Absolute Value of Absolutely Continuous Function is Absolutely Continuous

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

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

Then $\size f$ is absolutely continuous.

Proof
Let $\varepsilon$ 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:


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

By the Reverse Triangle Inequality, we have:


 * $\size {\map f {b_i} - \map f {a_i} } \ge \size {\size {\map f {b_i} } - \size {\map f {a_i} } }$

Therefore:


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

whenever:


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

Since $\varepsilon$ was arbitrary:


 * $\size f$ is absolutely continuous.