# Let $f:\Omega\to\textbf{R}^{m}$ be a function. Then $f$ is measurable if and only if $f^{-1}(B)$ is measurable for every open box $B$.

Let $$\Omega$$ be a measurable subset of $$\textbf{R}^{n}$$, and let $$f:\Omega\to\textbf{R}^{m}$$ be a function. Then $$f$$ is measurable if and only if $$f^{-1}(B)$$ is measurable for every open box $$B$$.

My solution

Let us prove the implication $$(\Rightarrow)$$ first.

If $$f$$ is measurable, the $$f^{-1}(V)$$ is measurable for every open subset $$V\subseteq\textbf{R}^{m}$$. In particular, since the open box $$B$$ is open, we conclude that $$f^{-1}(B)$$ is measurable for every open box $$B$$.

Let us prove the implication $$(\Leftarrow)$$ now.

Let us consider an open subset $$V\subseteq\textbf{R}^{m}$$. Thus we can express $$V$$ as a countable union of open boxes $$(B_{j})_{j\in J}$$. Thus we get \begin{align*} V = \bigcup_{j\in J}B_{j} \Rightarrow f^{-1}(V) = f^{-1}\left(\bigcup_{j\in J}B_{j}\right) = \bigcup_{j\in J}f^{-1}(B_{j}) \end{align*}

Since $$f^{-1}(B_{j})$$ is measurable and countable union of measurable sets is measurable, the result holds.

Could someone please verify if the wording of my proof is satisfactory?