Let $(S, d)$ be a complete separable metric space and consider the corresponding Borel $\sigma$-algebra $\mathcal{B}(S)$. Let $\mu$ and $\mu_n$, $n \in \mathbb{N}$, be probability measures on $\mathcal{B}(S)$ with finite first moments such that
$$
\lim_{n \rightarrow \infty} \mu_n(A) = \mu(A) \quad \forall A \in \mathcal{B}(S).
$$
According to this post, we have
$$
\lim_{n \rightarrow \infty} \int_S f \, d\mu_n = \int_S f \, d\mu
$$
for all bounded and Borel-measurable functions $f : S \rightarrow \mathbb{R}$. Since the latter is true for bounded and Lipschitz continuous functions, it further follows that $\mu_n$ converges weakly to $\mu$. I am further interested in the following question:
Is it true that
$$
\lim_{n \rightarrow \infty} \int_S d ( x, x_0 ) \mu_n (dx) = \int_S d (x, x_0) \mu (dx), \tag{1}
$$
for some $x_0 \in S$. In other words, do we have the convergence of the corresponding moments?
Moreover, it is known that weak convergence together with $(1)$ is equivalent to the convergence with respect to the Wasserstein metric. Hence the question could be formulated as:
Does set-wise convergence of probability measures imply convergence with respect to the Wasserstein metric?
Best Answer
No.
Take $S = \mathbb{N}$. Let $\mu_n = \frac{1}{n} \delta_n + \frac{n-1}{n} \delta_0$ as the measure which puts mass $1/n$ at the point $n$, and the rest at 0. Note that the first moment of each $\mu_n$ is $1$ (when centered at 0; the case of a different center $x_0$ I leave as an exercise).
Let $\mu = \delta_0$. I claim that $\mu_n \to \mu$ setwise. If $0 \in A$ then $\mu_n(A) \ge \frac{n-1}{n}$, so $\mu_n(A) \to 1 = \mu(A)$. If $0 \notin A$ then $\mu_n(A) \le \frac{1}{n}$ so $\mu_n(A) \to 0 = \mu(A)$. But the first moment of $\mu$ is not 1 but 0.