Let $B$ be a Boolean algebra, $S(B)$ the set of ultrafilters on $B$, and $\lambda(a)=\{U\in S(B):a\in U\}$ for every $a\in B$. Then the set $S(B)$ equipped with the topology for which $\{\lambda(a):a\in B\}$ is an open base is called the Stone space of $B$.
Suppose now that $B$ is a $\sigma$-algebra and that I want to construct the Stone space of $B$. Shall I now define $S(B)$ as the set of $\sigma$-complete ultrafilters on $B$, or shall I still define $S(B)$ as the set of ultrafilters on $B$?
A $\sigma$-algebra is a special case of a Boolean-algebra, so the Stone space (and its definition) stay exactly the same.
The space $S(B)$ will have other properties that reflect the $\sigma$-completeness of the algebra we start with, of course. I think Halmos' text has parts of it that deal with that.