I'm following a lecture note to prove that a monotonic function is differentiable almost everywhere. Below is an exposition of my understanding. I have no question, but I'm very happy to receive your suggestion.
Given a function $f:(a, b) \rightarrow \mathbb{R}$, its lower and upper derivatives are defined respectively as $$ \begin{aligned} & \underline{D} f(x)=\liminf _{y \rightarrow x} \frac{f(y)-f(x)}{y-x}, \\ & \overline{D} f(x)=\limsup _{y \rightarrow x} \frac{f(y)-f(x)}{y-x} . \end{aligned} $$
Clearly, $\underline{D} f \leq \overline{D} f$. Also, $f$ is differentiable at $x$ IFF $\underline{D} f(x)=\overline{D} f(x)$ is a finite value.
Lemma Let $f: [a, b] \rightarrow \mathbb{R}$ be non-decreasing. Then $$ \lambda^* \{x \in(a, b): \overline{D} f(x)>\alpha\} \leq \frac{f(b)-f(a)}{\alpha} \quad \forall \alpha>0. $$ Here $\lambda^*$ be the Lebesgue outer measure.
Proof Let $E_\alpha := \{x \in(a, b): \overline{D} f(x)>\alpha\}$. For each $(x, \delta) \in E_\alpha \times \mathbb R_{>0}$, there is $0 < |h| < \delta$ such that $$ \frac{f(x+h)-f(x)}{h} \ge \alpha. $$
It follows that the collection $\mathcal I$ of closed intervals with $$ \mathcal I := \left \{ [c, d] \subset (a, b) : \frac{f(d)-f(c)}{d-c} \ge \alpha \right \} $$ is a Vitali cover of $E_\alpha$. Fix $\varepsilon >0$. By Vitali's covering lemma, there is a finite subcollection $([c_i, d_i])_{i=1}^n$ of pairwise disjoint intervals in $\mathcal I$ such that $$ \lambda^* \left ( E_\alpha \setminus \bigcup_{i=1}^n [c_i, d_i] \right ) < \varepsilon. $$
Because $f$ is non-decreasing, $$ \begin{align} \lambda^* \left ( \bigcup_{i=1}^n [c_i, d_i] \right ) &=\sum_{i=1}^n (d_i-c_i) \\ &\le \frac{1}{\alpha} \sum_{i=1}^n (f(d_i)-f(c_i)) \\ &\le \frac{f(b) - f(a))}{\alpha}. \end{align} $$
Hence $$ \begin{align} \lambda^* (E_\alpha) &\le \lambda^* \left ( E_\alpha \setminus \bigcup_{i=1}^n [c_i, d_i] \right ) + \lambda^* \left ( \bigcup_{i=1}^n [c_i, d_i] \right ) \\ &\le \varepsilon + \frac{f(b) - f(a))}{\alpha}. \end{align} $$
The claim then follows by taking the limit $\varepsilon \downarrow 0$. $\tag*{$\blacksquare$}$
Corollary Let $f:[a, b] \to \mathbb{R}$ be non-decreasing. Then $$ \lambda\{x \in(a, b): \overline{D} f(x)=\infty\}=0 . $$ Here $\lambda$ be the Lebesgue measure.
Now comes our main theorem.
Theorem Let $I$ be an interval in $\mathbb{R}$ and $f: I \rightarrow \mathbb{R}$ monotonic. Then $f$ is differentiable $\lambda$-a.e. on $I$.
Proof It suffices to consider $I= [a, b]$ and $f$ is non-decreasing. Fix $\varepsilon>0$. For $s, t \in \mathbb{R}$ with $s<t$,
- define $A_{s, t}=\{x \in(a, b): \underline{D} f(x)<s <t < \overline{D} f(x)\}$,
- pick an open subset $O$ of $\mathbb R$ such that $A_{s, t} \subset O \subset (a, b)$ and $\lambda(O)<\lambda^*\left(A_{s, t}\right)+\varepsilon$,
- let $$ \mathcal{I}_{s, t} = \left\{[c, d] \subseteq O: \frac{f(d)-f(c)}{d-c} \leq s\right\} . $$
As in the proof of the Lemma, $\mathcal I_{s, t}$ is a Vitali cover of $A_{s, t}$. Since $\lambda^*\left(A_{s, t}\right) \leq b-a<\infty$, by Vitali's covering theorem there is a finite subcollection of pairwise disjoint intervals in $\mathcal{I}_{s, t}$, denoted by $\left(\left[c_i, d_i\right]\right)_{i=1}^n$, such that $$ \lambda^*\left(A_{s, t} \backslash \bigcup_{i=1}^n\left[c_i, d_i\right]\right)<\varepsilon . $$
By the Lemma, $$ \lambda^*\left(A_{s, t} \cap\left[c_i, d_i\right]\right) \leq \frac{f\left(d_i\right)-f\left(c_i\right)}{t} \leq \frac{s}{t}\left(d_i-c_i\right) \quad \forall i = 1, \ldots, n. $$
Hence $$ \begin{aligned} \lambda^*\left(A_{s, t}\right) & <\varepsilon+\frac{s}{t} \sum_{i=1}^n\left(d_i-c_i\right) \\ & \leq \varepsilon+\frac{s}{t} \lambda(O) \\ & \leq \varepsilon+\frac{s}{t}\left(\lambda^*\left(A_{s, t}\right)+\varepsilon\right) \end{aligned} $$
Taking $\varepsilon \downarrow 0$ gives $\lambda^*\left(A_{s, t}\right) \leq \frac{s}{t} \lambda^*\left(A_{s, t}\right)$, which implies $\lambda\left(A_{s, t}\right)=0$. Note that $f$ is not differentiable at $x$ if and only if there exist $s, t \in \mathbb{Q}$ with $$ \underline{D} f(x)<s<t<\overline {D} f(x) . $$ Thus $f$ is not differentiable only on the set $\bigcup_{s, t \in \mathbb{Q}} A_{s, t}$, which is a countable union of $\lambda$-null sets, and is hence $\lambda$-null. Hence $f$ is differentiable $\lambda$-a.e. $\tag*{$\blacksquare$}$
Proposition Let $f:[a, b] \rightarrow \mathbb{R}$ be non-decreasing, and let $$ g(x)= \begin{cases} f^{\prime}(x) & \text{if } f \text { differentiable at } x, \\ 0 & \text {otherwise }. \end{cases} $$ Then $g \geq 0$, and $$ \int_a^b g(x) \mathrm d x \leq f(b)-f(a). $$
Proof By Theorem, $f$ is differentiable $\lambda$-a.e. We define $$ g_n(x)= \begin{cases} \frac{f\left(x+\frac{1}{n}\right)-f(x)}{\frac{1}{n}} & \text{if } a \leq x \leq b-\frac{1}{n}, \\ 0 & \text{if } b-\frac{1}{n}<x \leq b. \end{cases} \quad \forall x \in [a, b]. $$
Then $g_n$ are well-defined and converge pointwise to $g$ for $\lambda$-a.e. $x \in[a, b]$. By Fatou's lemma, $$ \begin{aligned} \int_a^b g (x) \mathrm d x & \leq \liminf _{n \to \infty} \int_a^{b-1/n} g_n (x) \mathrm d x \\ & = \liminf _{n \to \infty} n \left [ \int_{b-1/n}^b f (x) \mathrm d x - \int_a^{a+1/n} f(x) \mathrm d x \right ] \\ &\le \liminf _{n \to \infty} n \left ( \frac{f(b)}{n} -\frac{(a)}{n} \right ) \\ &= f(b) - f(a). \end{aligned} $$
$\tag*{$\blacksquare$}$
