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We know the following fundamental property of functions of class $\mathcal C^{1}$:

Let $S$ be open in $\mathbb R^{n}$;let $f:S\rightarrow \mathbb R^{n}$ be a function of class $\mathcal C^{1}$.If the subset $A$ of $S$ has Lebesgue measure zero in $\mathbb R^{n}$,then the set $f(A)$ also has Lebesgue measure zero in $\mathbb R^{n}.(\star)$

If we replace Lebesgue measure zero by Jordan region (i.e. Jordan measurable set) in $(\star)$, the last argument may not be correct. I want to find some counterexamples to confirm the last argument is wrong.

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Here is a counterexample: Let $S:=\{(x,y)\in{\mathbb R}^2\>|\>-1<x<1\}$ and consider $$f: \quad(x,y)\mapsto\bigl({\rm artanh\,}x,0\bigr)\ .$$ The set $A:=\ ]{-1},1[\>\times\{0\}\subset S$ has Jordan measure $0$, but $f(A)={\mathbb R}\times\{0\}$ is not Jordan measurable.

But if you assume in addition that the set $A$ is compact then you will be able to prove that $f(A)$ has Jordan measure $0$. This follows from global Lipschitz estimates for $f$ valid on $A$.

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