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 real functions.

Then $f + g$ is absolutely continuous.

Proof

Let $\epsilon$ be a positive real number.

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

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

we have:

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

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

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

we have:

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

Let:

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

Then, for all sets 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:

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

and:

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

We then have:

\(\ds \sum_{i \mathop = 1}^n \size {\map {\paren {f + g} } {b_i} - \map {\paren {f + g} } {a_i} }\)

\(=\)

\(\ds \sum_{i \mathop = 1}^n \size {\paren {\map f {b_i} - \map f {a_i} } + \paren {\map g {b_i} - \map g {a_i} } }\)

\(\ds \)

\(\le\)

\(\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } + \sum_{i \mathop = 1}^n \size {\map g {b_i} - \map g {a_i} }\)

Triangle Inequality for Real Numbers

\(\ds \)

\(<\)

\(\ds \frac \epsilon 2 + \frac \epsilon 2\)

\(\ds \)

\(=\)

\(\ds \epsilon\)

whenever:

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

Since $\epsilon$ was arbitrary:

$f + g$ is absolutely continuous.

$\blacksquare$