Product of Absolutely Continuous Functions is Absolutely Continuous

Theorem
Let $a, b$ be real numbers with $a < b$.

Let $f, g : \closedint a b \to \R$ be absolutely continuous functions.

Then $f \times g$ is absolutely continuous.

Proof
From Absolutely Continuous Function is Continuous:


 * $f$ and $g$ are continuous.

From Closed Real Interval is Compact:


 * $\closedint a b$ is compact.

Therefore, by Continuous Function on Compact Subspace of Euclidean Space is Bounded:


 * $f$ and $g$ are bounded.

That is, there exists $M_f, M_g \in \R_{> 0}$ such that:


 * $\size {\map f x} \le M_f$
 * $\size {\map g x} \le M_g$

for all $x \in \closedint a b$.

Let $\closedint {a_1} {b_1}, \dotsc, \closedint {a_n} {b_n} \subseteq \closedint a b$ be a collection of disjoint closed real intervals.

Then:

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


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

we have:


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

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


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

we have:


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

Let:


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

Then, whenever:


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

We have both:


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

and hence:

Since $\varepsilon$ was arbitrary, we have:


 * $f \times g$ is absolutely continuous.