Equivalence of two properties of a measure space Let $(X,M,\mu)$ be a measure space. Why are the following statements equivalent?
i) There exists a sequence of pairwise disjoint measurable sets $\{A_{n}\}$ in $M$ , each of finite measure, such that $\mu(A) = \sum_{n=1}^{\infty} \mu(A \cap A_{n})$ for every measurable set $A \in M$.
ii) $\mu$ is a countable sum of pairwise mutually singular finite measures.

(ii) $\implies$ (i)
Suppose $\mu = \sum_{n=1}^\infty \mu_n$, where $\mu_i \perp \mu_j$ for all $i \neq j$.  By definition of "mutually singular," there exist pairwise disjoint $A_n \in M$ such that $\mu_n(A) = \mu_n(A \cap A_n)$ for all $A \in M$.  Therefore, $\mu(A_k) = \sum_{n=1}^\infty \mu_n(A_k) = \mu_k(A_k) < \infty$, so that each $A_n$ has finite $\mu$-measure.  Moreover, $\mu(A) = \sum_{n=1}^\infty \mu_n(A_n) = \sum_{n=1}^\infty \mu_n(A \cap A_n) = \sum_{n=1}^\infty \mu(A \cap A_n)$, where in the last step we used the fact that $\mu(A \cap A_n) = \mu_n(A \cap A_n)$ for all $n \in N$.
(i) $\implies$ (ii)
Let $\mu_n$ be a measure concentrated on $A_n \in M$, normalized so that $\mu_n(A_n) = 1$.  Then $\mu(A) = \sum_{n=1}^\infty \mu(A\cap A_n) = \sum_{n=1}^\infty \mu_n(A)$.
(EDIT following Nate's comment: We can define $\mu_n(A) = \mu(A \cap A_n)$.)
