Let $(X,\mu)$ be a measure space and $f_1,\cdots,f_d\in L^\infty(X)$.

Set $$g:=\displaystyle\sum_{k=1}^d|f_k|^2\;\;\text{and}\;\;c:=\|g\|_\infty.$$

Let $\sigma:=\{a_1,b_1,\cdots,a_d,b_d\}$ be such that $a_i,b_i\in \mathbb{Q}_+$ for all $i$. Set $$S_\sigma=\left\{x \in X;\; \left[\Re(f_k(x))\right]^2>a_k,\; \left[\Im(f_k(x))\right]^2>b_k,\;\;k=1,\cdots,d\right\}.$$ and $$\mathfrak{F}=\left\{\{a_1,b_1,\cdots,a_d,b_d\}\subset \mathbb{Q}_+^{2d};\;\;\sum_{k=1}^d (a_k+b_k) > c-\varepsilon,\;\forall \varepsilon>0\right\}.$$

By (1), we have $A_\varepsilon \subset \bigcup_{\sigma \in \mathfrak{F}} S_\sigma,\;\forall \varepsilon>0$ with $A_\varepsilon=\left\{x\in X;\; g(x)>c-\varepsilon\right\}.$

Since $\mu(A_\varepsilon)>0$ and $\mathfrak{F}$ is countable, then there exists $\sigma_0\in \mathfrak{F}$ such that $\mu(S_{\sigma_0})>0$. Why by subdivising $S_{\sigma_0}$, we can find $S\subset S_{\sigma_0}$ such that $\mu(S)>0$ and $\Re(f_j),\;\Im(f_j)$ keep a constant sign in $S$ for all $j\in\{1,\cdots,d\}$?

My attempt We can subdivise $S_{\sigma_0}$ as $$S_{\sigma_0}=B_1\sqcup B_2 \sqcup B_3 \sqcup B_4,$$ with

$B_1:=\{x\in X;\;\Re(f_1(x))\ge 0,\;\Im(f_1(x)) \ge 0\}$;

$B_2:=\{x\in X;\;\Re(f_1(x))\ge 0,\;\Im(f_1(x)) < 0\}$;

$B_3:=\{x\in X;\;\Re(f_1(x))<0,\;\Im(f_1(x)) \ge 0\}$;

$B_4:=\{x\in X;\;\Re(f_1(x))<0,\;\Im(f_1(x)) < 0\}$.

So there exists $k_0\in \{1,\cdots,4\}$ such that $\mu(B_{k_0})>0$. But I'm facing difficulties to end the proof.


Define for $k\in \left\{1,\dots,d\right\}$ the sets

$B_{1,k}:=\{x\in X;\;\Re(f_k(x))\geqslant 0,\;\Im(f_k(x)) \ge 0\}$;

$B_{2,k}:=\{x\in X;\;\Re(f_k(x))\geqslant 0,\;\Im(f_k(x)) < 0\}$;

$B_{3,k}:=\{x\in X;\;\Re(f_k(x))<0,\;\Im(f_k(x)) \geqslant 0\}$;

$B_{4,k}:=\{x\in X;\;\Re(f_k(x))<0,\;\Im(f_k(x)) < 0\}$.

For $v=\left(v_1,\dots,v_d\right)\in\left\{1;2;3;4\right\}^d$, define $E_v:= \bigcap_{k=1}^dB_{v_k,k}$. The family $\left(E_v\right)_{v\in\left\{1;2;3;4\right\}^d}$ is pairwise disjoint and its union is $S_{\sigma_0}$ hence there is some $v\in\left\{1;2;3;4\right\}^d$ such that the measure of $E_v$ is positive. On $E_v$, the functions $\Re\left(f_k\right)$ and $\Im\left(f_k\right)$ have a constant sign for each fixed $k$.


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