Let $(X,\mu)$ be a measure space and $f_1,\cdots,f_d\in L^\infty(X)$.
Set $$g:=\displaystyle\sum_{k=1}^d|f_k|^2\;\;\text{and}\;\;c:=\|g\|_\infty.$$ and $$A_\varepsilon=\left\{x\in X;\;\displaystyle\sum_{k=1}^d \left[\Re(f_k(x))\right]^2+\left[\Im(f_k(x))\right]^2>c-\varepsilon\right\}.$$
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 $$\mathfrak{F}=\left\{\{a_1,b_1,\cdots,a_d,b_d\}\subset \mathbb{Q}_+^{2d};\;\;\sum_{k=1}^d (a_k+b_k) > c-\varepsilon,\;\forall \varepsilon>0\right\}.$$
I want to show that $$A_\varepsilon \subset \bigcup_{\sigma \in \mathfrak{F}} S_\sigma,\;\forall \varepsilon>0\;?$$
If $x\in A_\varepsilon$ such that $\left[\Re(f_i(x))\right]$ and $\left[\Im(f_i(x))\right]$ are non-zero, then we can choose non negative rational numbers $a_i$ and $b_i$ such that $$ 0\lt \left[\Re(f_i(x))\right]^2- a_i\lt \frac{ \varepsilon}{2d} \mbox{ and } 0\lt \left[\Im(f_i(x))\right]^2 -b_i \lt \frac{ \varepsilon}{2d} .$$
Hence we can set $\sigma:=\{a_1,b_1,\cdots,a_d,b_d\}$. Then $\sigma$ belongs to $\mathfrak{F}$ and $x$ belongs to $S_\sigma$.
The problem if $x\in A_\varepsilon$ such that $\left[\Re(f_i(x))\right]=0$ or $\left[\Im(f_i(x))\right]=0$.
In the latter case the inclusion may fail because if $x$ belongs to some $S_\sigma$ then the real part and imaginary part of $f_i(x)$ are both positive. But $x\in A_\varepsilon$ do not impose this. In order to guarantee the inclusion, we should impose large inequalities instead of strict, that is, define
$$S_\sigma=\left\{x \in X;\; \left[\Re(f_k(x))\right]^2\geqslant a_k,\; \left[\Im(f_k(x))\right]^2\geqslant b_k,\;\;k=1,\cdots,d\right\}.$$ Then choose $a_i$ and $b_i$ such that $$ 0\leqslant \left[\Re(f_i(x))\right]^2- a_i\lt \frac{ \varepsilon}{2d} \mbox{ and } 0\leqslant \left[\Im(f_i(x))\right]^2 -b_i \lt \frac{ \varepsilon}{2d} . $$