Does this phenomenon regarding volumes of images of small balls hold uniformly?

Question

Let $\Omega \subseteq \mathbb{R}^n$ be a nice domain with smooth boundary (say a ball), and let $f:\Omega \to \mathbb{R}^n$ be smooth. Set $\Omega_0=\{ x \in \Omega \, | \, \det df_x =0 \} $

In this answer, zhw proves that if $x \in \Omega_0$, and $B(r)$ is an Euclidean ball of radius $r$ centered at $x$, then $\lim_{r \to 0}\frac{m(f(B(r))}{m(B(r))} =0$.

My question is whether this claim holds uniformly in the center of the balls:

For every $r$, let $x(r) \in \Omega_0$, and suppose that $B_{x(r)}(r)$ (the ball with radius $r$ centered at $x_i$) is contained in $\Omega$. (so $f$ is defined on it).

I also assume that $d(x(r),\partial \Omega) \ge \epsilon$ for some positive $\epsilon$.

Is it true that $\lim_{r \to 0}\frac{m(f(B_{x(r)}(r)))}{m(B_{x(r)}(r))} =0$?

The $x(r)$ can change with $r$.

I think that this question might be connected to the question of uniform differentiability, which in general does not hold for non-convex domains.

Answer 1

The claim is true, but the following argument might not be what you are looking for, since it uses facts that are stronger than (a form of) Sard's theorem, which I think is what you actually want to prove.

First, note that $K := \{x \in \Omega : d(x, \partial \Omega) \geq \epsilon/2 \}$ is a compact subset of $\Omega$ (I assume that $\Omega$ is bounded). By your assumptions, you have $B_{x(r)}(r) \subset K$ for $r < \epsilon/2$. Since $|\det D f|$ is continuous on $\Omega$, it is uniformly continuous on $K$. Hence, for $\delta > 0$, there is $0<\epsilon' < \epsilon/2$ such that $|\det Df(y)| \leq \delta$ for all $y \in B_{x(r)}(r)$ and $r < \epsilon'$. Here, I used that $x(r) \in \Omega_0$.

Next, the area formula from geometric measure theory (see for example the book by Evans and Gariepy) shows that \begin{align*} \delta \cdot m(B(r)) & \geq \int_{B_{x(r)}(r)} |\det Df (y)| d y \\ & = \int \# \big(B_{x(r)}(r) \cap f^{-1}(\{z\}) \big) d z \\ & \geq \int_{f(B_{x(r)}(r))} d z = m(f(B_{x(r)}(r))) \end{align*} for all $0 < r < \epsilon'$, proving your claim.