Notation: Let $(E,d)$ be a Polish space (separable and complete w.r.t. $d$). We say that a probability measure $\nu$ has finite first moment, if $$ \int_E d(x,x_0)\nu(dx)<\infty $$ for some $x_0 \in E$ and then, by the triangle inequality, $$ \int_E d(x,y_0)\nu(dx)\le\int_E d(x,x_0)\nu(dx)+d(x_0,y_0)<\infty $$ for all $y_0\in E$.
Question: Let $(\mu_n)_{n\in\mathbb{N}}, \mu$ be probability measures on $E$ with finite first moment. Suppose $$\int_E d(x, x_0)\mu_n(dx) \overset{n\to\infty}{\longrightarrow} \int_E d(x,x_0)\mu(dx) \qquad (*)$$ for some $x_0\in E$ and $\mu_n \rightharpoonup \mu$ (i.e. $\int_E f d\mu_n \longrightarrow \int_E fd\mu$ for all bounded, continuous $f:E\to\mathbb{R}$). Does $(*)$ then hold for all $x_0\in E$?
Own work: Let $y_0\in E$. Then
$$ \int_E d(x,y_0)\mu_n(dx)\le\int_E d(x,x_0)\mu_n(dx)+d(x_0,y_0),$$
so $\limsup_n \int_E d(x,y_0)\mu_n(dx) \le \int_E d(x,x_0)\mu(dx)+d(x_0,y_0)$. Since $d(x,y_0) \ge d(x,x_0)-d(x_0,y_0)$, we also $\int_E d(x,x_0)\mu(dx)-d(x_0,y_0)\le \liminf_n \int_E d(x,y_0)\mu_n(dx)$, which is a dead-end (?)
Different approach:
$$ \int_E d(x,y_0)\mu_n(dx) - \int_E d(x,y_0)\mu(dx) \le \int_E d(x,x_0)\mu_n(dx) - \int_E d(x,y_0)-d(x_0,y_0)\mu(dx), $$
and now I'd like to continue with $d(x,y_0)-d(x_0,y_0)\ge d(x, x_0)$, but the opposite is true. I know this should be easy but I don't see it, can someone help?
[The question has been edited after I posted thsi answer]
This is false. For a counter-example take $\mu_n=\delta_{-1}, \mu=\delta_1$ and $x_0=0$. The conclusion fails for $y_0=1$.
[Note that $\int f(x) \mu (dx)=f(z)$ if $\mu=\delta_z$].