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 $\nu(X_1\times\cdots\times X_{i1}\times\cdot\times X_{i+1}\times\cdots\times X_n)$). Is it known whether a there exists a $\nu$ with nonatomic marginals arbitrarily close to $\mu$?

$\begingroup$ Without loss of generality, each of your $X_i$ can be taken to be the Hilbert cube $[0,1]^\omega$, in which case $X = [0,1]^\omega$ also, so it would be sufficient to show the following: every probability measure on the Hilbert cube can be approximated by measures with nonatomic marginals. That should be easier because the Hilbert cube has nice structure. In particular, it might follow from the statement for the finitedimensional cube $[0,1]^n$ and the Kolmogorov extension theorem. $\endgroup$ – Nate Eldredge Mar 10 '15 at 15:10
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 nonatomic Borel probability measures on X is weak$^*$ dense in the set of all Borel probability measures on $X$. In particular, any deltameasure on $X$ can be approximated by nonatomic measures. It implies that in the case of a product space $X=X_1\times\dots\times X_n$ any deltameasure on $X$ can be approximated by products of nonatomic measures (so that, in particular, all their marginals are nonatomic). Now, in turn, an arbitrary measure on $X$ can be approximated by finite convex combinations of deltameasures on $X$.
I believe the answer is yes. Although it was not specifically stated in the question, let us assume that each $X_i$ has no isolated points.
Since $\mu$ has only finitely many marginals, and each marginal has only countably many atoms, we may write $\mu = \sum_i a_i \, \mu_i$, where, for each $i$, $a_i > 0$, $\mu_i$ is a probability measure, and either
 the marginals of $\mu_i$ have no atoms, or
 some marginal of $\mu_i$ is a single atom $\delta_x$.
For each $i$, we will show there is a sequence $\{\mu_{i,k}\}$ of good probability measures that converges to $\mu_i$. Then $\nu_k = \sum_i a_i \, \mu_{i,k}$ is also good, for each $k$. Also, we have $\nu_k \to \mu$. (I am not familiar with the Prokhorov metric, but $\{\nu_k\}$ converges to $\mu$ in the weak topology, which I believe gives the same topology on the space of probability measures.)
To complete the proof, we may assume the marginal of $\mu$ on $X_n$ is a single atom $\delta_x$ at some $x \in X_n$. That is, we may write $\mu = \mu' \times \delta_{x}$, where $\mu'$ is a probability measure on $X_1 \times \cdots \times X_{n1}$. By induction on $n$, we may approximate $\mu'$ by a probability measure $\nu'$, whose marginals have no atoms. Also, we may approximate $\delta_x$ by an atomless probability measure $\nu''$ on $X_n$. Then $\nu' \times \nu''$ approximates $\mu$, and its marginals have no atoms.
Some additional details of the proof can be found below.
Write the atomic part of the $X_1$marginal of $\mu$ as $\sum_j a_j \delta_{x_j}$, where the $x_j$'s are distinct and $a_j \ge 0$. For each $j$, let $$\nu_j'(E) = \mu \Bigl( E \cap \bigl( \{x_j\} \times X_2 \times \cdots X_n \bigr) \Bigr),$$ and write $\nu_j' = a_j \nu_j$, where $\nu_j$ is a probability measure. Then we have $\mu = \mu^{(1)} + \sum_j a_j \nu_j$, where the $X_1$marginal of $\mu^{(1)}$ has no atoms, and the $X_1$marginal of $\nu_j$ is $\delta_{x_j}$ (or $a_j = 0$).
Applying a similar argument, we may write $\mu^{(1)} = \mu^{(2)} + \sum_j b_j \nu^2_j$, where both the $X_1$ and $X_2$marginals of $\mu^{(2)}$ are atomless, and the $X_2$marginal of $\nu^2_j$ is an atom (or $b_j = 0$). Continuing in this way, we write $$ \mu = \mu^{(n)} + \sum_{i=1}^n \sum_j a_{i,j} \mu_{i,j} ,$$ where all marginals of $\mu^{(n)}$ are atomless, and, for all $i$ and $j$, the $X_i$marginal of $\mu_{i,j}$ is an atom (or $a_{i,j} = 0$). Deleting the terms with $a_{i,j} = 0$ yields the desired decomposition of $\mu$.
We have $\nu_k = \sum_i a_i \mu_{i,k}$. Given a continuous function $f \colon X \to \mathbb{R}$ and $\epsilon > 0$, assume, for simplicity, that $\f\_\infty \le 1$. Since $\sum_i a_{i,k} = 1$, there is some $M$, such that $\sum_{i > M} a_{i,k} < \epsilon$. Then, since $\mu_{i,k} \to \mu_i$, we have
$\left \int_X f \, d\mu  \int_X f \, d\nu_k \right$
$\le \left \int_X f \, d\mu  \sum_{i=1}^M a_i \int_X f \, d\mu_{i,k} \right + \epsilon$
$\to \left \int_X f \, d\mu  \sum_{i=1}^M a_i \int_X f \, d\mu_{i} \right + \epsilon$
$\le \left \int_X f \, d\mu  \sum_{i=1}^\infty a_i \int_X f \, d\mu_{i} \right + 2\epsilon$
$= 0 + 2\epsilon $.
So $\nu_k \to \mu$ weakly.
Suppose $\nu'_i \to \mu'$ on $X_1 \times \cdots \times X_{n1}$ and $\nu''_i \to \delta_x$ on $X_n$. Let $f$ be a continuous function on $X_1 \times \cdots \times X_{n1}$ and let $g$ be a continuous function on $X_n$. Then $$\int_X (f \times g) \, d(\nu'_i \times \nu''_i) = \nu'_i(f) \cdot \nu''_i(g) \to \mu'(f) \cdot \delta_x(g) .$$ Since functions of the form $f \times g$ span a dense subspace of the continuous functions on $X$, this implies $\nu'_i \times \nu''_i \to \mu' \times \delta_x$.