Let $(\mathcal{X}, \Sigma)$ be a Polish metric space, endowed with the Borel $\sigma$-algebra. Let $\mathscr P$ be the space of probability measures on $\mathcal X$ and $\mathscr P^1$ be defined as
$$\mathscr P^1=\{\pi\in\mathscr P:\mathbb E_\pi[d(x_0, X)]<+\infty\}\,,$$
for some $x_0\in\mathcal X$.
Then it is well known that if $\mu$ and $\nu$ are measures in $\mathscr P^1$
$$\mathscr W(\mu, \nu) = \sup_{f\in\mathrm{Lip}(1)}(\mathbb E_\mu[f(X)]-\mathbb E_\nu[f(X)])\,,$$
where $\mathscr W$ is the 1-Wasserstein distance.
Can something be said when $\mu$ and $\nu$ are not required to be in $\mathscr P^1$, but are just in $\mathscr P$?
Of course now it might be that $\mathbb E_\mu[f(X)]$ and $\mathbb E_\nu[f(X)]$ are not finite. It might even be that $f$ is not integrable. But can we still say that if $f$ is $1$-Lipschitz then $$\mathscr W(\mu, \nu) \geq \mathbb{E}_\mu[f(X)]-\mathbb{E}_\nu[f(X)]\,,$$ for all $\mu$ and $\nu$ such that $\mathbb E_\mu[|f(X)|]$ and $\mathbb E_\nu[|f(X)|]$ are finite?
I think that the answer is yes, and it is not hard to show it. We have that if $f\in L^1(\mu)\cap L^1(\nu)$ is 1-Lipschitz, then, given any coupling $\pi$ with marginals $\mu$ and $\nu$ we have $$\mathbb E_\mu[f(X)]-\mathbb E_\nu [f(X)] = \mathbb E_{(X, X')\sim\pi} [f(X)-f(X')]\leq \mathbb E_{(X, X')\sim\pi} [d(X,X')]\,.$$ Taking the inf wrt $\pi$ we conclude.