Definition: Let $(X,\mu)$ be a measure space and $\phi=(\phi_1,\cdots,\phi_d)\in L^{\infty}(X)$. The joint essential range of $\phi$ is the set $\mathcal{C}(\phi)$ which is consisting of all $z = (z_1,\cdots,z_d)\in \mathbb{C}^d$ such that for every $\varepsilon>0$ $$\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)-z_i|<\varepsilon \right\}\right)>0 .$$
Assume that $$r=\max\left\{\sum_{i=1}^d|z_i|^2; (z_1,\cdots,z_d)\in \mathcal{C}(\phi)\right\}.$$ i.e. $$ r:=\max\left\{\sum_{i=1}^d|z_i|^2\,;\;\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)-z_i|<\varepsilon \right\}\right)>0,\;\text{for every}\;\varepsilon>0\right\}.$$ I want to understand why the following two facts hold? $$\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)|^2>r \right\}\right)=0 ,$$ and $$ \int_X\left(\displaystyle\sum_{k=1}^d|\phi_k|^2\right)|f|^2d\mu\leq r \int_X|f|^2d\mu.$$
Note that the two facts figure in this paper (page 6 and 7).
Observe that by definition, $ z=(z_1,\ldots,z_d)\notin \mathcal{C}(\phi), $ if and only if there exists a neiborhood $U_z$ of $z$ such that $$ \mu\left(\{t\in X\;|\;\phi(t)\in U_z\}\right) =0. $$ We can see that if $w\in U_z$, then $w$ has a neighborhood that satisfies the above condition, namely $U_z$. This implies that $w\notin\mathcal{C}(\phi)$ for all $w\in U_z$, i.e. $U_z \subset \Bbb C^d\setminus \mathcal{C}(\phi).$ From this, it follows that $$\Bbb C^d\setminus\mathcal{C}(\phi)=\bigcup_{z\notin\mathcal{C}(\phi)}U_z.$$ We can see that $\mathcal{C}(\phi)$ is closed, and since $r:=\max\limits_{\lambda \notin \cup_z U_z}|\lambda|^2=\max\limits_{\lambda \in \mathcal{C}(\phi)}|\lambda|^2$ is a bounded constant, it is contained in $$B(0,\sqrt{r})=\{(z_1,z_2,\ldots,z_d)\in\Bbb C^d\;|\;|z|^2=\sum_j |z_j|^2\le r\}.$$ This establishes that $\mathcal{C}(\phi)$ is compact. Since $\Bbb C^d\setminus\mathcal{C}(\phi)$ is a second-countable space as a subspace of $\Bbb C^d$, there exists a countable family $\{z_i\}$ such that $$ \Bbb C^d\setminus\mathcal{C}(\phi)=\bigcup_{i\in\Bbb N}U_{z_i}. $$ This gives $$\begin{eqnarray} \mu\left(\left\{t\in X\;\big|\;\sum_k |\phi_k(t)|^2>r\right\} \right)&\le& \mu\left(\left\{t\in X\;\big|\;\phi(t)\notin\mathcal{C}(\phi)\right\} \right)\\&\le&\mu\left(\bigcup_{i\in\Bbb N}\left\{t\in X\;\big|\;\phi(t)\in U_{z_i}\right\} \right)\\ &\le&\sum_{i\in\Bbb N}\mu\left(\left\{t\in X\;\big|\;\phi(t)\in U_{z_i}\right\} \right)=0. \end{eqnarray}$$ This in turn implies that $$ \sum_k |\phi_k(t)|^2\le r $$ for $\mu$-almost every $t\in X$. This gives $$ \int_X\left(\displaystyle\sum_{k}|\phi_k(t)|^2\right)|f(t)|^2d\mu(t)\leq r \int_X|f(t)|^2d\mu(t). $$