Let $X$ be a metric space, and $\mathscr{B}$ the $\sigma$algebra generated by open sets of $X$. Can we find a countable dense subsets of the metric space $(\mathscr{B},d)$ with the metric $d(A,B)=m(A\Delta B)$ where $m$ is any Borel probability measure on the measurable space $(X,\mathscr{B})$?

1$\begingroup$ I don't think this is a metric: $d([0,1],(0,1))=0$, if we work with the usual Lebesgue measure on $\mathbb R$. Do you identify the sets such that $m(A\triangle B)=0$? (So the space would be the space of equivalence classes rather than the space of all Borel sets?) $\endgroup$– Martin SleziakApr 29, 2014 at 7:28

$\begingroup$ Even after identifying equivalence classes, the answer might be no. If $\mathfrak{c}$ is a realvalued measurable cardinal, we can use the discrete metric and put an atomless probability measure on all subsets. The resulting measure will be as far as possible from separable by the GitikShelah theorem. But the answer will be yes if the metric space is separable. $\endgroup$– Michael GreineckerApr 29, 2014 at 7:42

$\begingroup$ @Michael: why don´t you turn your comment into an answer? $\endgroup$– Ramiro de la VegaApr 29, 2014 at 15:56

1$\begingroup$ Note that by identifying sets with their indicator functions, your metric space $(\mathcal{B},d)$ embeds as a total subset of $L^1(X,m)$. So I think your question is equivalent to asking about the separability of $L^1(X,m)$ where $X$ is metrizable. The answers to this question seem to apply here. $\endgroup$– Nate EldredgeApr 29, 2014 at 21:37

$\begingroup$ Would you like to give some more detail? I know that $(X,\mathscr{B})$ is separable if and only if the space $L^{1}(X,\mathscr{B},m)$ is separable ,but I donot know when $L^{1}(X,\mathscr{B},m)$ is separable. $\endgroup$– Yee NeilApr 30, 2014 at 0:55
1 Answer
As Martin Sleziak has pointed out in a comment, one could only a get a pseudometric space this way and has to quotient out the null sets to obtain a proper metric. The question of separability is the same in either case.
If the space one started with is a separable metric space, then the Borel $\sigma$algebra is generated by a countable base of the topology and (equivalence classes of) elements of the countable algebra generated by the basis form a countable dense subset.
If the original metric space was not separable, the answer might well be no. Under set theoretic assumptions widely considered to be consistent, there exists an atomless probability measure on the powerset of some set and hence on the Borel$\sigma$algebra of some discrete metric space. Using the GitikShelah Theorem, one can show that there must be an uncountable family of independent events with nontrivial probability in such a probability space. One can then show that for every countable family of measurable events, an event in this family must be independent of it. So no countable family of events can be dense under the pseudometric you gave.
Spelling this argument out, takes some effort. One can find the relevant material in the humongous measure theory treatise of Fremlin. The GitikShelah Theorem can be found there at 543E. That this implies that there exists an uncountable family of independent nontrivial events follows from Maharam's representation theorem, which figures prominently in his third volume. That this means one (actually, most) of these events are independent of a countable family of measurable sets follows now from 272Q.
Maybe, someone can give a more direct answer.

$\begingroup$ Thanks for your detail answer,I think I have got it.Maybe the space $X$ needs to be separable. $\endgroup$– Yee NeilApr 30, 2014 at 0:19