Absolutely Continuous Real Function is Uniformly Continuous

Theorem
Let $I \subseteq \R$ be a real interval.

Let $f : I \to \R$ be an absolutely continuous function.

Then $f$ is uniformly continuous.

Proof
Let $\epsilon$ be a positive real number.

Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:


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

we have:


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

Consider specifically the case $n = 1$.

From the absolute continuity of $f$, we have that whenever $a \le x \le y \le b$ and:


 * $y - x < \delta$

we have:


 * $\size {\map f x - \map f y} < \epsilon$

Notice however that:


 * $\size {\map f x - \map f y} = \size {\map f y - \map f x}$

We therefore have:


 * $\size {\map f x - \map f y} < \epsilon$

when:


 * $x - y < \delta$

as well.

So, in fact, for all $x, y \in \closedint a b$ with:


 * $\size {x - y} < \delta$

we have:


 * $\size {\map f x - \map f y} < \epsilon$

Since $\epsilon$ was arbitrary:


 * $f$ is uniformly continuous.