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Given a set $A \subset \mathbb{R}^n$, this is called d-rectifiable if it can be covered by a countable union of images of lipshitz functions from $\mathbb{R}^d $ to $ \mathbb{R}^n $ and a $\mathcal{H}^d$ null set. So given $A_j = f_j (\mathbb{R}^d)$ with $f_j$ Lipshitz, one has that A is d-rectifiable if

$$ A \subset \cup_{j=1} ^\infty A_j \cup A_0 \text{ where } \mathcal{H}^d (A_0)=0 .$$

Now, consider $d=1$ and $n=2$ for simplicity. If I consider a countable number of "bad points", I wouldn't really need that $A_0$ in the definition. The only relevant case is when the cardinality of $A_0$ is more than countable.

Now the question is the following. I know there is the structure theorem by Federer-Fleming which tells you that you can decompose any $\mathcal{H}^d $- measurable set of finite measure in a rectifiable part and in a purely unrectifiable part. Let's call this set $A$, the rectifiable part $A^{rect}$ and the unrectifiable part $A^{unr}$. Given the definition above, the only relevant case is when $\mathcal{H}^d (A^{unr}) >0$, otherwise one could consider it to be the null set in the definition, making $A$ rectifiable.

The question is: why did we choose to define rectifiability that way and not without $A_0$? I mean, one could take $A^{rect}$ to be the union of the $A_j$s above, while discarding the $A_0$ in the unrectifiable part. It somehow seems to even make more sense, you could partition a set $A$ in a good part which is ALL contained in unions of images of Lipshitz functions, and in a bad part which contains all the rest. Is this just a convention or there's some good reason to have this definition? I guess it has to do with the fact of wanting to have some "compactness" in the class of rectifiable sets, meaning that a perturbation by a null set don't make you go out of the class, but maybe there's a deeper reason.

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  • $\begingroup$ This is something I should know better, but I think the reason is that for measure-theoretic stuff like this you're going to get a "left over" $d$-measure zero set, and some $d$-measure zero sets cannot be covered by a countable union of $d$-measure zero images of Lipschitz maps. Regarding this last part, I believe the latter are countable unions of what I called "cone free sets" near the end of this sci.math post, which are first Baire category sets, and so we're looking at $d$-measure zero sets (continued) $\endgroup$
    – Dave L Renfro
    Commented Apr 19, 2023 at 18:45
  • $\begingroup$ that are also first category. However, there exist $d$-measure zero sets that are not first category -- even more, ${\mathbb R}^d$ can be expressed as the union of a $d$-measure zero set and a first category set. Incidentally, being cone free is a stronger smallness notion than nowhere dense -- a cone free set is lower porous (see end of this MSE answer), so their countable unions are even $\sigma$-lower porous (hence have Hausdorff dimension less than $d).$ $\endgroup$
    – Dave L Renfro
    Commented Apr 19, 2023 at 18:51
  • $\begingroup$ @DaveLRenfro yes, some d-measure zero sets cannot be covered by a countable union of Lipshitz graphs (I think a product of suitable Cantor sets does the trick), however the issue I am having is another: since every set can be decomposed in a rectifiable part and in an unrectifiable one, why did they choose to keep this null set in the rectifiable part instead of the unrectifiable one? My guess is that you want to have some sort of compactness which you wouldn't have if you dropped this condition. For example, sets of finite perimeter wouldn't have a rectifiable reduced boundary (I think) $\endgroup$
    – tommy1996q
    Commented Apr 19, 2023 at 22:31
  • $\begingroup$ I think here is a counterexample to sets of locally finite perimeter having rectifiable reduced boundary for your definition. Consider the even extension of $y = x^{1/2}$ to $[-1, 1]$. Its graph $\gamma$ has finite perimeter (you could even give the problem of computing its length to a slightly competent calc student). But the germ of $\gamma$ at $(0, 0)$ probably cannot be covered by Lipschitz curves. $\endgroup$
    – Aidan Backus
    Commented Apr 20, 2023 at 3:14
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    $\begingroup$ I think there are simple practical reasons. Rectifiable sets are often used to integrate wrt. the corresponding Hausdorff-measure, so having an extra null-set does not matter. Also a common way to prove rectifiability is by exhaustion, i.e. cover 1% (in measure) of the set by a Lipschitz map and then iterate. This procedure naturally leaves over a null-set. $\endgroup$
    – mlk
    Commented Apr 20, 2023 at 7:48

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If $f:\mathbb{R}^n\to\mathbb{R}^m$, $m<n$ is Lipschitz continuous, then for almost all $x\in\mathbb{R}^m$, the set $f^{-1}(x)$ is ($n-m$)-rectifiable according to the definition that includes the null set $A_0$. (Such sets are often called countably $(n-m)$-rectifiable). This is a nice generalization of the Sard theorem for Lipschitz mappigns. This is one of the main reasons why the definition is stated the way it is stated. When you remove the set $A_0$ from the definition you will have much less natural and generic examples of rectifiable sets.

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