Is the space of probability measure dense in the space of probability measure with density?

Question

Let $\mathcal{P}_2(\mathbb{R}^n)$ be the set of all probability measures with finite second moment. Let $\mathcal{P}_2^*(\mathbb{R}^n)$ be the subset of $\mathcal{P}_2(\mathbb{R}^n)$ such that any element $\mu \in \mathcal{P}_2^*(\mathbb{R}^n)$ can be represented by $\mu=\nu(x)dx$ where $\mu \geq0$. Consider $\mathcal{P}_2(\mathbb{R}^n)$ and $\mathcal{P}_2^*(\mathbb{R}^n)$ with Wasserstein 2-distance, is $\mathcal{P}_2^*(\mathbb{R}^n)$ dense in $\mathcal{P}_2(\mathbb{R}^n)$? Thank you!!

Answer 1

Yes.

Let $\varepsilon\in(0,1)$. Let $X$ be a random variable following $\mu$, and $X_0$ be a fixed square-integrable random variable independent of $X$ with density $p_0$. You can easily find disjoint measurable sets $A_1,\cdots,A_n$ each with diameter smaller than $\varepsilon$ such that $\mu(A_1\cup\cdots\cup A_n)\ge1-\varepsilon$. Let $A_0=\mathbb R^n\backslash(A_1\cup\cdots\cup A_n)$, and for $i\in\{1,\cdots,n\}$ let $X_i$ be a random variable independent of $X$ and uniformly distributed on $A_i$. Therefore $X_i$ has a density, which we denote $p_i$. Then define $$ Y=\sum_{i=0}^nX_i1_{\{X\in A_i\}}. $$

The random variable $Y$ follows a distribution $\mu'$ which has a density, namely $\sum_{i=0}^n\mu(A_i)p_i$. And $$ \begin{align*} \mathcal W_2^2(\mu,\mu')&\le\mathbb E[\vert X-Y\vert^2]\\ &=\mathbb E[1_{\{X\in A_0\}}\vert X-X_0\vert^2]+\sum_{i=1}^n\mathbb E[1_{\{X\in A_i\}}\vert X-X_i\vert^2]\\&\le\Vert X-X_0\Vert_{L^2}\varepsilon+\varepsilon^2, \end{align*} $$ which proves the density of $\mathcal P_2^*(\mathbb R^n)$ in $\mathcal P_2(\mathbb R^n)$.