Lebesgue's Theorem for the Differentiability of Monotone Functions

# Lebesgue's Theorem for the Differentiability of Monotone Functions

On the Upper and Lower Derivatives of Real-Valued Functions page we defined the upper and lower derivatives of a real-valued function $f$ defined on an open interval $I$ by:

(1)
\begin{align} \quad \overline{D}f(x) = \limsup_{h \to 0} \frac{f(x + h) - f(x)}{h} \quad \mathrm{and} \quad \underline{D}f(x) = \liminf_{h \to 0} \frac{f(x + h) - f(x)}{h} \end{align}

We said that a function $f$ is differentiable at $x \in I$ if the upper and lower derivatives of $f$ at $x$ are finite and equal.

We then proved an important result. If $f$ is an increasing function on $[a, b]$ then for all $\alpha \in \mathbb{R}$ with $\alpha > 0$ we have that:

(2)
\begin{align} \quad m^*( \{ x \in (a, b) : \overline{D} f(x) \geq \alpha \}) \leq \frac{1}{\alpha} [f(b) - f(a)] \end{align}
(3)
\begin{align} \quad m^*( \{ x \in (a, b) : \overline{D} f(x) = \infty \}) = 0 \end{align}

We use these results to prove an extremely important theorem called Lebesgue's theorem for the differentiability of monotone functions. This result tells us that every monotone function $f$ (increasing or decreasing) defined on an open interval $(a, b)$ is differentiable almost everywhere on $(a, b)$.

 Theorem 1 (Lebesgue's Theorem for the Differentiability of Monotone Functions): Let $f$ be a monotone function on $(a, b)$ (bounded or unbounded). Then $f$ is differentiable almost everywhere on $(a, b)$. • Proof: Let $f$ be an increasing function on the interval $(a, b)$ .
• It suffices to prove the result for when $(a, b)$ is a bounded open interval. If $(a, b)$ is an unbounded, we can write $\displaystyle{(a, b) = \bigcup_{n=1}^{\infty} (a_n, b_n)}$ where $\{ (a_n, b_n) \}_{n=1}^{\infty}$ is a sequence of bounded open intervals such that $(a_n, b_n) \subseteq (a_{n+1}, b_{n+1})$ for all $n \in \mathbb{N}$.
• Thus if $f$ is differentiable almost everywhere on each bounded open interval $(a_n, b_n)$ there exists a set $E_n \subseteq (a_n, b_n)$ such that $m(E_n) = 0$ and $f$ is differentiable on all of $(a_n, b_n) \setminus E_n$. So $f$ is differentiable on all of $\bigcup_{n=1}^{\infty} (a_n, b_n) \setminus E_n = (a, b) \setminus \bigcup_{n=1}^{\infty} E_n$. But a countable union of sets of Lebesgue measure $0$ has Lebesgue measure $0$, so $f$ is differentiable almost everywhere on $(a, b)$.
• Now recall that $f$ is said to be differentiable at $x \in (a, b)$ if the upper and lower derivatives of $f$ are finite and equal. So the set of points for which $f$ is not differentiable on $(a, b)$ is:
(4)
\begin{align} \quad \{ x \in (a, b) : \overline{D} f(x) = \infty \} \cup \{ x \in (a, b) : \infty > \overline{D} f() > \underline{D} f(x) \} \end{align}
• We have already proven that if $f$ is an increasing function then $m^*(\{ x \in (a, b) : \overline{D} f(x) = \infty \}) = 0$. So we need to show that $m^*(\{ x \in (a, b) : \infty > \overline{D} f() > \underline{D} f(x) \}) = 0$.
• For each $\alpha, \beta \in \mathbb{Q}$, by letting $E_{\alpha, \beta} = \{ x \in (a, b) : \underline{D}f(x) < \beta < \alpha < \overline{D}f(x) < \infty \}$, we can write the above set as the following union:
(5)
\begin{align} \quad \{ x \in (a, b) : \infty > \overline{D} f(x) > \underline{D} f(x) \} &= \bigcup_{\alpha, \beta \in \mathbb{Q}} \{ x \in (a, b): \underline{D}f(x) < \beta < \alpha < \overline{D}f(x) < \infty \} \\ &= \bigcup_{\alpha, \beta \in \mathbb{Q}} E_{\alpha, \beta} \end{align}
• So we need to show that $m^*(E_{\alpha, \beta}) = 0$ for each $\alpha, \beta \in \mathbb{Q}$.
• Let $\epsilon > 0$ be given. For a fixed $\alpha, \beta \in \mathbb{R}$ with $\beta < \alpha$ let $E = E_{\alpha, \beta}$. Then there exists an open set $O'$ with $E \subseteq O'$ such that $m^*(O') \leq m^*(E) + \epsilon$. Let $O = O' \cap (a, b)$. Then $O$ is also an open set, and is such that $E \subseteq O \subseteq (a, b)$ and such that:
(6)
\begin{align} \quad m^*(O) \leq m^*(O') < m^*(E) + \epsilon \end{align}
• Define the collection of closed intervals $\mathcal I$ by:
(7)
\begin{align} \quad \mathcal I = \{ [c, d] \subseteq O : f(d) - f(c) < \beta (d - c) \} \end{align}
• Then $\mathcal I$ is a Vitali cover of $E$. (This can be proven in a similar manner to a part of the proof mentioned at the top of this page). Since $E \subseteq (a, b)$ and we're assuming that $(a, b)$ is a bounded open interval we have that $m(E) < \infty$. So by the Vitali cover theorem, since $m(E) < \infty$ and $\mathcal I$ is a Vitali cover of $E$ there exists a finite collection of mutually disjoint closed intervals $\{ [c_1, d_1], [c_2, d_2], ..., [c_n, d_n] \}$ in $\mathcal I$ such that:
(8)
\begin{align} \quad m^* \left ( E \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) < \epsilon \quad (\dagger) \end{align}
• Now:
(9)
\begin{align} \quad \sum_{k=1}^{n} [f(d_k) - f(c_k)] < \sum_{k=1}^{n} \beta [d_k - c_k] < \beta m^*(O) < \beta [m^*(E) + \epsilon] \quad (*) \end{align}
• Now note that for each $k \in \{ 1, 2, ..., n \}$ we have that $E \cap [c_k, d_k] \subseteq \{ x \in [c_k, d_k] : \overline{D} f(x) \geq |\alpha| \}$. So for each $k$ we have
(10)
\begin{align} \quad m^*(E \cap [c_k, d_k]) \leq m^* (\{ x \in [c_k, d_k] : \overline{D} f(x) \geq |\alpha| \}) \leq \frac{1}{|\alpha|} [f(d_k) - f(c_k)] \quad (**) \end{align}
• We write $E = \left ( E \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) \cup \left ( \bigcup_{k=1}^{n} [c_k, d_k] \cap E \right )$ so that then:
(11)
\begin{align} \quad m^*(E) &= m^* \left ( E \setminus \bigcup_{k=1}^{n} [c_k, d_k] \right ) + m^* \left ( \bigcup_{k=1}^{n} [c_k, d_k] \cap E \right ) \\ & \overset{(\dagger)} < \epsilon + \sum_{k=1}^{n} m^* ([c_k, d_k] \cap E) \quad (\mathrm{since \: the \:} [c_k, d_k]'s \: \mathrm{are \: mutually \: disjoint} \\ & \overset{(**)} < \epsilon + \sum_{k=1}^{n} \frac{1}{\alpha} [f(d_k) - f(c_k)] \\ & \overset{(*)} < \epsilon + \frac{\beta}{|\alpha|} (m^*(E) + \epsilon) \end{align}
• Since $\epsilon > 0$ is arbitrary we get that $m^*(E) \leq \frac{\beta}{|\alpha|} m^*(E)$. Note that if $m^*(E) > 0$ then this inequality implies that $1 \leq \frac{\beta}{|\alpha|}$ and so $|\alpha| \leq \beta$ so $\alpha \leq \beta$ which is contradiction since $\beta < \alpha$. Therefore we must have that $m^*(E) = 0$.
• Hence $f$ is differentiable almost everywhere on $(a, b)$. $\blacksquare$