# Why finite condition is needed in absolute continuity for a function?

I heard that absolute continuity for measure $$\nu$$ w.r.t. to $$\mu$$ where $$\nu$$ is a signed measure and $$\mu$$ is a positive measure, is as follows.

If $$\nu (E) = 0$$ for every $$E \in \mathcal M$$ for which $$\mu (E) = 0$$

Here, $$\mathcal M$$ is $$\sigma$$-algebra.

 But I know that for real line, absolutely continuous for function $$f$$ is as follows.

For all $$\epsilon > 0$$ , there exist $$\delta$$ such that whenever finite disjoint intervals $$(a_k , b_k )$$, $$\sum_k (b_k - a_k ) < \delta$$ then $$\sum _{k}|f(b_k)-f(a_k)|< \epsilon$$ satisfied.

 I think the signed measure can be considered as a function. Then I wonder why finite condition is needed in the definition of absolute continuous for a function.

Moreover, I saw a theorem that $$\nu$$ is absolute continuous w.r.t. to $$\mu$$ iff $$\forall \epsilon >0 \exists \delta$$ s.t. $$\mu(E) < \delta$$ implies $$|\nu(E)| < \epsilon$$.

Thus, I think, at real line, if I make $$E = \cup_{k\in A} (a_k , b_k )$$ for index set $$A$$, $$\nu$$ is absolute continuous iff $$\forall \epsilon >0 \exists \delta$$ s.t. $$\sum_k (b_k - a_k) < \delta$$ implies $$|\sum_k ( \nu(b_k) - \nu(a_k) )| < \epsilon$$ .

Regarding $$\nu$$ as a function $$f$$ , I think finite condition is not needed.



Can anyone can explain (not need to be rigorously) about finite condition in absolutely continuous?