Suppose we have a field of sets $\mathcal F$ such that no infinite union of members of $\mathcal F$ belong to it.
Let $m$ be any finitely additive measure on $\mathcal F$, then $m$ is automatically countably additive on $\mathcal F$. (I have seen this in a paper) but would like to know how this is the case??
I assume here that you mean: no infinite union of distinct members of $\mathcal F$ belongs to $\mathcal F$.
Let it be that $F_{n}\in\mathcal{F}$ for $n=1,2,\dots$ are disjoint, and that $F:=\cup_{n\in\mathbb{N}}F_{n}\in\mathcal{F}$. Then some $k\in\mathbb{N}$ must exist with $n>k\Rightarrow F_{n}=\emptyset$. (If not then $F$ is a infinite union of distinct members of $\mathcal{F}$ so cannot belong to it). Then $F=\cup_{n=1}^{k}F_{n}$ and by additivity $m\left(F\right)=\sum_{n=1}^{k}m\left(F_{n}\right)=\sum_{n=1}^{\infty}m\left(F_{n}\right)$. The last equation is a consequence of $m\left(\emptyset\right)=0$.