Is there a measurable space that cannot be covered by countable union of disjoint finite sets? Take some set $X$ with a non-counting measure $\mu$ attached to it. Is it possible that $X=\cup_{\alpha\in A}S_\alpha$ an uncountable disjoint union where $\mu(S_\alpha)=c$ some fixed finite number for all $\alpha$.
Certainly $\mathbb R=\cup_{r\in\mathbb R}\{r\}$ and $\mu(r)=1$ for each $r$ if we take counting measure, so I wish to exclude that possibility.
As far as I know, this is not possible with $\mathbb R$ and Lebesgue measure. I.e. we cannot break apart $\mathbb R$ into an uncountable disjoint union of sets of positive Lebesgue measure.

Take $X=\mathbb{R}\times[0,1]$ with $\mu(A)=\sum_{r\in\mathbb{R}} \lambda(A_r)$, where $A_r=\{s\in[0,1]:(r,s)\in A\}$ and $\lambda$ is Lebesgue measure on $[0,1]$.  Then $X$ is the disjoint union of the sets $S_r=\{r\}\times[0,1]$ for each $r\in\mathbb{R}$, each of which has measure $1$.
More generally, given any family of measure spaces, you can take their disjoint union with the obvious induced measure.  If you had an uncountable family of measure spaces, each of total measure $c$, then the disjoint union is a disjoint union of subsets of measure $c$.
