Absolutely Continuous Real Function is Uniformly Continuous
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Theorem
Let $I \subseteq \R$ be a real interval.
Let $f: I \to \R$ be an absolutely continuous real function.
Then $f$ is uniformly continuous.
Proof
Let $\epsilon$ be a positive real number.
Since $f$ is absolutely continuous, there exists real $\delta > 0$ such that for all collections of pairwise 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:
- $\ds \sum_{i \mathop = 1}^n \size {\map f {b_i} - \map f {a_i} } < \epsilon$
Consider specifically the case $n = 1$.
Suppose that $a \le x \le y \le b$ and:
- $0 \le y - x < \delta$
so that:
- $\size {x - y} < \delta$
From the absolute continuity of $f$ we then have:
- $\size {\map f y - \map f x} < \epsilon$
so:
- $\size {\map f x - \map f y} < \epsilon$
Now suppose that $y < x$ and:
- $\size {x - y} < \delta$
then:
- $0 < x - y < \delta$
and:
- $\size {\map f x - \map f y} < \epsilon$
So, in fact, for all $x, y \in \closedint a b$ with:
- $\size {x - y} < \delta$
we have:
- $\size {\map f x - \map f y} < \epsilon$
Since $\epsilon$ was arbitrary:
- $f$ is uniformly continuous.
$\blacksquare$