Nonatomic probability measures 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 $\mu$?
 A: 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 $X$.    
A: 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 $X_1 \times \cdots \times X_{n-1}$. 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 $X_1 \times \cdots \times X_{n-1}$ and $\nu''_i \to \delta_x$ on $X_n$. Let $f$ be a continuous function on $X_1 \times \cdots \times X_{n-1}$ 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 $\nu'_i \times \nu''_i \to \mu' \times \delta_x$.
