[proofplan] We test the determinants against a compactly supported smooth function and use the divergence structure of the determinant. For smooth maps, expansion of the determinant along one row and the Piola identity rewrite the tested determinant as an integral involving one undifferentiated component of the map and an $(n-1)$-minor of its gradient. We then extend this identity to Sobolev maps by local mollification. Finally, Rellich compactness gives strong local convergence of the undifferentiated component, while weak continuity of lower-order minors gives [weak convergence](/page/Weak%20Convergence) of the cofactor factor, and the product pairing passes to the limit. [/proofplan] [step:Handle the one-dimensional case directly] If $n=1$, then $W^{1,1}(U;\mathbb R)=W^{1,n}(U;\mathbb R^n)$ and \begin{align*} \det \nabla u_j=\partial_{x_1}u_j. \end{align*} Weak convergence $u_j\rightharpoonup u$ in $W^{1,1}(U)$ implies $\partial_{x_1}u_j\rightharpoonup \partial_{x_1}u$ in $L^1(U)$. Hence, for every $\phi\in C_c^\infty(U)$, \begin{align*} \lim_{j\to\infty}\int_U \phi(x)\det \nabla u_j(x)\,d\mathcal L^1(x)=\int_U \phi(x)\det \nabla u(x)\,d\mathcal L^1(x). \end{align*} This proves the theorem when $n=1$. For the rest of the proof assume $n\ge 2$. [/step] [step:Rewrite the smooth determinant using one component and cofactors] Let $\phi\in C_c^\infty(U)$ be fixed. Choose an [open set](/page/Open%20Set) $V\subset U$ with Lipschitz boundary such that \begin{align*} \operatorname{supp}\phi\subset V \end{align*} and $\overline V$ is compact in $U$. For a matrix $A\in\mathbb R^{n\times n}$, let $\operatorname{cof}A\in\mathbb R^{n\times n}$ denote the cofactor matrix, with entry $(\operatorname{cof}A)_{\alpha i}$ equal to the signed determinant of the minor obtained by deleting row $\alpha$ and column $i$. Let \begin{align*} v:V\to\mathbb R^n \end{align*} be a map in $C^\infty(\overline V;\mathbb R^n)$. Write $v_\alpha:V\to\mathbb R$ for its $\alpha$-th component, and write $\nabla v:V\to\mathbb R^{n\times n}$ for its weak gradient, which agrees with the classical Jacobian matrix. Expanding the determinant along the first row gives \begin{align*} \det \nabla v(x)=\sum_{i=1}^n \partial_{x_i}v_1(x)(\operatorname{cof}\nabla v(x))_{1i}. \end{align*} The Piola identity for smooth gradients states that \begin{align*} \sum_{i=1}^n \partial_{x_i}(\operatorname{cof}\nabla v)_{1i}=0 \end{align*} pointwise in $V$. Therefore \begin{align*} \det \nabla v=\sum_{i=1}^n \partial_{x_i}\big(v_1(\operatorname{cof}\nabla v)_{1i}\big) \end{align*} in $V$. Multiplying by $\phi$ and integrating by parts over $V$, with no boundary term because $\operatorname{supp}\phi\subset V$, yields \begin{align*} \int_U \phi(x)\det \nabla v(x)\,d\mathcal L^n(x)=-\sum_{i=1}^n\int_V v_1(x)\partial_{x_i}\phi(x)(\operatorname{cof}\nabla v(x))_{1i}\,d\mathcal L^n(x). \end{align*} [guided] The purpose of this step is to move one derivative away from the component $v_1$ and onto the [test function](/page/Test%20Function) $\phi$. This is what turns the determinant, which is a product of $n$ weakly convergent derivative factors, into a pairing between a strongly convergent undifferentiated factor and a weakly convergent cofactor factor. First define precisely the algebraic object being used. For $A\in\mathbb R^{n\times n}$, the cofactor matrix $\operatorname{cof}A\in\mathbb R^{n\times n}$ has entry $(\operatorname{cof}A)_{\alpha i}$ equal to the signed determinant of the matrix obtained from $A$ by deleting row $\alpha$ and column $i$. For a smooth map \begin{align*} v:V\to\mathbb R^n, \end{align*} the gradient $\nabla v:V\to\mathbb R^{n\times n}$ is the matrix-valued map whose $(\alpha,i)$ entry is $\partial_{x_i}v_\alpha$. Expanding $\det \nabla v(x)$ along the first row gives \begin{align*} \det \nabla v(x)=\sum_{i=1}^n \partial_{x_i}v_1(x)(\operatorname{cof}\nabla v(x))_{1i}. \end{align*} The Piola identity for smooth gradients says that each row of the cofactor matrix is divergence-free. For the first row this is \begin{align*} \sum_{i=1}^n \partial_{x_i}(\operatorname{cof}\nabla v)_{1i}=0. \end{align*} Combining the row expansion with the product rule gives \begin{align*} \sum_{i=1}^n \partial_{x_i}\big(v_1(\operatorname{cof}\nabla v)_{1i}\big)=\sum_{i=1}^n \partial_{x_i}v_1(\operatorname{cof}\nabla v)_{1i}+v_1\sum_{i=1}^n \partial_{x_i}(\operatorname{cof}\nabla v)_{1i}. \end{align*} The second term is zero by the Piola identity, so \begin{align*} \det \nabla v=\sum_{i=1}^n \partial_{x_i}\big(v_1(\operatorname{cof}\nabla v)_{1i}\big). \end{align*} Now multiply by $\phi$. Since $\phi\in C_c^\infty(U)$ and $\operatorname{supp}\phi\subset V$, [integration by parts](/theorems/210) over $V$ produces no boundary contribution: \begin{align*} \int_U \phi(x)\det \nabla v(x)\,d\mathcal L^n(x)=-\sum_{i=1}^n\int_V v_1(x)\partial_{x_i}\phi(x)(\operatorname{cof}\nabla v(x))_{1i}\,d\mathcal L^n(x). \end{align*} This is the exact identity we will pass to Sobolev maps and then to the limit $j\to\infty$. [/guided] [/step] [step:Extend the divergence identity to Sobolev maps by local mollification] Let \begin{align*} w:V\to\mathbb R^n \end{align*} belong to $W^{1,n}(V;\mathbb R^n)$. Choose an open set $V_0$ with \begin{align*} \operatorname{supp}\phi\subset V_0 \end{align*} and $\overline{V_0}\subset V$. Let $(\eta_\varepsilon)_{\varepsilon>0}$ be a [standard mollifier](/page/Standard%20Mollifier) on $\mathbb R^n$, and define \begin{align*} w_\varepsilon:V_0\to\mathbb R^n \end{align*} by $w_\varepsilon=\eta_\varepsilon*w$ on $V_0$, for all sufficiently small $\varepsilon>0$ after extending $w$ locally from $V$ to a function in $W^{1,n}(\mathbb R^n;\mathbb R^n)$ on a neighbourhood of $\overline{V_0}$. Then \begin{align*} w_\varepsilon\to w \quad \text{in } W^{1,n}(V_0;\mathbb R^n). \end{align*} For matrices $A,B\in\mathbb R^{n\times n}$, the polynomial nature of the determinant and cofactor maps gives constants $C_n,c_n>0$ such that \begin{align*} |\det A-\det B|\le C_n(|A|+|B|)^{n-1}|A-B| \end{align*} and \begin{align*} |\operatorname{cof}A-\operatorname{cof}B|\le c_n(|A|+|B|)^{n-2}|A-B|. \end{align*} Applying the first estimate with $A=\nabla w_\varepsilon(x)$ and $B=\nabla w(x)$, then using Hölder's inequality with exponents $n/(n-1)$ and $n$, gives \begin{align*} \det \nabla w_\varepsilon\to \det \nabla w \quad \text{in } L^1(V_0). \end{align*} When $n>2$, apply the second estimate and Hölder's inequality with the exponents \begin{align*} \frac{n}{n-2} \quad \text{and} \quad n. \end{align*} When $n=2$, use the immediate Lipschitz estimate for the cofactor map. In both cases this gives \begin{align*} \operatorname{cof}\nabla w_\varepsilon\to \operatorname{cof}\nabla w \quad \text{in } L^{\frac{n}{n-1}}(V_0;\mathbb R^{n\times n}). \end{align*} The smooth identity applied to $w_\varepsilon$ may be written with the right-hand integral over $V_0$, because $\operatorname{supp}\phi\subset V_0$. Since $\operatorname{supp}\partial_{x_i}\phi\subset\operatorname{supp}\phi\subset V_0$, the same integral is unchanged if its domain is enlarged from $V_0$ to $V$. Passing to the limit in the left-hand side by the $L^1(V_0)$ convergence of determinants and in the right-hand side by $w_{\varepsilon,1}\to w_1$ in $L^n(V_0)$ together with the cofactor convergence in $L^{n/(n-1)}(V_0)$ yields \begin{align*} \int_U \phi(x)\det \nabla w(x)\,d\mathcal L^n(x)=-\sum_{i=1}^n\int_V w_1(x)\partial_{x_i}\phi(x)(\operatorname{cof}\nabla w(x))_{1i}\,d\mathcal L^n(x). \end{align*} The right-hand side is finite because $w_1\in L^n(V)$, $\partial_{x_i}\phi\in L^\infty(V)$, and $(\operatorname{cof}\nabla w)_{1i}\in L^{\frac{n}{n-1}}(V)$. [/step] [step:Extract the two convergence inputs needed in the cofactor identity] Since $u_j\rightharpoonup u$ in $W^{1,n}(U;\mathbb R^n)$, the restrictions satisfy \begin{align*} u_j|_V\rightharpoonup u|_V \quad \text{in } W^{1,n}(V;\mathbb R^n). \end{align*} By local Rellich compactness on the bounded Lipschitz set $V$, every subsequence of $(u_j)$ has a further subsequence converging strongly to $u$ in $L^n(V;\mathbb R^n)$. Hence the whole sequence satisfies \begin{align*} u_j\to u \quad \text{in } L^n(V;\mathbb R^n). \end{align*} For each $i\in\{1,\dots,n\}$, the function \begin{align*} x\mapsto (\operatorname{cof}\nabla u_j(x))_{1i} \end{align*} is an $(n-1)\times(n-1)$ minor of the Jacobian matrix $\nabla u_j(x)$. We apply the lower-order case of the [weak continuity of minors](/theorems/8750), namely [citetheorem:8750], on the bounded open set $V$ with target dimension $m=n$, minor order $r=n-1$, and Sobolev exponent $p=n$. The hypotheses are satisfied: $1\le r=n-1\le n=\min\{m,n\}$, the restricted maps satisfy $u_j|_V\rightharpoonup u|_V$ in $W^{1,n}(V;\mathbb R^n)$, and $p=n>n-1=r$. Therefore the cited theorem applies to each specified $(n-1)\times(n-1)$ minor. The exponent in the weak convergence conclusion is \begin{align*} \frac{p}{r}=\frac{n}{n-1}. \end{align*} Hence \begin{align*} (\operatorname{cof}\nabla u_j)_{1i}\rightharpoonup(\operatorname{cof}\nabla u)_{1i} \quad \text{in } L^{\frac{n}{n-1}}(V). \end{align*} [/step] [step:Pass to the limit in the tested determinant] Define, for each $i\in\{1,\dots,n\}$ and each $j\in\mathbb N$, \begin{align*} a_{j,i}:V\to\mathbb R \end{align*} by \begin{align*} a_{j,i}(x)=u_{j,1}(x)\partial_{x_i}\phi(x). \end{align*} Define also \begin{align*} a_i:V\to\mathbb R \end{align*} by \begin{align*} a_i(x)=u_1(x)\partial_{x_i}\phi(x). \end{align*} Since $u_{j,1}\to u_1$ in $L^n(V)$ and $\partial_{x_i}\phi\in L^\infty(V)$, we have \begin{align*} a_{j,i}\to a_i \quad \text{in } L^n(V). \end{align*} The weak convergence of cofactors from the previous step and the strong convergence of $a_{j,i}$ imply \begin{align*} \lim_{j\to\infty}\int_V a_{j,i}(x)(\operatorname{cof}\nabla u_j(x))_{1i}\,d\mathcal L^n(x)=\int_V a_i(x)(\operatorname{cof}\nabla u(x))_{1i}\,d\mathcal L^n(x). \end{align*} Indeed, split the difference into \begin{align*} \int_V (a_{j,i}-a_i)(\operatorname{cof}\nabla u_j)_{1i}\,d\mathcal L^n(x)+\int_V a_i(x)\big((\operatorname{cof}\nabla u_j(x))_{1i}-(\operatorname{cof}\nabla u(x))_{1i}\big)\,d\mathcal L^n(x). \end{align*} The first term tends to $0$ by Hölder's inequality and boundedness of $(\operatorname{cof}\nabla u_j)_{1i}$ in $L^{\frac{n}{n-1}}(V)$. The second term tends to $0$ because $a_i\in L^n(V)$ and $(\operatorname{cof}\nabla u_j)_{1i}\rightharpoonup(\operatorname{cof}\nabla u)_{1i}$ in $L^{\frac{n}{n-1}}(V)$. Using the Sobolev cofactor identity first with $w=u_j$ and then with $w=u$, we obtain \begin{align*} \lim_{j\to\infty}\int_U \phi(x)\det \nabla u_j(x)\,d\mathcal L^n(x)=\int_U \phi(x)\det \nabla u(x)\,d\mathcal L^n(x). \end{align*} Since $\phi\in C_c^\infty(U)$ was arbitrary, this is precisely convergence in the sense of distributions on $U$. The proof is complete. [/step]