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Assume a real-valued function f whose domain is Lebesgue-measurable. If the f inverse of every real open set is measurable, then the f inverse of every real Borel set is also measurable.

I had an incorrect solution since I assumed that every Borel set is an open set and used this idea to prove the implication. How should I go about this?

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    The proof may be more or less complex depending on the theorems you know. There is a theorem that says if $\mathcal{F}$ is a sigma algebra on the domain of a function $f$ and if $f^{-1}(A) \in \mathcal{F}$ for all $A$ in some collection of sets $C$, then $f^{-1}(A) \in \mathcal{F}$ for all $A \in \sigma(C)$. – Michael Nov 26 '22 at 03:50
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    Following up on the comment by @Michael, a useful technique for this and similar results is sometimes called the good sets principle. See this answer and this google search. – Dave L. Renfro Nov 26 '22 at 08:55

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