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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?

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Yes, it is perfectly clear and correct! Well done!

Only the fact that every open subset is the countable union of open boxes might need a proof, if you haven't proven this fact before.

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  • $\begingroup$ Thanks for the comment! Such result has already been proven. $\endgroup$ – KnowledgeSeeker Jun 2 at 22:48

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