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?


Makes no difference. We can replace 'finite collection' by 'any collection' of disjoint intervals ...


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