I'm reading a proof of below theorem from Real Analysis by Andrew M. Bruckner, Judith B. Bruckner, and Brian S. Thomson.
Theorem 5.27 Let $f$ be absolutely continuous on $[a, b]$. Then (i) $f$ is continuous on $[a, b]$. (ii) $f$ is of bounded variation on $[a, b]$. (iii) For every set $E$ of Lebesgue measure zero in $[a, b], \lambda(f(E))=0$.
Proof. Condition (i) is immediate. To prove (ii), choose $\delta>0$ such that if $\left[a_1, b_1\right]$, $\left[a_2, b_2\right], \ldots,\left[a_n, b_n\right]$ is any finite collection of nonoverlapping closed intervals with $$ \sum_{k=1}^n\left(b_k-a_k\right)<\delta $$ then $$ \sum_{k=1}^n\left|f\left(b_k\right)-f\left(a_k\right)\right|<1 . $$ If $[c, d]$ is any interval in $[a, b]$ with $d-c<\delta$, then $V(f ;[c, d]) \leq 1$. Let $N \in \mathbb{N}$ with $N>(b-a) / \delta$. Partition $[a, b]$ into $N$ intervals $I_1, \ldots, I_N$ of equal length $(b-a) / N<\delta$. The variation of $f$ on each of these intervals is less than $1$, so $$ V(f ;[a, b]) \leq N<\infty. $$ as required.
To prove (iii), let $\varepsilon>0$. Choose $\delta>0$ such that, if $\left\{\left[c_k, d_k\right]\right\}$ is any finite or countable collection of nonoverlapping closed intervals in $[a, b]$ with $\sum_{k=1}^{\infty}\left(d_k-c_k\right)<\delta$, then $$ \sum_{k=1}^{\infty}\left|f\left(d_k\right)-f\left(c_k\right)\right|<\varepsilon. $$ Let $G=\bigcup_{k=1}^{\infty}\left(a_k, b_k\right)$ be an open set containing $E$ with $$ \lambda(G)=\sum_{k=1}^{\infty}\left(b_k-a_k\right)<\delta. $$ Now $$ f(E) \subset f(G) \subset f\left(\bigcup_{k=1}^{\infty}\left[a_k, b_k\right]\right) \subset \bigcup_{k=1}^{\infty}\left[f\left(c_k\right), f\left(d_k\right)\right] $$ where $c_k$ is a point in $\left[a_k, b_k\right]$ at which $f$ assumes its minimum and $d_k$ is a point where $f$ assumes its maximum. Thus $$ \lambda^*(f(E)) \leq \sum_{k=1}^{\infty}\left(f\left(d_k\right)-f\left(c_k\right)\right)<\varepsilon $$ because $\sum_{k=1}^{\infty}\left|d_k-c_k\right| \leq \delta$. Since $\varepsilon$ is arbitrary, $\lambda(f(E))=0$.
My understanding It seems to me there is a sloppy pint in the proof of part (iii). The definition of absolute continuity requires the intervals to be disjoint. However, we don't know if $\left\{\left(a_k, b_k\right)\right\}$ are disjoint and thus if $\left\{\left(c_k \wedge d_k, c_k \vee d_k\right)\right\}$ are. As such, we could not apply consequence of absolute continuity of $f$ on $\left\{\left[c_k \wedge d_k, c_k \vee d_k\right]\right\}$.
Could you confirm if my understanding is correct? Is there a simple fix for above proof?
