First, the definitions:
$f$ is AC on $E$ if $$\forall \epsilon >0\ \exists \delta >0\ \forall \{[a_k,b_k]\}_{k=1}^N \mbox{ such that }a_k,b_k \in E,\ \Sigma(b_k - a_k) <\delta : \Sigma| f(b_k) - f(a_k)| <\epsilon.$$
$f$ is GAC on $E$ if $E= \underset{n \in \mathbb{N}} {\biguplus} E_n$, $f$ is AC on $E_n\ \forall n$ and continuous on $E$.
My question is to prove that if $f$ is GAC on every measurable $E \subset I=[a,b]$ then $f$ AC on $I$.
I tried using the fact that it's GAC on $I$, and concluding that it's bounded variation on every $E_n$, thus can be written as two monotone functions, and eventually write $f$ as the sum of two (almost-lacking continuity) GAC monotone functions. Every monotone and GAC function is also AC. How can I use what's given to me and get continuity?? Any suggestions will be much appreciated.
I'm quite sure that John is wrong... as Umberto P. mentioned earlier, GAC on [a,b] implies GAC on every borel subset (take $[a,b]=\biguplus E_n$, $E\in B([a,b])$ s.t. $f$ is AC on each $E_n$, then it will also be AC on each $E\cap E_n$). So, the claim actually says that GAC on a segment (or whatever) implies AC. Which is absurd (take $x\sin(\frac{1}{x})$).