## 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 $$t\in X$$.