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The definition that Leon Simon gives (lect. of Geom. Measure theory) for n-rectifiability implies that every subset of an n-rectifiable set is rectifiable. To quote: a set $M\subset \mathbb{R}^{n+k}$ is said to be countably $n$-rectifiable if $M\subset M_0\cup_{j=1}^{\infty} F_j(\mathbb{R}^n)$ where $H_n(M_0)=0$ and $F_j$ are Lipschitz functions.

For $k=0$ one could take $id_{\mathbb{R}^n}:\mathbb{R}^n\rightarrow\mathbb{R}^n$ as the Lipschitz map and hence every subset $M$ of $\mathbb{R}^n$ is $n$-rectifiable. Can that be, really? No condition of Lebesgue measurability, nothing? Shouldn't one be able to integrate over $M$, first of all, as that what the word rectifiability is about? The definition of Simon seems to me in harmony with the definition that Federer gives 3.2.14. Now, when Federer uses the symbol $\int_A$ he definitely means that $A$ is measurable,2.4.10. So shouldn't rectifiability say from the beginning we are dealing with Hausdorff measurable sets (in this case Lebesgue)?

Let's say that it is self-understood that $M$ is measurable. Shouldn't there be another condition (other than measurability), for n-rectifibility in $\mathbb{R}^n$, something that guarantees the existence of a tangent space almost everywhere?

I understand the use of Lipschitz functions as one wants to avoid "square filling curves" but it seems to me something is missing. Please help me understand.

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In the case where $k=0$ and the Lipschitz map is the identity, your definition requires $M$ to be a subset of some $M_0 \subseteq \mathbb{R}^n$ such that $H_n(M_0) = 0$. So you don't get that every subset of $\mathbb{R}^n$ is $n$-rectifiable, you get that every subset of a set in $\mathbb{R}^n$ of Lebesgue measure $0$ is $n$-rectifiable. This seems to just correspond to the familiar fact that Lebesgue measure is complete, i.e. every subset of a set of measure $0$ is measurable. –  Paul Siegel May 24 '12 at 19:34
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@Paul: With the identity the definition reads $M\subset M_0\cup \mathbb{R}^n = \mathbb{R}^n$. –  Tapio Rajala May 24 '12 at 23:07
    
The definition of $n$-rectifiable set in $\mathbb{R}^n$ gives the existence of a tangent space almost everywhere. In this case it is just the space $\mathbb{R}^n$ at density points. As mentioned by Benoît, the case $k=0$ is not that interesting. –  Tapio Rajala May 24 '12 at 23:16
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3 Answers

I think you are true that the definition is void when $k=0$, but this only means it is an interesting for $k>0$ only. It even seems to me that this notion makes sense mostly when $n$ is the Hausdorff dimension of the considered set; the important theorem to bear in mind is a decomposition result which I may not remember very precisely, but that roughly says that any (closed ?) set of dimension $n$ is the union of a $n$-rectifiable set and a $n$-totally unrectifiable set (which means a set that has small intersection with every $n$-rectifiable set). The classical example of totally unrectifiable set is the four-corner Cantor set, which has Hausdorff dimension $1$ but meets every Lipschitz curve along a set of null one-dimensional measure.

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Hi Benoit! The way the structure theorem you are referring to is the "Federer-Besicovitch structure theorem" which, in the version I know, says that any Borel set with finite $m$-Hausdorff measure can be written as the union of a $m$-rectifiable set and a purely $m$-unrectifiable set (i.e. it does not contain any rectifiable bit). This part is trivial, but what is difficult is characterizing unrectifiable sets as those which project onto almost all of the $m$-planes with $H^m$-measure zero. –  Otis Chodosh May 25 '12 at 4:33
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I think that the answer to the question is basically that one usually considers rectifiable sets in the sense you give and further specifies that they are measurable.

On the other hand, I do not think you technically need measurability for a.e. tangent planes. Recall that your definition of rectifiability is equivalent to demanding that up to a set of measure zero, the set is contained in a countable union of $C^1$-manifolds. One way to get at the tangent plane of a rectifiable set is to say that a point $x \in S$ has a tangent plane $T_xM_i$ when $x\in M_i$ as long as for $x\in M_j$ for some other $j$ has $T_xM_j= T_xM_i$. Then one would have to check that this is defined fo $H^m$-a.e. for $x\in S$, but this follows from a sort of transversality argument, intuitively because $M_i$ and $M_j$ cannot intersect very often if they do not have the same tangent planes at their intersections. So, this argument should go through without reference to the measurability of $S$ at all.

This said, as far as I know, people work with rectifiable sets which are also measurable, so it is just an issue of symantics whether or not it is included in the definition.

Rectifiable (measurable) sets can still be very bad though. For example, if you take circles inside $B(0,1)\subset \mathbb{R}^2$, which are centered at the rational points and have radii which are square summable, then the union of these is a $1$-rectifiable set, because it is obviously a countable union of (smooth) $1$-manifolds. However it is dense in the ball! On the other hand, it has tangent planes almost everywhere!

So, rectifiability does not really avoid "square filling things" in this sense.

If it helps, an example of a non-rectifiable set is as follows. Take a triangle. Remove a regular hexagon inscribed in the triangle (side length 1/3 that of the triangle). This gives three triangles. Continue doing this and then take the intersection of all of these. This is a purely $1$-rectifiable set, roughly because if you project on to an axis perpendicular to a base of the triangle, then you get a cantor set, of $H^1$-measure zero and it is not hard to see that a $1$-rectifiable set must have at most one direction with this property (this set has three).


This is in response to your comment on your answer:

You ask why your original definition and this one are the same. Let me list three equivalent definitions of $k$-rectifiability:

(1) Up to a set of measure zero $S \subset \cup_{i=1}^\infty M_i$ where the $M_i$ are $C^1$ $k$-dimensional submanifolds.

(2) Up to a set of measure zero $S \subset \cup_{i=1}^\infty F_i(\mathbb{R}^k)$ for $F:\mathbb{R}^k\to \mathbb{R}^N$ (where $\mathbb{R}^N$ is the ambient space).

(3) Up to a set of measure zero $S\subset \cup_{i=1}^\infty \text{image}(F_i)$ where the $F_i :T_i \to \mathbb{R}^N$ is $C^1$ and $T_i$ is a $k$-plane in $\mathbb{R}^N$ where if we denote $\Pi_{T_i}$ by the projection $\mathbb{R}^N\to T_i$, we have that $\Pi_{T_i} \circ F_i = Id$.

The equivalence of (1) and (2) is in Leon Simon's book, I believe. It uses a.e. differentiability of Lipschitz functions and a Sard type theorem. To understand (3), notice that the bit about the projection just says that the images of $F_i$ are just graphs over the $T_i$. Graphs are $C^1$ submanifolds, so this (3) => (1). On the other hand any $C^1$ manifold is locally the graph over its tangent plane, so we can slice up a decomposition as in (1) into countably many such graphs, implying (3).

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Thank you Otis, but the way I understand the definition of Allard is that the set $S$, in your notation, is equal (up to a set of measure $0$) to the countable union of images of Borel subsets (if you want) of $\mathbb{R}^k$ in R^n. The main difference is that it is one thing to say it is a subset and another thing to say they are equal. Am I wrong? My understanding of what rectifiability is exactly your definition if you replace $\subset$ by $=$ and add that $S$ is $\mathcal{H}^k$-measurable of finite Hausdorff measure. That is how Pitts defines rectifiability. –  Daniel May 25 '12 at 17:44
    
I guess I found the answer I was looking for. If $S= S_0\bigcup_{i} F_i(mathbb{R}^k)$ is a k-rectifiable set in the sense of Allard and $T\subset S$ is another $\mathcal{H}^k$ measurable subset then one can write $T:= T\cap S_0\cup F_i(F_i^{-1}(T))$. So the definitions are equivalent indeed. Pfiu! Thank you for the answers. –  Daniel May 25 '12 at 18:20
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Thank you for the answers. I would like to insist a bit on the fuzziness of this notion of rectifiability. Needless to say, I am obviously far from being an expert in Geometric Measure Theory.

Another forefather of the field, Allard gives the following definition in "On the first variation of a varifold", Annals of Mathematics, Second Series 95 (3): 417–491.

A subset $A$ of $U$ ($U$ an open subset of \mathbb{R}^n) is $(\mathcal{H}^k,k)$ rectifiable if $\mathcal{H}^k(A)<\infty$ and there are corresponding to each $i=1,2,\ldots$ a member $T_i\in {\bf{G}}(n,k)$ (the grassmannian of $k$-subspaces in $\mathbb{R}^n$) and a continuously differentiable mapping $F_i:T_i\rightarrow \mathbb{R}^n$ with $T_i\circ F_i={\bf{1}_{T_i}}$ ($T_i$ denotes as well the orthogonal projection onto the subspace $T_i$) such that: $\mathcal{H}^k(A\setminus\bigcup_{i=1}^{\infty} image (F_i))$.

Wouldn't you agree that this definition of Allard is different than the one given by Federer? If the answer to this question is yes, then it is not anymore a matter of semantics I believe. We would have different notions of rectifiability in the literature and different notions of integral varifolds, integral currents, etc. By the way, Allard points to the definition that Federer gives.

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Let me point out that I do not understand why Allard includes the condition $T_i\circ F_i={\bf{1}_{T_i}}$. In particular for $k=n$ it means that all $F_i$ equal the identity map and so $A$ would be just $\bR^n$ up to a set of measure zero. Maybe, this is indeed his intention. –  daniel May 25 '12 at 9:28
    
\bR^n above is of course $\mathbb{R}^n$. –  daniel May 25 '12 at 9:28
    
another typo in the third paragraph of my second comment: it should be "such that $\mathcal{H}^k(A\setminus \bigcup_{i=1}^{\infty}~image(F_i))=0$." –  daniel May 25 '12 at 9:34
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Hi daniel, it is more customary to edit your own question in order to write this instead of posting a response. This definition is equivalent. He is basically requiring that the set is contained inside a countable union of $C^1$-graphs. Do you see why this is equivalent? I'm pretty sure that any definition of rectifiability you come across in any recent literature will be equivalent to the one you know, but it is certainly possible to give a lot of different definitions of rectifiability! –  Otis Chodosh May 25 '12 at 14:03
    
I do not see the equivalence, sorry! I would certainly hope that all definitions are equivalent. If you can point me to a reference or bother to explain in more detail, I would really appreciate it. –  Daniel May 25 '12 at 14:55
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