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 $\pi.$