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Suppose $A$ is a set of finite measure. Is it possible that $A$ can be an uncountable union of disjoint subsets of $A$, each of which has positive measure?

Mykie
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    The answer is yes if you allow the sets to be nonmeasurable (so that they have positive outer measure as opposed to measure). I gave a reference for this in a comment on Qiaochu's answer. – Jonas Meyer Oct 27 '10 at 21:43

2 Answers2

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No. Suppose $A$ is an uncountable disjoint union of measurable subsets $A_i, i \in I$ with positive measure. Then $I$ is a countable union of the sets of indices $i$ such that $\mu(A_i) > \frac{1}{n}, n \in \mathbb{N}$, so it follows that one of these sets must be uncountable. In particular a countable union of some subcollection of the $A_i$ has arbitrarily large measure; contradiction.

Qiaochu Yuan
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    Just pointing this out, but Qiaochu's answer actually has nothing to do with measure theory. He was pointing out the fact that the sum of an uncountable set of positive numbers can never be finite (although for this statement to make sense, you would need to first define what it means to be sum an uncountable set of numbers). – user1736 Oct 27 '10 at 19:46
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    @user1736: It is not true that it has nothing to do with measure theory. The answer uses countable additivity (or at least finite additivity) on disjoint measurable sets. Sets of finite measure can be decomposed into uncountably many disjoint nonmeasurable sets. See for example http://www.emis.de/cgi-bin/JFM-item?46.0294.01 (which I haven't read, and found referenced at http://www.mathkb.com/Uwe/Forum.aspx/math/32728/Cardinality-of-equivalence-classes-and-measure). – Jonas Meyer Oct 27 '10 at 21:28
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    Yeah, I guess I was assuming that he already had that part down and was just figuring out why the sum would have to diverge. In any case, Qiaochu's answer seemed to assume that the OP was aware of countable addivity as well, so I was trying to say that the actual reasoning/logic Qiaochu employed would be equally useful towards similar problems in an entirely different context. Sorry about the mistake. – user1736 Oct 28 '10 at 07:43
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    The difficultly I was having was showing that an uncountable sum of positive numbers would diverge. – Mykie Oct 28 '10 at 16:52
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    This "uncountable sum" thing may be a good way to state the problem intuitively but I think you should avoid it; the terminology presupposes that such a thing is well-defined in the first place. – Qiaochu Yuan Oct 28 '10 at 17:30
  • A sum over an uncountable indexing set, is the supremum of sums over all finite subsets. At least, that is the convention I have been taught. – Jonathan Beardsley Apr 14 '11 at 20:18
  • @Jbeardz- yes the definition for uncountable sum if correct but how do you approach the problem using this definition? – user24367 Oct 12 '11 at 16:41
  • Wouldn't it be better to say that a finite union of the $A_i$ has arbitrarily large measure? Countable additivity is not needed here. – bof Dec 18 '14 at 16:30
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Take any collection of disjoint sets of positive measure. There can be only a finite number of them having measure $>1$, having measure $>\tfrac{1}{2}$, ..., having measure $>\tfrac{1}{n}$, ... So, how many sets can be in a countable union of finite collections of sets?

Vadim
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