Let $S$ be a finite set of points in $\mathbb{R}^{d}$, $c(s) \in [0,1]$ such that $\sum_{s \in S} c(s) = 1$, $\rho$ continuous and non-vanishing probability distribution on $[0,1]^{d}$ and $\mu $ measure with respect to $\rho$, i.e. $ \mu (X) = \int_{X} \rho (x) \ dx$. Let us suppose that there exists an assignment $ \text{L} \colon [0,1]^d \to S$ such that it minimizes
$$ \int_{ [0,1]^{d} } \rho (x) \ \delta^2 (x, \text{L} (x)) \ d x $$
over all possible assignments that satisfy $ \mu ( \text{L}^{-1} (s) ) = c(s)$ for all $s \in S$, where $\delta$ is the Euclidean distance in $\mathbb{R}^{d}$.
I would like to prove that for all $s,t \in S, \ s \not = t$ there exists a hyperplane $\alpha$ orthogonal to $(t-s)$ such that $ \mu ( \alpha_{ts} \cap \text{L}^{-1} (s) ) = 0 $ and $ \mu ( \alpha_{st} \cap \text{L}^{-1} (t)) = 0 $, where $ \alpha_{ts}$ is the half-space bounded by $\alpha$ and containing $\alpha + (t-s)$, and $\alpha_{st}$ is the complementary half-space.
At first, I considered proving this constructively by describing the hyperplane, but since I don't know how this assignment looks like, I think this could be proven by contradiction. So, let us assume that for all hyperplanes orthogonal to $(t-s)$ it holds: $ \mu ( \alpha_{ts} \cap \text{L}^{-1} (s) ) >0 $ or $\mu ( \alpha_{st} \cap \text{L}^{-1} (t)) >0 $. In case, where both $ \mu ( \alpha_{ts} \cap \text{L}^{-1} (s) ) >0 $ or $\mu ( \alpha_{st} \cap \text{L}^{-1} (t)) >0 $, we can simply rearrange points that are not assigned efficiently and get better assignment, which is a contradiction.
However, I don't know how to prove it in case, where only one of the measures is positive. I would appreciate any advice.