Why $A_\varepsilon \subset \bigcup_{\sigma \in \mathfrak{F}} S_\sigma,\;?$ 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
$$\mathfrak{F}=\left\{\{a_1,b_1,\cdots,a_d,b_d\}\subset \mathbb{Q}_+^{2d};\;\;\sum_{k=1}^d (a_k+b_k) > c-\varepsilon,\;\forall \varepsilon>0\right\}.$$

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


Let $x\in A_\varepsilon$. Choose non negative rational numbers $a_i$ and $b_i$ such that $$ 0\lt     \left[\Re(f_i(x))\right]^2-  a_i\lt \frac{ \varepsilon}{2d}   \mbox{ and }   0\lt     \left[\Im(f_i(x))\right]^2 -b_i \lt \frac{ \varepsilon}{2d}          .$$
Let $\sigma:=\{a_1,b_1,\cdots,a_d,b_d\}$. Then $\sigma$ belongs to $\mathfrak{F}$ and $x$ belongs to $S_\sigma$. 
