Sum of Absolutely Continuous Functions is Absolutely Continuous
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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.
$\blacksquare$