# Show continuously differentiable image of Jordan measurable set is Jordan measurable

Let $G:U \to V$ be a continuously differentiable and injective mapping with $U,V \subset \mathbb{R}^n$ Then I have to show first that if $K \subset U$ a compact set has zero content then so does $G(K)$, and secondly that if $T$ is Jordan measurable and $\bar T \subset U$ then $G(T)$ is also measurable.

For the first part I know $K \subset \cup_i B_i$ for some finite collection of boxes whose total volume can be arbitrarily small. Then if I look at the image of each $B_i$ under $G$ I thought I could similarly bound their volume by continuity of $G$. Is this the right idea? For the second part I am not sure because i wanted to use the first part to show that the boundary of $G(T)$ is zero content since the boundary of $T$ is, but I realized the boundary of $T$ can be mapped somewhere in the interior of $G(T)$...

Edit: The definition of Jordan measurable I know is that the boundary has zero content and the set is bounded

• For the second part: try instead to understand what would be the preimage of the interior of $G(T)$ Jan 31, 2018 at 23:05
• why don't you give the bounty as it is expiring very soon? Feb 5, 2018 at 21:24

$\DeclareMathOperator{\diam}{diam} \DeclareMathOperator{\vol}{vol} \DeclareMathOperator{\itr}{int} \DeclareMathOperator{\cl}{cl}$For the first part you are on the right track.

Continuity alone is not enough to bound the volume of the image of each box $B_i$ under $G$ by the volume of $B_i$. Here you should use the fact that a $C^1$ function on a compact set is Lipschitz, that is, $G$ is Lipschitz, say with constant $L$. Then $\diam(G(B_i)) \le L \diam(B_i)$. Without loss of generality assume that these boxes are cubes. Therefore

$$\vol(G(B_i)) = \left( \frac{\diam(G(B_i))}{\sqrt{n}} \right)^{\frac{1}{n}} \le \left( \frac{L \diam(B_i)}{\sqrt{n}} \right)^{\frac{1}{n}} = L^{\frac{1}{n}} \vol(B_i)$$

We are done.

Now for the second part you are also on the right track, you just need to ensure that $G(\partial T) \subset \partial G(T)$.

Note that since $G$ is continuous then $G(\cl{T}) \subset \cl{G(T)}$ and the preimage of open sets are open. So $G(\partial T) \subset G(\cl{T}) \subset \cl{G(T)}$. Moreover $G^{-1}(\itr G(T))$ is open and $G^{-1}(\itr G(T)) \subset G^{-1}(G(T))$, hence $G^{-1}(\itr G(T)) \subset \itr G^{-1}(G(T))$.

By the injectivity of $G$ it follows that $G^{-1}(G(T)) = T$. Then $G^{-1}(\itr G(T)) \subset \itr T$, whence $\itr G(T) \subset G(\itr T)$.

This last inclusion implies that $G(\partial T) \bigcap \itr G(T) \subset G(\partial T) \bigcap G(\itr T) = \emptyset$, because $\partial T \bigcap \itr T = \emptyset$ and $G$ is injective.

Putting things together we have $G(\partial T)\subset \cl{G(T)} = \partial G(T) \bigcup \itr G(T)$ and $G(\partial T) \bigcap \itr G(T) = \emptyset$. Therefore $G(\partial T)\subset \partial G(T)$ as we wanted to show.

• Why WLOG they are cubes? Feb 2, 2018 at 22:48
• @Garou Because any $n$-rectangle can be approximated (and hence its volume too) by a finite collection of $n$-cubes. Given a rectangle $R$, start with any collection of cubes, say $C_1$, covering it. Then you split each of these cubes into $2^n$ equal subcubes (by splitting each edge in half) and take $C_2$ to be the ones that intersects $R$ and so on...Thus $C_k$ is a sequence of "cube coverings" of $R$ and $A_k = \bigcup_{S \in C_k} S$ is a decresing sequence of sets. Therefore $\lim_{k \to \infty} A_k = \liminf_{k \to \infty} A_k = R$. Feb 2, 2018 at 23:51