Let $\phi: D_t\to D_x$ be a diffeomorphism of an open set $D_t\subset \mathbb{R}^n$ onto a set $D_x \subset \mathbb{R}^n $ of the same type. Then the following assertions hold.
a) If $E_t\subset D_t$ is a set of (Lebesgue) measure zero, its image $\phi(E_t) \subset D_x$ is also a set of measure zero.
b)If a set $E_t$ contained in $D_t$ along with its closure $\bar E_t$ has Jordan measure zero, its image $\phi(E_t) = E_x $ is contained in $D_x$ along with its closure and also has (Jordan) measure zero.
c) If a (Jordan) measurable set $E_t$ is contained in the domain $D_t$ along with its closure $\bar E_t$, its image $E_x = \phi(E_t)$ is Jordan measurable and $\bar E_x \subset D_x$.
The lemma above is from $\S$11.5 of Mathematical Analysis by Zorich. The proof of the lemma based on the fact that every open subset in $\mathbb{R}^n$ can be represented as the union of a countable number of closed intervals (no two of which have any interior points in common).
According to a) we can conclude that: a diffeomorphism maps the set of (Lebesgue) measure zero to the set of (Lebesgue) measure zero.
In b) and c) the Lemma restrict the set $E_t$ to satisfying $\bar E_t\subset D_t$.
But I wanna know if b) and c) still holds for sets $E_t$ that $\bar E_t$ is not subset of $D_t$($D_t\setminus \bar E_t\neq \varnothing$)? So that we can say : a diffeomorphism maps the set of (Jordan) measure zero to the set of (Jordan) measure zero, and maps a Jordan measurable set to a Jordan measurable set. I guess this is not true, so could you give a counterexample, where a diffeomorphism maps a set of (Jordan) measure zero to a set not of (Jordan) measure zero ,or maps a Jordan measurable set to a Jordan unmeasurable set?