The two definitions alluded to in the title can be found here: <http://en.wikipedia.org/wiki/Separable_sigma_algebra> (one is that the $\sigma$-algebra is countably generated, the other is pretty much the standard usage of the word separable wrt the semi-metric given by the measure). Why are they equivalent?



Separable sigma-algebra: equivalence of two definitions
The two notions are not equivalent. Indeed, they are not
equivalent even when one considers completing the measure
by adding all null sets with respect to any countably
generated $\sigma$-algebra. Nevertheless, the forward
implication holds.


First, let me explain the forward implication. Suppose that
$S$ is a $\sigma$-algebra generated by a countable
subfamily $S\_0$ and $\mu$ is a finite measure defined on
$S$. The semi-metric on $S$ is defined by
$d(A,B)=\mu(A\triangle B)$. Let $S\_1$ be the collection of
finite Boolean combinations of sets in $S\_0$. This is a
countable family, and I claim it is dense in the
semi-metric. To see this, let $S\_2$ be the closure of $S\_1$
in the semi-metric, that is, the sets $A\in S$ that are
approximable by sets in $S\_1$, in the sense that for any
$r\gt 0$ there is $B\in S\_1$ such that $d(A,B)\lt r$. Note
that $S\_2$ contains $S\_1$ and is closed under complement
since the measure was finite. I claim it is also closed
under countable unions: if each $A\_n$ is approximable by
$B\_n$ to within $r/2^n$, then $\cup\_n A\_n$ is approximated
by $\cup\_n B\_n$ to within $r$, and so one may find an
approximating finite union. So $S\_2$ is actually a
$\sigma$-algebra, and since it contains $S\_0$, it follows
that $S\_2=S$. That is, every set in $S$ is approximable by
sets in $S\_1$, and so $S\_1$ is a countable dense set in the
semi-metric, as desired.


Let's turn now to the reverse implication, which is not
generally true. The easiest counterexample for this in the
strict sense of the question is to let $X$ be a set of size
continuum and $S=P(X)$, the full power set of $X$. This is
a $\sigma$-algebra, but it is easily seen not to be
countably generated on cardinality grounds. Fix any $p\in
X$ and let $\mu$ be the measure placing mass $1$ at $p$ and
0 mass outside {p}. In this case, the family {emptyset, X}
is dense in the semi-metric, since every subset is
essentially empty or all of $X$, depending on whether it
contains $p$. So the semi-metric is separable, but the
$\sigma$-algebra is not countably generated.


Note that in this counterexample, the $\sigma$ algebra is
obtained from the counting measure on {p} by adding a large
cardinality set of measure $0$ and taking the completion.
Similar counterexamples can be obtained by adding such
large cardinality set of measure $0$ to any space and
taking the completion.


At first, I thought incorrectly that one could address the
issue by considering the completion of the measure, and
showing that the $\sigma$-algebra would be contained within
the completion of a countably generated $\sigma$-algebra.
But I now realize that this is incorrect, and I can provide
a counterexample even to this form of the equivalence.


To see this, consider the filter $F$ of all sets $A\subset
\omega\_1$ that contain a closed unbounded set of countable
ordinals. This is known as the *club filter*, and it is
closed under countable intersection. The corresponding
ideal $NS$ consists of the *non-stationary sets*, those
that omit a club, and these are closed under countable
union. It follows that the collection $S=F\cup NS$, which
are the sets measured by a club set, forms a
$\sigma$-algebra. The natural measure $\mu$ on $S$ gives
every set in $F$ measure $1$ and every set in $NS$ measure
$0$. This is a countably additive 2-valued measure on $S$.
Note that every set in $S$ has measure $0$ or $1$; in
particular, there are no disjoint positive measure sets. It
follows that the family {emptyset,$\omega\_1$} is dense in
the semi-metric, since every set in $S$ either contains or
omits a club set, and hence either agrees with emptyset or
with the whole set on a club. Thus, the semi-metric is
separable. But for any countable subfamily $S\_0\subset S$,
we may intersect the clubs used to decide the members of
$S\_0$, and find a single club set $C\subset\omega\_1$ that
decides every member of $S\_0$, in the sense that every
member of $S\_0$ either contains or omits $C$. This feature
is preserved under complements, countable unions and
intersections, and therefore $C$ decides every member of
the $\sigma$-algebra generated by $S\_0$. The completion of
the measure on the $\sigma$-algebra generated by $S\_0$ is
therefore contained within the principal filter generated
by $C$ together with its dual ideal. This is not all of
$S$, since there are club sets properly contained within
$C$, such as the set of limit points of $C$. Thus, this is
a measure space that has a separable semi-metric, but the
$\sigma$-algebra is not contained in the completion of any
countably generated $\sigma$-algebra.


