Let $B_R$ be a closed ball of radius $R$ in the space $\mathbb{R}^d$. As the title suggests I have this feeling that the set of functions
$$S:= \left\lbrace f:\mathbb{R}^d \to \mathbb{R} \hspace{2mm}| \hspace{2mm} \text{supp} f\subset B_R, \int_{\mathbb{R}^d}f(x)dx = 1, f\geq 0\right\rbrace$$
is complete with respect to the Wasserstein metric, which is defined by
$$W_1(f,g) = \inf_{\pi\in \Pi(f,g)}\int_{\mathbb{R}^{2d}}\|x-y\|_{l^1}\pi(x,y)dxdy$$
and $\Pi(f,g)$ is the set of transference plans between $f$ and $g$, i.e. for any $\phi\in C_b(\mathbb{R}^d; \mathbb{R})$ we have
$$\int_{\mathbb{R}^{2d}}\phi(x)\pi(x,y)dxdy = \int_{\mathbb{R}^d}\phi(x)f(x)dx$$ $$\int_{\mathbb{R}^{2d}}\phi(y)\pi(x,y)dxdy = \int_{\mathbb{R}^d}\phi(y)g(y)dy.$$
This was taken as a fact by my supervisor but I don't think it is trivial to show. Does it follow from the alternate way of writing the Wasserstein distance
$$W_1(f,g) = \sup_{\|\phi\|_{\text{Lip}}\leq 1}\left|\int_{\mathbb{R}^d}\phi(x)f(x)dx - \int_{\mathbb{R}^d}\phi(y)g(y)dy\right|$$
in terms of the bounded Lipschitz functions? Then somehow showing that the set $S$ is a closed subset of the (complete) set of probability functions under the Wasserstein metric? Any help or proof would be greatly appreciated!