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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    $\begingroup$ The basic idea is to subtract out the atoms, as you've indicated. However, there are still a couple of gaps in your approach. First, you have to show that there are at most countably many atoms. Second, you should explain why $\mu$ is purely atomic on the union of the atoms. $\endgroup$ – user83827 Aug 8 '11 at 17:11
  • $\begingroup$ Measure $\mu$ is finite, every atom has a positive measure. Thus there exist at most countable number of atoms. Let $E \in \cal{S}$, $E \subset B$. Then $E=\bigcup_{n=1}^\infty E\cap B_n$. In order to show that $\mu$ is purely atomic on $B$ I consider two cases. If $\mu(E)=0$ it is nothing to prove. Assume that $\mu(E)>0$. Since $B_n$ is an atom, so its subset $E\cap B_n$ is either an atom or has the measure zero. Let $M=\{n \in N: E\cap B_n \ is \ an \ atom \}$. Then $M$ is nonempty and $E=\bigcup_{n\in M} (E\cup B_n)$ -sum of atoms. $\endgroup$ – Richard Aug 8 '11 at 19:00
  • $\begingroup$ I don't see why what you said implies there are at most countably many atoms. I can build continuum-many subsets of $[0,1]$ of measure $1/2$ whose pairwise symmetric differences are non-null. At some point you should use the atomicity (?) of the atoms. I'm sure you have the right idea, but just aren't writing out all the details. Maybe you should submit an answer to your own question? $\endgroup$ – user83827 Aug 8 '11 at 20:46
  • $\begingroup$ Sorry, I wrote wrong. If measure is finite then there is at most countably many pairwise disjoint measurable subsets with positive measure. In particular there is at most many pairwise disjoint atoms. $\endgroup$ – Richard Aug 9 '11 at 8:52
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    $\begingroup$ @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. $\endgroup$ – George Lowther Sep 22 '11 at 23:40

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

  • $\begingroup$ In that article, they proved that $\mu$ can be decomposed into a purely atomic measure and a nonatomic measure. If $X$ can be decomposed in to $B$ and $C$ as Richard wants, then $\mu|_B$ is purely atomic and $\mu|_C$ is nonatomic, so we have the decomposition in that article. My question is how to show the conversed side, if $\mu = \mu_1 + \mu_2$ where $\mu_1$ purely atomic, $\mu_2$ nonatomic, then how to find $B$ and $C$? $\endgroup$ – Du Phan May 8 '14 at 10:42
  • $\begingroup$ @DuPhan $B=\operatorname{supp}(\mu_1)$, $C=\operatorname{supp}(\mu_2)$ $\endgroup$ – Norbert May 8 '14 at 11:51
  • $\begingroup$ 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. $\endgroup$ – Du Phan May 8 '14 at 14:28

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