# Converse on the rectifiability of products of rectifiable sets

Let $1\leq k\leq m$ and $1\leq l\leq n$ fixed integers, $\mathscr{H}^k$ the $k$ dimensional Hausdorff measure and $E\subset \mathbb{R}^m$. We say that :

(1) $E$ is $k$ rectifiable if there exists $C\subset \mathbb{R}^k$ bounded and a Lipschitz function $f:C\rightarrow\mathbb{R}^m$ such that $E=f(C)$.
(2) $E$ is countably $k$ rectifiable if $E$ is the countable union of $k$ rectifiable sets.
(3) $E$ is countably $(\mathscr{H}^k,k)$ rectifiable is there exists a countably $k$ rectifiable subset $K\subset \mathbb{R}^n$ such that $\mathscr{H}^k(E\setminus K)=0$.
(4) $E$ is $(\mathscr{H}^k,k)$ rectifiable is $E$ is countably $(\mathscr{H}^k,k)$ rectifiable and $\mathscr{H}^k(E)<\infty$.

Theorem 3.2.23. If $E\subset \mathbb{R}^n$ is $k$-rectifiable Borel and $F\subset \mathbb{R}^n$ is $(\mathscr{H}^l,l)$ rectifiable Borel, then $E\times F$ is $(\mathscr{H}^{k+l},k+l)$ rectifiable and $\mathscr{H}^{k+l}\llcorner(E\times F)=(\mathscr{H}^k\llcorner E)\times (\mathscr{H}^l \llcorner F)$ (where $\llcorner$ is the restriction of measures).

Furthermore, this is false in general if $k$ rectifiable is replaced by $(\mathscr{H}^k,k)$ rectifiable.

My question : does any form of converse holds? More precisely for fixed $1\leq k\leq m$ and $1\leq l\leq n$, if $E\subset \mathbb{R}^m$ is $(\mathscr{H}^k,k)$ rectifiable Borel such that for all (or some "large" class of) $(\mathscr{H}^l,l)$ rectifiable Borel set $F\subset \mathbb{R}^n$, $E\times F$ is $(\mathscr{H}^{k+l},k+l)$ rectifiable, is $E$ is countably $k$ rectifiable (or even the stronger $E$ is $k$ rectifiable)? One could imagine more complicated statements where one would have to enjoy this property for all integers $l$ and $n$ such that $1\leq l\leq n$.

Reference : Herbert Federer, Geometric Measure Theory, Springer-Verlag, 1969, 2.10.29, 3.2.14, 3.2.23.

• Maybe I am looking at this the wrong way, but the difference in the two definitions is whether you allow "a few" corners or intersections of two Lipschitz images; I don't see why a statement of that kind would hold considering that we only look at things up to measure zero we can't conclude pointwise statements. I am not sure if you are familiar with rectifiability, but just FYI what people refer to as "$k$-rectifiability" nowadays is your definition (3) (Federer's terminology is not widely used anymore), see Wikipedia. – Silvia Ghinassi Mar 24 '17 at 15:10