Let $f$ be a continuous function on a set $E$. Is it always true that $f^{-1}(A)$ is always measurable if $A$ is measurable?
Is this correct?
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1What $\sigma$-algebras? It's true for the Borel $\sigma$-algebras, but for unrelated $\sigma$-algebras, there's no reason for a continuous function to be measurable. – Daniel Fischer Oct 17 '13 at 21:27
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Would you please give me the proof? – Yang Oct 17 '13 at 21:33
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2The usual counterexample is to take the Cantor Function $\phi$ and define $\psi(x)=x+\phi(x)$. $\psi$ is one-to-one and maps the measure-zero Cantor set onto a set $A$ of positive measure. Now consider the inverse of $\psi$ and a non-measurable subset of $A$. c.f. Gelbaum and Olmsted, Counterexamples in Analysis, example 8.16. – David Mitra Oct 17 '13 at 22:09
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1The above of course is meant for the case where "measurable" means "Lebesgue measurable". – David Mitra Oct 17 '13 at 22:16
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Hey David,thank you. I know we should use that Cantor function. Actually I searched that book but I can't find example 8.16. – Yang Oct 17 '13 at 23:49
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5Does this answer your question? Pre-image of a measurable set A is always measurable? – user0 Feb 20 '21 at 19:30
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Negative answer by David Mitra.
Take the Cantor Function $\phi$ and define $\psi(x)=x+\phi (x)$. The function $\psi$ is one-to-one and maps the measure-zero Cantor set onto a set $E$ of positive measure. Let $B$ be a non-measurable subset of $E$. (Every set of positive measure contains a nonmeasurable set.) The set $A=\psi^{-1}(B)$ is a subset of the Cantor set, and is therefore measurable. This gives a counterexample ($ f=\psi^{-1}$).
c.f. Gelbaum and Olmsted, Counterexamples in Analysis, example 8.16.
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