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Let $\mu$ be a positive finite measure in some $\sigma$-field $\cal{S}$ of subsets of $X$. How find a decomposition $X=B \cup C$ on the disjoint sum of two measurable sets $B, C$ such that the measure $\mu|_B$ is atomless (i.e. it does not contain atoms) and $\mu|_{C}$ is purely atomic (i.e every measurable subset of $C$ of positive measure is the sum of finite on countable many atoms), where, for fixed $Z \in \cal{S}$, $\mu|_Z (E)=\mu(E \cap Z)$ for every $E \in \cal{S}$.

I think that if there exist a finite number of disjoint atoms, we can find decomposition in the following way.
Let $B_1$ be an arbitrary atom (if there exists),and assume that we have defined atoms $B_1,...,B_n$ such that $B_n \subset X\setminus \bigcup_{i=1}^{n-1}B_i$. We denote by $B_{n+1}$ an atom contained in $X\setminus \bigcup_{i=1}^{n}B_i$ if there exist. In the case when there is a finite number atoms this sequence is finite. We take $B=\bigcup B_n$, $C=X\setminus B$ and we have desired decomposition.

I don't sure, how to find the decomposition in the case when there exist a countable number of disjoints atoms? Does it suffice to take $B=\bigcup_{n=1}^\infty B_n$, $C=X\setminus B$?

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    @Richard: Just normal induction is needed. Choose the atom $B_{n+1}\subseteq X\setminus\bigcup_{i=1}^nB_i$ to maximize $\mu(B_{n+1})$. The maximum is achieved, because there can only be finitely many atoms of measure at least any given \epsilon > 0. Note that $\mu(B_n)\to0$, so you must eventually include every atom.2011-09-22

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This article answers your question even for the more general situation when the space is $\sigma$-finite and shows uniqueness of the desired decomposition.

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    Thank for your answer. I cannot find the general definition of support for a $\sigma$-finite measure in the internet (unless $\mu$ is Borel). But after re-read this article, I figure it out. Because $\mu_1$ is singular to $\mu_2$, there exists $B\subset X$ such that $\mu(X) = \mu_1(B)$ and $\mu_2(B) = 0$, then put $C = X - B$. In general, when $X$ is $\sigma$-finite, write $X$ as a countable disjoint union of finite measure subsets. Then we have a required decomposition. It is unique up to a difference of null sets.2014-05-08