This is an exercise from Royden, Flitzpatrick's Real Analysis (4th edition), chapter 6, Q.N. 39.

Let $f:[a,b]\to \mathbb R$ be an increasing function. I have to show that $f$ is absolutely continuous iff for every $\varepsilon >0$ there exists $\delta >0$ such that for every measurable subset $E$ of $[a,b]$ with $m(E)<\delta$, we have $m^*(f(E))<\varepsilon$.

I proceed like this: Let $f$ be increasing and absolutely continuous. Let $\varepsilon >0$. Thus there exists $\delta >0$ such that $\delta$ satisfies the requirements for absolute continuity. Let $E$ be a measurable subset of $[a,b]$ with $m(E)<\delta$. Thus there exists a pairwise disjoint countable collection of open intervals $\{(a_k,b_k):k=1,2,\ldots,n\}$ such that $E\subset \bigcup\limits_{k=1}^{\infty}(a_k,b_k)$ and $\sum\limits_{k=1}^{\infty}|b_k-a_k|<\delta$. Since $f$ is increasing, therefore $f((a_k,b_k))\subset (f(a_k),f(b_k))$ for all $k\in \mathbb{N}$. Thus $f(E)\subset f(\bigcup\limits_{k=1}^{\infty}(a_k,b_k))\subset \bigcup\limits_{k=1}^{\infty} (f(a_k),f(b_k))$ and so $m^*(f(E))\leq \sum\limits_{k=1}^{\infty}|f(b_k)-f(a_k)|<\varepsilon$ as $f$ is absolutely continuous.

Conversely, let the given condition hold. Let $\varepsilon >0$ and let $\delta$ work for it. Let $\{(a_n,b_n):n\in \mathbb{N}\}$ be a pairwise disjoint collection of open intervals such that $\sum\limits_{n=1}^{\infty}|b_n-a_n|<\delta$. Then $E=\bigcup\limits_{n=1}^{\infty}(a_n,b_n)$ is a Lebesgue measurable set with $m(E)=\sum\limits_{n=1}^{\infty}|b_n-a_n|<\delta$. Therefore by hypothesis, $m^*(f(E))<\varepsilon$.

Problem starts here. Unless $f$ is continuous $f(a_k,b_k)$ and $(f(a_k),f(b_k))$ may not be same. Then how to proceed. Please help.


The statement is false without assuming $f$ is continuous. For instance, let $f:[0,2]\to\mathbb{R}$ be given by $f(x)=x$ if $x\leq 1$ and $f(x)=x+1$ if $x>1$. Then it is easy to see that for any measurable $E\subseteq[0,2]$, $f(E)$ is measurable and $m(f(E))=m(E)$ (split $E$ into $E\cap [0,1]$ and $E\cap(1,2]$). So, the given condition holds, with $\delta=\epsilon$, but $f$ is not absolutely continuous.


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