I am trying to understand Remark 4.29 in Pavlov's paper. He defines measures as follows.
Definition 4.28. A (complex infinite) measure on an enhanced measurable space $(X,M,N)$ is a map $\mu:M' →\mathbf{C}$, where $M' \subset M$ is an ideal of $M$ such that $N \subset M'$, the map $\mu$ vanishes on $N$, and the ideal $M'$ satisfies the following saturation condition: for any countable family $\{m_i\}_{i\in I}$ of disjoint elements of $M'$ such that the sum $\sum_{i\in I}\mu(m'_i)$ converges absolutely for any family $\{m'_i\}_{i\in I}$ satisfying $m'_i \subset m_i$ and $m'_i \in M'$, we have \begin{equation} \bigcup_{i\in I}m_i\in M',\qquad \mu\bigg(\bigcup_{i\in I}m_i\bigg)=\sum_{i\in I}\mu(m_i). \end{equation} We say that $\mu$ is a finite measure if $M' = M$.
And he says that his definition of finite measures can be simplified in Remark 4.29:
a finite measure on an enhanced measurable space $(X,M,N)$ is a countably additive map $M \to \mathbf{C}$ that vanishes on $N$. Countable additivity means that for any countable family $\{m_i\}_{i∈I}$ of disjoint elements of $M$ the sum $\sum_{i\in I}\mu(m_i)$ converges absolutely and $\mu(\bigcup_{i\in I}m_i)=\sum_{i\in I}\mu(m_i)$.
I don't know why these definitions are equivalent. Perhaps it is a elementary problem, but I couldn't prove it. In particular, I do not see how Definition 4.28 implies the simplified definition in Remark 4.29 when $M' = M$.
