[proofplan] The nondegeneracy hypothesis first turns the image measure $\nu = u_\#\lambda$ into a measure absolutely continuous with respect to [Lebesgue measure](/page/Lebesgue%20Measure), so Brenier theory applies in both directions. The forward Brenier map gives the monotone factor $T = \nabla\varphi$ transporting $\lambda$ to $\nu$, while the reverse Brenier map $R = \nabla\psi$ transports $\nu$ back to $\lambda$. The standard essential inverse relation for the forward and reverse Brenier maps then lets us define $s := R \circ u$, prove $s_\#\lambda = \lambda$, and recover $u = T \circ s$ almost everywhere. Uniqueness is asserted only for the monotone factor, because it is exactly the unique Brenier map from $\lambda$ to $\nu$. [/proofplan] [step:Verify absolute continuity and finite second moments for the two transport measures] The measure $\lambda$ is a probability measure because $\mathcal{L}^n(\Omega) \in (0,\infty)$. Since $\Omega$ is bounded, there is a constant $M_\Omega > 0$ such that $|x| \leq M_\Omega$ for every $x \in \Omega$, after changing $\Omega$ on a set of $\lambda$-measure zero if necessary. Hence \begin{align*}\int_\Omega |x|^2 \, d\lambda(x) \leq M_\Omega^2 < \infty.\end{align*} Define $\nu := u_\#\lambda$, meaning that for every Borel set $A \subset \mathbb{R}^n$, \begin{align*} \nu(A) = \lambda(u^{-1}(A)). \end{align*} Since $u \in L^2(\Omega,\lambda;\mathbb{R}^n)$, the change-of-variables identity for pushforward measures gives \begin{align*}\int_{\mathbb{R}^n} |y|^2 \, d\nu(y) = \int_\Omega |u(x)|^2 \, d\lambda(x) < \infty.\end{align*} Thus both $\lambda$ and $\nu$ have finite second moments. It remains to check absolute continuity of $\nu$ with respect to $\mathcal{L}^n$. Let $N \subset \mathbb{R}^n$ be a Borel set with $\mathcal{L}^n(N)=0$. By nondegeneracy of $u$, \begin{align*} \nu(N) = \lambda(u^{-1}(N)) = 0. \end{align*} Therefore $\nu \ll \mathcal{L}^n$. Also $\lambda \ll \mathcal{L}^n$ because $\lambda$ is a normalized restriction of $\mathcal{L}^n$. [guided] We need Brenier theory in both directions, so we must verify its measure-theoretic hypotheses for both measures. First, $\lambda$ is a probability measure because it is obtained by normalizing $\mathcal{L}^n$ on the bounded [open set](/page/Open%20Set) $\Omega$, and the assumption $\mathcal{L}^n(\Omega)>0$ makes the normalization meaningful. The source measure $\lambda$ has finite second moment. Since $\Omega$ is bounded, choose $M_\Omega > 0$ such that $|x| \leq M_\Omega$ for $\lambda$-a.e. $x \in \Omega$. Then \begin{align*} \int_\Omega |x|^2 \, d\lambda(x) \leq \int_\Omega M_\Omega^2 \, d\lambda(x) = M_\Omega^2 < \infty. \end{align*} Now define the image measure $\nu := u_\#\lambda$. This means that for every Borel set $A \subset \mathbb{R}^n$, \begin{align*} \nu(A) = \lambda(u^{-1}(A)). \end{align*} The second moment of $\nu$ is exactly the $L^2$ norm of $u$ with respect to $\lambda$. Indeed, applying the defining integration identity for pushforward measures to the Borel function $y \mapsto |y|^2$ gives \begin{align*} \int_{\mathbb{R}^n} |y|^2 \, d\nu(y) = \int_\Omega |u(x)|^2 \, d\lambda(x). \end{align*} The right-hand side is finite because $u \in L^2(\Omega,\lambda;\mathbb{R}^n)$. The nondegeneracy assumption is used precisely to prove $\nu \ll \mathcal{L}^n$. Let $N \subset \mathbb{R}^n$ be a Borel set with $\mathcal{L}^n(N)=0$. Since $u$ is nondegenerate, the preimage $u^{-1}(N)$ has $\lambda$-measure zero. Therefore \begin{align*} \nu(N) = \lambda(u^{-1}(N)) = 0. \end{align*} This proves $\nu \ll \mathcal{L}^n$. Finally, $\lambda \ll \mathcal{L}^n$ follows directly from the formula $\lambda = \mathcal{L}^n|_\Omega/\mathcal{L}^n(\Omega)$. [/guided] [/step] [step:Apply Brenier theory in the forward direction to obtain the monotone factor] We apply the Brenier theorem (citing a result not yet in the wiki: Brenier Theorem) to the probability measures $\lambda$ and $\nu$ on $\mathbb{R}^n$. The source measure $\lambda$ is absolutely continuous with respect to $\mathcal{L}^n$, and both measures have finite second moments by the previous step. Hence there exists a proper lower semicontinuous convex function $\varphi: \mathbb{R}^n \to (-\infty,\infty]$ such that the Borel map \begin{align*}T: \mathbb{R}^n \to \mathbb{R}^n\end{align*} is given by $T = \nabla\varphi$ $\lambda$-a.e. and satisfies \begin{align*} T_\#\lambda = \nu. \end{align*} Moreover, Brenier uniqueness gives that $T$ is unique $\lambda$-a.e. among optimal transport maps from $\lambda$ to $\nu$ for the quadratic cost, and in particular unique as the Brenier map from $\lambda$ to $\nu$. [/step] [step:Use the reverse Brenier map as an essential inverse] We apply the reverse Brenier theorem and inverse-map statement (citing a result not yet in the wiki: Brenier Inverse Map Theorem) to the pair $(\nu,\lambda)$. The source measure $\nu$ is absolutely continuous with respect to $\mathcal{L}^n$, and both $\nu$ and $\lambda$ have finite second moments. Therefore there exists a proper lower semicontinuous convex function $\psi: \mathbb{R}^n \to (-\infty,\infty]$ such that the Borel map \begin{align*}R: \mathbb{R}^n \to \mathbb{R}^n\end{align*} is given by $R = \nabla\psi$ $\nu$-a.e. and satisfies \begin{align*} R_\#\nu = \lambda. \end{align*} We regard $\lambda$ as a probability measure on $\mathbb{R}^n$ supported on $\Omega$. Under the standard conjugate normalization, the reverse Brenier potential may be chosen as the convex conjugate potential $\varphi^*$ up to an additive constant, so the reverse map $R$ is the Brenier map induced by $\varphi^*$ on the set where $\varphi^*$ is differentiable. The Brenier inverse-map theorem then applies because the source measures in both directions, namely $\lambda$ and $\nu$, are absolutely continuous with respect to $\mathcal{L}^n$ and both measures have finite second moments. It gives the essential inverse identities \begin{align*}R(T(x)) = x\end{align*} for $\lambda$-a.e. $x \in \Omega$, and \begin{align*}T(R(y)) = y\end{align*} for $\nu$-a.e. $y \in \mathbb{R}^n$. Here $\partial\varphi(x) := \{y \in \mathbb{R}^n : \varphi(z) \geq \varphi(x) + y \cdot (z - x) \text{ for all } z \in \mathbb{R}^n\}$ denotes the subdifferential of $\varphi$ at $x$, and $\varphi^*: \mathbb{R}^n \to (-\infty,\infty]$ denotes the convex conjugate defined by $\varphi^*(y) := \sup_{x \in \mathbb{R}^n} \bigl(x \cdot y - \varphi(x)\bigr)$. The inverse relation is the usual consequence of Fenchel equality for convex conjugates on the two optimal graph plans: at points where the Brenier maps are single-valued, $y \in \partial\varphi(x)$ is equivalent to $x \in \partial\varphi^*(y)$. [/step] [step:Define the rearrangement and prove that it preserves $\lambda$] Fix a point $x_0 \in \Omega$. Define the Borel map \begin{align*}s: \Omega \to \Omega\end{align*} by \begin{align*}s(x) := R(u(x))\end{align*} for $\lambda$-a.e. $x \in \Omega$, and set $s(x) = x_0$ on the $\lambda$-null exceptional set if necessary to make it everywhere defined as a Borel map into $\Omega$. We prove that $s_\#\lambda=\lambda$. Let $A \subset \Omega$ be a Borel set. By the definition of $s$, then the definition of pushforward measure, and finally $u_\#\lambda=\nu$, \begin{align*} s_\#\lambda(A) = \lambda(s^{-1}(A)). \end{align*} Since $s = R \circ u$ $\lambda$-a.e., \begin{align*} \lambda(s^{-1}(A)) = \lambda(u^{-1}(R^{-1}(A))). \end{align*} Using $\nu = u_\#\lambda$, \begin{align*} \lambda(u^{-1}(R^{-1}(A))) = \nu(R^{-1}(A)). \end{align*} Since $R_\#\nu=\lambda$, \begin{align*} \nu(R^{-1}(A)) = \lambda(A). \end{align*} Therefore $s_\#\lambda(A)=\lambda(A)$ for every Borel set $A \subset \Omega$, so $s_\#\lambda=\lambda$. [/step] [step:Recover $u$ by composing the measure-preserving factor with the monotone factor] Let \begin{align*}E := \{y \in \mathbb{R}^n : T(R(y)) = y\}.\end{align*} The essential inverse relation gives $\nu(\mathbb{R}^n \setminus E)=0$. Since $\nu = u_\#\lambda$, \begin{align*}\lambda(u^{-1}(\mathbb{R}^n \setminus E)) = \nu(\mathbb{R}^n \setminus E) = 0.\end{align*} Thus for $\lambda$-a.e. $x \in \Omega$, the point $u(x)$ belongs to $E$, and hence \begin{align*}T(s(x)) = T(R(u(x))) = u(x).\end{align*} Therefore \begin{align*}T \circ s = u\end{align*} $\lambda$-a.e. on $\Omega$. [/step] [step:Record the uniqueness statement for the monotone factor only] The map $T$ constructed above is the Brenier map from $\lambda$ to $\nu$. Since $\lambda \ll \mathcal{L}^n$ and both $\lambda$ and $\nu$ have finite second moments, Brenier uniqueness implies that any other gradient of a proper lower semicontinuous convex function transporting $\lambda$ to $\nu$ agrees with $T$ $\lambda$-a.e. This proves the asserted uniqueness of the monotone factor. The construction $s=R\circ u$ gives the canonical rearrangement factor whenever the reverse Brenier map $R$ is taken as an essential inverse for $T$. The theorem does not assert uniqueness of $s$ among all $\lambda$-measure-preserving maps satisfying $u=T\circ s$; it asserts existence of such an $s$ together with uniqueness of the monotone Brenier factor. [/step]