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](https://math.stackexchange.com/questions/3078303/joint-essential-range-of-varphi-in-l-inftyx-mud#3078456) 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 $t\in X$.



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


