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?

  • $\begingroup$ The answer to the question in your link is invalid. $\endgroup$ Commented Apr 29, 2023 at 7:49

1 Answer 1

  • 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.

  • $\begingroup$ Thanks for your answer , instead for $f$ being Lebesgue Measurable is true that $X$ could be any subset of $\mathbb{R} ?$ $\endgroup$ Commented Apr 29, 2023 at 8:01
  • $\begingroup$ and another question ,sorry if I am bothering you your example is correct and you are completely right . In the link the guy says the first set is open , but it is open even if I change the set X with another set not borel will still be open and so i will arrive to the same conclusion , that is my problem . What is the missing point here ? $\endgroup$ Commented Apr 29, 2023 at 8:23
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    $\begingroup$ For your first question, yes as far as $X$ is Lebesgue measurable. For your second question, since the Borel $\sigma -$algebra is generated by open sets, open sets are Borel sets, so no chance to find an open set that is not a Borel set... @Dsrksidemath $\endgroup$
    – Surb
    Commented Apr 29, 2023 at 8:56

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