Let $Q:=[0,1]^2$ and suppose we have a horizontal strip $S=\{\alpha\le y\le\beta\}\cap Q$ inside it (with $0\le\alpha<\beta\le 1$).
We want to cover $S$ in an accurate way with vertical bands: given any $\epsilon>0$ we want to find a finite collection of parallelograms $P_1,\dots,P_n$, each having a side in $[0,1]\times\{0\}$ and the opposite one in $[0,1]\times\{1\}$, in such a way that
- $S\subseteq\cup_{i=1}^n P_i$
- $|\cup P_i\setminus S|<\epsilon$ (the wasted area is small).
How can this be done? This is somewhat counterintuitive as clearly in any such collection the $P_i$'s must overlap a lot in the complement of $S$.
The picture shows an example of a cover of the strip (blue) with parallelograms (red):
Motivation: I am trying to prove Davies' theorem, which states that such a good cover is possible if $S$ is any compact subset of $Q$ (but this statement reduces easily to the case where $S$ is a horizontal strip).
This is proved in this paper, which unfortunately I can't access.
Update: In fact, one can also reduce to the case of $\alpha=0$, so that the complement of the strip consists of only one region (i.e. $Q\cap\{y>\beta\}$).
Davies' proof, although elementary, is too long to type here. He controls the leaking area by ensuring that the intersection of every horizontal line which is not too close to the horizontal strip $S$ with the union of the parallelograms has small length so by Fubini the leakage area is small too. Lemma 1 proves the existence of this sort of covering for what Davies calls admissible triangles (triangles with horizontal base whose other two sides when extended meet the horizontal sides of the square in the interior). If this doesn't make sense, send me your email and I will send the paper to you.