[proofplan] The proof has two compactness ingredients. The uniform second-moment bound gives tightness of every discrete time slice, so we first extract a diagonal subsequence converging narrowly at a fixed countable dense set of times. The discrete action estimate gives quantitative Wasserstein bounds between nearby discrete representatives; passing these bounds to the dense-time limits gives a $W_2$-continuous extension. Finally, the same partition estimates give lower semicontinuity of the metric action, which identifies the limit as an $AC^2$ curve and gives the stated derivative bound. [/proofplan]

[step:Extract narrow limits on a countable dense set of times]Let $D\subset[0,T]$ be a countable dense set containing $0$ and $T$. For each $j\in\mathbb N$, define the discrete index map \begin{align*} k_j:[0,T]&\to\{0,\dots,N_j\} \end{align*} by setting $k_j(0):=0$ and, for $t\in(0,T]$, \begin{align*} k_j(t):=\min\{k\in\{0,\dots,N_j\}: t\le k\tau_j\}. \end{align*} Then $\bar\rho_j(t)=\rho_{j,k_j(t)}$ for every $t\in[0,T]$. We claim that the family \begin{align*} \mathcal K:=\{\rho_{j,k}:j\in\mathbb N,\ 0\le k\le N_j\} \end{align*} is narrowly relatively compact. Indeed, for every $R>0$ and every $\mu\in\mathcal K$, Markov's inequality gives \begin{align*} \mu(\mathbb R^n\setminus B(0,R))\le \frac{1}{R^2}\int_{\mathbb R^n}|x|^2\,d\mu(x)\le \frac{M_T}{R^2}. \end{align*} Thus $\mathcal K$ is tight. By Prokhorov's compactness theorem on the Polish space $\mathbb R^n$, $\mathcal K$ is narrowly relatively compact. Enumerate $D$ as $(q_\ell)_{\ell\in\mathbb N}$. Applying narrow relative compactness successively to the sequences $(\rho_{j,k_j(q_\ell)})_{j\in\mathbb N}$ and using a diagonal extraction, we obtain a subsequence, still denoted by $(j_m)_{m\in\mathbb N}$, and probability measures $(\rho_q)_{q\in D}\subset\mathcal P(\mathbb R^n)$ such that \begin{align*} \rho_{j_m,k_{j_m}(q)}\to\rho_q \end{align*} narrowly as $m\to\infty$ for every $q\in D$. The second-moment bound and lower semicontinuity of the map $\mu\mapsto \int_{\mathbb R^n}|x|^2\,d\mu(x)$ under narrow convergence imply \begin{align*} \int_{\mathbb R^n}|x|^2\,d\rho_q(x)\le M_T, \end{align*} so $\rho_q\in\mathcal P_2(\mathbb R^n)$ for every $q\in D$.[/step]

[guided]The first obstacle is compactness in space. We are not trying to prove $W_2$-compactness of the whole family, because a uniform second-moment bound alone gives tightness and narrow compactness, not compactness in the $W_2$ topology. So we work in the narrow topology first. Fix a countable dense set $D\subset[0,T]$ containing $0$ and $T$. For a time $t\in[0,T]$, the piecewise-constant curve $\bar\rho_j$ takes the value at the first grid point lying at or to the right of $t$. We encode this by defining \begin{align*} k_j(t):=\min\{k\in\{0,\dots,N_j\}: t\le k\tau_j\}, \end{align*} with $k_j(0)=0$. Hence $\bar\rho_j(t)=\rho_{j,k_j(t)}$. Now define the full collection of all discrete measures by \begin{align*} \mathcal K:=\{\rho_{j,k}:j\in\mathbb N,\ 0\le k\le N_j\}. \end{align*} We verify tightness. Let $R>0$ and let $\mu\in\mathcal K$. Since $\mu=\rho_{j,k}$ for some $j$ and $k$, the hypothesis gives \begin{align*} \int_{\mathbb R^n}|x|^2\,d\mu(x)\le M_T. \end{align*} Markov's inequality applied to the nonnegative function $x\mapsto |x|^2$ gives \begin{align*} \mu(\mathbb R^n\setminus B(0,R))=\mu(\{x\in\mathbb R^n:|x|^2\ge R^2\})\le \frac{1}{R^2}\int_{\mathbb R^n}|x|^2\,d\mu(x)\le \frac{M_T}{R^2}. \end{align*} Given $\varepsilon>0$, choosing $R>0$ such that \begin{align*} R>\left(\frac{M_T}{\varepsilon}\right)^{1/2} \end{align*} makes the right-hand side at most $\varepsilon$. Thus $\mathcal K$ is tight. Since $\mathbb R^n$ with its Euclidean topology is a Polish space, Prokhorov's compactness theorem implies that $\mathcal K$ is relatively compact for narrow convergence. We now use this compactness at countably many times. Enumerate $D$ as $(q_\ell)_{\ell\in\mathbb N}$. For $q_1$, the sequence $(\rho_{j,k_j(q_1)})_{j\in\mathbb N}$ has a narrowly convergent subsequence. Inside that subsequence, the sequence at $q_2$ has a narrowly convergent subsequence, and so on. Taking the usual diagonal subsequence gives indices $(j_m)_{m\in\mathbb N}$ and measures $(\rho_q)_{q\in D}$ such that \begin{align*} \rho_{j_m,k_{j_m}(q)}\to\rho_q \end{align*} narrowly for every $q\in D$. Finally, the limit measures have finite second moment. The function $x\mapsto |x|^2$ is lower semicontinuous and nonnegative, so the portmanteau lower semicontinuity property gives \begin{align*} \int_{\mathbb R^n}|x|^2\,d\rho_q(x)\le \liminf_{m\to\infty}\int_{\mathbb R^n}|x|^2\,d\rho_{j_m,k_{j_m}(q)}(x)\le M_T. \end{align*} Therefore $\rho_q\in\mathcal P_2(\mathbb R^n)$ for every $q\in D$.[/guided]

[step:Bound discrete increments between arbitrary grid representatives] For $0\le a\le b\le N_j$, define the local discrete action map \begin{align*} A_j:\{(a,b)\in\{0,1,\dots,N_j\}^2:0\le a\le b\le N_j\}\to[0,\infty) \end{align*} by \begin{align*} A_j(a,b):=\sum_{k=a}^{b-1}\frac{W_2^2(\rho_{j,k+1},\rho_{j,k})}{\tau_j}, \end{align*} where the empty sum is $0$. By the triangle inequality for $W_2$ and the Cauchy-Schwarz inequality for the finite sum over $k=a,\dots,b-1$, \begin{align*} W_2(\rho_{j,b},\rho_{j,a})\le \sum_{k=a}^{b-1}W_2(\rho_{j,k+1},\rho_{j,k}). \end{align*} Writing each summand as \begin{align*} W_2(\rho_{j,k+1},\rho_{j,k})=\tau_j^{1/2}\frac{W_2(\rho_{j,k+1},\rho_{j,k})}{\tau_j^{1/2}}, \end{align*} Cauchy-Schwarz gives \begin{align*} W_2^2(\rho_{j,b},\rho_{j,a})\le (b-a)\tau_j A_j(a,b). \end{align*} In particular, \begin{align*} W_2^2(\rho_{j,b},\rho_{j,a})\le C_T(b-a)\tau_j. \end{align*} If $s,t\in[0,T]$ with $s\le t$, then $k_j(s)\le k_j(t)$ and \begin{align*} (k_j(t)-k_j(s))\tau_j\le t-s+\tau_j. \end{align*} Consequently \begin{align*} W_2^2(\bar\rho_j(t),\bar\rho_j(s))\le C_T(t-s+\tau_j). \end{align*} [/step]

[step:Pass the discrete estimates to the dense-time limits] Let $q,r\in D$ with $q\le r$. Applying the preceding estimate to the subsequence $(j_m)_{m\in\mathbb N}$ gives \begin{align*} W_2^2(\rho_{j_m,k_{j_m}(r)},\rho_{j_m,k_{j_m}(q)})\le C_T(r-q+\tau_{j_m}). \end{align*} The measures $\rho_{j_m,k_{j_m}(q)}$ and $\rho_{j_m,k_{j_m}(r)}$ converge narrowly to $\rho_q$ and $\rho_r$, respectively. Since the cost $(x,y)\mapsto |x-y|^2$ is lower semicontinuous and nonnegative on $\mathbb R^n\times\mathbb R^n$, the Wasserstein distance is lower semicontinuous under narrow convergence of both marginals. Therefore \begin{align*} W_2^2(\rho_r,\rho_q)\le \liminf_{m\to\infty}W_2^2(\rho_{j_m,k_{j_m}(r)},\rho_{j_m,k_{j_m}(q)})\le C_T(r-q). \end{align*} Thus the map \begin{align*} D&\to(\mathcal P_2(\mathbb R^n),W_2) \end{align*} \begin{align*} q&\mapsto\rho_q \end{align*} is $1/2$-Hölder continuous with constant $C_T^{1/2}$. [/step]

[step:Extend the dense-time limit to a continuous Wasserstein curve]Since $(\mathcal P_2(\mathbb R^n),W_2)$ is complete, the uniformly continuous map $q\mapsto\rho_q$ from the dense subset $D\subset[0,T]$ has a unique continuous extension \begin{align*} \rho:[0,T]\to(\mathcal P_2(\mathbb R^n),W_2). \end{align*} For every $s,t\in[0,T]$ with $s\le t$, choose sequences $(q_i)_{i\in\mathbb N}$ and $(r_i)_{i\in\mathbb N}$ in $D$ such that $q_i\to s$, $r_i\to t$, and $q_i\le r_i$. Passing to the limit in \begin{align*} W_2^2(\rho_{r_i},\rho_{q_i})\le C_T(r_i-q_i) \end{align*} gives \begin{align*} W_2^2(\rho_t,\rho_s)\le C_T(t-s). \end{align*} This proves in particular that $\rho_t\in\mathcal P_2(\mathbb R^n)$ for every $t\in[0,T]$ and that $\rho$ is continuous with respect to $W_2$.[/step]

[guided]The dense-time estimates are strong enough to define the limiting curve at every time. The target space is $(\mathcal P_2(\mathbb R^n),W_2)$, and this metric space is complete. From the previous step, the map \begin{align*} D&\to(\mathcal P_2(\mathbb R^n),W_2) \end{align*} \begin{align*} q&\mapsto\rho_q \end{align*} satisfies \begin{align*} W_2(\rho_r,\rho_q)\le C_T^{1/2}|r-q|^{1/2} \end{align*} for all $q,r\in D$. Hence it is uniformly continuous on the dense subset $D$ of $[0,T]$. Completeness now gives a unique continuous extension \begin{align*} \rho:[0,T]\to(\mathcal P_2(\mathbb R^n),W_2). \end{align*} Concretely, if $t\in[0,T]$ and $(q_i)_{i\in\mathbb N}$ is any sequence in $D$ with $q_i\to t$, then $(\rho_{q_i})_{i\in\mathbb N}$ is a Cauchy sequence in $W_2$, hence converges in $\mathcal P_2(\mathbb R^n)$. The Hölder estimate also shows that the limit is independent of the approximating sequence, so it is meaningful to call this limit $\rho_t$. The same limiting argument preserves the estimate. If $s,t\in[0,T]$ with $s\le t$, choose sequences $(q_i)_{i\in\mathbb N}$ and $(r_i)_{i\in\mathbb N}$ in $D$ such that $q_i\to s$, $r_i\to t$, and $q_i\le r_i$. For every $i$, \begin{align*} W_2^2(\rho_{r_i},\rho_{q_i})\le C_T(r_i-q_i). \end{align*} Since $\rho_{q_i}\to\rho_s$ and $\rho_{r_i}\to\rho_t$ in $W_2$, the left-hand side converges to $W_2^2(\rho_t,\rho_s)$. The right-hand side converges to $C_T(t-s)$. Therefore \begin{align*} W_2^2(\rho_t,\rho_s)\le C_T(t-s). \end{align*} This gives a continuous curve in the correct Wasserstein space, but it is not yet the sharp $AC^2$ conclusion. The sharper action bound is obtained from partitions in the next step.[/guided]

[step:Prove the metric action bound from partition estimates]Let $P=\{0=t_0<t_1<\dots<t_L=T\}$ be a finite partition with all $t_i\in D$. For each $j$ and each $i\in\{1,\dots,L\}$, set \begin{align*} a_{j,i}:=k_j(t_{i-1}),\qquad b_{j,i}:=k_j(t_i). \end{align*} The intervals of indices $\{a_{j,i},\dots,b_{j,i}-1\}$ are disjoint as $i$ varies, up to empty intervals. From the discrete estimate with local action, \begin{align*} W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})\le (b_{j,i}-a_{j,i})\tau_j A_j(a_{j,i},b_{j,i}). \end{align*} Since \begin{align*} (b_{j,i}-a_{j,i})\tau_j\le t_i-t_{i-1}+\tau_j, \end{align*} we obtain \begin{align*} \frac{W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})}{t_i-t_{i-1}}\le \left(1+\frac{\tau_j}{t_i-t_{i-1}}\right)A_j(a_{j,i},b_{j,i}). \end{align*} Summing over $i$ and using disjointness of the index intervals gives \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})}{t_i-t_{i-1}} \le \left(1+\frac{\tau_j}{\delta_P}\right)C_T, \end{align*} where \begin{align*} \delta_P:=\min_{1\le i\le L}(t_i-t_{i-1})>0. \end{align*} Passing to the extracted subsequence and using lower semicontinuity of $W_2$ under narrow convergence at the points of $D$ yields \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{t_i},\rho_{t_{i-1}})}{t_i-t_{i-1}}\le C_T. \end{align*} Now let $P=\{0=t_0<t_1<\dots<t_L=T\}$ be an arbitrary finite partition. Choose partitions $P_m=\{0=t_{m,0}<t_{m,1}<\dots<t_{m,L}=T\}$ with each $t_{m,i}\in D$, $t_{m,i}\to t_i$, and $t_{m,i-1}<t_{m,i}$ for every $i$. The $W_2$-continuity of $\rho$ gives \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{t_i},\rho_{t_{i-1}})}{t_i-t_{i-1}} =\lim_{m\to\infty}\sum_{i=1}^L\frac{W_2^2(\rho_{t_{m,i}},\rho_{t_{m,i-1}})}{t_{m,i}-t_{m,i-1}} \le C_T. \end{align*} Thus every finite partition satisfies \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{t_i},\rho_{t_{i-1}})}{t_i-t_{i-1}}\le C_T. \end{align*} By the metric characterization of absolutely continuous curves with square-integrable metric derivative, a curve $\gamma:[0,T]\to X$ in a complete metric space is absolutely continuous of order two and satisfies \begin{align*} \int_0^T |\gamma'|^2(t)\,d\mathcal L^1(t)\le A \end{align*} whenever every finite partition $0=t_0<t_1<\dots<t_L=T$ satisfies \begin{align*} \sum_{i=1}^L\frac{d^2(\gamma(t_i),\gamma(t_{i-1}))}{t_i-t_{i-1}}\le A. \end{align*} Applying this characterization with $X=\mathcal P_2(\mathbb R^n)$, $d=W_2$, $\gamma=\rho$, and $A=C_T$ gives that $\rho$ is absolutely continuous of order two and \begin{align*} \int_0^T |\rho'|^2(t)\,d\mathcal L^1(t)\le C_T. \end{align*}[/step]

[guided]The goal of this step is to upgrade the Hölder continuity estimate to the sharper quadratic action estimate. Fix a finite partition $P=\{0=t_0<t_1<\dots<t_L=T\}$ whose points belong to $D$. For each $j\in\mathbb N$ and each $i\in\{1,\dots,L\}$, define the left and right grid indices \begin{align*} a_{j,i}:=k_j(t_{i-1}),\qquad b_{j,i}:=k_j(t_i). \end{align*} The index intervals $\{a_{j,i},\dots,b_{j,i}-1\}$ are disjoint as $i$ varies because the partition times are increasing and the map $k_j$ is nondecreasing. The local action estimate from the previous step gives \begin{align*} W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})\le (b_{j,i}-a_{j,i})\tau_j A_j(a_{j,i},b_{j,i}). \end{align*} Since the right-continuous grid representative of $t_i$ can overshoot $t_i$ by at most one time step, we have \begin{align*} (b_{j,i}-a_{j,i})\tau_j\le t_i-t_{i-1}+\tau_j. \end{align*} Dividing by $t_i-t_{i-1}>0$ gives \begin{align*} \frac{W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})}{t_i-t_{i-1}}\le \left(1+\frac{\tau_j}{t_i-t_{i-1}}\right)A_j(a_{j,i},b_{j,i}). \end{align*} Let \begin{align*} \delta_P:=\min_{1\le i\le L}(t_i-t_{i-1}). \end{align*} Summing over $i$ and using the disjointness of the index intervals, the local actions add to at most the full discrete action. Hence \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{j,b_{j,i}},\rho_{j,a_{j,i}})}{t_i-t_{i-1}}\le \left(1+\frac{\tau_j}{\delta_P}\right)C_T. \end{align*} Now pass to the extracted subsequence. At every point of $D$, the measures converge narrowly to the prescribed dense-time limit, and lower semicontinuity of $W_2$ under narrow convergence of both marginals yields \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{t_i},\rho_{t_{i-1}})}{t_i-t_{i-1}}\le C_T. \end{align*} It remains to remove the restriction that the partition points lie in $D$. Let $P=\{0=t_0<t_1<\dots<t_L=T\}$ be an arbitrary finite partition. Since $D$ is dense and contains the endpoints, choose partitions $P_m=\{0=t_{m,0}<t_{m,1}<\dots<t_{m,L}=T\}$ with $t_{m,i}\in D$ and $t_{m,i}\to t_i$ for every $i$. The curve $\rho$ is continuous in $W_2$, so each quotient converges, and therefore \begin{align*} \sum_{i=1}^L\frac{W_2^2(\rho_{t_i},\rho_{t_{i-1}})}{t_i-t_{i-1}}\le C_T. \end{align*} The metric characterization of absolutely continuous curves with square-integrable metric derivative now applies in the complete metric space $(\mathcal P_2(\mathbb R^n),W_2)$. It says that this uniform partition bound implies that $\rho$ is absolutely continuous of order two and that its metric derivative satisfies \begin{align*} \int_0^T |\rho'|^2(t)\,d\mathcal L^1(t)\le C_T. \end{align*}[/guided]

[step:Upgrade dense-time convergence to every time by a Lipschitz test argument]Fix $t\in[0,T]$ and let $\varphi:\mathbb R^n\to\mathbb R$ be bounded and Lipschitz. Denote its Lipschitz constant by $\operatorname{Lip}(\varphi)$ and its supremum norm by $\|\varphi\|_\infty$. For every $q\in D$, the triangle inequality gives \begin{align*} \left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\rho_t\right|\le \left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(q)\right|+\left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(q)-\int_{\mathbb R^n}\varphi\,d\rho_q\right|+\left|\int_{\mathbb R^n}\varphi\,d\rho_q-\int_{\mathbb R^n}\varphi\,d\rho_t\right|. \end{align*} Let $W_1$ denote the Wasserstein distance with cost $|x-y|$ on Borel probability measures on $\mathbb R^n$ with finite first moments. The first and third terms are controlled by the Kantorovich-Rubinstein bound \begin{align*} \left|\int_{\mathbb R^n}\varphi\,d\mu-\int_{\mathbb R^n}\varphi\,d\nu\right|\le \operatorname{Lip}(\varphi)W_1(\mu,\nu) \end{align*} and by $W_1\le W_2$. Using the discrete estimate and the continuity estimate for $\rho$, we obtain \begin{align*} \limsup_{m\to\infty}\left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\rho_t\right|\le 2\operatorname{Lip}(\varphi)C_T^{1/2}|t-q|^{1/2}. \end{align*} Since $D$ is dense in $[0,T]$, let $q\to t$ through points of $D$. This proves convergence against every bounded Lipschitz test function. It remains only to pass from bounded Lipschitz tests to bounded continuous tests. The uniform second-moment bound implies tightness of the family consisting of all $\bar\rho_{j_m}(t)$ and all $\rho_t$. Given a bounded continuous function $\psi:\mathbb R^n\to\mathbb R$ and $\varepsilon>0$, choose $R>0$ so large that all these measures assign mass at most \begin{align*} \frac{\varepsilon}{6\|\psi\|_\infty} \end{align*} to $\mathbb R^n\setminus B(0,R)$. By uniform continuity of $\psi$ on $\overline{B}(0,R)$, choose $\delta>0$ such that $|\psi(x)-\psi(y)|\le\varepsilon/3$ whenever $x,y\in\overline{B}(0,R)$ and $|x-y|\le\delta$. Choose a finite $\delta$-net $(x_i)_{i=1}^I$ in $\overline{B}(0,R)$ and define a Lipschitz function $\varphi_0:\overline{B}(0,R)\to\mathbb R$ by \begin{align*} \varphi_0(x):=\max_{1\le i\le I}\left(\psi(x_i)-\frac{\varepsilon}{3\delta}|x-x_i|\right). \end{align*} Then $|\varphi_0-\psi|\le\varepsilon/3$ on $\overline{B}(0,R)$. Extending $\varphi_0$ to $\mathbb R^n$ by the McShane extension formula and truncating to the interval $[-\|\psi\|_\infty,\|\psi\|_\infty]$, we obtain a bounded Lipschitz function $\varphi:\mathbb R^n\to\mathbb R$ such that $|\varphi|\le\|\psi\|_\infty$ and $|\varphi-\psi|\le\varepsilon/3$ on $\overline{B}(0,R)$. The tail estimate then gives \begin{align*} \limsup_{m\to\infty}\left|\int_{\mathbb R^n}\psi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\psi\,d\rho_t\right|\le \varepsilon. \end{align*} Because $\varepsilon>0$ is arbitrary, $\bar\rho_{j_m}(t)$ converges narrowly to $\rho_t$ for this fixed $t$. Since $t\in[0,T]$ was arbitrary, convergence holds for every time; in the language of the original assertion, one may take the exceptional set to be $E:=\varnothing$. Together with the action estimate from the previous step, this proves the asserted subsequential compactness and completes the proof.[/step]

[guided]The dense set $D$ was only a tool for extraction; the theorem needs convergence at actual times. Fix a time $t\in[0,T]$. To prove narrow convergence at $t$, it is enough first to prove convergence against bounded Lipschitz test functions and then use tightness to approximate bounded continuous test functions on a large compact ball. Let $\varphi:\mathbb R^n\to\mathbb R$ be bounded and Lipschitz. For $q\in D$, insert the dense-time value $q$ between $t$ and the limit. The triangle inequality gives \begin{align*} \left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\rho_t\right|\le \left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(q)\right|+\left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(q)-\int_{\mathbb R^n}\varphi\,d\rho_q\right|+\left|\int_{\mathbb R^n}\varphi\,d\rho_q-\int_{\mathbb R^n}\varphi\,d\rho_t\right|. \end{align*} The middle term tends to zero because $q\in D$ and the subsequence was constructed to converge narrowly at every point of $D$. Let $W_1$ denote the Wasserstein distance with cost $|x-y|$ on Borel probability measures on $\mathbb R^n$ with finite first moments. For the first term, the Kantorovich-Rubinstein estimate gives a $W_1$ bound, and $W_1\le W_2$ on probability measures with finite second moments. Therefore \begin{align*} \left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(q)\right|\le \operatorname{Lip}(\varphi)W_2(\bar\rho_{j_m}(t),\bar\rho_{j_m}(q)). \end{align*} The discrete increment estimate bounds this by a quantity whose limit superior is at most $\operatorname{Lip}(\varphi)C_T^{1/2}|t-q|^{1/2}$. The third term is handled in the same way using the $W_2$-continuity estimate for the limiting curve $\rho$. Hence \begin{align*} \limsup_{m\to\infty}\left|\int_{\mathbb R^n}\varphi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\varphi\,d\rho_t\right|\le 2\operatorname{Lip}(\varphi)C_T^{1/2}|t-q|^{1/2}. \end{align*} Because $D$ is dense, we can choose $q\in D$ arbitrarily close to $t$, forcing the right-hand side to zero. Thus convergence holds against every bounded Lipschitz test function. Now let $\psi:\mathbb R^n\to\mathbb R$ be bounded and continuous. The uniform second-moment bound gives a uniform tightness estimate for all measures under discussion, including the limit measures by lower semicontinuity of the second moment. Choose $R>0$ so that the mass outside $B(0,R)$ is uniformly small. On the compact set $\overline{B}(0,R)$, the function $\psi$ is uniformly continuous, so it can be approximated uniformly there by a bounded Lipschitz function $\varphi$ with $|\varphi|\le\|\psi\|_\infty$. Splitting the integrals over $B(0,R)$ and its complement gives \begin{align*} \limsup_{m\to\infty}\left|\int_{\mathbb R^n}\psi\,d\bar\rho_{j_m}(t)-\int_{\mathbb R^n}\psi\,d\rho_t\right|\le \varepsilon. \end{align*} Since $\varepsilon>0$ is arbitrary, the convergence is narrow at the fixed time $t$. The time $t$ was arbitrary, so the exceptional set is empty.[/guided]