I am trying to follow Evans' argument for Thm. 4.2.3 (i) in his book on Weak Convergence Methods for Nonlinear Partial Differential Equations. To keep things simple, let $B$ be the unit ball in $\mathbb{R}^2$, and suppose that $\{v_m\}_{m=1}^{\infty}\subset W^{1, 2}(B;\mathbb{R}^N)$ is a sequence of weak solutions to the following PDE: \begin{equation}\tag{1} -\triangle v_m=0\quad\text{in }B. \end{equation}
Suppose further that $v_m\rightharpoonup v$ in $W^{1, 2}(B;\mathbb{R}^N)$. Then we can pass to a sub-sequence, if necessary, such that $v_m$ converges to $v$ quasi-uniformly with respect to $p\text{-cap}$ for any $p\in [1, 2)$. That is, given $\delta>0$, there exists a relatively closed set $E_{\delta}\subset B$ such that \begin{equation} p\text{-cap}(B\setminus E_{\delta})<\delta\quad\text{and}\quad v_m\rightrightarrows v\text{ on }E_{\delta}. \end{equation}
Henceforth, fix $p\in[1, 2)$. The reduced defect measure $\theta$ is defined as \begin{equation} \theta(E)\equiv\limsup_{m\rightarrow\infty}\int_E|Dv_m-Dv|^2\ \mathrm{d}y\quad(E\subset B\text{ and Borel}). \end{equation}Given $\delta>0$, I want to show that $\theta(E_{\delta})=0$.
My work
Given $\delta>0$, there is a set $E_{\delta}$ on which we have uniform convergence of $v_m$ to $v$ and $p\text{-cap}(B\setminus E_{\delta})<\delta$. Thus, there exists a $\phi\in C_c^{\infty}(B)$ such that $\phi\geq 1$ on $B\setminus E_{\delta}$, so I used the test function \begin{equation} \varphi=(1-\phi)(v_m-v) \end{equation}in (1) to arrive at \begin{equation} \int_{E_{\delta}}|Dv_m|^2(1-\phi)\ \mathrm{d}y=\int_{E_{\delta}}D_iv_m^{\alpha} D_iv^{\alpha}(1-\phi)\ \mathrm{d}y+\int_BD_iv_m^{\alpha}D_i\phi(v_m-v)^{\alpha}\ \mathrm{d}y \end{equation}Upon passing to a further sub sequence, if necessary, we can assume that $v_m-v\rightarrow 0$ in $L^2$. The weak convergence of $Dv_m $ to $Dv$ in $L^2$ and the strong convergence of $D_i\phi(v_m-v)^{\alpha}$ to zero in $L^2$ imply that \begin{equation} \lim_{m\rightarrow\infty}\int_{E_{\delta}}|Dv_m|^2(1-\phi)\ \mathrm{d}y=\int_{E_{\delta}}|Dv|^2(1-\phi)\ \mathrm{d}y \end{equation}or \begin{equation} \limsup_{m\rightarrow\infty}\int_{E_{\delta}}|Dv_m-Dv|^2(1-\phi)\ \mathrm{d}y=0 \end{equation}I am not sure how to show $\theta(E_{\delta})=0$ from here.