Let $\mu$ and $\nu$ be Borel probability measures on $\mathbb{R}^n$, and suppose $\mu \ll \mathcal{L}^n$. Let $\phi:\mathbb{R}^n \to \mathbb{R}$ be a finite convex Brenier potential, and let $T:\mathbb{R}^n \to \mathbb{R}^n$ be the Brenier map transporting $\mu$ to $\nu$, so that $T(x)=\nabla \phi(x)$ for $\mu$-almost every point $x$ at which $\phi$ is differentiable. Then $\phi$ is a transport Alexandrov solution in the single-valued almost-everywhere sense:

\begin{align*} (\nabla \phi)_{\#}\mu=\nu. \end{align*}

Moreover, let $U,V\subset \mathbb{R}^n$ be open sets. Suppose $\mu=\rho_0\,\mathcal{L}^n\!\restriction_U$ and $\nu=\rho_1\,\mathcal{L}^n\!\restriction_V$, where $\rho_0:U\to(0,\infty)$ and $\rho_1:V\to(0,\infty)$ are measurable density representatives. If $\phi\in C^2(U)$ is convex, $\nabla\phi:U\to V$ is a diffeomorphism onto $V$, and

\begin{align*} (\nabla\phi)_{\#}(\rho_0\,\mathcal{L}^n\!\restriction_U)=\rho_1\,\mathcal{L}^n\!\restriction_V, \end{align*}

then

\begin{align*} \det D^2\phi(x)=\frac{\rho_0(x)}{\rho_1(\nabla\phi(x))} \end{align*}

for $\mathcal{L}^n$-almost every $x\in U$. If, in addition, $\rho_0$ and $\rho_1$ are continuous, then this identity holds for every $x\in U$.