Absolute Continuity

# Absolute Continuity

Definition: Let $f : [a, b] \to \mathbb{R}$. We say that $f$ is Absolutely Continuous on $[a, b]$ if for all $\epsilon > 0$ there exists a $\delta > 0$ such that if $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ is any finite collection of disjoint open subintervals of $[a, b]$ such that $\displaystyle{\sum_{i=1}^{n} (y_i - x_i) < \delta}$ we have that $\displaystyle{\sum_{i=1}^{n} |f(y_i) - f(x_i)| < \epsilon}$. |

Theorem 1: If $f : [a, b] \to \mathbb{R}$ is absolutely continuous on $[a, b]$ then $f$ is uniformly continuous on $[a, b]$. |

**Proof:**Let $f$ be an absolutely continuous function on $[a, b]$ and let $\epsilon > 0$. By the absolute continuity of $f$ there exists a $\delta > 0$ such that for any finite collection of disjoint open subintervals of $[a, b]$, $\{ (x_1, y_1), (x_2, y_2), ..., (x_n, y_n) \}$ with $\displaystyle{\sum_{i=1}^{n} (y_i - x_i) < \delta}$ we have that:

\begin{align} \quad \sum_{i=1}^{n} | f(y_i) - f(x_i) | < \epsilon \end{align}

- In particular, for all $\epsilon > 0$ there exists a $\delta > 0$ such that for any open subinterval $(x, y)$ of $[a, b]$ with $|y - x| = y - x < \delta$ we have that:

\begin{align} \quad |f(y) - f(x)| < \epsilon \end{align}

- Therefore $f$ is uniformly continuous on $[a, b]$. $\blacksquare$

Corollary 2: If $f : [a, b] \to \mathbb{R}$ is absolutely continuous on $[a, b]$ then $f$ is continuous on $[a, b]$. |

**Proof:**The absolute continuity of $f$ on $[a, b]$ implies that $f$ is uniformly continuous on $[a, b]$ which implies that $f$ is continuous on $[a, b]$. $\blacksquare$