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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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    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 $0$.