prove that any Jordan measurable set in $\mathbb{R}^n$ under Diffeomorphism stays Jordan measurable set

Question

prove that any Jordan measurable set in $\mathbb{R}^n$ under Diffeomorphism stays Jordan measurable set So what I thought is to split the set to bounded sets and show that for each the boundary is negligible. and we know that countable union of negligible sets is negligible. What I can't understand, is why can't a non boundary point in the Jordan set be mapped to a boundary in the output set. There might be a connection to this: Show continuously differentiable image of Jordan measurable set is Jordan measurable but Diffeomorphismness doesn't imply injectiveness.

Answer 1

The key facts are that (i) $A$ is Jordan measurable iff $\partial A$ is a null set, (ii) if $f$ is a homeomorphism then $f(\partial A) = \partial f(A)$ (show that $\overline{f(A)} = f(\overline{A})$), and (iii) a Lipschitz homeomorphism maps null sets to null sets.

First suppose $A$ is bounded & Jordan measurable, then since $f$ is a diffeomorphism it is Lipschitz on a compact rectangle containing $A$. Since $A$ is Jordan measurable we see that $\partial A $ is a null set. Then $f(\partial A) = \partial f(A)$ is a null set and hence $f(A)$ is Jordan measurable.

Now let $A_m = A \cap [m_1,m_1+1) \times \cdots \times [m_n,m_n+1)$ for $m \in \mathbb{Z}^n$. It is straightforward to show that $\partial (B \cap C) \subset \partial B \cup \partial C$, hence we see that the $A_m$ are Jordan measurable and so $\partial A_m$ is null.

I claim that $\partial A \subset \cup_m \partial A_m$. If this is true, then $\partial f(A) \subset \cup_m \partial f(A_m)$ and so $f(A)$ is Jordan measurable.

To demonstrate the claim, first suppose $x \in \partial A$, then I claim that $x \in \partial (A \cap B(x,1))$. To see this, suppose $a_n \in A, b_n \in A^c$ and $a_n \to x, b_n \to x$. Then for sufficiently large $n$ we have $a_n \in A \cap B(x,1)$ and $b_n \in A^c \subset (A \cap B(x,1))^c =A^c \cup B(x,1)^c $ and so $x \in \partial (A \cap B(x,1))$. Now let $I= \{ m | B(x,1) \cap A_m \ne \emptyset \}$ and hence $A \cap B(x,1) \subset \cup_{m \in I} A_m$ and so $\partial (A \cap B(x,1)) \subset \cup_{m \in I} \partial A_m \subset \cup_m \partial A_m$. Hence $\partial A \subset \cup_m \partial A_m$.