= Equality in the three-dimensional Loomis--Whitney inequality
{c}
For a <Euclidean body> $S\subseteq\mathbb R^3$, equality
$$
|S|^2=|S_{12}|\,|S_{13}|\,|S_{23}|
$$
in the <Loomis--Whitney inequality> forces $S$ to agree outside a <null set> with a <Cartesian product> $E_1\times E_2\times E_3$ of measurable <sets>. The equality conditions in the two <Cauchy-Schwarz inequalities> used in the proof give this product structure. Agreement outside a <null set> alone cannot specify which boundary <points> belong to $S$.
If $S$ is connected and a finite <union> of \b[closed], positive-volume <axis-parallel boxes>, equality does imply that $S$ is exactly one closed <axis-parallel box>. Each <coordinate projection of a Euclidean body> onto one axis is then an interval. On the finite grid of box endpoints, the product structure forces every open cell inside the bounding <axis-parallel box> to belong to $S$. Since $S$ is a <closed set> and is the <closure> of its <interior>, it contains the whole bounding <axis-parallel box>.
Closedness matters. Consider
$$
S=(0,1)^3\ \cup\ \bigl((0,1/2]\times(0,1)\times(0,1]\bigr).
$$
This connected <set> is a <union> of two positive-volume <axis-parallel boxes>. Its volume and all three projected areas are one, so equality holds. But only part of its top face is present, and it is not a <Cartesian product> of intervals.
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