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

Claim 1. Define $G:I\to\mathbb{R}$ by $G(t)=g(t)+Lt$. If $x,y\in I$ with $x\le y$ then, by the Lipschitz hypothesis, $g(x)-g(y)\le|g(x)-g(y)|\le L|x-y|=L(y-x)$, hence

$$G(y)-G(x)=\bigl(g(y)-g(x)\bigr)+L(y-x)\ge0 .$$

Thus $G$ is nondecreasing on $I$. By the differentiation theorem for nondecreasing functions, the set $I\setminus D_{G}$ is null, where $D_{G}$ is the set of points of $I$ at which $G$ has a derivative.

We show $D_{G}\subseteq D$, which gives $I\setminus D\subseteq I\setminus D_{G}$ and hence, by the null-set claim, that $I\setminus D$ is null. Let $x\in D_{G}$ and put $c=G'(x)$. For $h\ne0$ with $x+h\in I$,

$$\frac{g(x+h)-g(x)}{h}=\frac{G(x+h)-G(x)-Lh}{h}=\frac{G(x+h)-G(x)}{h}-L .$$

Given a real $\varepsilon>0$, the definition of the derivative of $G$ at the interior point $x$ supplies a real $\delta>0$ such that the left-hand quotient above differs from $c$ by at most $\varepsilon$ whenever $0<|h|<\delta$ and $x+h\in I$; by the displayed identity the difference quotients of $g$ then differ from $c-L$ by at most $\varepsilon$. Hence $g$ has the derivative $g'(x)=c-L$ at $x$, so $x\in D$.

Claim 2. Let $x\in D$. For $h\ne0$ with $x+h\in I$ the Lipschitz hypothesis gives

$$\Bigl|\frac{g(x+h)-g(x)}{h}\Bigr|=\frac{|g(x+h)-g(x)|}{|h|}\le L .$$

Suppose $|g'(x)|>L$ and put $\varepsilon=|g'(x)|-L>0$. By the definition of the derivative there is a real $\delta>0$ such that every $h$ with $0<|h|<\delta$ and $x+h\in I$ has its difference quotient within $\varepsilon$ of $g'(x)$; and such an $h$ exists, because $I$ is open, so some real $\rho>0$ has $\{t:|t-x|<\rho\}\subseteq I$ and any $h$ with $0<|h|<\min\{\delta,\rho\}$ will do. Fix such an $h$, for which the difference quotient differs from $g'(x)$ by less than $\varepsilon$. Then, by the reverse triangle inequality for the absolute value (Properties of the Absolute Value in an Ordered Field), that quotient has absolute value greater than $|g'(x)|-\varepsilon=L$, contradicting the display. Hence $|g'(x)|\le L$.