Answer to Question #170522 in Real Analysis for Prathibha Rose

Question

Define absolutely continuous function on [a,b] .suppose f is absolutely continuous, prove that |f| is absolutely continuous.

Answer 1

A function $f:[a,b] \to \mathbb{R}$ is absolutely continuous on $[a,b]$ if for every $\epsilon >0 \, \exist \, \delta >0$ such that whenever a finite sequence of pairwise disjoint subintervals $(x_k,y_k)$ of $[a,b]$ with $x_k,y_k \in [a,b]$ satisfies $\sum_k (y_k -x_k) < \delta$ then $\sum_k |f(y_k)-f(x_k) | < \epsilon$

Suppose that f is absolutely continuous. This implies that the condition of the definition above has been satisfied.

By triangle inequality, we have that

$||f(y_k)| - |f(x_k)|| \leq |f(y_k)-f(x_k)|$

$\implies \sum_k||f(y_k)| - |f(x_k)|| \leq \sum_k |f(y_k)-f(x_k)| < \epsilon \\ \therefore \sum_k||f(y_k)| - |f(x_k)|| < \epsilon$

Hence |f| is absolutely continuous.