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.
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.