Measure of a set under diffeomorphism

Question

Let $\Omega,\Omega'\subset\mathbb{R}^n$ be bounded open sets. Let $F:\Omega\to\Omega'$ be a diffeomorphism. Let $A\subset\mathbb{R}^n$ be a compact measurable set such that $A\subset\Omega$ and $$\lambda(A)\ge\alpha,$$ where $\lambda$ is the $n$-dimensional Lebesgue measure and $\alpha>0$.

Is it possible to prove that $$\lambda(F(A))\ge C\alpha,$$ where $C>0$ depends just on $F$?

Answer 1

Solution supposing $\alpha$ constant.

Let be $C_n = \{x\in\Omega: |\det DF(x)| > 1/n\}$. $F$ diffeomorphism $\implies\Omega = \bigcup_{n = 1}^\infty C_n$. Take $n_0$ s.t. $\lambda(C_{n_0}) > \lambda(\Omega) - \alpha/2$.

If $\lambda(A)\ge\alpha$, $\lambda(C_{n_0}\cap A)\ge\alpha/2$. Using the cov theorem: $$ \lambda(F(A)) = \int_{F(A)}d\lambda = \int_{A}|\det DF|\,d\lambda\ge \int_{C_{n_0}\cap A}|\det DF|\,d\lambda\ge \alpha/2n_0 .$$

Answer 2

Example: $n=1$, $\Omega = \Omega' = (0,1)$ and $F(x) = x^2$. Then for $\alpha < 1/2$, let $A = [\alpha,2\alpha]$ to get $F(A) = [\alpha^2,4\alpha^2]$ so that $\lambda(A) = \alpha$ and $\lambda(F(A)) = 3\alpha^2$. Note $\lambda(F(A))/\lambda(A) = 3\alpha$ can be as small as we like.