[proofplan] We use Kantorovich duality for the compact semi-discrete quadratic transport problem. A dual maximizer gives finitely many target potentials $a_i$, and after setting $a_i=2w_i$, the corresponding contact sets are exactly the Laguerre cells $P_i(w)$. Complementary slackness gives an optimal plan supported on these contact sets. Since distinct sites have pairwise tie sets contained in affine hyperplanes, absolute continuity of $\mu_0$ makes the cells disjoint up to $\mu_0$-null sets; the prescribed atomic masses therefore force $\mu_0(P_i(w))=m_i$, and the tie-broken map is optimal. [/proofplan]

[step:Apply semi-discrete Kantorovich duality and choose optimal target potentials]Let \begin{align*} Y:=\{y_1,\dots,y_N\} \end{align*} and define the finite Borel measure $\nu$ on $Y$ by \begin{align*} \nu:=\sum_{i=1}^N m_i\delta_{y_i}. \end{align*} Since $\nu(Y)=\sum_{i=1}^N m_i=\mu_0(\Omega)$, the transport problem between $\mu_0$ and $\nu$ is balanced. We use Kantorovich duality with dual attainment for continuous costs on a compact source and a finite target (citing a result not yet in the wiki: Kantorovich duality with dual attainment for compact semi-discrete transport). Applied to the continuous cost $c:\Omega\times Y\to[0,\infty)$, $c(x,y)=|x-y|^2$, it gives a vector $a=(a_1,\dots,a_N)\in\mathbb R^N$ and a Borel function $\varphi:\Omega\to\mathbb R$ such that \begin{align*} \varphi(x)+a_i\le |x-y_i|^2 \end{align*} for every $x\in\Omega$ and every $i\in\{1,\dots,N\}$, and such that the dual value \begin{align*} \int_\Omega \varphi(x)\,d\mu_0(x)+\sum_{i=1}^N m_i a_i \end{align*} equals the minimum primal transport cost. Replacing $\varphi$ by its $c$-transform relative to the finite vector $a$ preserves admissibility and cannot decrease the dual value. Thus we may assume that $\varphi:\Omega\to\mathbb R$ is the continuous function \begin{align*} \varphi(x)=\min_{1\le i\le N}\bigl(|x-y_i|^2-a_i\bigr). \end{align*} Define $w=(w_1,\dots,w_N)\in\mathbb R^N$ by \begin{align*} w_i:=\frac{a_i}{2} \end{align*} for every $i\in\{1,\dots,N\}$.[/step]

[guided]The target measure has only finitely many atoms, so the dual potential on the target is just a finite vector. We define \begin{align*} Y:=\{y_1,\dots,y_N\} \end{align*} and \begin{align*} \nu:=\sum_{i=1}^N m_i\delta_{y_i}. \end{align*} The equality \begin{align*} \nu(Y)=\sum_{i=1}^N m_i=\mu_0(\Omega) \end{align*} verifies that source and target have the same total mass, which is the basic feasibility condition for transport plans. We now invoke Kantorovich duality with dual attainment for continuous costs on a compact source and a finite target (citing a result not yet in the wiki: Kantorovich duality with dual attainment for compact semi-discrete transport). The hypotheses are satisfied: $\Omega$ is compact, $Y$ is finite, and the function $c:\Omega\times Y\to[0,\infty)$ defined by \begin{align*} c(x,y):=|x-y|^2 \end{align*} is continuous. Therefore there are real numbers $a_1,\dots,a_N$ and a Borel function $\varphi:\Omega\to\mathbb R$ satisfying \begin{align*} \varphi(x)+a_i\le |x-y_i|^2 \end{align*} for every $x\in\Omega$ and every $i\in\{1,\dots,N\}$, and the dual value \begin{align*} \int_\Omega \varphi(x)\,d\mu_0(x)+\sum_{i=1}^N m_i a_i \end{align*} equals the minimal primal transport cost. Why can we take $\varphi$ to be a minimum of finitely many functions? The admissibility inequalities force \begin{align*} \varphi(x)\le |x-y_i|^2-a_i \end{align*} for every $i$. Hence \begin{align*} \varphi(x)\le \min_{1\le i\le N}\bigl(|x-y_i|^2-a_i\bigr). \end{align*} Replacing $\varphi$ by the right-hand side keeps every inequality valid and can only increase the integral against the nonnegative measure $\mu_0$. Since the original pair was already optimal, the replacement is also optimal. Thus we may assume \begin{align*} \varphi(x)=\min_{1\le i\le N}\bigl(|x-y_i|^2-a_i\bigr). \end{align*} Finally, the theorem defines cells with the convention $|x-y_i|^2-2w_i$, so we set \begin{align*} w_i:=\frac{a_i}{2} \end{align*} for every $i\in\{1,\dots,N\}$.[/guided]

[step:Identify the dual contact sets with the Laguerre cells] For each $i\in\{1,\dots,N\}$, define the contact set $C_i\subset\Omega$ by \begin{align*} C_i:=\{x\in\Omega:\varphi(x)+a_i=|x-y_i|^2\}. \end{align*} Using the formula for $\varphi$, we have $x\in C_i$ if and only if \begin{align*} |x-y_i|^2-a_i\le |x-y_j|^2-a_j \end{align*} for every $j\in\{1,\dots,N\}$. Since $a_i=2w_i$, this is equivalent to \begin{align*} |x-y_i|^2-2w_i\le |x-y_j|^2-2w_j \end{align*} for every $j\in\{1,\dots,N\}$. Hence \begin{align*} C_i=P_i(w) \end{align*} for every $i\in\{1,\dots,N\}$. Because $\varphi$ is the minimum of the $N$ functions $x\mapsto |x-y_i|^2-a_i$, every $x\in\Omega$ belongs to at least one $C_i$. Therefore \begin{align*} \Omega=\bigcup_{i=1}^N P_i(w). \end{align*} [/step]

[step:Show that pairwise tie sets have zero source measure] For $i,j\in\{1,\dots,N\}$ with $i\ne j$, define the tie set $H_{ij}\subset\Omega$ by \begin{align*} H_{ij}:=\{x\in\Omega: |x-y_i|^2-2w_i=|x-y_j|^2-2w_j\}. \end{align*} Expanding the squares gives, for $x\in H_{ij}$, \begin{align*} 2x\cdot(y_j-y_i)=|y_j|^2-|y_i|^2+2w_j-2w_i. \end{align*} Since $y_i\ne y_j$, the vector $y_j-y_i$ is nonzero, so $H_{ij}$ is contained in the intersection of $\Omega$ with an affine hyperplane of $\mathbb R^n$. Every affine hyperplane in $\mathbb R^n$ has $\mathcal L^n$-measure zero. Since $\mu_0$ is absolutely continuous with respect to $\mathcal L^n\!\restriction_\Omega$, it follows that \begin{align*} \mu_0(H_{ij})=0. \end{align*} Define the total tie set $H\subset\Omega$ by \begin{align*} H:=\bigcup_{\substack{1\le i, j\le N, i\ne j}} H_{ij}. \end{align*} This is a finite union of $\mu_0$-null sets, hence \begin{align*} \mu_0(H)=0. \end{align*} On $\Omega\setminus H$, the sets $P_1(w),\dots,P_N(w)$ are pairwise disjoint. [/step]

[step:Use complementary slackness to force the prescribed cell masses] Let $\gamma$ be an optimal transport plan between $\mu_0$ and $\nu$, whose existence follows from compactness of $\Omega$, finiteness of $Y$, and lower semicontinuity of the continuous cost (citing a result not yet in the wiki: existence of optimal transport plans for compact lower semicontinuous cost). Since $\gamma$ has first marginal $\mu_0$ and second marginal $\nu$, define finite Borel measures $\gamma_i$ on $\Omega$ by \begin{align*} \gamma_i(A):=\gamma(A\times\{y_i\}) \end{align*} for every $A\in\mathcal B(\Omega)$ and every $i\in\{1,\dots,N\}$. Then \begin{align*} \gamma_i(\Omega)=m_i \end{align*} and \begin{align*} \sum_{i=1}^N \gamma_i=\mu_0. \end{align*} Complementary slackness for the optimal primal-dual pair says that \begin{align*} \varphi(x)+a_i=|x-y_i|^2 \end{align*} for $\gamma$-a.e. $(x,y_i)\in\Omega\times Y$ (citing a result not yet in the wiki: complementary slackness in Kantorovich duality). Therefore $\gamma_i$ is concentrated on $C_i=P_i(w)$, and hence \begin{align*} m_i=\gamma_i(\Omega)=\gamma_i(P_i(w))\le \mu_0(P_i(w)). \end{align*} Summing over $i$ gives \begin{align*} \sum_{i=1}^N m_i\le \sum_{i=1}^N \mu_0(P_i(w)). \end{align*} Since the cells cover $\Omega$ and overlap only inside the $\mu_0$-null set $H$, finite additivity gives \begin{align*} \sum_{i=1}^N \mu_0(P_i(w))=\mu_0(\Omega). \end{align*} Using $\sum_{i=1}^N m_i=\mu_0(\Omega)$, all the preceding inequalities must be equalities. Thus \begin{align*} \mu_0(P_i(w))=m_i \end{align*} for every $i\in\{1,\dots,N\}$. [/step]

[step:Verify that the tie-broken map pushes the source measure to the prescribed atomic target] For each $i\in\{1,\dots,N\}$, define the tie-broken cell $A_i\subset\Omega$ by \begin{align*} A_i:=\{x\in\Omega:I_w(x)=i\}. \end{align*} Because each $P_i(w)$ is Borel and $I_w$ is defined by the least-index rule, each $A_i$ is Borel. Moreover, \begin{align*} A_i\subset P_i(w) \end{align*} and \begin{align*} P_i(w)\setminus A_i\subset H. \end{align*} Since $\mu_0(H)=0$, we obtain \begin{align*} \mu_0(A_i)=\mu_0(P_i(w))=m_i. \end{align*} The preimage of $\{y_i\}$ under $T_w$ is exactly $A_i$. Therefore, for every $i\in\{1,\dots,N\}$, \begin{align*} (T_w)_\#\mu_0(\{y_i\})=\mu_0(T_w^{-1}(\{y_i\}))=\mu_0(A_i)=m_i. \end{align*} Since $Y=\{y_1,\dots,y_N\}$ is finite, this proves \begin{align*} (T_w)_\#\mu_0=\sum_{i=1}^N m_i\delta_{y_i}. \end{align*} [/step]

[step:Prove optimality of the induced transport map from dual equality] Define the transport plan $\gamma_{T_w}$ on $\Omega\times Y$ by \begin{align*} \gamma_{T_w}:=(\operatorname{id}_\Omega,T_w)_\#\mu_0, \end{align*} where $\operatorname{id}_\Omega:\Omega\to\Omega$ is the identity map. By the pushforward computation above, $\gamma_{T_w}$ has marginals $\mu_0$ and $\nu$. For $\mu_0$-a.e. $x\in A_i$, we have $x\in P_i(w)=C_i$, and therefore \begin{align*} \varphi(x)+a_i=|x-y_i|^2. \end{align*} Using the partition $\Omega=\bigcup_{i=1}^N A_i$ and the identities $\mu_0(A_i)=m_i$, we compute \begin{align*} \int_{\Omega\times Y} |x-y|^2\,d\gamma_{T_w}(x,y)=\sum_{i=1}^N\int_{A_i}|x-y_i|^2\,d\mu_0(x). \end{align*} By the contact identity on $A_i$, \begin{align*} \sum_{i=1}^N\int_{A_i}|x-y_i|^2\,d\mu_0(x)=\sum_{i=1}^N\int_{A_i}\bigl(\varphi(x)+a_i\bigr)\,d\mu_0(x). \end{align*} Since the $A_i$ partition $\Omega$ and $\mu_0(A_i)=m_i$, the right-hand side equals \begin{align*} \int_\Omega \varphi(x)\,d\mu_0(x)+\sum_{i=1}^N m_i a_i. \end{align*} This is the optimal Kantorovich dual value, which equals the optimal primal transport value by duality. Hence $\gamma_{T_w}$ is an optimal transport plan. Since $\gamma_{T_w}$ is induced by the map $T_w$, the map $T_w$ is an optimal transport map from $\mu_0$ to $\nu$ for the quadratic cost. [/step]