Density in Wasserstein space

I am wondering whether the following result is true:

Let $\mathcal W_p(\mathbb R^d)$ be the Wasserstein space of order $p$ and let $\eta$ and $\gamma$ be two probability measures in $\mathcal W_p(\mathbb R^d)$, such that supp($\eta$) $\subset$ supp($\gamma$). Then there is a sequence of probability measures $(\eta_n)$ that converges to $\eta$ in $\mathcal W_p(\mathbb R^d)$ and such that $\eta_n$ is absolutely continuous w.r.t. $\gamma$, with $\frac{d\eta_n}{d\gamma}$ being bounded.

It seems to be a well-known result but I was not able to prove it nor to find a reference. Any help and suggestion are appreciated!

• What do you mean by $\frac{d \eta_n}{d \gamma}$ being bounded''? Do you mean that for each $n$ separately, or uniformly in $n$? The latter is not possible. – A Blumenthal Jan 8 '18 at 19:08
• I mean it is bounded for each n – Ryan Jan 8 '18 at 19:13

First approximate $\eta$ with finite convex combinations of $\delta$-measures at points from $\mathrm{supp}\gamma$, then approximate each of these $\delta$-measures with measures with constant density with respect to $\gamma$ (just take the normalized restrictions of $\gamma$ to small balls).
• Thanks! Could you provide a little bit of details? I am not sure how to do the approximation in second moment, when supp($\eta$) is unbounded. – Ryan Jan 8 '18 at 20:19
• So you are not assuming that the measures have a finite $p$-th moment? – R W Jan 9 '18 at 3:34
• Sorry I forgot in the statement I use $p$ instead of 2. I do not think the value of $p$ matters though. It seems that one can show $\int \phi(x) d\mu_n \rightarrow \int \phi(x) d\mu$ for any continuous function $\phi$. Am I correct? – Ryan Jan 9 '18 at 4:54
• Who are $\mu_n$? Convergence in the Wasserstein $p$-distance is equivalent to weak convergence plus convergence of the $p$-th moments. – R W Jan 9 '18 at 7:52
• what i meant was that one should be able to find a sequence $(\mu_n)$ that converges to $\mu$ in the above sense and such that each $\frac{\mu_n}{\gamma}$ is bounded. – Ryan Jan 9 '18 at 16:49