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In a measure space, let's call a measurable subset atomless wrt the measure, if it does not have an atomic subset. In particular a measurable subset with zero measure is atomless.

There may be measurable subsets that are neither atomic nor atomless, for instance, the union of atomic subset(s) and atomless subset(s) with positive measure(s).

  1. I was wondering if the converse is true, i.e. if a measurable subset is neither atomic nor atomless, then must it be the union of atomic subset(s) and atomless subset(s) with positive measure(s)?
  2. I think the previous question is equivalent to whether any measurable subset can be partitioned into atomic subset(s) and atomless subset(s)?

Thanks and regards!

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See math.stackexchange.com/questions/56327/…. An answer hasn't yet been posted, but it may be helpful. –  Jonas Meyer Jan 14 '12 at 4:57
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1 Answer 1

  1. It is at least true if the measure is $\sigma$-finite. This guarantees that the measure space has at most countably many atomic subsets, so the union of them is measurable, and the complement of that union has positive measure and is atomless. (I don't see a reason why the union of all of the atomic subsets would have to be measurable in the non-$\sigma$-finite case, nor do I know of a counterexample.)
  2. In the $\sigma$-finite case at least, you can apply the same reasoning. Restricting the measure to the $\sigma$-algebra you get by intersecting each of the sets in the original $\sigma$-algebra with some particular measurable set gives you another $\sigma$-finite measure space. Then you can use a method similar to the above to get the indicated partition. Or, just intersect with the partition of the whole space.
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+1 Thanks, Jonas! In another post, I was wondering why σ-finite measures occur to you and others so naturally. –  Tim Jan 14 '12 at 6:46
    
About 1, why, when sigma finite, (1) "the measure space has at most countably many atomic subsets"? (2) "the complement of that union has positive measure and is atomless"? –  Tim Jan 14 '12 at 8:20
    
(1) Here is an outline: First consider the finite case. Show that distinct atoms are disjoint. Show that if there are uncountably many atoms, then there is a positive integer $n$ and infinitely many atoms with measure greater than $1/n$. This is impossible in the finite case, so there are only countably many atoms. In the general $\sigma$-finite case, the space is a countable union of sets of finite measure, and each of those only has countably many atoms, and this implies that the whole space has only countably many atoms. –  Jonas Meyer Jan 14 '12 at 9:02
    
(2) Rereading your question I may have been thinking of the wrong hypothesis there, but the positive measure part would follow under the assumption that the space is not merely a union of atoms, and regardless the atomless part follows immediately from the definition. An atom in a restricted subspace would also be an atom in the whole space, and we already removed all of those. –  Jonas Meyer Jan 14 '12 at 9:04
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