Suppose $\Omega\subset \mathbb R^N$ is bounded and lipschitz boundary. Suppose $u_n, u\in H^1(\Omega)$ such that $u_n\to u$ weakly in $H^1$. Then can I conclude that $$\liminf_{n\to\infty}\int_{\Omega\setminus T_n} |\nabla u_n|^2dx\geq \int_\Omega |\nabla u|^2dx$$ or not? (where $T_n\subset \Omega$ such that $|T_n|\to 0$).

I know if $\nabla u_n$ is $L^2$ equi-integrable then this is a trivial result. But what if I don't have equi-integrability? Can this still be true?

Thank you!

PS: Actually what I need is a weaker version of this. In my problem I only require $T_n:=\{x\in\Omega, \text{dist}(x,\partial\Omega)<1/n\}$, i.e., just around the boundary, and $\max{(\|T[u_n]\|_{L^\infty(\partial\Omega)},\|T[u]\|_{L^\infty})}<\infty$. But if in general is true is much better!

This is more of a problem relating to weak convergence in $L^{2}$. Since $f_{n}\to f$ weakly in $H^{1}\left(\Omega\right)$, we know $g_{n}:=\nabla f_{n}\to\nabla f=:g$ weakly in $L^{2}\left(\Omega\right)$.
Now, we require the following fact: If $x_{n}\rightharpoonup x$ weakly and $y_{n}\to y$ in norm, then $\left(x_{n}\right)_{n}$ is bounded (lets say by $C>0$), hence \begin{eqnarray*} \left|\left\langle x_{n},y_{n}\right\rangle -\left\langle x,y\right\rangle \right| & \leq & \left|\left\langle x_{n},y_{n}\right\rangle -\left\langle x_{n},y\right\rangle \right|+\left|\left\langle x_{n},y\right\rangle -\left\langle x,y\right\rangle \right|\\ & \leq & C\cdot\left\Vert y_{n}-y\right\Vert +\left|\left\langle x_{n},y\right\rangle -\left\langle x,y\right\rangle \right|\\ & \xrightarrow[n\to\infty]{} & 0. \end{eqnarray*}
I leave it to you to verify that $g\cdot\chi_{\Omega\setminus T_{n}}\to g$ in $L^{2}\left(\Omega\right)$, where $\chi_{\Omega\setminus T_{n}}$ is the characteristic function/indicator function of $\Omega\setminus T_{n}$.
Canceling $\left\Vert g\right\Vert _{L^{2}}$ (the claim is trivial if $\left\Vert g\right\Vert _{L^{2}}=0$), we get $$\left\Vert g\right\Vert _{L^{2}}\leq\liminf_{n\to\infty}\left\Vert \chi_{\Omega\setminus T_{n}}\cdot g_{n}\right\Vert _{L^{2}},$$ which (by squaring) yields exactly the desired estimate.