[Edited 10/22/15: there were embarrassing errors in my original answer, which were pointed out in another question; I've corrected these, and I've written up further details in a blog post.]
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 (19621952), 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.
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 $\Gamma$. (In fact, Rokhlin's definition is given for a measure space, not just a measured $\sigma$-algebra, and requires that $\Gamma$ separate points of the space.) Since we're already using the word "separable" for (S), let's say that in this case $(S,\mu)$ is one-sided countably generated mod zero, and denote this property by (CG1CG0+). 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 Another natural condition would be that $(S,\mu)$$S$ itself is one-sided separable if it has a countable subset $\Gamma$countably generated, that is not only dense w.r.t. $\rho$ but also has the property, that for every $A\in S$ and$S = \sigma(\Gamma)$ for some countable $\epsilon>0$ there exists$\Gamma$; call this $B\in \Gamma$ such(CG), and not that it applies to the Borel $A\subset B$ and$\sigma$-algebra on $\mu(B\setminus A) < \epsilon$; we denote this property by$[0,1]$, but not the Lebesgue $\sigma$-algebra. The latter satisfies the weaker condition (S1CG0+). Similarly, say $(S,\mu)$and hence is also countably generated mod zero if it has, meaning that there is a countable subset $\Gamma$$\Gamma \subset S$ such that for every $A\in S$, there exists $B\in \sigma(\Gamma)$ such that $\rho(A,B)=0$$\mu(A \bigtriangleup 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(In the previous version of this answer I got carried away and also defined (S1), one-sided separability, to be the property that every $A\in S$ and $\epsilon>0$ there exists $B\in \Gamma$ such that $A\subset B$ and $\mu(B\setminus A) < \epsilon$. As was correctly pointed out by Rina Shora and Nik Weaver on another question, this fails to hold for even the most standard examples.)
It is immediate that (CG), and similarly $\Rightarrow$ (S1CG0+) implies$\Rightarrow$ (SCG0). It was shown in, and [JDH] shows that (CGCG0) implies$\Rightarrow$ (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 $(S,\mu)$(it is non-atomicwritten with (has no atoms), then(CG) in fact all four definitions are equivalent. To see thismind, observe that any of the four implybut works just as well for (SCG0), and). In fact one can also show that (S) in turn implies that there is a $\sigma$-algebra isomorphism from$\Rightarrow$ $(S,\mu)$ 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.(CG0) Intuitively, one expects that if $E$ is an atom(details are in $(S,\mu)$, then there should be athis blog post) $\sigma$-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 $\nu$separability is a point mass with total weight $\mu(E)$. In particular, this requires that there exists a setequivalent to being countably generated $F\subset E$ such thatmod zero $\mu(F) = \mu(E)$-- 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 befirst two implications above are strict (mod zero) isomorphic to a point mass even ifthe Lebesgue (S) holds.
The one$\sigma$-sided conditionsalgebra satisfies (S1CG0+) andbut not (CG1CG) serve to fill this gap, and let us deal appropriately with the atomic pieces. Indeed, if either$\sigma$-algebra 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 endLebesgue subsets of the day, we see that (CG1) and (S1) are equivalent, and imply both (CG) and (S). Furthermore,$[0,1]$ with measure 0 or 1 satisfies (CGCG0) impliesbut not (SCG0+), 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 notFinally, because otherwise I doubt that [Ro] would introduce the extraanother related condition would be that $A\subset B$. However, I do not know a counterexample.
Comments on the proof of (A). For$S$ is contained in the proofcompletion of a countably generated $\sigma$-algebra; this property lies somewhere between (ACG), one must use and (CG1CG0) or, but its relation to (S1CG0+) 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 younot clear 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 $\mu(\bigcup_n B_n) < \epsilon$me. (For
The example in [JDH] involving the club filter shows that (S1S) one takes $S' = \Gamma$, while for(and hence (CG1CG0) 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 zeronot enough to imply this condition.
