Characterising $\sigma$-algebras as posets

Question

A $\sigma$-algebra is defined as a set $X$ together with a subset $\Sigma$ of the power set $\mathcal{P}(X)$, such that

1. $X\in \Sigma$
2. $\Sigma$ is closed under complementation
3. $\Sigma$ is closed under countable unions

Since $\Sigma$ is a set of sets, it can be regarded as a poset under the inclusion order. It seems as if it should be possible to give an equivalent characterisation of a $\sigma$-algebra as a poset with a certain set of properties. (Fixing such a poset would define $\Sigma$, and then $X$ would be defined as the set of atoms of the poset.)

My question is, what properties must a poset have in order to be equivalent to a $\sigma$-algebra in this sense? Is it as simple as saying the poset must be a complemented distributive lattice (i.e. a Boolean algebra) with countable joins? Or are some additional conditions needed?

Answer 1

The relevant reference for your question is Sikorski's Boolean Algebras, section 24.

Your characterization will not hold in general, and an example is to take the quotient of the Borel sets of reals modulo the ideal of countable sets (it fails to satisfy an infinitary distributive law that holds for $\sigma$-algebras). It is sufficient (by 24.4 op. cit.) to add that the Boolean algebra is atomic (every element has an atom below), and this is necessary for countably generated ones.

The general result is 24.1 in Sikorski's, and here is one of the characterizations given there.

The following are equivalent:

1. $B$ is isomorphic to $\sigma$-algebra.
2. $B$ is a Boolean algebra with countable joins, and for every $a\in B\smallsetminus\{0\}$ there exists maximal filter $F\subset B$ closed under countable meets such that $a\in F$.