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If $f$ is a continuous function on $X$ and E a Lebesgue measurable set, can we conclude that $f^{-1}(E)$ is measurable?

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    Actually the example in the link provided by Chris is used to proof that "measurable" is really more general than "Borel": there is a measurable set $B$ such that its inverse image by a continuous map is not measurable so is not Borel. The same function in that post serves to show that measurability is not a topological property.2012-09-28

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