As we know, Lebesgue measurable function f is defined as all preimages of the open sets are Lebesgue measurable. So I'm wondering if this definition guarantee that the preimage of any Lebesgue measurable set is Lebesgue measurable.
2026-05-06 01:20:02.1778030402
any countereample in measurable function
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No - if $f$ is injective and sends any positive-measure set $B$ to null set $A$, then fix some nonmeasurable subset $C$ of $B$; the image $D$ of $C$ is a subset of $A$, which is null, and hence measurable. So $D$ is measurable with non-measurable preimage.
This is the only obstacle that can arise: if $f$ is measurable and the $f$-preimage of any null set is null, then the $f$-preimage of any measurable set is measurable (use the fact that measurable sets differ from a $G_\delta$ set by a null set).