Building the Wasserstein space by pushforwards

Question

Let $\mathbb{R}^d$ denote the $d$-dimensional Euclidean space, $\mathcal{W}_2(\mathbb{R}^d)$ denote the $2$-Wasserstein space with respect to the $d$-dimensional Euclidean space $\mathbb{R}^d$. Let $L^2(\mathbb{R}^d)$ denote the Bochner space of all Borel functions $f:\mathbb{R}^d\rightarrow \mathbb{R}^d$ satisfying $\int \, \lVert f(x)\rVert^2dx<\infty$.

Let $\mathcal{X}\subseteq \mathcal{W}_2(\mathbb{R}^d)$ consist of all measures ${\nu}$ for which there is some $f\in L^2(\lambda,\mathbb{R}^d)$ satisfying: $$ {\nu}=f_{\#}\lambda $$ where $\lambda$ is the uniform measure on $[0,1]^d$.

How is $\mathcal{X}$ related to $\mathcal{W}_2(\mathbb{R}^d)$? Is $\mathcal{X}$ a dense subset of $\mathcal{W}_2(\mathbb{R}^d)$?

Answer 1

Here is an alternative answer, which is weaker than @Benoît Kloeckner's strong form and only asserts density.

Yes, your set is dense. Indeed, $\mathcal X$ contains finitely atomic measures, which are obviously weakly-$\ast$ dense in the Wasserstein space and therefore also dense in distance (it is well known that the Wasserstein distance $W_2$ metrize the weak-$\ast$ convergence, modulo minor decay conditions at infinity given by convergence of the second moments).

To see that $\mathcal X$ contains finitely atomic measures, fix an arbitrary integer $N$ and take any $\nu=\sum_{i=1}^N \nu_i\delta_{x_i}\in \mathcal W_2(\mathbb R^d)$. Choose now any partition of your unit cube $[0,1]^d=\bigcup_{i=1}^1 C_i$ into $N$ disjoint sets with corresponding measures $|C_i|=\nu_i$. Then the piecewise constant map $f$ defined as $f(x)=x_i$ for $x\in C_i$ pushes forward $\lambda$ to $\nu$.

Answer 2

Perhaps another simple argument that $\mathcal{X}$ is indeed equal to $\mathcal{W}_2(\mathbb{R}^d$): Starting from the fact that there is a bimeasurable bijection $b : \mathbb{R}^d \rightarrow \mathbb{R}$. (c.f. here).

For any $\nu \in \mathcal{W}_2(\mathbb{R}^d)$, let $\tilde\nu := b_{\#}\nu$, and thus for the inverse $a$ of $b$ it holds $\nu = a_{\#} \tilde\nu$. Let further $\mathcal{U}$ be the uniform measure on $(0, 1)$ and note that $\mathcal{U} = (\pi_1)_{\#} \lambda$, where $\pi_1$ is the projection onto the first coordinate. Let $Q$ be the quantile function of $\tilde\nu$ and thus $\tilde\nu = Q_{\#} \mathcal{U}$. Putting things together, $\nu = (a \circ Q \circ \pi_1)_{\#} \lambda$, where $a \circ Q \circ \pi_1 \in L^2(\lambda)$ follows through this equality.