[proofplan] The proof is the smooth Jacobian formula for a pushforward measure. In the forward direction, the pushforward identity gives the density of $\nu$ on the open image $T(U)$ after applying the smooth change-of-variables formula, and the fact that $T_\#\mu$ is supported on $T(U)$ gives the image condition. The determinant identity is first obtained almost everywhere and then upgraded to a pointwise identity by continuity. Conversely, the determinant equation and the image condition allow us to compute $T_\#\mu$ on arbitrary Borel sets and identify it with $\nu$. [/proofplan] [step:Record the inverse map and the sign of the Jacobian determinant] Let \begin{align*} S:T(U)\to U \end{align*} denote the inverse diffeomorphism of $T$. Since $T(x)=\nabla \phi(x)$ and $\phi \in C^2(U)$, the Jacobian matrix of $T$ at $x$ is \begin{align*} JT_x=D^2\phi(x). \end{align*} Because $T$ is a diffeomorphism, $JT_x$ is invertible for every $x \in U$, hence $\det D^2\phi(x)\neq 0$. Since $\phi$ is convex and $C^2$, the Hessian matrix $D^2\phi(x)$ is positive semidefinite for every $x \in U$. Therefore all eigenvalues of $D^2\phi(x)$ are nonnegative, and invertibility forces them to be positive. Thus \begin{align*} \det D^2\phi(x)>0 \end{align*} for every $x \in U$. [/step] [step:Derive the density equation on the image from the pushforward identity] Assume $T_\#\mu=\nu$. Let $B \subset T(U)$ be a Borel set. By the definition of pushforward measure, \begin{align*} \nu(B)=T_\#\mu(B)=\mu(T^{-1}(B)). \end{align*} Using the definition of $\mu$, this becomes \begin{align*} \nu(B)=\int_{T^{-1}(B)} \rho_0(x)\,d\mathcal{L}^n(x). \end{align*} The set $T^{-1}(B)$ is Borel because $T:U\to V$ is continuous, and the function $x\mapsto \rho_0(x)\mathbb{1}_{T^{-1}(B)}(x)$ is nonnegative and Borel measurable. Apply the smooth change-of-variables formula to the $C^1$ diffeomorphism $T:U\to T(U)$ with the substitution $y=T(x)$, equivalently $x=S(y)$. Under this substitution, \begin{align*} d\mathcal{L}^n(x)=\left|\det JS_y\right|\,d\mathcal{L}^n(y). \end{align*} Since $JS_y=(JT_{S(y)})^{-1}$ and $\det JT_{S(y)}=\det D^2\phi(S(y))>0$, we have \begin{align*} \left|\det JS_y\right|=\frac{1}{\det D^2\phi(S(y))}. \end{align*} Therefore \begin{align*} \nu(B)=\int_B \frac{\rho_0(S(y))}{\det D^2\phi(S(y))}\,d\mathcal{L}^n(y). \end{align*} On the other hand, by the definition of $\nu$, \begin{align*} \nu(B)=\int_B \rho_1(y)\,d\mathcal{L}^n(y). \end{align*} Hence \begin{align*} \int_B \rho_1(y)\,d\mathcal{L}^n(y)=\int_B \frac{\rho_0(S(y))}{\det D^2\phi(S(y))}\,d\mathcal{L}^n(y) \end{align*} for every Borel set $B \subset T(U)$. By the uniqueness of densities with respect to [Lebesgue measure](/page/Lebesgue%20Measure), it follows that \begin{align*} \rho_1(y)=\frac{\rho_0(S(y))}{\det D^2\phi(S(y))} \end{align*} for $\mathcal{L}^n$-almost every $y \in T(U)$. [/step] [step:Upgrade the almost everywhere identity to a pointwise identity on $U$] Define \begin{align*} G:U\to \mathbb{R}, \qquad G(x)=\rho_1(T(x))\det D^2\phi(x)-\rho_0(x). \end{align*} The map $G$ is continuous because $\rho_0$, $\rho_1$, $T$, and $x\mapsto \det D^2\phi(x)$ are continuous. Let $N\subset T(U)$ be the Borel exceptional set on which the density identity may fail, so $\mathcal{L}^n(N)=0$. The smooth change-of-variables formula applied to the nonnegative Borel function $\mathbb{1}_N\circ T$ gives $\mathcal{L}^n(T^{-1}(N))=0$, because the corresponding integral over $N$ is zero. For every $x\in U\setminus T^{-1}(N)$, the identity at $y=T(x)$ gives $G(x)=0$. Hence $G(x)=0$ for $\mathcal{L}^n$-almost every $x \in U$. If there were a point $x_0 \in U$ with $G(x_0)\neq 0$, continuity would give an open ball $B(x_0,r)\subset U$ on which $G$ has constant nonzero sign. Such a ball has positive $\mathcal{L}^n$-measure, contradicting $G=0$ almost everywhere. Therefore $G(x)=0$ for every $x \in U$, which is exactly \begin{align*} \det D^2\phi(x)=\frac{\rho_0(x)}{\rho_1(T(x))}=\frac{\rho_0(x)}{\rho_1(\nabla\phi(x))}. \end{align*} [/step] [step:Show that no target mass lies outside the image] Since $T(U)\subset V$, the preimage of $V\setminus T(U)$ under $T:U\to V$ is empty: \begin{align*} T^{-1}(V\setminus T(U))=\varnothing. \end{align*} Using $T_\#\mu=\nu$ and the definition of pushforward measure, \begin{align*} \nu(V\setminus T(U))=T_\#\mu(V\setminus T(U))=\mu(\varnothing)=0. \end{align*} This proves the image condition in the forward direction. [/step] [step:Use the determinant equation and image condition to recover the pushforward measure] Conversely, assume \begin{align*} \det D^2\phi(x)=\frac{\rho_0(x)}{\rho_1(T(x))} \end{align*} for every $x \in U$, and assume \begin{align*} \nu(V\setminus T(U))=0. \end{align*} Let $B \subset V$ be a Borel set. Since $T^{-1}(B)=T^{-1}(B\cap T(U))$, the definition of $\mu$ gives \begin{align*} T_\#\mu(B)=\mu(T^{-1}(B))=\int_{T^{-1}(B\cap T(U))}\rho_0(x)\,d\mathcal{L}^n(x). \end{align*} The set $T^{-1}(B\cap T(U))$ is Borel because $T:U\to V$ is continuous, and the integrand $x\mapsto \rho_0(x)\mathbb{1}_{T^{-1}(B\cap T(U))}(x)$ is nonnegative and Borel measurable. Apply the smooth change-of-variables formula to the $C^1$ diffeomorphism $T:U\to T(U)$ with $y=T(x)$ and $x=S(y)$. Since $d\mathcal{L}^n(x)=1/\det D^2\phi(S(y))\,d\mathcal{L}^n(y)$, we obtain \begin{align*} T_\#\mu(B)=\int_{B\cap T(U)}\frac{\rho_0(S(y))}{\det D^2\phi(S(y))}\,d\mathcal{L}^n(y). \end{align*} The determinant equation applied at $S(y)$ gives \begin{align*} \det D^2\phi(S(y))=\frac{\rho_0(S(y))}{\rho_1(y)}. \end{align*} Hence \begin{align*} \frac{\rho_0(S(y))}{\det D^2\phi(S(y))}=\rho_1(y). \end{align*} Therefore \begin{align*} T_\#\mu(B)=\int_{B\cap T(U)}\rho_1(y)\,d\mathcal{L}^n(y)=\nu(B\cap T(U)). \end{align*} The image condition gives $\nu(B\setminus T(U))\leq \nu(V\setminus T(U))=0$, so \begin{align*} \nu(B)=\nu(B\cap T(U))+\nu(B\setminus T(U))=\nu(B\cap T(U)). \end{align*} Thus $T_\#\mu(B)=\nu(B)$ for every Borel set $B\subset V$, and consequently $T_\#\mu=\nu$. [guided] We now prove the reverse implication carefully. The goal is to show equality of the two finite Borel measures $T_\#\mu$ and $\nu$ on $V$. By definition, this means proving \begin{align*} T_\#\mu(B)=\nu(B) \end{align*} for every Borel set $B\subset V$. Fix a Borel set $B\subset V$. Since the map $T:U\to T(U)$ only takes values in $T(U)$, points of $B$ outside $T(U)$ have no preimage. Therefore \begin{align*} T^{-1}(B)=T^{-1}(B\cap T(U)). \end{align*} Using the definition of the pushforward measure and then the definition of $\mu$, we get \begin{align*} T_\#\mu(B)=\mu(T^{-1}(B))=\int_{T^{-1}(B\cap T(U))}\rho_0(x)\,d\mathcal{L}^n(x). \end{align*} Now we use the smooth change-of-variables formula for the diffeomorphism $T:U\to T(U)$. The substitution is $y=T(x)$, and the inverse substitution is $x=S(y)$, where $S:T(U)\to U$ is the inverse diffeomorphism. The relevant integrand is nonnegative and Borel measurable because $T$, $S$, and $\rho_0$ are continuous and $B\cap T(U)$ is Borel. Since $JT_x=D^2\phi(x)$, the diffeomorphism hypothesis gives that $D^2\phi(x)$ is invertible for every $x\in U$. Convexity of the $C^2$ function $\phi$ gives that $D^2\phi(x)$ is positive semidefinite, so invertibility forces all eigenvalues to be positive. Therefore $\det D^2\phi(x)>0$ for every $x\in U$, and the measure transformation is \begin{align*} d\mathcal{L}^n(x)=\frac{1}{\det D^2\phi(S(y))}\,d\mathcal{L}^n(y). \end{align*} Applying this to the previous integral gives \begin{align*} T_\#\mu(B)=\int_{B\cap T(U)}\frac{\rho_0(S(y))}{\det D^2\phi(S(y))}\,d\mathcal{L}^n(y). \end{align*} This is where the Monge-Ampere equation enters. The assumed determinant equation holds at every point of $U$, so in particular it holds at $S(y)\in U$ for every $y\in T(U)$. Thus \begin{align*} \det D^2\phi(S(y))=\frac{\rho_0(S(y))}{\rho_1(T(S(y)))}. \end{align*} Because $S$ is the inverse of $T$, we have $T(S(y))=y$. Hence \begin{align*} \det D^2\phi(S(y))=\frac{\rho_0(S(y))}{\rho_1(y)}. \end{align*} Since $\rho_0$ and $\rho_1$ are positive, division is legitimate, and therefore \begin{align*} \frac{\rho_0(S(y))}{\det D^2\phi(S(y))}=\rho_1(y). \end{align*} Substituting this into the changed-variable integral yields \begin{align*} T_\#\mu(B)=\int_{B\cap T(U)}\rho_1(y)\,d\mathcal{L}^n(y)=\nu(B\cap T(U)). \end{align*} It remains to compare $\nu(B\cap T(U))$ with $\nu(B)$. The set $B$ decomposes as the disjoint union of $B\cap T(U)$ and $B\setminus T(U)$. The image condition says that the whole complement $V\setminus T(U)$ has zero $\nu$-measure, so its subset $B\setminus T(U)$ also has zero $\nu$-measure: \begin{align*} \nu(B\setminus T(U))\leq \nu(V\setminus T(U))=0. \end{align*} By finite additivity on disjoint Borel sets, \begin{align*} \nu(B)=\nu(B\cap T(U))+\nu(B\setminus T(U))=\nu(B\cap T(U)). \end{align*} Combining the two identities gives \begin{align*} T_\#\mu(B)=\nu(B). \end{align*} Since $B\subset V$ was arbitrary, the measures agree on all Borel subsets of $V$, which proves $T_\#\mu=\nu$. [/guided] [/step]