in this question Borel Measurability of a function with countable discontinuity points. $X$ is a Borel set , but I have $2$ questions $1)$ if $X$ is every subset of $\mathbb{R}$ I think the theorem is also valid we only used preimage of open set is an open set and countability so I don't see the point of being Borel . $2)$ And I also think that the function also Lebesgue measurable since borel sigma algebra is contained in lebesgue sigma algebra , so open set are also measurable in lebesgue sigma algebra . Are my statements correct?
2026-03-25 17:26:14.1774459574
regarding measurability of functions
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For $f$ being Borel Measurable, $X$ must be a Borel set. Indeed, take a (Lebesgue) measurable set $N$ set that is not a Borel set. Then, $f:N\to \mathbb R$ defined by $f(x)=1$ is continuous, However, $f^{-1}(\{1\})=N$ which is not Borel, and thus, $f$ is not Borel measurable.
Indeed, if $f$ is Borel measurable, then it's Lebesgue measurable.