This question is from the book The Integrals of Lebesgue, Denjoy, Perron, and Henstock. page 141..
This involves a more general type of absolute continuity:
For a gauge $\delta:[a,b]\to \mathbb{R}^+$ . We say that $([v,u],\xi)$ is $\delta$-fine provided $$\xi\in [u,v]\subseteq (\xi-\delta(\xi),\xi+\delta(\xi)).$$
Given a Function $F:[a,b]\to\mathbb{R}$.
We say the $F$ is $AC_\delta$ on $E\subseteq [a,b]$ provided for every $\epsilon>0$, there exists a gauge $\delta:[a,b]\to \mathbb{R}^+$ and a constant $\eta>0$ such that if $D=\{[u_i,v_i],\xi\}_i^n$ is a collection of non-overlapping $\delta$-fine intervals such that with $\displaystyle\sum_{i=1}^n (v_i-u_i)<\eta$ we have $$|\sum F(v)-F(u)|<\epsilon.$$
$F$ is $ACG_{\delta}$ if $[a,b]=\bigcup_{n=1}^\infty X_n$ such that $F$ is $AC_\delta$ on each $X_n$.
Given $E\subseteq [a,b]$ with $\mu(E)=0$ and $F$ is $ACG_\delta$ on $[a,b]$. Why can we express $E=\displaystyle\bigcup_{i=1}^\infty E_i$ with $E_i$'s pairwise disjoint and that $F$ is $AC_\delta$ on each $E_i$.
I'm stuck on finding the close form on this $E_i$ in terms of $X_i$. I think im missing something.
Any insight is highly appreciated.
I think this formula for $E_i's$ holds. Let $E_1=X_1$ and set $E_n= X_n\setminus\displaystyle\bigcup_{i=1}^{n-1} X_i$ for $n\geq 2$. $E_i$ are disjoint. Also, Since $F$ $AC_\delta$ on any finite union of sets $X_n$ and any subsets of those finite unions, it follows that $F$ is $AC_\delta$ each $E_n$ and that $[a,b]=\displaystyle\bigcup_{i=1}^\infty E_i$.