I have a question regarding the example of a set that is Lebesgue measurable, but not a Borel set on $\mathbb{R}$ given in https://www.math3ma.com/blog/lebesgue-but-not-borel. Wouldn't it be the case that we have $N = f(f^{-1}(N)) = (f^{-1})^{-1}(f^{-1}(N))$ measurable following this analysis? We namely have that $f^{-1}$ is continuous, hence measurable and the inverse mappings of Lebesgue measurable sets under $f^{-1}$ are hence Lebesgue measurable..
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2$\begingroup$ The preimage of a measurable set under a continuous function need not be measurable. $\endgroup$– Umberto P.Oct 15, 2021 at 17:13
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$\begingroup$ Why? All continuous functions are Lebesgue measurable right? $\endgroup$– SBMSOct 15, 2021 at 17:18
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$\begingroup$ Pre-images of open sets are measurable. $\endgroup$– Umberto P.Oct 15, 2021 at 17:39
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$\begingroup$ Ah, thanks, I see it now. $\endgroup$– SBMSOct 15, 2021 at 17:53
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Umberto P. answered it in the comments:
The preimage of a measurable set under a continuous function need not be measurable. Pre-images of open sets are measurable.
