[proofplan]
Write the transport map as $T=\nabla\phi$. The pushforward identity says that integrating a bounded continuous [test function](/page/Test%20Function) on $V$ against $\nu$ is the same as integrating its pullback by $T$ against $\mu$. Since $T$ is a diffeomorphism, the classical change-of-variables formula rewrites the $\nu$-integral over $V$ as an integral over $U$ with Jacobian factor $|\det DT|$. Convexity makes $D^2\phi$ positive semidefinite, while the diffeomorphism assumption makes its determinant nonzero, so the absolute Jacobian is $\det D^2\phi$. Finally, testing against functions of the form $h \circ T^{-1}$ yields pointwise equality of the continuous densities.
[/proofplan]
[step:Translate the pushforward identity into test-function integrals]
Define the map
\begin{align*}
T: U &\to V
\end{align*}
by $T(x)=\nabla\phi(x)$. Let $f:V\to\mathbb{R}$ be a bounded [continuous function](/page/Continuous%20Function). Since $T_\#\mu=\nu$, the defining property of pushforward measure gives
\begin{align*}
\int_V f(y)\, d\nu(y) = \int_U f(T(x))\, d\mu(x).
\end{align*}
Using the density definitions of $\mu$ and $\nu$, this becomes
\begin{align*}
\int_V f(y)\rho_1(y)\, d\mathcal{L}^n(y) = \int_U f(T(x))\rho_0(x)\, d\mathcal{L}^n(x).
\end{align*}
[guided]
We first convert the measure-theoretic hypothesis into an integral identity. Define
\begin{align*}
T: U &\to V
\end{align*}
by $T(x)=\nabla\phi(x)$. The assumption $T_\#\mu=\nu$ means that the distribution of points after applying $T$ to the measure $\mu$ is exactly $\nu$. Equivalently, for every bounded continuous function $f:V\to\mathbb{R}$,
\begin{align*}
\int_V f(y)\, d\nu(y) = \int_U f(T(x))\, d\mu(x).
\end{align*}
Now we insert the definitions of the measures. Since $\nu=\rho_1\,d\mathcal{L}^n$ on $V$ and $\mu=\rho_0\,d\mathcal{L}^n$ on $U$, the preceding equality becomes
\begin{align*}
\int_V f(y)\rho_1(y)\, d\mathcal{L}^n(y) = \int_U f(T(x))\rho_0(x)\, d\mathcal{L}^n(x).
\end{align*}
This is the identity that contains the transport information. The rest of the proof rewrites the left-hand side over the same domain $U$ so that the two densities can be compared directly.
[/guided]
[/step]
[step:Apply change of variables to the diffeomorphism $T$]
The map $T:U\to V$ is a $C^1$ diffeomorphism by hypothesis. Applying the classical change-of-variables formula for diffeomorphisms to the substitution $y=T(x)$ gives
\begin{align*}
\int_V f(y)\rho_1(y)\, d\mathcal{L}^n(y) = \int_U f(T(x))\rho_1(T(x))|\det DT_x|\, d\mathcal{L}^n(x).
\end{align*}
Because $\phi\in C^2(U)$, the Jacobian matrix of $T=\nabla\phi$ at $x$ is the Hessian matrix $D^2\phi(x)$. Since $\phi$ is convex, $D^2\phi(x)$ is positive semidefinite for every $x\in U$. Since $T$ is a diffeomorphism, $DT_x$ is invertible for every $x\in U$, so $\det D^2\phi(x)\ne 0$. Therefore every eigenvalue of $D^2\phi(x)$ is nonnegative and their product is nonzero, hence
\begin{align*}
|\det DT_x| = \det D^2\phi(x).
\end{align*}
Thus, for every bounded continuous $f:V\to\mathbb{R}$,
\begin{align*}
\int_U f(T(x))\rho_1(T(x))\det D^2\phi(x)\, d\mathcal{L}^n(x) = \int_U f(T(x))\rho_0(x)\, d\mathcal{L}^n(x).
\end{align*}
[/step]
[step:Compare the two continuous densities pointwise on $U$]
Define
\begin{align*}
g:U &\to \mathbb{R}
\end{align*}
by
\begin{align*}
g(x)=\rho_1(T(x))\det D^2\phi(x)-\rho_0(x).
\end{align*}
The function $g$ is continuous because $\rho_0$, $\rho_1$, $T$, and $x\mapsto \det D^2\phi(x)$ are continuous. We claim that $g(x)=0$ for every $x\in U$.
Let $h:U\to\mathbb{R}$ be any continuous function with compact support. Define
\begin{align*}
f:V &\to \mathbb{R}
\end{align*}
by $f(y)=h(T^{-1}(y))$. Since $T^{-1}:V\to U$ is continuous and $h$ is continuous with compact support, $f$ is bounded and continuous. Substituting this $f$ into the previous identity gives
\begin{align*}
\int_U h(x)g(x)\, d\mathcal{L}^n(x)=0.
\end{align*}
If there were $x_0\in U$ with $g(x_0)>0$, continuity of $g$ would give a radius $r>0$ such that $\overline{B}(x_0,r)\subset U$ and $g(x)>g(x_0)/2$ for every $x\in B(x_0,r)$. Choose a continuous function $h:U\to[0,1]$ with compact support contained in $B(x_0,r)$ and with $h(x_0)=1$. Then $h g\ge 0$ on $U$ and $h g>0$ on some smaller ball around $x_0$, so
\begin{align*}
\int_U h(x)g(x)\, d\mathcal{L}^n(x)>0,
\end{align*}
contradicting the identity above. The same argument applied to $-g$ excludes $g(x_0)<0$. Therefore $g=0$ on $U$.
[/step]
[step:Divide by the positive target density]
The equality $g=0$ gives, for every $x\in U$,
\begin{align*}
\rho_1(T(x))\det D^2\phi(x)=\rho_0(x).
\end{align*}
Since $T(x)=\nabla\phi(x)\in V$ and $\rho_1:V\to(0,\infty)$ is positive, division by $\rho_1(T(x))$ is valid. Hence, for every $x\in U$,
\begin{align*}
\det D^2\phi(x)=\frac{\rho_0(x)}{\rho_1(\nabla\phi(x))}.
\end{align*}
This is the claimed smooth Monge-Ampere equation.
[/step]
