[proofplan] The converse direction follows from the dynamic Wasserstein estimate: an admissible solution of the continuity equation has transport length bounded by the time integral of its $L^2(\mu_t)$ velocity norm. If this norm is square-integrable in time, the resulting metric derivative is also square-integrable, so the curve is $2$-absolutely continuous. For the existence direction, one approximates the $2$-absolutely continuous curve by piecewise constant-speed optimal transport interpolations, extracts weak limits of the associated momentum measures under a uniform quadratic-action bound, and uses lower semicontinuity of the quadratic action to obtain a limiting Borel velocity satisfying the continuity equation and the estimate by $|\mu'|$. [/proofplan]

[step:Define the velocity norm and record the dynamic estimate]For a Borel map \begin{align*} v:(0,1)\times\mathbb R^n\to\mathbb R^n, \end{align*} define the function \begin{align*} g:(0,1)\to[0,\infty] \end{align*} by \begin{align*} g(t):=\left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2}. \end{align*} The converse part uses the dynamic estimate obtained from the Benamou-Brenier formula [citetheorem:9556]: if $\mu$ is narrowly continuous in $\mathcal P_2(\mathbb R^n)$, $v$ is Borel, $(\mu_t,v_t)$ solves the distributional continuity equation, and $g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, then for every $0\le s\le t\le 1$, \begin{align*} W_2(\mu_s,\mu_t)\le \int_s^t g(r)\,d\mathcal L^1(r). \end{align*} The hypotheses of this estimate are satisfied in the converse direction: narrow continuity is assumed, the continuity equation is assumed in the displayed distributional sense, and the square-integrability hypothesis on $g$ implies $g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$ because $\mathcal L^1((0,1))=1$.[/step]

[guided]We first isolate the quantity that measures how much mass is being moved at time $t$. For the given Borel field \begin{align*} v:(0,1)\times\mathbb R^n\to\mathbb R^n, \end{align*} we define \begin{align*} g:(0,1)\to[0,\infty] \end{align*} by \begin{align*} g(t):=\left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2}. \end{align*} This is the $L^2(\mu_t)$ norm of the velocity field at time $t$. The theorem assumes \begin{align*} \int_0^1 g(t)^2\,d\mathcal L^1(t)<\infty, \end{align*} so $g$ belongs to $L^2((0,1),\mathcal B((0,1)),\mathcal L^1)$. Since $\mathcal L^1((0,1))=1$, the Cauchy-Schwarz inequality gives \begin{align*} \int_0^1 g(t)\,d\mathcal L^1(t)\le \left(\int_0^1 g(t)^2\,d\mathcal L^1(t)\right)^{1/2}, \end{align*} and hence $g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$. The key external input is the dynamic estimate obtained from the Benamou-Brenier formula [citetheorem:9556]. It says that whenever $\mu:[0,1]\to\mathcal P_2(\mathbb R^n)$ is narrowly continuous, $v$ is Borel, the pair $(\mu_t,v_t)$ solves \begin{align*} \int_0^1\int_{\mathbb R^n}\partial_t\varphi(t,x)\,d\mu_t(x)\,d\mathcal L^1(t)+\int_0^1\int_{\mathbb R^n}\nabla_x\varphi(t,x)\cdot v_t(x)\,d\mu_t(x)\,d\mathcal L^1(t)=0 \end{align*} for every $\varphi\in C_c^\infty((0,1)\times\mathbb R^n)$, and $g\in L^1$, then \begin{align*} W_2(\mu_s,\mu_t)\le \int_s^t g(r)\,d\mathcal L^1(r) \end{align*} for every $0\le s\le t\le 1$. Why is this the right estimate? The continuity equation says that the measure changes only by the flux $v_t\mu_t$. The integral of the $L^2(\mu_t)$ norm of this flux controls the length of the path in the Wasserstein metric. All hypotheses required by the dynamic estimate are present in the converse part of the theorem: narrow continuity, the displayed distributional continuity equation, Borel measurability of $v$, and the integrability of $g$.[/guided]

[step:Deduce absolute continuity and the metric derivative inequality] Assume the hypotheses of the converse direction. By the dynamic estimate, for every $0\le s\le t\le 1$, \begin{align*} W_2(\mu_s,\mu_t)\le \int_s^t g(r)\,d\mathcal L^1(r). \end{align*} Since $g\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, this is the defining estimate for absolute continuity of $\mu$ as a curve in the metric space $(\mathcal P_2(\mathbb R^n),W_2)$. The pointwise estimate below gives $|\mu'|\le g$ for $\mathcal L^1$-a.e. $t$, and the hypothesis $g\in L^2((0,1),\mathcal B((0,1)),\mathcal L^1)$ therefore implies $|\mu'|\in L^2((0,1),\mathcal B((0,1)),\mathcal L^1)$. Thus $\mu$ is $2$-absolutely continuous. By the definition of the metric derivative, for every $t\in(0,1)$ at which the Lebesgue differentiation theorem applies to $g$ and the metric derivative exists, \begin{align*} |\mu'|(t)=\lim_{h\to0,\ t+h\in[0,1]}\frac{W_2(\mu_{t+h},\mu_t)}{|h|}\le \lim_{h\to0,\ t+h\in[0,1]}\frac{1}{|h|}\int_{\min\{t,t+h\}}^{\max\{t,t+h\}}g(r)\,d\mathcal L^1(r)=g(t). \end{align*} The metric derivative exists for $\mathcal L^1$-a.e. $t$ for absolutely continuous metric curves, and the Lebesgue differentiation theorem applies to $g$ for $\mathcal L^1$-a.e. $t$. Therefore \begin{align*} |\mu'|(t)\le \left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2} \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. [/step]

[step:Approximate the absolutely continuous curve by optimal short-time interpolations] Assume now that $\mu:[0,1]\to\mathcal P_2(\mathbb R^n)$ is $2$-absolutely continuous in $(\mathcal P_2(\mathbb R^n),W_2)$. Let $m:(0,1)\to[0,\infty)$ denote the metric derivative of $\mu$, so $m(t)=|\mu'|(t)$ for $\mathcal L^1$-a.e. $t$. Since $\mu$ is $2$-absolutely continuous, $m\in L^2((0,1),\mathcal B((0,1)),\mathcal L^1)$. For each $k\in\mathbb N$, let $\mathcal P_k$ be the uniform partition of $[0,1]$ with nodes $t_{k,j}:=j2^{-k}$ for $j\in\{0,\dots,2^k\}$. For each $j\in\{0,\dots,2^k-1\}$, choose an optimal coupling between $\mu_{t_{k,j}}$ and $\mu_{t_{k,j+1}}$, and let \begin{align*} \mu^k:[0,1]\to\mathcal P_2(\mathbb R^n) \end{align*} be the constant-speed displacement interpolation on each interval $[t_{k,j},t_{k,j+1}]$. On that interval let \begin{align*} v^k:(t_{k,j},t_{k,j+1})\times\mathbb R^n\to\mathbb R^n \end{align*} be a Borel velocity field realizing the corresponding constant-speed geodesic in the continuity equation. This construction is justified by the Benamou-Brenier representation of constant-speed Wasserstein geodesics [citetheorem:9556], applied on each interval to the endpoint measures $\mu_{t_{k,j}}$ and $\mu_{t_{k,j+1}}$. Since there are only finitely many intervals for each fixed $k$, choosing one optimal dynamical pair on each interval and pasting these Borel fields defines a Borel field $v^k$ on $(0,1)\times\mathbb R^n$. For $\mathcal L^1$-a.e. $t\in(t_{k,j},t_{k,j+1})$ one has \begin{align*} \left(\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\right)^{1/2}=\frac{W_2(\mu_{t_{k,j}},\mu_{t_{k,j+1}})}{t_{k,j+1}-t_{k,j}}. \end{align*} Using absolute continuity of $\mu$ on the interval $[t_{k,j},t_{k,j+1}]$ gives \begin{align*} W_2(\mu_{t_{k,j}},\mu_{t_{k,j+1}})\le \int_{t_{k,j}}^{t_{k,j+1}}m(r)\,d\mathcal L^1(r). \end{align*} Therefore the actions of these approximating curves are controlled by the metric derivative. Define the conditional average \begin{align*} M_km:(0,1)\to[0,\infty] \end{align*} by \begin{align*} M_km(t):=\frac{1}{t_{k,j+1}-t_{k,j}}\int_{t_{k,j}}^{t_{k,j+1}}m(r)\,d\mathcal L^1(r) \end{align*} for $t\in(t_{k,j},t_{k,j+1})$. On each interval $(t_{k,j},t_{k,j+1})$ the constant-speed identity and the preceding endpoint estimate give \begin{align*} \int_{t_{k,j}}^{t_{k,j+1}}\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_{t_{k,j}}^{t_{k,j+1}}\left(M_km(t)\right)^2\,d\mathcal L^1(t). \end{align*} Summing over $j$ and applying Jensen's inequality on each partition interval gives \begin{align*} \int_0^1\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_0^1 \left(M_km(t)\right)^2\,d\mathcal L^1(t)\le \int_0^1 m(t)^2\,d\mathcal L^1(t)<\infty. \end{align*} Hence the approximating curves have uniformly bounded quadratic action. We shall also need the localized form of this estimate. If $0<a<b<1$, let $I_k(a,b)$ be the union of all partition intervals $(t_{k,j},t_{k,j+1})$ that meet $(a,b)$. Then \begin{align*} \int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_{I_k(a,b)}\left(M_km(t)\right)^2\,d\mathcal L^1(t)\le \int_{I_k(a,b)}m(t)^2\,d\mathcal L^1(t). \end{align*} Since the endpoints of $I_k(a,b)$ converge to $a$ and $b$, and $m^2\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$, absolute continuity of the Lebesgue integral gives \begin{align*} \limsup_{k\to\infty}\int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} [/step]

[step:Extract a limiting momentum and pass the continuity equation to the limit]For each $k\in\mathbb N$, define the positive measure $\sigma_k$ on $(0,1)\times\mathbb R^n$ by \begin{align*} \sigma_k(B):=\int_0^1\int_{\mathbb R^n}\mathbb 1_B(t,x)\,d\mu_t^k(x)\,d\mathcal L^1(t) \end{align*} for every Borel set $B\subset(0,1)\times\mathbb R^n$. Define the vector-valued Radon measure $\omega_k$ componentwise by \begin{align*} (\omega_k)_i(B):=\int_B (v_t^k(x))_i\,d\sigma_k(t,x) \end{align*} for $i\in\{1,\dots,n\}$ and every Borel set $B\subset(0,1)\times\mathbb R^n$ for which the integral is finite. This is locally finite: if $K\subset(0,1)\times\mathbb R^n$ is compact, then Cauchy-Schwarz gives \begin{align*} |\omega_k|(K)\le \left(\int_K |v_t^k(x)|^2\,d\sigma_k(t,x)\right)^{1/2}\left(\sigma_k(K)\right)^{1/2}\le \left(\int_0^1\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\right)^{1/2}, \end{align*} because $\sigma_k(K)\le\sigma_k((0,1)\times\mathbb R^n)=1$. The last quantity is bounded independently of $k$ by the preceding step. The pairs $(\sigma_k,\omega_k)$ solve the distributional continuity equation associated to $(\mu^k,v^k)$. We use the compactness and lower semicontinuity theorem for continuity-equation measures with bounded quadratic action. The precise external result is: if $\sigma_k$ are nonnegative Radon measures on $(0,1)\times\mathbb R^n$, $\omega_k$ are vector-valued Radon measures, $\sigma_k\overset{*}{\rightharpoonup}\sigma$ against compactly supported continuous functions, $\omega_k$ have uniformly bounded total variation on every compact set, and \begin{align*} \sup_k\int_{(0,1)\times\mathbb R^n}\left|\frac{d\omega_k}{d\sigma_k}(t,x)\right|^2\,d\sigma_k(t,x)<\infty, \end{align*} then a subsequence satisfies $\omega_k\overset{*}{\rightharpoonup}\omega$ locally as vector-valued Radon measures, $\omega\ll\sigma$, and for every open set $O\subset(0,1)\times\mathbb R^n$, \begin{align*} \int_O\left|\frac{d\omega}{d\sigma}(t,x)\right|^2\,d\sigma(t,x)\le \liminf_{k\to\infty}\int_O\left|\frac{d\omega_k}{d\sigma_k}(t,x)\right|^2\,d\sigma_k(t,x). \end{align*} Here $d\omega_k/d\sigma_k=v^k$, so the global quadratic-action bound and the compact-variation estimate above verify the hypotheses except for $\sigma_k\overset{*}{\rightharpoonup}\sigma$. We now verify this remaining convergence. Let $\sigma$ be the measure on $(0,1)\times\mathbb R^n$ defined by \begin{align*} \sigma(B):=\int_0^1\int_{\mathbb R^n}\mathbb 1_B(t,x)\,d\mu_t(x)\,d\mathcal L^1(t). \end{align*} For every compactly supported continuous function $\psi:(0,1)\times\mathbb R^n\to\mathbb R$, the map \begin{align*} F_\psi:[0,1]\times\mathcal P_2(\mathbb R^n)\to\mathbb R \end{align*} defined by \begin{align*} F_\psi(t,\nu):=\int_{\mathbb R^n}\psi(t,x)\,d\nu(x) \end{align*} is continuous when $\mathcal P_2(\mathbb R^n)$ is given the narrow topology. The interpolating curves $\mu^k$ converge uniformly in $W_2$ to $\mu$: if $t\in[t_{k,j},t_{k,j+1}]$, then the triangle inequality, the constant-speed property of $\mu^k$, and absolute continuity of $\mu$ give \begin{align*} W_2(\mu_t^k,\mu_t)\le W_2(\mu_t^k,\mu_{t_{k,j}})+W_2(\mu_{t_{k,j}},\mu_t)\le W_2(\mu_{t_{k,j}},\mu_{t_{k,j+1}})+\int_{t_{k,j}}^{t_{k,j+1}}m(r)\,d\mathcal L^1(r). \end{align*} The right-hand side is bounded by twice the integral of $m$ over a dyadic interval of length $2^{-k}$, and these integrals converge uniformly to $0$ because $m\in L^1((0,1),\mathcal B((0,1)),\mathcal L^1)$ and the Lebesgue integral is absolutely continuous. Thus $\mu_t^k\to\mu_t$ narrowly for every $t\in[0,1]$. Since \begin{align*} \left|\int_{\mathbb R^n}\psi(t,x)\,d\mu_t^k(x)\right|\le \|\psi\|_\infty \end{align*} for every $t$ and $k$, dominated convergence gives $\sigma_k\overset{*}{\rightharpoonup}\sigma$. The compactness theorem therefore provides a subsequence, not relabelled, and a vector-valued Radon measure $\omega$ on $(0,1)\times\mathbb R^n$ such that \begin{align*} \omega_k\overset{*}{\rightharpoonup}\omega \end{align*} against compactly supported continuous vector fields. Passing to the limit in the distributional continuity equation is now justified term by term: the time-derivative term converges by $\sigma_k\overset{*}{\rightharpoonup}\sigma$, and the flux term converges by $\omega_k\overset{*}{\rightharpoonup}\omega$. Hence \begin{align*} \int_0^1\int_{\mathbb R^n}\partial_t\varphi(t,x)\,d\mu_t(x)\,d\mathcal L^1(t)+\int_{(0,1)\times\mathbb R^n}\nabla_x\varphi(t,x)\cdot d\omega(t,x)=0 \end{align*} for every $\varphi\in C_c^\infty((0,1)\times\mathbb R^n)$.[/step]

[guided]The compactness step has two separate tasks. First, we must show that the spacetime mass measures converge. Second, we must extract a weak limit of the vector-valued momentum measures. The quadratic action estimate is what controls the momenta. For each $k\in\mathbb N$, define \begin{align*} \sigma_k(B):=\int_0^1\int_{\mathbb R^n}\mathbb 1_B(t,x)\,d\mu_t^k(x)\,d\mathcal L^1(t) \end{align*} for Borel $B\subset(0,1)\times\mathbb R^n$. The momentum measure is defined component by component: \begin{align*} (\omega_k)_i(B):=\int_B (v_t^k(x))_i\,d\sigma_k(t,x),\qquad i\in\{1,\dots,n\}. \end{align*} Why is this a locally finite vector measure? If $K\subset(0,1)\times\mathbb R^n$ is compact, then Cauchy-Schwarz gives \begin{align*} |\omega_k|(K)\le \left(\int_K |v_t^k(x)|^2\,d\sigma_k(t,x)\right)^{1/2}\left(\sigma_k(K)\right)^{1/2}. \end{align*} Since $\sigma_k$ is obtained by integrating probability measures in time, $\sigma_k(K)\le 1$. Therefore \begin{align*} |\omega_k|(K)\le \left(\int_0^1\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\right)^{1/2}, \end{align*} and the right-hand side is uniformly bounded by the action estimate from the preceding step. The external compactness input we use is the compactness and lower semicontinuity theorem for continuity-equation measures with bounded quadratic action. It requires: $\sigma_k\overset{*}{\rightharpoonup}\sigma$ as nonnegative Radon measures, local uniform boundedness of $|\omega_k|$, and a uniform bound on \begin{align*} \int\left|\frac{d\omega_k}{d\sigma_k}\right|^2\,d\sigma_k. \end{align*} Here $d\omega_k/d\sigma_k=v^k$, so the last condition is exactly the quadratic-action bound. The local boundedness of $|\omega_k|$ was just proved by Cauchy-Schwarz. It remains to prove $\sigma_k\overset{*}{\rightharpoonup}\sigma$. Define \begin{align*} \sigma(B):=\int_0^1\int_{\mathbb R^n}\mathbb 1_B(t,x)\,d\mu_t(x)\,d\mathcal L^1(t). \end{align*} Take a compactly supported continuous test function $\psi:(0,1)\times\mathbb R^n\to\mathbb R$. We need to prove \begin{align*} \int_0^1\int_{\mathbb R^n}\psi(t,x)\,d\mu_t^k(x)\,d\mathcal L^1(t)\to \int_0^1\int_{\mathbb R^n}\psi(t,x)\,d\mu_t(x)\,d\mathcal L^1(t). \end{align*} For fixed $t$, the measures $\mu_t^k$ converge narrowly to $\mu_t$. Indeed, if $t\in[t_{k,j},t_{k,j+1}]$, then \begin{align*} W_2(\mu_t^k,\mu_t)\le W_2(\mu_t^k,\mu_{t_{k,j}})+W_2(\mu_{t_{k,j}},\mu_t). \end{align*} The first term is bounded by $W_2(\mu_{t_{k,j}},\mu_{t_{k,j+1}})$ because $\mu^k$ moves along a constant-speed geodesic between these endpoints. The second term is bounded by the absolute-continuity estimate for $\mu$. Hence \begin{align*} W_2(\mu_t^k,\mu_t)\le 2\int_{t_{k,j}}^{t_{k,j+1}}m(r)\,d\mathcal L^1(r), \end{align*} up to replacing the interval by the same dyadic cell containing $t$. These cell integrals tend uniformly to $0$ because $m\in L^1$. Thus $\mu_t^k\to\mu_t$ narrowly for each $t$. The integrands are bounded by $\|\psi\|_\infty$, so dominated convergence proves $\sigma_k\overset{*}{\rightharpoonup}\sigma$. The compactness theorem now gives a subsequence and a vector-valued Radon measure $\omega$ with $\omega_k\overset{*}{\rightharpoonup}\omega$. Since the continuity equation is tested only against compactly supported smooth functions, weak convergence is sufficient to pass to the limit: the term containing $\partial_t\varphi$ uses convergence of $\sigma_k$, and the term containing $\nabla_x\varphi$ uses convergence of $\omega_k$. Therefore the limiting pair satisfies \begin{align*} \int_0^1\int_{\mathbb R^n}\partial_t\varphi(t,x)\,d\mu_t(x)\,d\mathcal L^1(t)+\int_{(0,1)\times\mathbb R^n}\nabla_x\varphi(t,x)\cdot d\omega(t,x)=0 \end{align*} for every $\varphi\in C_c^\infty((0,1)\times\mathbb R^n)$.[/guided]

[step:Represent the limiting momentum by a Borel velocity] Let $\sigma$ be the positive measure on $(0,1)\times\mathbb R^n$ defined by \begin{align*} \sigma(B):=\int_0^1\int_{\mathbb R^n}\mathbb 1_B(t,x)\,d\mu_t(x)\,d\mathcal L^1(t) \end{align*} for every Borel set $B\subset(0,1)\times\mathbb R^n$. The lower semicontinuity part of the compactness theorem for quadratic action implies that $\omega$ is absolutely continuous with respect to $\sigma$. Hence the Radon-Nikodym theorem gives a Borel map \begin{align*} v:(0,1)\times\mathbb R^n\to\mathbb R^n \end{align*} such that \begin{align*} \omega(B)=\int_B v_t(x)\,d\mu_t(x)\,d\mathcal L^1(t) \end{align*} for every Borel set $B\subset(0,1)\times\mathbb R^n$ for which the integral is defined componentwise. Substituting this representation of $\omega$ into the limiting continuity equation yields \begin{align*} \int_0^1\int_{\mathbb R^n}\partial_t\varphi(t,x)\,d\mu_t(x)\,d\mathcal L^1(t)+\int_0^1\int_{\mathbb R^n}\nabla_x\varphi(t,x)\cdot v_t(x)\,d\mu_t(x)\,d\mathcal L^1(t)=0 \end{align*} for every $\varphi\in C_c^\infty((0,1)\times\mathbb R^n)$. Thus $(\mu_t,v_t)$ solves the continuity equation. [/step]

[step:Use localized lower semicontinuity to obtain the pointwise velocity bound]The lower semicontinuity theorem from the preceding step applies on open spacetime sets. Fix $0<a<b<1$ and $R>0$, and set \begin{align*} O_{a,b,R}:=(a,b)\times B(0,R). \end{align*} Since $O_{a,b,R}$ is open, lower semicontinuity gives \begin{align*} \int_a^b\int_{B(0,R)}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \liminf_{k\to\infty}\int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t). \end{align*} The localized estimate proved for the dyadic interpolations gives \begin{align*} \limsup_{k\to\infty}\int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} Combining the two inequalities yields \begin{align*} \int_a^b\int_{B(0,R)}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} Letting $R\to\infty$ and applying the monotone convergence theorem to the nonnegative functions $\mathbb 1_{B(0,R)}(x)|v_t(x)|^2$ gives \begin{align*} \int_a^b\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t) \end{align*} for every $0<a<b<1$. Define \begin{align*} h:(0,1)\to[0,\infty] \end{align*} by \begin{align*} h(t):=\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x). \end{align*} The preceding interval inequality says that the absolutely continuous measure $h(t)\,d\mathcal L^1(t)$ is dominated by the absolutely continuous measure $m(t)^2\,d\mathcal L^1(t)$ on every interval $(a,b)\subset(0,1)$. By the monotone class theorem, the same domination holds on every Borel subset of $(0,1)$. Hence the Radon-Nikodym densities satisfy \begin{align*} h(t)\le m(t)^2 \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. Since $m(t)=|\mu'|(t)$ for $\mathcal L^1$-a.e. $t$, taking square roots gives \begin{align*} \left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2}\le |\mu'|(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$. The velocity $v$ constructed above is Borel, solves the continuity equation with $\mu$, and satisfies the required pointwise $L^2(\mu_t)$ bound. Together with the converse direction already proved, this completes the proof.[/step]

[guided]The pointwise bound is obtained by first proving an inequality on every time interval and then converting that interval inequality into a statement about densities. Fix $0<a<b<1$ and $R>0$, and define \begin{align*} O_{a,b,R}:=(a,b)\times B(0,R). \end{align*} The lower semicontinuity theorem applies to this open set and gives \begin{align*} \int_a^b\int_{B(0,R)}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \liminf_{k\to\infty}\int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t). \end{align*} This is the precise place where the compactness theorem's lower-semicontinuity conclusion is used. The right-hand side is controlled by the localized dyadic estimate already proved: \begin{align*} \limsup_{k\to\infty}\int_a^b\int_{\mathbb R^n}|v_t^k(x)|^2\,d\mu_t^k(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} Thus \begin{align*} \int_a^b\int_{B(0,R)}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} We introduced $B(0,R)$ only because lower semicontinuity is stated for open spacetime sets and compactly supported tests. To recover the full spatial integral, let $R\to\infty$. The functions $\mathbb 1_{B(0,R)}(x)|v_t(x)|^2$ increase pointwise to $|v_t(x)|^2$, so the monotone convergence theorem gives \begin{align*} \int_a^b\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t). \end{align*} Now define \begin{align*} h:(0,1)\to[0,\infty] \end{align*} by \begin{align*} h(t):=\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x). \end{align*} The interval inequality says \begin{align*} \int_a^b h(t)\,d\mathcal L^1(t)\le \int_a^b m(t)^2\,d\mathcal L^1(t) \end{align*} for every $0<a<b<1$. Intervals generate the Borel sets of $(0,1)$, and both sides define absolutely continuous measures, so the same domination holds for every Borel set. Therefore the Radon-Nikodym densities satisfy $h(t)\le m(t)^2$ for $\mathcal L^1$-a.e. $t$. Since $m(t)=|\mu'|(t)$ almost everywhere, we obtain \begin{align*} \left(\int_{\mathbb R^n}|v_t(x)|^2\,d\mu_t(x)\right)^{1/2}\le |\mu'|(t) \end{align*} for $\mathcal L^1$-a.e. $t\in(0,1)$.[/guided]