Constant Real Function is Absolutely Continuous

Theorem

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

Let $f : I \to \R$ be an constant real function.

Then $f$ is absolutely continuous.

Proof

Let $\delta, \epsilon$ be positive real numbers.

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

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

Since $f$ is constant, for all $i \in \set {1, 2, \ldots, n}$ we have:

$\size {\map f {b_i} - \map f {a_i} } = 0$

so:

$\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } = 0 < \epsilon$

Since $\epsilon$ was arbitrary:

$f$ is absolutely continuous.

$\blacksquare$