# Rectifiable sets and approximate tangent planes#

Rectifiable sets are the measure theoretic counterpart to manifolds.

Definition 22 (Rectifiable set)

A $$\mathcal H^n$$-measurable set $$E\subset \mathbb R^m$$ is $$n$$-rectifiable if there exist Lipschitz $$f_i \colon \mathbb R^n \to \mathbb R^m$$ such that

$\mathcal H^n\left(E\setminus \bigcup_{i\in\mathbb N} f_i(\mathbb R^n)\right)=0.$

We will show that $$n$$-rectifiable sets possess a unique approximate $$n$$-dimensional tangent plane at almost every point.

Given $$V\in G(m,n)$$, $$a\in \mathbb R^n$$ and $$0<s<1$$ define the cone around $$V$$ centred at $$a$$ with aperture $$s$$ as

$C(a,V,s)=\{x\in \mathbb R^n : \operatorname{dist}(x-a, V)< s\|x-a\|\}.$

Definition 23 (Approximate tangent plane)

Let $$A\subset \mathbb R^m$$ and $$a\in A$$. A $$V\in G(m,n)$$ is an approximate tangent plane to $$A$$ at $$a$$ if

(22)#$\limsup_{r\to 0} \frac{\mathcal H^n(A \cap B(a,r))}{r^n}>0$

and, for every $$0<s<1$$,

(23)#$\lim_{r\to 0} \frac{\mathcal H^n(A \cap B(a,r) \setminus C(a,V,s))}{r^n}=0.$

Rademacher’s theorem gives a candidate for the approximate tangent plane to a rectifiable set. There are three steps required to prove that the derivative is indeed an approximate tangent plane: show that the derivative has full rank at almost every point; prove the density condition (22); and show that the sets from other parametrisations of the rectifiable set do not destroy the approximation by a tangent plane at almost every point.

The second and third steps follow from the results of the previous section. For the first step we use the following.

Lemma 7 (Easy Sard’s theorem)

If $$f\colon \mathbb R^n \to \mathbb R^m$$ is Lipschitz then

$\mathcal H^n(\{f(x): \operatorname{rank} Df(x)<n\})=0.$

Proof. Let $$L=\operatorname{Lip}f$$. Fix $$0<R<\infty$$, $$\delta,\epsilon>0$$ and let

$A=\{x\in B(x,R): \operatorname{rank} Df(x)<n\}.$

For $$x\in A$$ let

$W_x = f(x)+Df(x)(\mathbb R^n).$

Then for sufficiently small $$0<r_x<\delta$$,

$f(B(x,r_x))\subset B(f(x),Lr_x) \cap \{y: \operatorname{dist}(y,W_x)< \epsilon r_x\}.$

Since $$\operatorname{rank}Df(x)<n$$, the set on the right hand side can be covered by $$(L/\epsilon)^{n-1}$$ cubes of side length $$\epsilon r_x$$.

Since $$A$$ is covered by balls of the form $$B(x,r_x/5)$$, there exists a disjoint collection of balls $$B(x_i,r_i/5)$$ such that $$A$$ is covered by the union of the $$B(x_i,r_i)$$. Then

$f(A) \subset f\left(\bigcup_{i\in\mathbb N}B(x_i,r_i)\right) \subset \bigcup_{i\in\mathbb N} f(B(x_i,r_i)).$

By the previous argument, each factor of the right hand side is covered by $$(L/\epsilon)^{n-1}$$ cubes of side length $$\epsilon r_i$$. Thus

$\mathcal H^n_{2\delta}(f(A)) \leq \sum_{i\in\mathbb N} \left(\frac{L}{\epsilon}\right)^{n-1} (\epsilon r_i)^{n-1} = L^{n-1}\epsilon \sum_{i\in\mathbb N}r_i^n.$

However, the $$B(x_i,r_i/5)$$ are disjoint subsets of $$B(0,R+\delta)\subset \mathbb R^n$$ and so

$\sum_{i\in\mathbb N}\left(\frac{r_i}{5}\right)^n \leq (R+\delta)^n.$

Since $$\epsilon>0$$ is arbitrary, this implies that $$\mathcal H^n_{2\delta}(f(A))=0$$ and hence $$\mathcal H^n(f(A))=0$$. Taking a countable union over $$R\to\infty$$ completes the proof.

Lemma 8

Let $$f\colon \mathbb R^n\to \mathbb R^m$$ be Lipschitz and $$S\subset \mathbb R^n$$. Suppose that there exists a $$\delta>0$$ such that, for each $$x,y\in S$$,

$\|f(x)-f(y)\|\geq \delta \|x-y\|.$

Then $$f(S)$$ has a unique approximate tangent plane at $$\mathcal H^n$$ almost every point.*

Proof. By Lemma 7, we may suppose $$\operatorname{rank}Df(x)=n$$ for every $$x\in S$$. By Lemma 5, we may suppose $$\Theta^{*,n}(E,f(x))>0$$ for every $$x\in S$$. Fix $$x\in S$$ and $$0<s<1$$. There exists $$\epsilon>0$$ such that

$\|f(y)-f(x) -Df(x)(y-x)\| < s\|y-x\|$

for all $$y\in B(x,\epsilon)\cap S$$. Moreover, if $$y\in S\setminus B(x,\epsilon)$$ then

$\|f(y)-f(x)\| \geq \delta \epsilon.$

That is, if $$a=f(x)$$ and $$b=f(y)$$ with $$\|a-b\|\leq \delta\epsilon$$ and $$V=a+Df(x)(\mathbb R^n)$$,

$\operatorname{dist}(b-a, V) < s\|y-x\| \leq s\|b-a\|/\delta.$

Therefore, $$V$$ is an approximate tangent plane to $$f(S)$$ at $$a$$.

This approximate tangent plane is unique at any density point $$x$$ of $$S$$. Indeed, if $$V'\neq V$$, let $$v\in V\setminus V'$$ and let $$0<s<1$$ be such that $$C(f(x),\mathbb Rv,s)\cap C(f(x),V',s)= \{0\}$$. Since $$\operatorname{rank}Df(x)=n$$ and $$x$$ is a density point of $$S$$, for sufficiently small $$r>0$$ there exists $$y\in S \cap B(x,r)$$ such that $$B(f(y),s r)\cap C(f(x),V',s) = \emptyset$$ and

$\frac{\mathcal H^n(B(f(y),s r)\cap S)}{r^n} \geq \delta s.$

In particular, $$V'$$ is not an approximate tangent to $$f(S')$$ at $$x$$.

Theorem 14

Let $$E\subset \mathbb R^m$$ be $$n$$-rectifiable with $$\mathcal H^n(E)<\infty$$. Then for $$\mathcal H^n$$-a.e. $$x\in E$$, $$E$$ has a unique approximate tangent plane at $$x$$.

Proof. Let $$f\colon \mathbb R^n\to \mathbb R^m$$ be one of the Lipschitz functions as in the definition of a rectifiable set and let $$S=f^{-1}(E)$$. It suffices to prove that $$E$$ has a unique approximate tangent plane at $$f(x)$$ for $$\mathcal L^n$$-a.e. $$x\in S$$.

By Lemma 7, we may suppose that $$\dim Df(\mathbb R^n)=n$$ for all $$x\in S$$. Fix such an $$x$$ and let $$0<\epsilon<\|Df(x)^{-1}\|/2$$. There exists $$\delta>0$$ such that

$\|f(y)-f(x) -Df(x)(y-x)\| < \epsilon\|y-x\|$

for all $$y\in B(x,\delta)\cap S$$. In particular, by the triangle inequality,

$\|f(y)-f(x) \| > \epsilon\|y-x\|/2.$

Therefore, the sets

$S_\epsilon:=\{x\in G: \|f(y)-f(x) \| > \epsilon \|y-x\|\ \forall y\in B(x,\epsilon)\}$

are Borel and monotonically increase to $$S$$ as $$\epsilon\to 0$$. Therefore it suffices to prove the result for $$\mathcal L^n$$-a.e. $$x$$ in some fixed $$S_\epsilon$$. Cover $$S_\epsilon$$ by finitely many balls $$B_1,B_2,\ldots,B_N$$ of radius $$\epsilon$$. It suffices to prove the result for $$\mathcal L^n$$-a.e. $$x$$ in some fixed $$S':=S_\eta\cap B_i$$.

However, $$S'$$ satisfies the hypotheses of Lemma 8 and so $$f(S')$$ has a unique approximate tangent at $$\mathcal H^n$$ almost every point. To see that this tangent is a unique approximate tangent to $$E$$ at $$\mathcal H^n$$ almost every point, we simply use Lemma 6: for $$\mathcal H^n$$-a.e. $$x\in f(S')$$, $$\Theta^{*,n}(E\setminus f(S'),x)=0$$.

In Theorem 16 we will see that the converse to Theorem 14 holds.

For $$V\in G(n,m)$$, write $$\pi_V$$ for the orthogonal projection onto $$V$$ and equip $$G(n,m)$$ with the metric $$d(V,W)=\|\pi_V-\pi_W\|$$. We will consider $$\mathcal L^n$$ on an element of $$G(n,m)$$.

Lemma 9

Let $$f\colon \mathbb R^m \subset \mathbb R^n \to \mathbb R^m$$ be Lipschitz and let $$S\subset \mathbb R^n$$ satisfy $$\mathcal L^n(S)>0$$. For $$\epsilon>0$$ suppose that there exists an invertible linear $$L\colon \mathbb R^n\to \mathbb R^m$$ such that, for all $$x,y \in S$$,

$\|f(x)-f(y)-L(x-y)\| < \frac{\epsilon}{\|L^{-1}\|}\|x-y\|.$

Then for any $$V\in G(n,m)$$ with $$\|(\pi_V|_{L(\mathbb R^n)})^{-1}\|^{-1} \geq 2\epsilon$$, $$\mathcal L^n(\pi_V(f(S)))>0$$.*

Proof. For any $$V\in G(n,m)$$,

$\|\pi_V(f(x)-f(y)) - \pi_V( L(x-y))\| < \epsilon\|L^{-1}\|^{-1}\|x-y\|$

and so, if $$\|(\pi_V|_{L(\mathbb R^n)})^{-1}\|^{-1} \geq 2\epsilon$$,

$\begin{split}\|\pi_V(f(x)-f(y))\| &\geq \|\pi_V( L(x-y))\| - \epsilon\|L^{-1}\|^{-1}\|x-y\|\\ &\geq \|(\pi_V|_{L(\mathbb R^n)})^{-1}\|^{-1}\|L(x-y)\|-\epsilon\|L(x-y)\|\\ &\geq \epsilon \|L(x-y)\|\\ &\geq \epsilon \|L^{-1}\|\|x-y\|.\end{split}$

Thus $$\pi_V\circ f$$ has Lipschitz inverse on $$S$$ and hence $$\mathcal L^n(\pi_V(f(S)))>0$$ by Example 5.

Corollary 2

Let $$E\subset \mathbb R^m$$ be $$n$$-rectifiable with $$\mathcal H^n(E)>0$$. Then there exists $$W\in G(m-n,m)$$ such that $$\pi_V(E)>0$$ for all $$V\in G(n,m)$$ with $$V\cap W=\{0\}$$.*

Remark 5

The set of $$V$$ that satisfy the conclusion of Corollary 2 is very large; try some examples in reasonable dimensions.

Proof. Since $$E\subset \mathbb R^m$$ is rectifiable with $$\mathcal H^n(E)>0$$, there exists a Lipschitz $$f\colon \mathbb R^n\to \mathbb R^m$$ with $$\mathcal H^n(E\cap f(\mathbb R^n))>0$$. In particular, $$S:= f^{-1}(E)$$ satisfies $$\mathcal L^n(S)>0$$. By Lemma 7, for $$\mathcal L^n$$-a.e. $$x\in S$$, $$Df(x)$$ is injective.

Fix $$\epsilon>0$$ For $$M>0$$, the set of invertible $$L\colon \mathbb R^n\to \mathbb R^m$$ with $$\|L^{-1}\|<M$$ may be covered by countably many sets of diameter $$\epsilon/M$$. Varying $$M\in\mathbb N$$, we see that $$\mathcal L^n$$ almost all of $$S$$ is covered by countably many sets of the form

$\{x\in S: \|Df(x)-L\|< \epsilon/2\|L^{-1}\|\}.$

Moreover, each of these sets may be covered by countably many sets of the form

$\{x\in S: \|f(x)-f(y)-L(x-y)\| < \epsilon\|x-y\|/\|L^{-1}\|\ \forall y\in B(x,\epsilon)\}.$

Finally, these sets may be covered by countably many sets of diameter $$\epsilon$$. Therefore, for each $$j\in\mathbb N$$ there exists $$S_j\subset S$$ and invertible $$L_j\colon \mathbb R^n\to\mathbb R^m$$ such that, for all $$x,y\in S_j^\epsilon$$,

$\|f(x)-f(y)-L_j(x-y)\| < \epsilon\|x-y\|/\|L_j\|^{-1},$

and $$\mathcal L^n(S\setminus \bigcup_{j\in \mathbb N}S_j)=0$$.

Since $$\mathcal L^n(S)>0$$, there exists $$j\in \mathbb N$$ with $$\mathcal L^n(S_j)>0$$. Then $$S_j$$ satisfies the hypotheses of Lemma 9 and so $$\mathcal L^n(\pi_V(S))\geq \mathcal L^n(\pi_V(S_j))>0$$ for all $$V\in G(n,m)$$ with $$\|(\pi_V|_{L_j(\mathbb R^n)})^{-1}\|^{-1} \geq 2\epsilon$$. Let $$L_\epsilon=L_j$$. Repeat this for each $$i\in\mathbb N$$ with $$\epsilon =1/i$$. The set $$G(n,m)$$ is compact and so we may suppose that $$L_{1/i}(\mathbb R^n) \to W\in G(n,m)$$. The only $$V\in G(n,m)$$ for which $$\mathcal L^n(\pi_V(S))=0$$ satisfy $$\|(\pi_V|_{L_{1/i}(\mathbb R^n)})^{-1}\|^{-1} < 2/i$$ and hence $$\|(\pi_V|_{W})^{-1}\|^{-1} < 2/i$$ for each $$i\in \mathbb N$$. That is, $$V\cap W^{\perp} \neq \{0\}$$ as required.

## Exercises#

Example 46

Let $$X$$ be a metric space, $$Y\subset X$$ and $$f\colon Y \to \mathbb R$$ $$L$$-Lipschitz. Define $$\tilde f\colon X \to \mathbb R$$ by

$\tilde f(x)= \sup\{f(y)-Ld(x,y):y\in Y\}.$
1. Show that $$\tilde f$$ is an $$L$$-Lipschitz extension of $$f$$ to $$X$$. This is called the McShane–Whitney extension theorem

2. If $$f\colon Y\to \mathbb R^n$$ is $$L$$-Lipschitz, show that there is a $$\sqrt{n}L$$-Lipschitz extension of $$f$$ to $$X$$.

3. The following example shows that the vector valued extension cannot have the same Lipschitz constant in general: Let

$Y=\{(-1,1),(1,-1),(1,1)\} \subset \ell_\infty^2$

and define

$f(-1,1)=(-1,0), \quad f(1,-1)=(1,0), \quad f(1,1)=(0,\sqrt{3}).$

Show that $$f$$ is 1-Lipschitz but has no 1-Lipschitz extension to $$Y\cup \{(0,0)\}$$.

4. However, the Kirszbraun extension theorem states that any Lipschitz map between any two Hilbert spaces may be extended whilst preserving the Lipschitz constant.

Example 47

1. Let $$E\subset \mathbb R^m$$ be $$n$$-rectifiable. Show that $$\mathcal H^n|_E$$ is $$\sigma$$-finite.

2. Show that Theorem 14 may not be true if $$E$$ does not satisfy $$\mathcal H^n(E)<\infty$$.