I'm late to the party, but here's my two cents. References in what follows are to
- [Ha] P.R. Halmos, Measure Theory, Springer, 1950.
- [Ro] V.A. Rokhlin, On the fundamental ideas of measure theory, Transl. AMS, Series 1, No. 10 (1962), 1-54. (The original Russian article is from 1949. A pdf of the English translation is presently available here.)
- [JDH] will denote the answer already given to this question by Joel David Hamkins.
Consider a measured $\sigma$-algebra $(S,\mu)$. Assume that $\mu$ is normalised to have total weight 1, and that $S$ is complete (contains all subsets of null sets).
In [Ha], $(S,\mu)$ is said to be separable if it has a countable subset that is dense w.r.t. the metric $\rho(A,B) = \mu(A\bigtriangleup B)$. We denote this property by (S).
In [Ro], $(S,\mu)$ is said to be separable if it has a countable subset $\Gamma$ such that for every $A\in S$, there exists $B\in \sigma(\Gamma)$ such that $A\subset B$ and $\mu(B\setminus A) = 0$. Here $\sigma(\Gamma)$ is the $\sigma$-algebra generated by $\Gamma$. Since we're already using the word "separable" for (S), let's say that in this case $(S,\mu)$ is one-sided countably generated, and denote this property by (CG1). To keep terminology manageable, we won't explicitly say "mod zero", but this is understood, and thus we need to specify "one-sided" because of the restriction that $A\subset B$, which means that the "mod zero" only applies to the outer approximation, whereas the inner must be exact.
So that's two conditions. Let's round it out by saying that $(S,\mu)$ is one-sided separable if it has a countable subset $\Gamma$ that is not only dense w.r.t. $\rho$ but also has the property that for every $A\in S$ and $\epsilon>0$ there exists $B\in \Gamma$ such that $A\subset B$ and $\mu(B\setminus A) < \epsilon$; we denote this property by (S1). Similarly, say $(S,\mu)$ is countably generated if it has a countable subset $\Gamma$ such that for every $A\in S$ there exists $B\in \sigma(\Gamma)$ such that $\rho(A,B)=0$.
Now we have four conditions: two of them involve approximations from the outside, while the other two allow arbitrary approximations. Clearly (CG1) implies (CG), and similarly (S1) implies (S). It was shown in [JDH] that (CG) implies (S), but the converse is not true.
So far this is just a summary of what others have already said here. Here's the new bit.
Equivalence when non-atomic. Recall that an atom is a set $E\in S$ such that $\mu(E)>0$ and every subset $A\subset E$ has either $\mu(A)=0$ or $\mu(A)=\mu(E)$. If $(S,\mu)$ is non-atomic (has no atoms), then in fact all four definitions are equivalent. To see this, observe that any of the four imply (S), and that (S) in turn implies that there is a $\sigma$-algebra isomorphism from $(S,\mu)$ to the Lebesgue sets on the unit interval equipped with Lebesgue measure [Ha, Sec. 41, Theorem C]. Since all four properties hold for the Lebesgue space, we are done.
Atomic pieces. Intuitively, one expects that if $E$ is an atom in $(S,\mu)$, then there should be a $\sigma$-algebra map $(S|_E, \mu|_E) \to (T,\nu)$ that is a mod zero isomorphism, where $T$ is a $\sigma$-algebra with only two elements ($\emptyset$ and a single point) and $\nu$ is a point mass with total weight $\mu(E)$. In particular, this requires that there exists a set $F\subset E$ such that $\mu(F) = \mu(E)$ and every null set $A\subset E$ has $A\cap F = \emptyset$. The example in [JDH] shows that this need not always be the case, and that an atomic space need not be (mod zero) isomorphic to a point mass even if (S) holds.
The one-sided conditions (S1) and (CG1) serve to fill this gap and let us deal appropriately with the atomic pieces. Indeed, if either of these properties hold, then one can show the following:
(A) There exist atoms $E_n \in S$ such that $S|_{E_n}$ is the trivial $\sigma$-algebra for every $n$, and $S|_{(\bigcup_n E_n)^c}$ is isomorphic to the Lebesgue sets on an interval of length $1 - \sum_n \mu(E_n)$.
So at the end of the day, we see that (CG1) and (S1) are equivalent, and imply both (CG) and (S). Furthermore, (CG) implies (S), and the converse is true if $(S,\mu)$ is non-atomic, but may fail if it has atoms.
I don't know if (CG) is equivalent to (CG1) and (S1). I suspect it is not, because otherwise I doubt that [Ro] would introduce the extra condition that $A\subset B$. However, I do not know a counterexample.
Comments on the proof of (A). For the proof of (A), one must use (CG1) or (S1) to take an atom $E\in S$ and produce a subset $F\subset E$, $F\in S$ such that $F\cap A = \emptyset$ for all $A\in S$ with $\mu(A)=0$. The key to this is that both conditions allow you to produce a countable family $S'\subset S$ such that for every $\epsilon>0$ and every null set $A$, we have $A\subset \bigcup_n B_n$ for some $B_n \in S'$ with $\mu(\bigcup_n B_n) < \epsilon$. (For (S1) one takes $S' = \Gamma$, while for (CG1) one takes $S'$ to be the set of all finite intersections of sets in $\Gamma$ and their complements.) Then because $E$ is an atom, we can conclude that $\mu(B_n \cap E) = 0$ for all $n$, and so the collection of all null sets in $E$ can be covered by a countable collection of null sets, whose union therefore has measure zero.