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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?

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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.2010-10-27

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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.

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    I understand. Thanks2010-10-27
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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).2010-10-27
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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).2010-10-27
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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.2010-10-28
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    The difficultly I was having was showing that an uncountable sum of positive numbers would diverge.2010-10-28
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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.2010-10-28
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    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.2011-04-14
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    @Jbeardz- yes the definition for uncountable sum if correct but how do you approach the problem using this definition?2011-10-12
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    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.2014-12-18
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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?

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Is following argument correct?

Let the cardinality of set of natural numbers be N. Let A be collection of uncountable disjoint subsets. Let us arrange these subsets in descending order according to the value of measure $\mu$ on each subset. Let us take first N of these subsets. Denote measures on each of these sets as $\mu_1,\mu_2....\mu_N$. If $\mu(A)$ is finite each of these values should be finite. Suppose all if these values greater then $0$, the the sum $\Sigma_{1\le i\le N}\mu_i$ cannot converge to a finite value (A sum of sequence of positive real numbers converges only if there the $n_{th}$ term goes to $0$ as $n \to \infty$ ). This is in contradiction to our assumption. Therefore we can say that only countable number of these values are greater than $0$.