Prove that $\mu \left(\left\{t\in X\,;\;\sum_{i=1}^d|\phi_i(t)|^2>r \right\}\right)=0$

Let $ (X,\mu)$ be a measure space and $ \phi=(\phi_1,\cdots,\phi_d)\in L^{\infty}(X)$ .

Let $ $ r=\max\left\{\sum_{i=1}^d|z_i|^2; (z_1,\cdots,z_d)\in \mathcal{C}(\phi)\right\},$ $ where $ \mathcal{C}(\phi)$ 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 .$ $

Why $ $ \sum_{i=1}^d |\phi_i(t)|^2\le r $ $ for $ \mu$ -almost every $ t\in X$ .