[proofplan] We reduce the non-uniform cover inequality to the [Uniform Cover Inequality](/theorems/2609) by exploiting a monotonicity property special to unions of unit cubes: for $B \subseteq A \subseteq [n]$, the projection satisfies $|K_B| \leq |K_A|$. Given a $k$-cover (where each element belongs to at least $k$ sets), we trim it to a uniform $k$-cover by replacing each set $A_j$ with an appropriate subset $A_j' \subseteq A_j$ so that every element belongs to exactly $k$ of the $A_j'$. The monotonicity property ensures $|K_{A_j'}| \leq |K_{A_j}|$, so the right-hand side does not increase. The Uniform Cover Inequality then gives the result. [/proofplan] [step:Establish the monotonicity $|K_B| \leq |K_A|$ for $B \subseteq A$ when $K$ is a union of unit cubes] Suppose $K = \bigcup_{\alpha \in \Lambda} (a_\alpha + [0,1)^n)$ is a union of translates of the unit cube $[0,1)^n$, where the $a_\alpha \in \mathbb{Z}^n$ are distinct lattice points (we may assume integer translates since overlapping unit cubes with integer translates tile without affecting the union's measure, and non-integer translates can be handled by noting that $K$ decomposes into integer-translate copies). For $A \subseteq [n]$, the projection $K_A$ is itself a union of translates of the unit cube $[0,1)^A$ in $\mathbb{R}^A$: \begin{align*} K_A = \bigcup_{\alpha \in \Lambda} \bigl((a_\alpha)_A + [0,1)^A\bigr), \end{align*} where $(a_\alpha)_A = ((a_\alpha)_i)_{i \in A}$ denotes the projection of the lattice point $a_\alpha$ onto the coordinates indexed by $A$. The measure is $|K_A| = |\{(a_\alpha)_A : \alpha \in \Lambda\}|$, the number of distinct projected lattice points (since each unit cube contributes measure $1$ and distinct projected lattice points produce disjoint cubes in $\mathbb{R}^A$). Now let $B \subseteq A$. The map $(a_\alpha)_A \mapsto (a_\alpha)_B$ is a surjection from the set of distinct projected lattice points in $\mathbb{R}^A$ onto those in $\mathbb{R}^B$ (every $B$-projection of a point of $K$ is obtained by further projecting some $A$-projection). Therefore \begin{align*} |K_B| = |\{(a_\alpha)_B : \alpha \in \Lambda\}| \leq |\{(a_\alpha)_A : \alpha \in \Lambda\}| = |K_A|. \end{align*} The inequality holds because the image of a surjection has cardinality at most that of its domain. [guided] Why does this monotonicity hold for unions of unit cubes but not for general open bodies? For a union of unit cubes, the projection measure $|K_A|$ counts the number of distinct lattice points when projected onto the $A$-coordinates. Projecting further (from $A$ to $B \subseteq A$) can only merge lattice points, never create new ones, so the count cannot increase. For a general body, this fails. Consider a thin strip in $\mathbb{R}^2$ along the line $x_2 = x_1$: the body has small area, but its projections onto each coordinate axis are large. Adding more coordinates to the projection set (going from $B = \{1\}$ to $A = \{1, 2\}$) gives the body itself, which can have smaller measure than the one-dimensional projection. The unit-cube structure prevents this because it forces a discrete, grid-aligned geometry where projections are monotone in the coordinate set. [/guided] [/step] [step:Trim the $k$-cover to a uniform $k$-cover without increasing the product] Let $\mathcal{A} = \{A_1, \ldots, A_r\}$ be a $k$-cover of $[n]$, so every element $i \in [n]$ belongs to at least $k$ of the sets $A_j$. For each $i \in [n]$, let $d_i := |\{j : i \in A_j\}| \geq k$ denote the number of sets containing $i$. We construct a uniform $k$-cover $\mathcal{A}' = \{A_1', \ldots, A_r'\}$ with $A_j' \subseteq A_j$ for each $j$, such that every element $i \in [n]$ belongs to exactly $k$ of the $A_j'$. The construction proceeds greedily: for each element $i$ with $d_i > k$, remove $i$ from $d_i - k$ of the sets containing it (choosing arbitrarily which $d_i - k$ sets to remove $i$ from). After processing all elements, each $i$ belongs to exactly $k$ of the modified sets. More precisely, define $A_j' \subseteq A_j$ for each $j$ by: for each $i \in [n]$, among the $d_i$ indices $j$ with $i \in A_j$, keep $i$ in exactly $k$ of them and remove $i$ from the remaining $d_i - k$. This produces sets $A_j' \subseteq A_j$ with $|\{j : i \in A_j'\}| = k$ for all $i \in [n]$, i.e., a uniform $k$-cover. Since $A_j' \subseteq A_j$ for each $j$, the monotonicity established in the first step gives \begin{align*} |K_{A_j'}| \leq |K_{A_j}| \quad \text{for all } j = 1, \ldots, r. \end{align*} Therefore \begin{align*} \prod_{j=1}^r |K_{A_j'}| \leq \prod_{j=1}^r |K_{A_j}|. \end{align*} [guided] The trimming procedure is straightforward: each element $i$ that appears in "too many" sets (more than $k$) is removed from the excess sets. The order of removal does not matter -- what matters is that the final family $\mathcal{A}'$ has $A_j' \subseteq A_j$ for all $j$ and is a uniform $k$-cover. The monotonicity $|K_{A_j'}| \leq |K_{A_j}|$ is the ingredient that makes the trimming work. Replacing $A_j$ with a subset $A_j' \subseteq A_j$ can only decrease the projection measure (for unions of unit cubes), so the product $\prod |K_{A_j}|$ does not increase. For a general body, the projection onto a smaller coordinate set could be larger, and the trimming would potentially increase the right-hand side, invalidating the argument. [/guided] [/step] [step:Apply the Uniform Cover Inequality to conclude] The family $\mathcal{A}' = \{A_1', \ldots, A_r'\}$ is a uniform $k$-cover of $[n]$. By the [Uniform Cover Inequality](/theorems/2609) applied to $K$ and $\mathcal{A}'$: \begin{align*} |K|^k \leq \prod_{j=1}^r |K_{A_j'}|. \end{align*} Combining with the bound from the trimming step: \begin{align*} |K|^k \leq \prod_{j=1}^r |K_{A_j'}| \leq \prod_{j=1}^r |K_{A_j}|. \end{align*} [guided] The structure of the argument is a two-step reduction: 1. **Trim** the $k$-cover to a uniform $k$-cover, using the unit-cube monotonicity to control the right-hand side. 2. **Apply** the Uniform Cover Inequality to the resulting uniform cover. The non-uniform cover inequality thus follows immediately from the uniform version, provided one has the monotonicity $|K_B| \leq |K_A|$ for $B \subseteq A$. This monotonicity is the sole reason the unit-cube hypothesis appears in the theorem statement. [/guided] [/step]