= Solution
A multiset $\mathcal C$ of subsets of $[n]$ is a <uniform cover> of multiplicity $k$ when each coordinate occurs in exactly $k$ members. The <uniform covers theorem> states that every <Euclidean body> $S\subseteq\mathbb R^n$ satisfies
$$
\boxed{|S|^k\leq\prod_{A\in\mathcal C}|S_A|,}
$$
where $S_A$ is the <coordinate projection of a Euclidean body> onto the coordinates in $A$.
We prove it by induction on $n$. Split the cover into $\mathcal C^-$, whose members omit $n$, and $\mathcal C^+$, whose members contain $n$. Exactly $k$ members lie in $\mathcal C^+$. For a last-coordinate value $x$, let $S(x)\subseteq\mathbb R^{n-1}$ be the corresponding slice. Removing $n$ from the members of $\mathcal C^+$ and retaining the members of $\mathcal C^-$ gives a $k$-uniform cover of $[n-1]$. The inductive hypothesis and <Fubini's theorem> give
$$
\begin{aligned}
|S|
&=\int |S(x)|\,dx\\
&\leq
\prod_{A\in\mathcal C^-}|S_A|^{1/k}
\int\prod_{A\in\mathcal C^+}|S(x)_{A\setminus\{n\}}|^{1/k}\,dx.
\end{aligned}
$$
Apply <Hölder's inequality> to the $k$ factors in the integral. Since
$$
\int|S(x)_{A\setminus\{n\}}|\,dx=|S_A|,
$$
we obtain $|S|\leq\prod_{A\in\mathcal C}|S_A|^{1/k}$, which is the result after taking the $k$th power. The one-dimensional base case is immediate.
The <Bollobas--Thomason box theorem> states that for every body $S\subseteq\mathbb R^n$ there is an <axis-parallel box> $B$ such that
$$
\boxed{|B|=|S|\quad\text{and}\quad |B_A|\leq|S_A|\text{ for every }A\subseteq[n].}
$$
An <irreducible uniform cover> cannot be decomposed into two smaller uniform covers. There are only finitely many such covers of $[n]$: encode a cover by its multiplicity vector in $\mathbb N^{2^n}$ and apply the <Dickson lemma>.
Choose a componentwise minimal positive array $(x_A)_{A\subseteq[n]}$ satisfying
$$
x_A\leq|S_A|,
\qquad
|S|^k\leq\prod_{A\in\mathcal C}x_A
$$
for every irreducible $k$-uniform cover $\mathcal C$, together with
$$
x_A\leq\prod_{i\in A}x_{\{i\}}.
$$
The actual projection volumes are feasible by the <uniform covers theorem>, and finiteness gives a minimal array. Every uniform cover is a disjoint union of irreducible ones, so its cover inequality also holds for this array.
Minimality implies that, for each coordinate $i$, some tight uniform-cover inequality can be chosen whose cover contains the singleton $\{i\}$. Indeed, either such an inequality already blocks decreasing $x_{\{i\}}$, or a tight product inequality $x_A=\prod_{j\in A}x_{\{j\}}$ does; in the latter case take a tight cover containing $A$ and replace that occurrence of $A$ by its singleton coordinates. Let these tight covers be $\mathcal C_i$, of multiplicities $k_i$, and let $K=\sum_i k_i$. Their multiset union is a $K$-uniform cover. Removing one copy of every singleton leaves a $(K-1)$-uniform cover, so comparison of its cover inequality with the product of all the tight equalities yields
$$
\prod_{i=1}^n x_{\{i\}}\leq|S|.
$$
The singleton cover gives the reverse inequality, hence
$$
\prod_i x_{\{i\}}=|S|.
$$
For any $A\subseteq[n]$, the one-uniform cover consisting of $A$ and the singletons $\{i\}$ for $i\notin A$ now gives $x_A\geq\prod_{i\in A}x_{\{i\}}$. The defining product inequality gives the reverse bound. Thus all these quantities are equal. Taking the side lengths of $B$ to be $x_{\{1\}},\ldots,x_{\{n\}}$ proves the theorem.
Finally suppose that the proper body $S\subseteq\mathbb R^3$ satisfies
$$
|S|^2=|S_{12}|\,|S_{13}|\,|S_{23}|.
$$
This is equality in the three-dimensional <Loomis--Whitney inequality>. In its proof, equality must hold in both applications of <Cauchy-Schwarz inequality>. Their equality conditions force the three projection indicators to factor through one-dimensional measurable sets $E_1,E_2,E_3$, and force
$$
S=E_1\times E_2\times E_3
$$
up to a set of <Lebesgue measure> zero; this is <Equality in the three-dimensional Loomis--Whitney inequality>.
Because $S$ is connected, each one-coordinate projection is connected and hence is an interval. Because $S$ is a finite union of positive-volume <axis-parallel boxes>, a proper difference between $S$ and the product of those three intervals would contain a positive-volume rectangular cell in a common finite subdivision. That would contradict equality up to measure zero. Consequently the equality is exact and
$$
\boxed{S\text{ is an axis-parallel box}.}
$$
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