Let $(X, S, \mu)$ be a space with a positive measure and let ${\cal A}$ be a subfamily of ${\cal S}$ consisting of sets with finite measure.
I wish to prove that if
$$
K:=\sup_{B\in {\cal A}}\mu(B) <\infty
$$
then there exists a sequence $(B_n)$ of pairwise disjoint sets from $\{A\}$ such that $\mu(B_n) \rightarrow K$.
I know that by the property of supremum there is a sequence $(C_n)$ from ${\cal A}$ such that $\mu(C_n)\rightarrow K$. That sequence need not be pairwise disjoint. If we take instead of $(C_n)$ the sequence $(B_n)$, where $B_1=C_1$, $B_n=C_n \setminus (C_1 \cup...\cup C_{n-1})$ for $n\geq 2$, then sets $B_n$ are from ${\cal A}$, are pairwise disjoint with measures $\mu(B_n) \leq \mu(C_n)$, but I do not see whether or not $\mu(B_n)\rightarrow K$.
Edit.
As $K<\infty$ and $\neq 0$ it is not true.
What about the case $K=\infty$?
Answer
This doesn't seem to be true. It can only hold when $K=0$.
Suppose $B_n$ is such that $\mu(B_n)\rightarrow K$. Now assume that $\mu(B_k) = x\neq 0$ for some $k$. Then for any $n\neq k$ we have $B_n\cup B_k\in \mathcal{A}$ and therefore $\mu(B_n) +\mu(B_k) = \mu(B_n\cup B_k) \leq K$. So it follows that $B_n \leq K-x$. But this contradicts our assumption that $\mu(B_n)\to K$.
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