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

Question

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 $t\in X$.

Observe from this answer 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 $$\Bbb C^d\setminus\mathcal{C}(\phi)=\bigcup_{z\notin\mathcal{C}(\phi)}U_z.$$

Since $\Bbb C^d$ is a second-countable space, 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 $t\in X$.

Answer 1

It suffices to prove that, given $\rho>t$, $g(t) \leqslant \rho$ for almost all $t$, where $g$ is your sum of squares. Assume the contrary. Then the $\phi$-preimage of the set $\Omega=\{(z_1,\dots,z_d):\sum z_i^2 >\rho\} $ has positive measure. The set $\Omega$ is a countable union of compact sets lying in $\Omega$. Thus there exists such a compact set $K$ whose $\phi$-preimage has a positive measure. If any point of $K$ has a neighborhood whose preimage has zero measure, take a finite subcover to get a contradiction. In other words, $K$ contains a point in $C(\phi) $. Contradiction.