I have a Borel probability measure $\pi$ on $\mathbb{R}^{n+1}$ such that $\pi\_1=\mu\_1, \ldots, \pi\_{n+1}=\mu\_{n+1}$ for some fixed Borel probability measures $\mu\_1, \ldots, \mu\_{n+1}$ (where each $\mu\_i$ is absolutely continuous with respect to Lebesgue measure). I want to construct a sequence of probability measures $\pi^{(k)}$ such that $\pi^{(k)}$ converges weakly to $\pi$ and has the same marginals as $\pi,$ that is, $\pi^{(k)}\_j=\mu\_j$ for $j=2,\ldots, n+1.$


Now of course, I can take $\pi^{(k)}=\pi.$ But, I am trying to prove some inequality where I need $\pi^{(k)}$ to be absolutely continuous with respect to Lebesgue measure (on $\mathbb{R}^n$). 


Now I have a twofold goal: I can make $\pi^{(k)}$ absolutely continuous but then in the process I end up messing with the marginals. Can someone point it out to me if it is possible at all to cook up such a sequence $\pi^{(k)}$ which is absolutely continuous with respect to Leb, and has the fixed marginals and which weakly converges to $\pi.$



Absolutely continuous coupling of probability measures
Let me formulate and prove it in greater generality (which actually makes your question easier). Let $X$ be a metric space, and $\mu$ be a probability measure on $X\times X$ (for simplicity I consider the product of two copies of $X$ only; the general case is precisely the same). You want to obtain a sequence of measures $\nu\_n$ on $X\times X$ which


(1) have the same marginals $\mu\_1,\mu\_2$ as $\mu$;


(2) are absolutely continuous with respect to the product measure $\mu\_1\times\mu\_2$;


(3) weakly converge to $\mu$.


For each $n$ take a countable partition of $X$ into measurable sets $X^i$ with diameter $\le 1/n$ (presuming the space $X$ is such that partitions like this exist for any $n$) and denote by $\mu\_\epsilon^i$ the normalized restriction of the measure $\mu\_\epsilon$ to $X^i$. Then put 
$$
\nu\_n = \sum\_{i,j} \mu(X^i\times X^j) \mu\_1^i\times \mu\_2^j \;.
$$ 


