenever $0<|h|<\eta_{0}$ and $x+h\in(a,b)$. For real $\eta$ with $0<\eta\le\eta_{0}$ let

$$Q(\eta)=\{q(h):0<|h|<\eta,\ x+h\in(a,b)\},$$

a nonempty subset of $[0,M]$; let $m(\eta)$ be its greatest lower bound and $\bar{m}(\eta)$ its least upper bound, which exist by the least upper bound property and by Existence of the Infimum of a Nonempty Subset of $\mathbb{R}$ Bounded Below. As $\eta$ decreases $Q(\eta)$ shrinks, so $m$ does not decrease and $\bar m$ does not increase; both stay in $[0,M]$. Let $\alpha$ be the least upper bound of $\{m(\eta):0<\eta\le\eta_{0}\}$ and $\beta$ the greatest lower bound of $\{\bar m(\eta):0<\eta\le\eta_{0}\}$; then $0\le\alpha\le\beta\le M$, since $m(\eta)\le\bar m(\eta)$ for every $\eta$ and $m(\eta)\le\bar m(\eta')$ for all $\eta,\eta'$ by comparing at $\min\{\eta,\eta'\}$.

Suppose $\alpha<\beta$. By The Rational Numbers are Dense in the Real Numbers choose rationals $u,v$ with $\alpha<u<v<\beta$; then $0\le u$. For every $\eta$ with $0<\eta\le\eta_{0}$ we have $m(\eta)\le\alpha<u$, so $u$ is not a lower bound of $Q(\eta)$ and there is $h$ with $0<|h|<\eta$ and $q(h)<u$, that is $\Delta J^{h}<u|J^{h}|$; and $\bar m(\eta)\ge\beta>v$, so there is $h'$ with $0<|h'|<\eta$ and $\Delta J^{h'}>v|J^{h'}|$. Hence $x\in E_{u,v}$, contrary to the choice of $x$. Therefore $\alpha=\beta$; call this common value $c$.

Finally, let $\varepsilon>0$. There is $\eta_{1}$ with $m(\eta_{1})>c-\varepsilon$ and $\eta_{2}$ with $\bar m(\eta_{2})<c+\varepsilon$, since $c$ is the least upper bound of the $m(\eta)$ and the greatest lower bound of the $\bar m(\eta)$. Put $\delta=\min\{\eta_{1},\eta_{2},x-a,b-x\}$, a positive real because $a<x<b$. If $0<|h|<\delta$ and $x+h\in I$, then $|h|<\min\{x-a,b-x\}$ forces $x+h\in(a,b)$, so $q(h)$ belongs to $Q(\eta_{1})$ and to $Q(\eta_{2})$ and therefore satisfies $c-\varepsilon<m(\eta_{1})\le q(h)\le\bar m(\eta_{2})<c+\varepsilon$. Thus for every real $\varepsilon>0$ there is a real $\delta>0$ such that the difference quotient of $g$ at $x$ differs from $c$ by less than $\varepsilon$ for all $h$ with $0<|h|<\delta$ and $x+h\in I$; that is, $g$ has a derivative $g'(x)=c$ at the interior point $x$ of $I$, so $x\in D$.

Therefore $(a,b)\setminus D$ is contained in the union of $\bigcap_{M}G_{M}$ and of the sets $E_{u,v}$ over the countably many pairs of rationals with $0\le u<v$, all of which are null; by the null-set claim it is null. By Step 0, $I\setminus D$ is null.