TITLE: The volume of the image of a map with vanishing Jacobian is zero
QUESTION [7 upvotes]: 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 \} $

Is there an elementary way to prove that $m(f(\Omega_0))=0$? ($m$ is the Lebesgue measure).

I know that this follows from the (co)area formula or Sard's theorem*, but is there a way that avoids them? 
Edit:
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 $\frac{m(f(B(r))}{m(B(r))} \to 0$.
Since $\Omega_0$ is compact, we can proceed as follows:
Cover $\Omega_0$ by finitely many balls $B_i$ with radius $r$, centered around points that belong $\Omega_0$. (The centers depend on $r$)Then,
$$ m(f(\Omega_0)) \le \sum_i m(f(B_i))=\sum_i \frac{m(f(B_i))}{m(B_i)}m(B_i).$$

If we could prove that $\frac{m(f(B_i))}{m(B_i)} \to 0$  when $r \to 0$ uniformly in $i$, then we could get 
  $$
m(f(\Omega_0)) \le o(1) \sum_i m(B_i)=o(1) \sum_i m(B(r)).
$$
  Since we can cover $\Omega_0$ by $N(r)$ balls of radius $r$ with centers in $\Omega_0$, where $N(r) \le c \frac{1}{m(B(r))}$, we could conclude that
  $m(f(\Omega_0)) \le o(1)$, so it must be zero.

The problem is that I am not sure if $\frac{m(f(B_{x_i(r)}(r)))}{m(B_{x_i(r)}(r))} $ converges to zero independently of $i$. I asked about this separately here.

*If I am not mistaken, then Sard's theorem implies that almost every $y \in \mathbb R^n$ is a regular value of $f$- so it doesn't have a preimage in $\Omega_0$.
**I don't know an elementary proof even in the case where $\Omega_0=\Omega$. That is, even if we assume the the domain where the Jacobian vanishes is "open and nice" I don't know if its trivial. In general, $\Omega_0$ is an arbitrary closed set, which might be pretty complicated.

REPLY [2 votes]: This is quite well-known. You do not need to have proven area (or coarea) formula in order to prove this. The result follows from basic covering arguments. It will be included in a soon to be publicized paper of mine with Piotr Hajlasz. I am using Hausdorff measure below, but that should not be any issue.
Lemma: Let $f:R^n \to R^N$ be $L$-Lipschitz. Let
$$
E_k=\{ x\in R^n:\, \text{$Df(x)$ exists and $rank \, Df(x)=k$.}\},
\quad
0\leq k\leq n.
$$
Then every $x\in E_k$ has the following property:
For every integer $m\geq 1$, there is $r=r(x,m)>0$ such that $f(B(x,r))$ can be covered by $m^k$ balls, each of radius $\frac{2\sqrt{k}Lr}{m}$.
Proof sketch: Given $m$, there is a threshold $\delta>0$ such that if $r < \delta$ then
$$
 \|f(y)-f(x) - Df(x)(y-x)\| \lesssim \frac{1}{m} \|y-x\| \leq \frac{r}{m}  \, \quad \text{for all $y \in B(x,r)$.}
$$
where $\lesssim$ means up to a constant that depends of $k$ and the Lipschitz constant of $f$. This says that the image $f(B(x,r)$ is within $\lesssim r/m$ thin strip around the image $Df(x)(B(x,r))$. Now the image of the derivative is contained in a $k$-dimensional affine space. So, we may cover this image by $\lesssim m^k$ many balls of diameters $\lesssim r/m$. As a result $f(x,r)$ is also covered by those same balls except after doubling/trippling their radii.
Theorem (Sard-type): Suppose $f\colon R^n \to R^N$, is Lipschitz. Let $E$ be the set of all $x$ where $Df(x)$ exists and $rank \, Df(x) <n$. Then $\mathcal{H}^n(f(E))=0$.
Proof sketch: If we use the lemma then we can estimate the Hausdorff content as follows (the whole point is we compute $n$-dim measure and $k<n$):
$$
\mathcal{H}^n_\infty(f(B(x,r))) \lesssim \sum_{covering\, of \, f(B(x,r)) \, by \, C} (diam \, C)^n \lesssim m^k \times (r/m)^n = r^n m^{k-n}
$$
Now by a Vitali covering we find covering of $E$ with $B(x,r_i)$ such that $B(x,r_i/5)$ are disjoint. Then summing the inequality above over this collection produces (notice that $\sum_i r_i^n$ is bounded by $5^n$ times Lebesgue measure of $E$.)
$$
\mathcal{H}^n_\infty(f(E)) \lesssim  m^{k-n} m(E)
$$
But $m$ is arbitrary and $k<n$. $\mathcal{H}^n_\infty(f(E)) = 0$ from which as a general rule it follows that $\mathcal{H}^n(f(E))=0$.