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 real 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 pairwise disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:

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

we have:

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

Consider specifically the case $n = 1$.

Suppose that $a \le x \le y \le b$ and:

$0 \le y - x < \delta$

so that:

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

From the absolute continuity of $f$ we then have:

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

so:

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

Now suppose that $y < x$ and:

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

then:

$0 < x - y < \delta$

and:

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

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.

$\blacksquare$