# Preimages of Jordan-measurable sets

When is the preimage of a Jordan-measurable set Jordan-measurable?

In particular, is continuity sufficient? Piecewise continuity with finitely many pieces?

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Continuity of a function is not sufficient to guarantee that the preimage of a Jordan measurable set is Jordan measurable. Note that for bounded sets Jordan measurability is the same as saying that the boundary has Lebesgue measure $0$. Consider the Devil's ski slope homeomorphism, $f:[0,1]\to[0,2]$, that maps the standard Cantor set, $C$, to a fat Cantor set of measure $1$. So the inverse function $f^{-1}:[0,2]\to[0,1]$ maps the fat Cantor set of measure $1$ to the standard cantor set. So $(f^{-1})^{-1}(C)=f(C)$ is the fat Cantor set of measure $1$. The boundary of the standard Cantor set is itself which has measure $0$ and is hence Jordan measurable. The boundary of the fat Cantor set is itself but has positive measure and hence is not Jordan measurable.
One way to ensure that the preimage of a Jordan measurable set is Jordan measurable on bounded sets is to ensure that the function to which you are taking the preimage under is a homeomorphism whose inverse has the Luzin N property. More precisely if $f:A\to B$ is a homeomorphism, whose inverse has the Luzin N property, where $A,B\subset\mathbb{R}^{n}$ are open then the preimage of every bounded Jordan measurable subset, $E$, of $B$ whose boundary is contained in $B$ and is such that the boundary of $D=f^{-1}(E)$ is in $A$ is a bounded Jordan measurable subset of $A$. To make sure the homeomorphism has this property one can demand that the inverse be Lipschitz. Also if we demand $f$ be a $C^{1}$ diffeomorphism then $f^{-1}$ has the Luzin N property.
The reason I want these conditions is because if $E\subset B$ is Jordan measurable then I want $f^{-1}(\partial E)=\partial f^{-1}(E)$ and that measure $0$ sets are preserved under $f^{-1}$. That way the boundary of the preimage will have measure $0$.