Let $\mathcal{A}$ be a $\sigma$-algebra in $X$ and $f:X \to \mathbb{R}^N$ a measurable function. Prove that $G\subset \mathbb{R}^N$ open $\implies f^{-1}(G) \in\mathcal{A}$.
My attempet: If we write $f(x)=(f_1(x),\dots, f_N(x))$ where $f_i:X \to \mathbb{R}$, we have that $$ f_i=P_i \circ f$$ where $P_i:\mathbb{R}^N \to \mathbb{R}$ is the projection of the $i$-coordinate. Since $P_i$ is continuous and an open map it's true that $P_i(G)=A \subset \mathbb{R}$ open and then follows that $$f_i^{-1}(A)=f^{-1}(P^{-1}_i(A))). $$ By the description of the open sets in $\mathbb{R}$ we can write $A$ as an disjoint countable union of open intervals and maybe with some work we can prove by this that $f^{-1}_i(A) \in \mathcal{A}$. The real problem here is that it's not true that $P^{-1}_i(A)=G$. There is a way to make this argument work?
Any open set $G$ in $\mathbb R^{n}$ is a union of sets of the type $(a_1,b_1)\times (a_2,b_2)\times \cdots \times (a_n,b_n)$ with $a_i, b_i \in \mathbb Q$ for each $i$. Since $f_i^{-1}(a_i,b_i) \in \mathcal A$ for each $i$ it follows that $f^{-1}(G) \in \mathcal A$.