[proofplan] The hypotheses give exactly the narrow convergence and moment convergence needed to upgrade each sequence to convergence in the Wasserstein metric. Applying the characterization of $W_p$-convergence, we obtain $W_p(\mu_k,\mu)\to 0$ and $W_p(\nu_k,\nu)\to 0$. The desired convergence then follows from the triangle inequality for the metric $W_p$, applied twice to compare $W_p(\mu_k,\nu_k)$ with $W_p(\mu,\nu)$. [/proofplan]

[step:Upgrade narrow convergence and moment convergence to Wasserstein convergence]Because $\mu_k \to \mu$ narrowly and \begin{align*} \int_X d(x,x_0)^p\,d\mu_k(x) \to \int_X d(x,x_0)^p\,d\mu(x), \end{align*} the characterization of Wasserstein convergence by narrow convergence and convergence of $p$-moments applies to the sequence $(\mu_k)_{k \in \mathbb N}$ and gives \begin{align*} W_p(\mu_k,\mu) \to 0. \end{align*} Likewise, because $\nu_k \to \nu$ narrowly and \begin{align*} \int_X d(x,x_0)^p\,d\nu_k(x) \to \int_X d(x,x_0)^p\,d\nu(x), \end{align*} the same characterization applies to the sequence $(\nu_k)_{k \in \mathbb N}$ and gives \begin{align*} W_p(\nu_k,\nu) \to 0. \end{align*} Here we are using the characterization theorem for $W_p$-convergence on $\mathcal P_p(X)$: narrow convergence together with convergence of the $p$-moment about one, equivalently every, base point is equivalent to convergence in $W_p$ (citing a result not yet in the wiki: Characterization of Wasserstein convergence by narrow convergence and convergence of p-moments).[/step]

[guided]We first convert the stated hypotheses into the form that interacts directly with the Wasserstein metric. The relevant result is the characterization of $W_p$-convergence on $\mathcal P_p(X)$: for probability measures in $\mathcal P_p(X)$ on a Polish [metric space](/page/Metric%20Space), convergence in $W_p$ is equivalent to narrow convergence plus convergence of the $p$-moments about one, equivalently every, base point. For the sequence $(\mu_k)_{k \in \mathbb N}$, the theorem's hypotheses give both required inputs. First, $\mu_k \to \mu$ narrowly. Second, for the chosen point $x_0 \in X$, \begin{align*} \int_X d(x,x_0)^p\,d\mu_k(x) \to \int_X d(x,x_0)^p\,d\mu(x). \end{align*} Since every $\mu_k$ and $\mu$ lies in $\mathcal P_p(X)$, the measures have finite $p$-moment, so the characterization theorem applies. Therefore \begin{align*} W_p(\mu_k,\mu) \to 0. \end{align*} The same verification applies to the sequence $(\nu_k)_{k \in \mathbb N}$. The hypotheses give $\nu_k \to \nu$ narrowly and \begin{align*} \int_X d(x,x_0)^p\,d\nu_k(x) \to \int_X d(x,x_0)^p\,d\nu(x). \end{align*} Since every $\nu_k$ and $\nu$ lies in $\mathcal P_p(X)$, the characterization theorem again applies, and we obtain \begin{align*} W_p(\nu_k,\nu) \to 0. \end{align*} This is the only place where the moment convergence assumption is used. Narrow convergence alone would not control how mass may escape to infinity, while convergence of the $p$-moments supplies exactly the missing tightness at order $p$.[/guided]

[step:Compare the two Wasserstein distances by the triangle inequality] Since $W_p$ is a metric on $\mathcal P_p(X)$, its triangle inequality gives \begin{align*} W_p(\mu_k,\nu_k) \le W_p(\mu_k,\mu) + W_p(\mu,\nu) + W_p(\nu,\nu_k). \end{align*} Using the symmetry of $W_p$, this becomes \begin{align*} W_p(\mu_k,\nu_k) - W_p(\mu,\nu) \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu). \end{align*} Applying the same triangle inequality with the roles of $(\mu_k,\nu_k)$ and $(\mu,\nu)$ interchanged gives \begin{align*} W_p(\mu,\nu) \le W_p(\mu,\mu_k) + W_p(\mu_k,\nu_k) + W_p(\nu_k,\nu). \end{align*} Again using symmetry, we obtain \begin{align*} W_p(\mu,\nu) - W_p(\mu_k,\nu_k) \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu). \end{align*} Combining the two one-sided estimates yields \begin{align*} \left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu). \end{align*} [/step]

[step:Pass to the limit in the comparison estimate] From the first step, \begin{align*} W_p(\mu_k,\mu) \to 0 \end{align*} and \begin{align*} W_p(\nu_k,\nu) \to 0. \end{align*} Therefore the right-hand side of \begin{align*} \left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \le W_p(\mu_k,\mu) + W_p(\nu_k,\nu) \end{align*} converges to $0$. Since the left-hand side is non-negative for every $k \in \mathbb N$, the squeeze principle gives \begin{align*} \left|W_p(\mu_k,\nu_k)-W_p(\mu,\nu)\right| \to 0. \end{align*} Equivalently, \begin{align*} W_p(\mu_k,\nu_k) \to W_p(\mu,\nu). \end{align*} This is the desired stability statement. [/step]