Sard's Theorem with whatever function

Question

A lot of staetments of Sard's Theorem require a certain order of smoothness to the function $f$. At least it is $C^1$.

I want to prove a more general version of this theorem which holds (clearly) only with $f : \mathbb{R}^n \to \mathbb{R}^n$ but $f$ could be literally what you want (no regularity assumption). The statement sounds like that:

Let $f : \mathbb{R}^n \to \mathbb{R}^n$ and \begin{align} Z &= \{ x \in \mathbb{R^n} | f \text{ is differentiable at $x$ and } \det(J f(x)) =0 \} \end{align} Then $\mathcal{L}^n (f(Z))=0$

Here $\mathcal{L}^n$ is the Lebesgue measure.

I proved the claim for $n=1$. I'm reporting it here:

Firstly assume wlog $f :[a,b] \to \mathbb{R}$.

Let $x \in Z$, then by definition of differentiability, $\forall \epsilon >0$ there exists $r_{\epsilon,x}>0$ such that $\forall y \in \mathbb{R}^n$ with $0 < \vert y-x\vert<r_{\epsilon,x}$, then $\Big\rvert \dfrac{f(y)-f(x)}{y-x} -0\Big\lvert < \epsilon$.

Thus $\vert f(x) - f(y) \vert < \epsilon |y-x| < r \epsilon $ for all $r \in (0,r_{\epsilon,x}) \qquad (1)$.

Let me call $I_{x,r} := [x-r,x+r] $ and $A_M:=f(Z) \cap [-M,M]$. I have $f(Z) = \bigcup_{M=1}^{\infty}A_M$ and $\mathcal{L}(f(I_{x,r}))< 2r\epsilon$ by $(1)$.

Claim: $\mathcal{L}(A_M)=0$

I need Vitali covering Lemma

Let $A \subset \mathbb{R}^n$ bounded and $\mathcal{B}$ a fine covering of $A$ (i.e. a family of balls such that for all $x \in A$, then $\inf \{r>0 | B(x,r) \in \mathcal{B} \} = 0$). Then there exists a subfamily $\mathcal{B}' \subset \mathcal{B}$ of disjoint balls (at most countable) such that $\mathcal{L}^n(A) < \mathcal{L}^n(\bigcup_{B \in \mathcal{B}'}B) + \epsilon$

Fix $\epsilon >0$. I choose $\mathcal{B} = \{[f(x) - r \epsilon,f(x)+r\epsilon] | x \in Z, r \in (0,r_{\epsilon,x}) \text{ such that if } y \in [x-r,x+r] \text{ then } \vert f(y)-f(x) \vert < r \epsilon \}$

Clearly this is a fine covering, hence, by Vitali, there exists a subfamily of disjoint intervals $I_i:= I_{f(x),\epsilon r_i}$ such that $\mathcal{L}(A_M) < \mathcal{L}(\bigcup I_i) + \bar{\epsilon}$.

Now, I take $\epsilon = \bar{\epsilon} = \dfrac{1}{100} \dfrac{\mathcal{L}(A_M)}{(1+b-a)}$.

I observe that $\mathcal{L}(\bigcup I_i) \leq 2 \sum \epsilon r_i \leq \epsilon (b-a)$.

We put all these stuffs together and we get $\mathcal{L}(A_M) < \dfrac{1}{100} \mathcal{L}(A_M)$ which implies $\mathcal{L}(A_M)=0$.

Finally $\mathcal{L}(f(Z)) \leq \sum \mathcal{L}(A_M)=0$

NOW, I want to simulate this proof for the cases $n \geq 2$ but the main isssue is that I have more than one direction to control.

My idea is to prove that for every $\epsilon >0$ and every $x \in Z$ there exists $r_{\epsilon,x} >0$ such that $f(B(x,r))$ is contained in a parallelepiped with all sides controlled by $r$ and one side controlled by $\epsilon r$ for every $r \in (0, r_{\epsilon,x})$. Then I will use Vitali and finally discover $\mathcal{L}^n(f(B(x,r)) < c \epsilon r^n$.

Wlog I can suppose that the first column of Jacobian matrix is full of zeros and this leads me to prove that along $x_1$-axis my image is controlled by $\epsilon r$, but what about the other axis?

Can you give me some hints or references?

Answer 1

Maybe this can work:

Fix $\epsilon > 0$

Let $Z_M := Z \cap \overline{B(x,M)}$ with $M>0$. Take $x \in Z_M$.

Since $f(Z) = \bigcup f(Z_M)$, it's sufficient to show $\mathcal{L}^n(f(Z_M)) = 0$.

By definition of differentiability we get:

for every $\epsilon_1 > 0$ there exists $r=r_{\epsilon_1,x}> 0$ such that, for every $y$ with $ 0< \Vert y-x \Vert < r$ then $\Vert f(y) - f(x) - Jf(x)(y-x) \Vert < \epsilon_1 \Vert (y-x) \Vert < \epsilon_1 r$.

Now, take the hyperplane passing through point $f(x)$ and direction Im$(Jf(x)$ and call it $H$. Observe that $d(f(y), H) \leq \Vert f(y) -( f(x) + Jf(x)(y-x)) \Vert < \epsilon_1 \Vert (y-x) \Vert < \epsilon_1 r$

By Taylor's formula we get $\Vert f(y) - f(x) \Vert \leq \Vert Jf(x) \Vert \Vert y-x \Vert \leq C_x r$ where $C_x$ is a constant depending on $x$.

Therefore, $f(B(x,r)) \subset \text{ "Parallelepiped with one side controlled by $2\epsilon_1r$ and the others by $2rC_x$"}$, as in the picture below.

This has to hold for every $\epsilon_1$, so, in particular, we can take $\epsilon_1 = \dfrac{\epsilon}{2^{n} C_x^{n-1}}$.

Thus we have for every $\epsilon >0$ and for every $x \in Z_M$ there exists $r_{\epsilon,x}>0$ such that for every $r \in (0, r_{\epsilon,x})$ we get $\mathcal{L}^{n \star}(f(B(x,r))) \leq \epsilon r^n$ ($\mathcal{L}^{n \star}$ is the outer measure).

Consider now this fine covering: \begin{equation} \mathcal{B} = \{B(x,r) \, | \, x \in Z_M, \, r \in (0, r_{\epsilon,x}) \} \end{equation} By Vitali there exists a subfamily $\{B_i = B(x_i,r_i)\}$ such that \begin{equation} \mathcal{L}^{n \star} (Z_M - \bigcup B_i) =0 \\ \mathcal{L}^{n \star} (\bigcup B_i) \leq \mathcal{L}^{n \star}(Z_M) + \epsilon \end{equation}

Now, recall this fact

Let $f : A \subset \mathbb{R}^n -> \mathbb{R}^n$ where $f$ is differentiable in $A$ and $A$ is a null set, then $f(A)$ is a null set.

Then \begin{equation} \mathcal{L}^{n \star} (f(Z_M)) \leq \mathcal{L}^{n \star}(f(Z_M - \bigcup B_i)) + \mathcal{L}^{n \star}(\bigcup (f(B_i))) \leq \mathcal{L}^{n \star}(\bigcup(f(B_i))) \leq \sum \mathcal{L}^{n \star}(f(B_i)) \leq \sum \epsilon r_i^n \leq \dfrac{\epsilon}{w_n}(\mathcal{L}^{n \star}(Z_M) + \epsilon) \end{equation}

Where the last inequality holds since $\sum \mathcal{L}^n(B_i) = w_n \sum r_i^n \leq \mathcal{L}^{n \star}(Z_M) + \epsilon$. This concludes the proof by arbitrariness of $\epsilon$.