The Lebesgue Integral of the Derivative of an Increasing Function

Theorem 1: Let $f$ be an increasing function on $[a, b]$. Then $f'(x) \geq 0$ (where it's defined) and $\displaystyle{\int_a^b f'(x) \: dx \leq f(b) - f(a)}$.

• Proof: Let $f$ be an increasing function on $[a, b]$ (and hence increasing on $(a, b)$). Then by Lebesgue's Theorem for the Differentiability of Monotone Functions, $f$ is differentiable almost everywhere on $(a, b)$ and so $f'(x)$ exists almost everywhere on $(a, b)$. For each $x \in (a, b)$ for which $f'(x)$ is defined, we have that the following limit exists and is finite:

(1)

\begin{align} \quad f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} \end{align}

• By letting $h = \frac{1}{n}$ and letting $n \to \infty$ we have that $h \to 0$ and so for each $x \in (a, b)$ for which $f'(x)$ is defined we have that:

(2)

\begin{align} \quad f'(x) = \lim_{n \to \infty} \frac{f \left ( x + \frac{1}{n} \right ) - f(x)}{\frac{1}{n}} \quad (*) \end{align}

• (We must take care as for certain $n$ that are too large, $f$ may not be defined as $x + \frac{1}{n}$. We consider $n$ sufficiently large so that $f \left ( x + \frac{1}{n} \right )$ is defined).

• Since $f$ is an increasing function, $f \left (x + \frac{1}{n} \right ) - f(x) \geq 0$ and clearly $\frac{1}{n} \geq 0$. Also note that $f'(x)$ is a pointwise limit of Lebesgue measurable functions, so $f'(x)$ is a Lebesgue measurable function.

• Now the functions in the limit at $(*)$ are all Lebesgue measurable and nonnegative, and as $n \to \infty$ these functions converge pointwise to $f'(x)$ almost everywhere on $[a, b]$. So by Fatou's Lemma for Nonnegative Lebesgue Measurable Functions we have that:

(3)

\begin{align} \quad \int_a^b f'(x) \: dx & \leq \liminf_{n \to \infty} \int_a^b \frac{f \left ( x + \frac{1}{n} \right ) - f(x)}{\frac{1}{n}} \: dx \\ & \leq \liminf_{n \to \infty} \left [ n \int_a^b f \left (x + \frac{1}{n} \right ) \: dx - n \int_a^b f(x) \: dx \right ] \end{align}

• Now we have that:

(4)

\begin{align} \quad n \int_a^b f \left (x + \frac{1}{n} \right ) \: dx - n \int_a^b f(x) \: dx &= n \left [ \int_{a + \frac{1}{n}}^{b + \frac{1}{n}} f(x) \: dx - \int_a^b f(x) \: dx \right ] \\ &= n \left [ \int_{a + \frac{1}{n}}^b f(x) \: dx + \int_b^{b+\frac{1}{n}} f(x) \: dx - \int_a^{a+\frac{1}{n}} f(x) \: dx - \int_{a+\frac{1}{n}}^{b} f(x) \: dx \right ] \\ &= n \left [ \int_b^{b+\frac{1}{n}} f(x) \: dx - \int_a^{a+\frac{1}{n}} f(x) \: dx \right ] \quad (**) \end{align}

• Since $f$ is increasing on $[a, b]$ we have that $f(a) \leq f(x)$ for all $x \in [a, b]$ and so:

(5)

\begin{align} \quad \frac{1}{n} f(a) = \int_a^{a+\frac{1}{n}} f(a) \: dx \leq \int_a^{a+\frac{1}{n}} f(x) \: dx \end{align}

• Therefore:

(6)

\begin{align} \quad - \int_a^{a+\frac{1}{n}} f(x) \: dx \leq -\frac{1}{n} f(a) \quad (\dagger) \end{align}

• Similarly, we have that $f(x) \leq f(b)$ for all $x \in [a, b]$ and so:

(7)

\begin{align} \quad \int_b^{b+\frac{1}{n}} f(x) \: dx \leq \int_b^{b+\frac{1}{n}} f(b) \: dx = \frac{1}{n} f(b) \quad (\dagger \dagger) \end{align}

• So from $(\dagger)$ and $(\dagger \dagger)$ and $(**)$ we get:

(8)

\begin{align} \quad n \int_a^b f \left (x + \frac{1}{n} \right ) \: dx - n \int_a^b f(x) \: dx & \leq n \left [ \frac{1}{n} f(b) - \frac{1}{n} f(a) \right ] = f(b) - f(a) \end{align}

• Since this holds for all adequate $n$ we have that:

(9)

\begin{align} \quad \int_a^b f'(x) \: dx \leq f(b) - f(a) \quad \blacksquare \end{align}