Set of Finite Measure: Uncountable disjoint subsets of non-zero measure 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?  
 A: 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.
A: 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 $0$.
A: 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?
