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Let $f: \mathbb{R} \to \mathbb{R}$ be continuous and $g: \mathbb{R} \to \mathbb{R}$ Lebesgue measurable. Is $f \circ g$ Borel measurable?

I think that the answer to this question is no. I know that the hypothesis implies that $f \circ g$ is Lebesgue measurable, but I am sure that there is a counter-example to show that $f \circ g$ is not Borel measurable, since the Borel sigma algebra is a proper subset of the Lebesgue sigma algebra.

Note: In my textbook, a function is measurable with respect to a sigma algebra $\mathcal{E}$ if and only if $[f > \alpha] = \{x : f(x) > \alpha\} \in \mathcal{E}$ for all $\alpha \in \mathbb{R}$.

My current idea is to find a Lebesgue measurable set $D$ that is not Borel measurable. Then define $g(x) = \chi_{_{D}}$, which is Lebesgue measurable, as $D$ is Lebesgue measurable. Then $f(x) = x$ is continuous and $f \circ g = g$, but $g$ is not Borel measurable.

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  • $\begingroup$ Does $X=\mathbb{R}$? $\endgroup$ – confused_wallet Mar 10 at 3:02
  • $\begingroup$ Yes, that was a typo. Fixed. $\endgroup$ – johnny133253 Mar 10 at 3:05
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    $\begingroup$ I think your idea is a good one! $\endgroup$ – confused_wallet Mar 10 at 3:07
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    $\begingroup$ Your current idea is correct. $\endgroup$ – Ramiro Mar 10 at 15:31

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