The volume of the image of a map with vanishing Jacobian is zero

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 \} $

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.

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*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.

Answer 1

Too long for a hint, too short for a complete answer. Let's explore an easy case: $n=2,$ $f(0)=0,$ and $Df(0)$ is singular. Suppose further $Df(0)(e_1)=e_1,$ $Df(0)(e_2)=0.$

Claim: $|f(rB)|/|rB| \to 0$ as $r\to 0^+.$ (Here $rB$ is the ball of radius $r$ centered at $0,$ and $|\,|$ denotes area measure.)

Proof: We have $f(x,y) = (x,0) + \epsilon(x,y),$ where $\epsilon(x,y)$ is the usual $o(|(x,y)|)$ error. Define $M_r = \sup_{rB} |\epsilon|.$ Then $M_r=o(r)$ as $r\to 0^+.$

Next observe

$$f(rB) \subset I(r+M_r)\times I(M_r).$$

Here $I_t$ is the interval $[-t,t]$ for $t>0.$ Thus

$$|f(rB)|\le |I(r+M_r)\times I(M_r)| = 2(r+M_r)\cdot 2M_r.$$

Therefore

$$\frac{|f(rB)|}{|rB|} \le \frac{2(r+M_r)\cdot 2M_r}{\pi r^2} \to 0.$$

This proves the the claim. I'll stop here for now.

Answer 2

The uniformity of the $o(1)$ comes from uniform continuity of the determinant on compact subsets.

Let me go through the first couple of steps assuming that $f$ is at least $C^1$. Let us consider a compact unit cube $\Omega$ (slightly easier than balls). The determinant $d_x=\det(df_x)$ is uniformly continuous on $\Omega$. Thus for $\epsilon>0$ there is $N$ so that chopping up the unit cube into smaller cubes of side $1/N$ we have that the determinant only varies by at most $\epsilon$ in each of the smaller cubes.

Now let us mark all small cubes that contains a point for which $d_x=0$. On such a marked cube $d_x$ does not exceed $\epsilon$ by the above, so the image of this small cube has volume at most $\epsilon /N^n$. As $f(\Omega_0)$ is contained in the union of the images of the marked small cubes we have $m(f(\Omega_0)) \leq N^n \epsilon /N^n = \epsilon$.

Edit: To prove the upper bound estimate for the volume we may use continuity of $df_x$ (not its determinant) and consider a point $p\in \Omega$ for which $\det df_p=0$. Let $C_r$ be a cube (or ball) of side $r$ containing $p$. The linear map $df_p$ takes $R^n$ into a subspace of dimension $m<n$. In particular $df_p$ maps the cube $C_r$ into an $m$-dimensional object of diameter $\leq K\times r$. Given $\epsilon>0$, continuity of $df_x$ implies that for $r$ small enough $f(C_r)$ will be contained in a slab of diameter $Kr(1+\epsilon)$ and transversal width at most $\epsilon r$. So the volume does not exceed $K' vol(C_r) \epsilon$. Now proceed as above using uniform continuity of $df$. (This argument is pretty much like in Rudin : Real and complex analysis, Thm 7.24, case B)

Regarding this upper bound on the volume of the image, there are theorems of various levels of difficulty. The "ultimate" version is probably H. Federer, Geometric Measure Theory [Thm 3.2.3]. A simplified version is that for a Lipschitz map $f:R^n\rightarrow R^n$ and a measurable set $A$ (a Lipschitz map is almost surely differentiable): $$ \int_A |\det(df_x)|\; dvol_x = \int_{R^n} \#(f^{-1}(y)) \; dvol_y\geq vol(f(A))$$ The proof is rather involved, though.

Answer 3

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 $\mathcal{H}^n(f(E))=0$.