Differentiable Function with Bounded Derivative is Absolutely Continuous

Theorem
Let $a, b$ be real numbers with $a < b$.

Let $f : \closedint a b \to \R$ be a continuous function.

Let $f$ be differentiable on $\openint a b$, with bounded derivative.

Then $f$ is absolutely continuous.

Proof
Since the derivative of $f$ is bounded, there exists some $M \in \R_{> 0}$ such that:


 * $\size {\map {f'} x} \le M$

for all $x \in \openint a b$.

Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals.

Note that for each $i \in \set {1, 2, \ldots, n}$:


 * $f$ is continuous on $\closedint {a_i} {b_i}$ and differentiable on $\openint {a_i} {b_i}$.

So, by the Mean Value Theorem, for each $i$ there exists some $\xi_i \in \openint {a_i} {b_i}$ such that:


 * $\map {f'} {\xi_i} = \dfrac {\map f {b_i} - \map f {a_i} } {b_i - a_i}$

We then have:

Let $\varepsilon$ be a positive real number.

Then for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ with:


 * $\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \frac \varepsilon M$

we have:


 * $\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \varepsilon$

Since $\varepsilon$ was arbitrary:


 * $f$ is absolutely continuous.