Let $E \subset \mathbb{R^n}$ be a lebesgue-measurable set and let $f,g : E \to \mathbb{R}$ be lebesgue-mesurable functions.
Define $\Phi: E \to \mathbb{R^2}$ by $$ \Phi (x)= \begin{pmatrix} f(x) \\ g(x) \\ \end{pmatrix}. $$ Then, prove that $\Phi^{-1}(U)$ is lebesgue-measurable set for open set $U \in \mathbb{R^2}$.
Let $U\in \mathbb{R^2}$ be open and I have to prove $$ \Phi^{-1}(U)=\left\{ x \in E \ \Bigg| \ \Phi (x)= \begin{pmatrix} f(x) \\ g(x) \\ \end{pmatrix} \in U \right\} $$ is open, but I cannot know what I should do and I'm stacked.
I would like you to give me some ideas.
$U$ is a countable union of sets of the form $A\times B$ where $A$ and $B$ are open in $\mathbb R$. Hence the inverse image is a countable union of sets of the form $f^{-1}(A)\cap g^{-1}(B)$.
[Given any point $(x,y) \in U$ there is an open rectangle $(a,b)\times (c,d)$ containing $(x,y)$, contained in $U$ such that $a,b,c,d $ are all rational numbers. This gives you a countable family of open sets of the form $A \times B$ ($A,B$ open) whose union is $U$].