It is known that for a compact metric space $X$ without isolated points the set of nonatomic Borel probability measures on $X$ is dense in the set of all Borel probability measures on $X$ (endowed with the Prokhorov metric). In particular if $X$ is a product space $X=X_1\times\cdots\times X_n$ (each $X_i$ a compact metric space), and given a measure $\mu$ on $(X,\mathcal B(X))$ ($\mathcal B(X)$ the Borel subsets of $X$), there is a nonatomic $\nu$ measure on $(X,\mathcal B(X))$ arbitrarily close to $\mu$. In general, $\nu$ need not have nonatomic marginal probability measures (here the marginal for the $i$-th factor is $\nu(X_1\times\cdots\times X_{i-1}\times\cdot\times X_{i+1}\times\cdots\times X_n)$). Is it known whether a there exists a $\nu$ with nonatomic marginals arbitrarily close to $\mu$?
[Math] Nonatomic probability measures
measure-theory
Related Solutions
Of course, there are many ways of metrizing the weak topology on $\mathcal M(\Omega)$ by using various tools of functional analysis. However, as it has already been pointed out by Dan, the most natural way is to use the transportation metric on the space of measures. [It is much more natural than the Prokhorov metric. I don't want to go into historical details here - they can be easily found elsewhere, but I insist that the transportation metric should really be related with the names of Kantorovich (in the first place) and his collaborator Rubinshtein]. Dan gives its dual definition in terms of Lipschitz functions, however its "transport definition" is actually more appropriate here. Let me remind it.
Given two probability measures $\mu_1,\mu_2$ on $\Omega$ $$ \overline d(\mu_1,\mu_2) = inf_M \int d(x_1,x_2) dM(x_1,x_2) \;, $$ where $d$ is the original metric on $\Omega$, and the infimum (which is in fact attained) is taken over all probability measures $M$ on $\Omega\times\Omega$ whose marginals ($\equiv$ coordinate projections) are $\mu_1$ and $\mu_2$. One should think about such measures as "transportation plans" between distributions $\mu_1$ and $\mu_2$, while the integral in the RHS of the definition is the "cost" of the plan $M$.
It is obvious that the above definition makes sense not just for probability measures, but for any two positive measures $\mu_1,\mu_2$ with the same mass. Moreover, $\overline d(\mu_1,\mu_2)$ actually depends on the difference $\mu_1-\mu_2$ only, so that one can think about it as a "weak norm" $$ |||\mu_1-\mu_2||| = \overline d(\mu_1,\mu_2) $$ of the signed measure $\mu_1-\mu_2$ (clearly, it is homogeneous with respect to multiplication by scalars).
Let now $\mu=\mu_1-\mu_2$ be an arbitrary signed measure, where $\mu_1,\mu_2$ are the components of its Hahn decomposition. The only reason why the definition of the weak norm does not work in this situation is that the measure $\mu$ need not to be "balanced" in the sense that the total masses $\|\mu_1\|$ and $\|\mu_2\|$ need not be the same any more. However, this can be easily repaired in the following way: extend the original space $\Omega$ to a new metric space $\Omega'$ by adding to it an "ideal point" $o$ and putting $d(\omega,o)=1$ for any $\omega\in\Omega$. Then the measure $$ \mu'=\mu - (\|\mu_1\|-\|\mu_2\|)\delta_o \;, $$ where $\delta_o$ is the unit mass at the point $o$, is now balanced, so that $|||\mu'|||$ is well defined. Therefore, one can extend the definition of the weak norm $|||\cdot|||$ to arbitrary signed measures $\mu$ by putting $$ |||\mu|||=|||\mu'||| \;. $$
It is now easy to see that the distance $|||\mu_1-\mu_2|||$, where $\mu_1,\mu_2$ are two arbitrary signed measures, metrizes the weak topology on $\mathcal M(\Omega)$.
Consider the set $\mathcal A$ of measurable subsets $A \subset X$, such that $\mu \mapsto \mu[A]$ is measurable. Obviously, $\mathcal A$ contains all open sets.
It's also easy to see that $\mathcal A$ contains an algebra of sets - say, sets $A \subset X$, such that their indicator is a pointwise limit of a sequence of continuous functions. Indeed, if $f_n \to \mathsf{1} [A]$ and $0 \le f_n \le 1$ (which can always be assumed, since we can replace $f_n$ by $\max(\min(f_n, 0), 1)$) then the function $\mu \mapsto \mu[A]$ is the pointwise limit of $\mu \mapsto \intop f_n d\mu$.
A similar argument to the above shows that $\mathcal{A}$ is closed under sequential limits of sets. And anything that contains an algebra and is closed under sequential limits contains a $\sigma$-algebra (I would call that a version of the monotone class theorem). Therefore, $\mathcal A \supset \mathcal{B} (X)$.
Best Answer
I think that Dave's argument (as well as the reference to the Hilbert cube) make this question more complicated than it actually is.
Let's take for a starting point the claim already formulated by the topicstarter: for a compact metric space $X$ without isolated points the set of non-atomic Borel probability measures on X is weak$^*$ dense in the set of all Borel probability measures on $X$. In particular, any delta-measure on $X$ can be approximated by non-atomic measures. It implies that in the case of a product space $X=X_1\times\dots\times X_n$ any delta-measure on $X$ can be approximated by products of non-atomic measures (so that, in particular, all their marginals are non-atomic). Now, in turn, an arbitrary measure on $X$ can be approximated by finite convex combinations of delta-measures on $X$.