[proofplan] The proof has two ingredients. First, the $L^p$-boundedness of the gradients gives finite $p$-moments for the generated Young measure and the barycentre identity $\nabla u(x)=\int_{\mathbb R^n}\xi\,d\nu_x(\xi)$ for a.e. $x\in U$. Second, once the barycentre is known, [Jensen's inequality](/theorems/9) applied pointwise to the probability measure $\nu_x$ gives the desired inequality. The growth assumption ensures that $\varphi$ is integrable from above against $\nu_x$, while convexity supplies an affine lower bound, so the right-hand side is a well-defined extended real number. [/proofplan]

[step:Use weak boundedness to obtain finite moments of the Young measure]Since $u_k\rightharpoonup u$ in $W^{1,p}(U;\mathbb R)$, the sequence $(u_k)_{k=1}^{\infty}$ is bounded in $W^{1,p}(U;\mathbb R)$. Hence the gradient sequence \begin{align*} \nabla u_k:U\to\mathbb R^n \end{align*} is bounded in $L^p(U;\mathbb R^n)$. Define \begin{align*} M:=\sup_{k\in\mathbb N}\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x). \end{align*} Then $M<\infty$. For each $R>0$, define the bounded [continuous function](/page/Continuous%20Function) $\psi_R:\mathbb R^n\to[0,R]$ by \begin{align*} \psi_R(\xi):=\min\{|\xi|^p,R\}. \end{align*} By the defining convergence property of the Young measure generated by $(\nabla u_k)$, applied to $\psi_R$, \begin{align*} \int_U\int_{\mathbb R^n}\psi_R(\xi)\,d\nu_x(\xi)\,d\mathcal L^n(x) = \lim_{k\to\infty}\int_U \psi_R(\nabla u_k(x))\,d\mathcal L^n(x). \end{align*} Since $\psi_R(\nabla u_k(x))\le |\nabla u_k(x)|^p$ for every $x\in U$ and every $k\in\mathbb N$, it follows that \begin{align*} \int_U\int_{\mathbb R^n}\psi_R(\xi)\,d\nu_x(\xi)\,d\mathcal L^n(x)\le M. \end{align*} Letting $R\to\infty$ and using the [monotone convergence theorem](/theorems/509) for the increasing family $(\psi_R)_{R>0}$ gives \begin{align*} \int_U\int_{\mathbb R^n}|\xi|^p\,d\nu_x(\xi)\,d\mathcal L^n(x)\le M. \end{align*} Therefore \begin{align*} \int_{\mathbb R^n}|\xi|^p\,d\nu_x(\xi)<\infty \end{align*} for $\mathcal L^n$-a.e. $x\in U$.[/step]

[guided]The first point is that the Young measure must have enough integrability for the expression involving $\varphi$ to make sense. [Weak convergence](/page/Weak%20Convergence) in $W^{1,p}(U;\mathbb R)$ implies boundedness in that [Banach space](/page/Banach%20Space), so the gradients are uniformly bounded in $L^p(U;\mathbb R^n)$. We record this by defining \begin{align*} M:=\sup_{k\in\mathbb N}\int_U |\nabla u_k(x)|^p\,d\mathcal L^n(x), \end{align*} and the weak boundedness gives $M<\infty$. The function $\xi\mapsto |\xi|^p$ is not bounded, so it is not one of the basic bounded continuous test functions in the Young-measure definition. To reach it, we truncate. For each $R>0$, define the map $\psi_R:\mathbb R^n\to[0,R]$ by \begin{align*} \psi_R(\xi):=\min\{|\xi|^p,R\}. \end{align*} This map is bounded and continuous, so the defining property of the generated Young measure applies: \begin{align*} \int_U\int_{\mathbb R^n}\psi_R(\xi)\,d\nu_x(\xi)\,d\mathcal L^n(x) = \lim_{k\to\infty}\int_U \psi_R(\nabla u_k(x))\,d\mathcal L^n(x). \end{align*} For every $k\in\mathbb N$ and every $x\in U$, the truncation satisfies \begin{align*} \psi_R(\nabla u_k(x))\le |\nabla u_k(x)|^p. \end{align*} Thus \begin{align*} \int_U\int_{\mathbb R^n}\psi_R(\xi)\,d\nu_x(\xi)\,d\mathcal L^n(x)\le M. \end{align*} As $R\to\infty$, the functions $\psi_R$ increase pointwise to $\xi\mapsto |\xi|^p$. The monotone convergence theorem therefore yields \begin{align*} \int_U\int_{\mathbb R^n}|\xi|^p\,d\nu_x(\xi)\,d\mathcal L^n(x)\le M. \end{align*} Since this double integral is finite, the inner integral is finite for $\mathcal L^n$-a.e. $x\in U$: \begin{align*} \int_{\mathbb R^n}|\xi|^p\,d\nu_x(\xi)<\infty. \end{align*} This is the pointwise finite-moment fact needed later when we integrate a function with $p$-growth.[/guided]

[step:Identify the weak limit gradient as the barycentre] By the barycentre property for $L^p$ Young measures under weak $L^p$ convergence, applied to the sequence \begin{align*} \nabla u_k:U\to\mathbb R^n, \end{align*} the weak limit of $(\nabla u_k)$ in $L^p(U;\mathbb R^n)$ agrees with the barycentre of the generated Young measure. Since $u_k\rightharpoonup u$ in $W^{1,p}(U;\mathbb R)$, we have \begin{align*} \nabla u_k\rightharpoonup \nabla u \quad \text{in } L^p(U;\mathbb R^n). \end{align*} Therefore, for $\mathcal L^n$-a.e. $x\in U$, \begin{align*} \nabla u(x)=\int_{\mathbb R^n}\xi\,d\nu_x(\xi). \end{align*} Here the vector integral is understood componentwise: for each $i\in\{1,\dots,n\}$, \begin{align*} \partial_{x_i}u(x)=\int_{\mathbb R^n}\xi_i\,d\nu_x(\xi) \end{align*} for $\mathcal L^n$-a.e. $x\in U$. The integrals are finite because $p>1$ and the previous step gives finite $p$-moments for $\nu_x$ for a.e. $x$. [/step]

[step:Check that the convex integrand is meaningful against each fibre measure] Because $\varphi:\mathbb R^n\to\mathbb R$ is finite and convex, it is continuous and Borel measurable. Also, a finite [convex function](/page/Convex%20Function) on $\mathbb R^n$ admits an affine lower support: there exist $a\in\mathbb R^n$ and $b\in\mathbb R$ such that \begin{align*} \varphi(\xi)\ge a\cdot \xi+b \end{align*} for every $\xi\in\mathbb R^n$. Define the positive and negative part maps $\varphi^+:\mathbb R^n\to[0,\infty)$ and $\varphi^-:\mathbb R^n\to[0,\infty)$ by \begin{align*} \varphi^+(\xi):=\max\{\varphi(\xi),0\} \end{align*} and \begin{align*} \varphi^-(\xi):=\max\{-\varphi(\xi),0\} \end{align*} for $\xi\in\mathbb R^n$. Then $\varphi=\varphi^+-\varphi^-$ and $|\varphi|=\varphi^++\varphi^-$. Let $x\in U$ be such that \begin{align*} \int_{\mathbb R^n}|\xi|^p\,d\nu_x(\xi)<\infty. \end{align*} Since $p>1$ and $\nu_x$ is a probability measure, the first moment is finite: \begin{align*} \int_{\mathbb R^n}|\xi|\,d\nu_x(\xi)<\infty. \end{align*} The upper growth assumption gives \begin{align*} \varphi(\xi)\le C(1+|\xi|^p), \end{align*} so the positive part $\varphi^+$ is $\nu_x$-integrable. The affine lower bound gives \begin{align*} \varphi^-(\xi)\le |a|\,|\xi|+|b|, \end{align*} so the negative part $\varphi^-$ is also $\nu_x$-integrable. Hence \begin{align*} \int_{\mathbb R^n}\varphi(\xi)\,d\nu_x(\xi) \end{align*} is a well-defined finite real number for $\mathcal L^n$-a.e. $x\in U$. [/step]

[step:Apply Jensen's inequality to each probability measure $\nu_x$] Let $x\in U$ be a point for which the finite moment conclusion, the barycentre identity, and the integrability conclusion above all hold. This is true for $\mathcal L^n$-a.e. $x\in U$. Since $\nu_x$ is a Borel probability measure on $\mathbb R^n$, since $\varphi:\mathbb R^n\to\mathbb R$ is convex and $\nu_x$-integrable, and since \begin{align*} \nabla u(x)=\int_{\mathbb R^n}\xi\,d\nu_x(\xi), \end{align*} [Jensen's inequality](/theorems/1977) for probability measures gives \begin{align*} \varphi(\nabla u(x)) = \varphi\left(\int_{\mathbb R^n}\xi\,d\nu_x(\xi)\right) \le \int_{\mathbb R^n}\varphi(\xi)\,d\nu_x(\xi). \end{align*} Since the exceptional set has $\mathcal L^n$-measure zero, the inequality holds for $\mathcal L^n$-a.e. $x\in U$. This proves the theorem. [/step]