I'm working with a covering lemma by Córdoba that states:
"Given a parallelepiped $R \subset \mathbb{R}^3$ with sides parallel to the coordinate axes, denote its side lenghts by $x_R, y_R$ and $z_R$ (the letter indicating the side's direction).
Let $B$ be a family of parallelepipeds in $\mathbb{R}^3$ satisfying
a) Their projections over the coordinate planes belong to some dyadic decomposition of them.
b) $x_R < x_{R'}$ and $y_R < y_{R'} \implies z_R \le z_{R'}$ for every $R, R' \in B$.
Then given $\{R_\alpha\} \subset B$ we can select a subfamily $\{R_j\}$ such that
$|\bigcup R_\alpha| \le c|\bigcup R_j|$
$\int_{\bigcup R_j} \exp (\sum \chi_{R_j}(x))dx \le c|\bigcup R_j| $."
I am trying to prove that, once this is proven, the same result holds without the hypothesis a).
I've tried using the one third trick but I don't see how that would work. I can't apply it to the parallelepipeds $R$ directly (they are not cubes, and their excentricities are uncontrolled), and if I apply it to each side separately, the resulting parallelepipeds could not belong to the same dyadic grid.
I've also tried to find some subsets that satisfy some sparseness condition but the way I see it that would mess up the result b).
Most sources I've found consider reductions to dyadic scenarios trivially valid but ultimately rely on one of these two things that I can't seem to make work. Does anybody know why the reduction works in this case?
I finally found a way around this so I'm posting it here in case someone ever comes through this with the same question.
The original goal is proving the above lemma without hypothesis $a)$. For that, one is supposed to prove the lemma with hypothesis $a)$ and then prove the generalized version.
The thing is, hypothesis $a)$ has apparently been wrongly stated all along. At least other sources talk about it as if it were something different, and this changes everything and makes the problem solvable. So I'm going to explain how.
Hypothesis $a)$ should have been "Their projections over the coordinate AXES belong to some dyadic decomposition of them". This lemma is now proved in a similar way (although some details are a bit uglier), but then the version without hypothesis $a)$ follows from the version with hypothesis $a)$ by (essentially) applying the one third trick to each side of the parallelepipeds.
I don't think I should get into more details here, but if you are reading this from the future and would like to know more you can contact me.