Sum of Absolutely Continuous Functions is Absolutely Continuous

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

Let $f, g : I \to \R$ be absolutely continuous functions.

Then $f + g$ is absolutely continuous.

Proof
Let $\varepsilon$ be a positive real number.

Since $f$ is absolutely continuous, there exists real $\delta_1 > 0$ such that for all collections of disjoint closed real intervals $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq I$ with:


 * $\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_1$

we have:


 * $\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \varepsilon 2$

Similarly, since $g$ is absolutely continuous, there exists real $\delta_2 > 0$ such that whenever:


 * $\displaystyle \sum_{i \mathop = 1}^n \paren {b_i - a_i} < \delta_2$

we have:


 * $\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \varepsilon 2$

Let:


 * $\delta = \map \min {\delta_1, \delta_2}$

Then, 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:


 * $\displaystyle \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \frac \varepsilon 2$

and:


 * $\displaystyle \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} } < \frac \varepsilon 2$

We then have:

whenever:


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

Since $\varepsilon$ was arbitrary:


 * $f + g$ is absolutely continuous.