Whether $\varphi(E)$ is a Jordan measurable set?

Definition: A set $S \subset \mathbb {R^{n}}$ is Jordan measurable if it is bounded in $\mathbb {R^{n}}$ and its boundary is a set of Lebesgue measure zero.

The following conclusion has been proved to be correct.

Let $\varphi : \Omega\subset \mathbb {R^{n}} \rightarrow \mathbb {R}$ be a $C^{1}$ function in the open set $\Omega$. Let $E$ be a compact Jordan measurable set with $E\subset \Omega$ . If $\varphi$ is a diffeomorphism on the interior of $E$, then $\varphi(E)$ is a compact Jordan measurable set.

I want to relax one condition from the above conclusion, replacing the condition

"Let $E$ be a compact Jordan measurable set with $E\subset \Omega$ "

with

"Let $E$ be a Jordan measurable set with $\overline{E}\subset \Omega$".

Do we have that $\varphi(E)$ is a Jordan measurable set?

By my intuition, $\varphi(E)$ may not be a Jordan measurable set. Then I try to find some examples to support my view, let $\varphi(x)=\tan(x)$, $E= \Omega=(0,\pi/2)$. $\overline{E}\not\subset \Omega$, although $\varphi(E)$ is not a Jordan measurable set and $\varphi$ is a diffeomorphism on the interior of $E$.

Unfortunately,since $\overline{E}\not\subset \Omega$, my example fails to satisfy the changed condition.I want to find a example strictly denying "$\varphi(E)$ is a Jordan measurable set". I would really appreciate it!

All above is copied from here.

• Are you sure about the fact that you state? What if $E$ is a usual Cantor set and $\varphi$ is such that $\varphi(E)$ is a fat Cantor set? I am not sure that such $\varphi$ can be chosen to be $C^1$ though. – erz Feb 18 '17 at 13:37
• @erz: This won't work, a $C^1$ map is Lipschitz and thus sends null sets to null sets. – Christian Remling Feb 18 '17 at 17:56
• You should accept my answer, or otherwise the question will be periodically bumped to the homepage. Forever. – erz May 20 '17 at 17:37

This is the proof for the case when $E$ is bounded and Jordan measurable, but not necessarily closed. Also the proof seems to work for any $C^1$ map between manifolds.

We know that $\partial E$ is a null set. Let $F$ be the set of critical points of $\varphi$ in $int~E$ and let $U=int~ E\backslash F$. Then $\overline{E}=\partial E\bigcup int~E=\partial E\bigcup F\bigcup U$.

From the continuity of $\varphi$ and compactness of $\overline{E}$ we have $\overline{\varphi(E)}=\overline{\varphi(\overline{E})}=\varphi(\overline{E})=\varphi(\partial E)\bigcup \varphi(F)\bigcup \varphi(U)$.

Note that $U$ is open and $\varphi$ is an open map on $U$, and so $\varphi(U)$ is open. Hence $\varphi(U)\subset int~\varphi(E)$, and so $\partial \varphi(E)=\overline{\varphi(E)}\backslash int~\varphi(E)\subset [\varphi(\partial E)\bigcup \varphi(F)\bigcup \varphi(U)]\backslash \varphi(U)\subset \varphi(\partial E)\bigcup \varphi(F)$.

As Christian Rempling observed, $\varphi$ has to be Lipshitz on compacts, and so it maps compact null sets into compact null sets, from where $\varphi(\partial E)$ is null, while by Sard's Theorem $\varphi(F)$ is null. Thus $\partial \varphi(E)$ is null, and so $\varphi(E)$ is Jordan measurable.

As for the unbounded case, consider $\Omega=\mathbb{R}$ and $E=\mathbb{N}$, which is clearly Jordan measurable. Let $\psi:\mathbb{N}\to\mathbb{Q}$ be a bijection. Since $\mathbb{N}$ is a discreet space, this map can be extended to a $C^1$ map $\varphi:\mathbb{R}\to\mathbb{R}$ (easy to see). But then $\varphi(E)=\mathbb{Q}$, and the latter is not Jordan measurable.

The last example however is kind of artificial, so I'd like to know what happens with nicer unbounded sets, say connected ones (and maps into manifolds, not merely into $\mathbb{R}$).

• The definition of a Jordan measurable set given by the OP includes the assumption that the set is bounded, so $\Bbb{N}$ is not Jordan measurable according to the definition of the OP. – PhoemueX Feb 19 '17 at 21:51
• @PhoemueX: you are right, but one may still be curious about sets with null boundary. – erz Feb 20 '17 at 1:26