Question is about the proof of the theorem 2 in section $6.4.1$ page $347$ of Evans's PDE (second edition). The goal is to prove;
Weak maximum principle for $c \geq 0$:. Asume $u \in C^2(\Omega)\cap C(\overline{\Omega}) $ and $c \geq 0$ in $\Omega$. If $L(u) \leq 0$ in $\Omega$ then $max_{\overline{\Omega}} u \leq max_{\delta\Omega} u^+$
(For simplicity lets assume $L(u) = -\Delta u + cu$)
The proof defines $V = \{x \in \Omega | u(x) > 0\}$. Then by the previous theorem (or in this simplified case $\Delta u \geq 0$ (sub-harmonic) in $V$) we have that $max_{\overline{V}} u = max_{\delta V} u$. How to conclude then that $max_{\overline{V}} u = max_{\delta V} u = max_{\delta\Omega} u^+$.
I don't understand how to obtain the equality $max_{\delta V} u= max_{\delta\Omega}u^+$ in the proof.
In the notations of Evans:
\begin{align*} \max_{\partial V} u&=\max\left\{\max_{\partial V\cap U} u,\max_{\partial V\cap \partial U} u\right\}\quad\text{(because $\partial{V}\subset\overline{U}$)}\\\\ &=\max_{\partial V\cap \partial U} u\quad\text{(because $u=0$ on $\partial V\cap U$)}\\\\ &=\max_{\partial U} u\quad\text{(because $u\leq 0$ on $\partial U\setminus \partial V$)}\\\\ &=\max_{\partial U} u^+\quad\text{(because $u^+=\max\{u,0\}$)}\\\\ \end{align*}