Lipschitz Continuous Real Function is Absolutely Continuous

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

Let $f$ be Lipschitz continuous.

Then $f$ is absolutely continuous.

Proof
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.

Hence, by Constant Real Function is Absolutely Continuous:


 * $f$ 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:

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.