If $(X,\mathcal F, \sigma)$ is any space with a measure and $\mathcal{A \subseteq F}$ is a countable chain then $\sigma(\bigcup \mathcal A)=\sup_{A \in \mathcal A} \sigma(A)$; this is well-known. Can there be a space $(X,\mathcal F, \sigma)$ where an uncountable chain $\mathcal {A \subseteq F}$ such that $\bigcup \mathcal A \in \mathcal F$ and $\sigma(\bigcup \mathcal A)\neq\sup_{A \in \mathcal A}\sigma(A)$ can be found?
I generally accept Axiom of Choice.
Consider $(\aleph_1, \mathcal P(\aleph_1), \sigma)$ where $\aleph_1$ is the smallest uncountable ordinal and $\sigma(E)=0$ if $E$ is countable, $\sigma(E)=\infty$ if $E$ is uncountable. Consider $$ \mathcal A = \bigl\{ A_x=\{y\in\aleph_1:y\leq x\}: x\in\aleph_1 \bigr\}. $$ Then each $A_x$ is countable therefore $$ \sup_{A\in\mathcal A} \sigma(A) = 0 \neq \infty = \sigma\left(\bigcup\mathcal A\right) . $$