Absolute Value of Absolutely Continuous Function is Absolutely Continuous
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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 $\epsilon$ 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:
- $\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
we have:
- $\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
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:
- $\ds \sum_{i \mathop = 1}^n \size {\size {\map f {b_i} } - \size {\map f {a_i} } } \le \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
whenever:
- $\ds \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta$
Since $\epsilon$ was arbitrary:
- $\size f$ is absolutely continuous.
$\blacksquare$