# 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 $$t\in X$$.

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 $$t\in X$$.

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.
• $\phi^{-1}(...)$ – Fedor Petrov Jan 20 at 16:27