Lipschitz Continuous Real Function is Absolutely Continuous

Theorem

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

By the definition of Lipschitz continuity, there exists $K \in \R$ such that:

$\size {\map f x - \map f y} \le K \size {x - y}$

for all $x, y \in I$.

If $K = 0$, then:

$\size {\map f x - \map f y} = 0$

for all $x, y \in I$.

In this case, $f$ is constant.

$f$ is absolutely continuous in the case $K = 0$.

Now, take $K > 0$.

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

We have:

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

$\le$

$\ds \sum_{i \mathop = 1}^n K \size {b_i - a_i}$

$\ds $

$=$

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

since $b_i > a_i$

Let $\epsilon$ be a positive real number.

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

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

we have:

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

Since $\epsilon$ was arbitrary:

$f$ is absolutely continuous in the case $K > 0$.

Hence the result.

$\blacksquare$