Equality in the three-dimensional Loomis--Whitney inequality

For a Euclidean body $S\subseteq {\mathbb{R}}^3$, equalityin 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 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. ConsiderThis 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.