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Equality in the three-dimensional Loomis--Whitney inequality forces a measurable body to agree up to a null set with a Cartesian product $E_1\times E_2\times E_3$. This follows from the equality conditions in the two Cauchy-Schwarz inequalities used in its proof. If the body is connected and is a finite union of positive-volume axis-parallel boxes, each $E_i$ is an interval and equality up to a null set upgrades to exact equality with one box.

Each nonnegative integrable $f_j$ depends on all coordinates except the $j$th. In dimension three, Cauchy-Schwarz inequality first in the third coordinate and then in the remaining pair proves the displayed estimate. Induction integrates the last coordinate by Hölder's inequality on the first $d-1$ factors, then applies Hölder's inequality to the remaining factor and the dimension-$(d-1)$ product. Taking indicators of coordinate projections of a Euclidean body gives the volume version of the Loomis--Whitney inequality.

Each coordinate projection $S_j$ is compact, hence a Lebesgue measurable set, because it is the continuous image of the compact set $K$. Its $(d-1)$-dimensional measure is finite. Choose any nonnegative integrable function $\eta$ on $\mathbb{R}$ with integral one, and defineIf $x\in K$, every projection ${\hat{x}}_j$ belongs to $S_j$. If $x\notin K$, $f(x)=0$. Thus for every $j$, $f(x)\le 1_{S_j}({\hat{x}}_j)=\int s_j({\hat{x}}_j,t)dt$, so the introductory estimate applies. Tonelli theorem gives $\Vert s_j{\Vert }_1={\lambda }_{d-1}(S_j)$, andSubstitute these values and raise to the $d$th power:This proves the geometric Loomis--Whitney inequality, including zero-volume sets or zero-measure projections, without any division by their measures.

The uniform covers theorem says that if a multiset $\mathcal{C}$ of coordinate subsets covers every coordinate exactly $k$ times, then a Euclidean body $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ is its coordinate projection of a Euclidean body. Work with bounded Borel bodies, whose projections are Lebesgue measurable sets; an empty-coordinate projection has measure one.

Prove the theorem by induction on dimension, with the one-dimensional case immediate. Slice at the last coordinate, writing $S_t\subseteq {\mathbb{R}}^{n-1}$. The subsets $A\setminus \{n\}$ still cover each remaining coordinate $k$ times, so induction givesFor $A$ not containing $n$, the projected slice is contained in $S_A$, giving a constant bound. For $A$ containing $n$, its measure is the slice measure $f_A(t)$ of $S_A$. Exactly $k$ members of $\mathcal{C}$ contain $n$, counted with multiplicity. Integrate and apply Hölder's inequality to those $k$ factors:Fubini's theorem identifies $\int f_A=|S_A|$, completing the proof. Exact covering is needed for real volumes, which may be smaller than one.

The Bollobas--Thomason box theorem states that there is an axis-parallel box $B$ with $|B|=|S|$ and $|B_A|\le |S_A|$ for every nonempty coordinate subset $A$. Use the logarithmic linear program for the box theorem: if its side lengths are $b_i=e^{x_i}$, maximize ${\sum }_i{x}_i$ subject toThe linear program is feasible by choosing all $x_i$ sufficiently negative and is bounded above by the full-set constraint. Its dual minimizes ${\sum }_A{\lambda }_A\log |S_A|$ subject to ${\lambda }_A\ge 0$ and ${\sum }_{A\ni i}{\lambda }_A=1$ for each $i$: precisely a fractional uniform cover.

The dual feasible polytope is nonempty, bounded and defined by integer coefficients, so an optimal extreme point has rational coordinates. Multiply its weights by a common denominator $k$. The resulting integer uniform cover and the uniform covers theorem giveThus the dual optimum is at least $\log |S|$; weight one on $A=[n]$ attains equality. Strong duality gives a primal optimum ${\sum }_i{x}_i=\log |S|$. The corresponding box has the required volume and projected-volume bounds, proving the theorem.

For the Loomis–Whitney theorem, take the box supplied by the Bollobas--Thomason box theorem. Each side length occurs in exactly $n-1$ projections omitting one coordinate, soThis is the Loomis--Whitney inequality.

For the three projection areas of a volume-one body, necessity is $1=|S{|}^2\le abc$. Conversely suppose $abc\ge 1$ and setThe box $[0,x]\times [0,y]\times [0,z]$ has projected areas $a,b,c$ and volume $V=\sqrt{abc}\ge 1$. Inside it retain the three slabsFor $t>0$, each projection is the full corresponding rectangle, since the slab in the omitted coordinate supplies every point of that projection. The volume is $V[1-(1-t{)}^3]$. Chooseto make the volume one. This is a compact connected union of three positive-thickness boxes. Hence the complete answer is

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 $S\subseteq {\mathbb{R}}^n$ satisfieswhere $S_A$ 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 giveApply Hölder's inequality to the $k$ factors in the integral. Sincewe obtain $|S|\le {\prod }_{A\in \mathcal{C}}|S_A{|}^{1/k}$, 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 thatAn 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 $(x_A{)}_{A\subseteq [n]}$ satisfyingfor every irreducible $k$-uniform cover $\mathcal{C}$, together withThe 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 yieldsThe singleton cover gives the reverse inequality, henceFor any $A\subseteq [n]$, the one-uniform cover consisting of $A$ and the singletons $\{i\}$ for $i\notin A$ now gives $x_A\ge {\prod }_{i\in A}{x}_{\{i\}}$. The defining product inequality gives the reverse bound. Thus all these quantities are equal. Taking the side lengths of $B$ to be $x_{\{1\}},\dots ,x_{\{n\}}$ proves the theorem.

Finally suppose that the proper body $S\subseteq {\mathbb{R}}^3$ satisfiesThis 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 forceup 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

Positive numbers $a,b,c$ are the areas of the three coordinate projections of a volume-one Euclidean body exactly when $abc\ge 1$. Necessity is the three-dimensional Loomis--Whitney inequality. For sufficiency take box lengths $x=\sqrt{ac/b}$, $y=\sqrt{ab/c}$ and $z=\sqrt{bc/a}$, and retain the union of the three slabs with first, second or third normalized coordinate at most $t$. All three projections are full rectangles, while volume is $\sqrt{abc}[1-(1-t{)}^3]$. Choosing $t=1-(1-1/\sqrt{abc}{)}^{1/3}$ gives volume one. The slabs have positive thickness, so the construction uses a body with interior rather than zero-volume additions.