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**.