[proofplan] We prove $|K|^k \leq \prod_{i=1}^r |K_{A_i}|$ by induction on $n$, the ambient dimension. For $n = 1$, the only uniform $k$-cover is the singleton $\{1\}$ repeated $k$ times, and equality holds. For $n \geq 2$, we peel off the last coordinate by splitting the cover into two families: $\mathcal{A}_- = \{A \in \mathcal{A} : n \notin A\}$ (sets not containing $n$) and $\mathcal{A}_+ = \{A \setminus \{n\} : A \in \mathcal{A}, n \in A\}$ (sets containing $n$, with $n$ removed). The induction hypothesis bounds $|K(x_n)|$ section-by-section, the containment $K(x_n)_A \subseteq K_A$ handles the $\mathcal{A}_-$ terms, and Holder's inequality with exponent $k$ (since $|\mathcal{A}_+| = k$) combined with Fubini's theorem handles the $\mathcal{A}_+$ terms. [/proofplan] [step:Base case: $n = 1$] When $n = 1$, the ground set is $[1] = \{1\}$. A uniform $k$-cover of $\{1\}$ must contain $k$ copies of $\{1\}$ (the only non-empty subset of $\{1\}$). The inequality reads \begin{align*} |K|^k \leq |K_{\{1\}}|^k. \end{align*} Since $K \subseteq \mathbb{R}^1$ and $K_{\{1\}} = K$, this is $|K|^k \leq |K|^k$, which holds with equality. [/step] [step:Split the cover into $\mathcal{A}_-$ and $\mathcal{A}_+$ by peeling off the last coordinate] Let $n \geq 2$ and let $\mathcal{A} = \{A_1, \ldots, A_r\}$ be a uniform $k$-cover of $[n]$. Define \begin{align*} \mathcal{A}_- &:= \{A \in \mathcal{A} : n \notin A\}, \\ \mathcal{A}_+ &:= \{A \setminus \{n\} : A \in \mathcal{A},\, n \in A\}. \end{align*} Since every element of $[n]$ belongs to exactly $k$ members of $\mathcal{A}$, the element $n$ belongs to exactly $k$ members. Therefore $|\mathcal{A}_+| = k$. For every $j \in [n-1]$, the element $j$ appears in exactly $k$ members of $\mathcal{A}$. Some of these members contain $n$ (and contribute $j$ to $\mathcal{A}_+$ after removing $n$) and the rest do not contain $n$ (and contribute $j$ to $\mathcal{A}_-$). Therefore $j$ belongs to exactly $k$ members of $\mathcal{A}_- \cup \mathcal{A}_+$. This shows that $\mathcal{A}_- \cup \mathcal{A}_+$ is a uniform $k$-cover of $[n-1]$. [guided] The splitting strategy is the natural one for an induction on dimension: we express the $n$-dimensional problem in terms of $(n-1)$-dimensional sections. The cover $\mathcal{A}$ splits into sets that do and do not involve the last coordinate. Removing the index $n$ from the sets in $\mathcal{A}$ that contain it produces $\mathcal{A}_+$, and the sets not containing $n$ form $\mathcal{A}_-$. Together, $\mathcal{A}_- \cup \mathcal{A}_+$ is a uniform $k$-cover of $[n-1]$, which is exactly what the induction hypothesis requires. The count $|\mathcal{A}_+| = k$ will be essential: it is the number of functions to which we apply Holder's inequality in the integration step. [/guided] [/step] [step:Establish the two key inequalities relating sections to projections] For $x_n \in \mathbb{R}$, define the section $K(x_n) := \{(x_1, \ldots, x_{n-1}) \in \mathbb{R}^{n-1} : (x_1, \ldots, x_{n-1}, x_n) \in K\}$. **Inequality (1): sections project inside full projections.** If $A \subseteq [n-1]$ (i.e., $n \notin A$), then for every $x_n \in \mathbb{R}$, \begin{align*} K(x_n)_A \subseteq K_A, \end{align*} and therefore $|K(x_n)_A| \leq |K_A|$. This holds because if $(x_i)_{i \in A} \in K(x_n)_A$, there exists a point in $K$ with those $A$-coordinates and last coordinate $x_n$, and projecting that point onto $A$ shows $(x_i)_{i \in A} \in K_A$. **Inequality (2): projections involving $n$ are integrals of section projections.** If $A \subseteq [n]$ with $n \in A$, then by Fubini's theorem, \begin{align*} |K_A| = \int_{\mathbb{R}} |K(x_n)_{A \setminus \{n\}}| \, d\mathcal{L}^1(x_n). \end{align*} This holds because $K_A = \{((x_i)_{i \in A \setminus \{n\}}, x_n) : (x_i)_{i \in A \setminus \{n\}} \in K(x_n)_{A \setminus \{n\}}\}$, so its measure decomposes as an integral over the $x_n$-coordinate of the measures of the section projections. [guided] These two inequalities are the workhorses of the induction. Inequality (1) says that fixing the last coordinate and projecting can only make the set smaller than projecting without fixing. This is a containment of sets: every point in $K(x_n)_A$ comes from a point in $K$ (with last coordinate $x_n$), and projecting that same point onto $A$ gives a point in $K_A$. Inequality (2) is Fubini's theorem: when the projection includes the last coordinate, the measure of the projection decomposes as $|K_A| = \int |K(x_n)_{A \setminus \{n\}}| \, d\mathcal{L}^1(x_n)$. The slice of $K_A$ at height $x_n$ is exactly $K(x_n)_{A \setminus \{n\}}$, and integrating the slice measures recovers the full measure. The distinction is: for sets $A$ not containing $n$, the projection $K_A$ does not depend on $x_n$ at all (it is the "worst case" over all $x_n$), giving an upper bound on each section's projection. For sets $A$ containing $n$, the projection $K_A$ decomposes as an integral over $x_n$, which is what we need to apply Holder's inequality. [/guided] [/step] [step:Apply the induction hypothesis section by section] By the induction hypothesis, the $(n-1)$-dimensional body $K(x_n) \subseteq \mathbb{R}^{n-1}$ satisfies the uniform cover inequality under the cover $\mathcal{A}_- \cup \mathcal{A}_+$ of $[n-1]$: \begin{align*} |K(x_n)|^k \leq \prod_{A \in \mathcal{A}_-} |K(x_n)_A| \cdot \prod_{A \in \mathcal{A}_+} |K(x_n)_A|. \end{align*} Taking the $k$-th root: \begin{align*} |K(x_n)| \leq \prod_{A \in \mathcal{A}_-} |K(x_n)_A|^{1/k} \cdot \prod_{A \in \mathcal{A}_+} |K(x_n)_A|^{1/k}. \end{align*} [/step] [step:Extract the $\mathcal{A}_-$ terms using the containment bound and integrate using Holder] Integrating over $x_n$ and using Fubini's theorem ($|K| = \int |K(x_n)| \, d\mathcal{L}^1(x_n)$): \begin{align*} |K| \leq \int_{\mathbb{R}} \prod_{A \in \mathcal{A}_-} |K(x_n)_A|^{1/k} \cdot \prod_{A \in \mathcal{A}_+} |K(x_n)_A|^{1/k} \, d\mathcal{L}^1(x_n). \end{align*} By inequality (1), for each $A \in \mathcal{A}_-$ (so $n \notin A$), we have $|K(x_n)_A| \leq |K_A|$ for every $x_n$. The factors $|K(x_n)_A|^{1/k}$ for $A \in \mathcal{A}_-$ are therefore bounded by the constants $|K_A|^{1/k}$ and can be extracted from the integral: \begin{align*} |K| \leq \prod_{A \in \mathcal{A}_-} |K_A|^{1/k} \cdot \int_{\mathbb{R}} \prod_{A \in \mathcal{A}_+} |K(x_n)_A|^{1/k} \, d\mathcal{L}^1(x_n). \end{align*} The remaining integral involves a product of $|\mathcal{A}_+| = k$ non-negative measurable functions of $x_n$. We apply Holder's inequality with exponents all equal to $k$ (the $k$-function version: for non-negative measurable functions $f_1, \ldots, f_k$, $\int f_1 \cdots f_k \, d\mathcal{L}^1 \leq \prod_{j=1}^k (\int f_j^k \, d\mathcal{L}^1)^{1/k}$). Writing $\mathcal{A}_+ = \{B_1, \ldots, B_k\}$, with $f_j(x_n) := |K(x_n)_{B_j}|^{1/k}$: \begin{align*} \int_{\mathbb{R}} \prod_{j=1}^{k} |K(x_n)_{B_j}|^{1/k} \, d\mathcal{L}^1(x_n) \leq \prod_{j=1}^{k} \left( \int_{\mathbb{R}} |K(x_n)_{B_j}| \, d\mathcal{L}^1(x_n) \right)^{1/k}. \end{align*} The Holder exponents are valid: we have $k$ functions each raised to the power $1/k$, and $\sum_{j=1}^k \frac{1}{k} = 1$. Each function $x_n \mapsto |K(x_n)_{B_j}|$ is non-negative and measurable (as a section-measure function of a bounded open set). [guided] This step is the heart of the proof. The $\mathcal{A}_-$ terms are handled by the pointwise bound from inequality (1): since $|K(x_n)_A| \leq |K_A|$ for $A \in \mathcal{A}_-$, these factors can be pulled out of the integral as constants. The $\mathcal{A}_+$ terms remain inside the integral and depend on $x_n$. There are exactly $k$ of them (since $n$ belongs to exactly $k$ members of the cover). Holder's inequality with $k$ equal exponents separates the integral of a product into a product of integrals. This is where the uniformity condition is consumed: if some elements appeared in more or fewer than $k$ of the covering sets, the number of factors in $\mathcal{A}_+$ would not match the Holder exponent, and the inequality would not balance. The version of Holder's inequality we use is: for non-negative measurable functions $f_1, \ldots, f_k$ and exponents $p_1 = \cdots = p_k = k$ (with $\sum 1/p_j = 1$), \begin{align*} \int \prod_{j=1}^k f_j \, d\mathcal{L}^1 \leq \prod_{j=1}^k \|f_j\|_{L^k} = \prod_{j=1}^k \left( \int f_j^k \, d\mathcal{L}^1 \right)^{1/k}. \end{align*} Applied with $f_j = |K(x_n)_{B_j}|^{1/k}$, the $L^k$ norm of $f_j$ is $(\int |K(x_n)_{B_j}| \, d\mathcal{L}^1)^{1/k}$. [/guided] [/step] [step:Apply Fubini to identify the integrals and conclude] By inequality (2), for each $B_j \in \mathcal{A}_+$ (where $B_j = A \setminus \{n\}$ for some $A \in \mathcal{A}$ with $n \in A$), we have \begin{align*} \int_{\mathbb{R}} |K(x_n)_{B_j}| \, d\mathcal{L}^1(x_n) = |K_{B_j \cup \{n\}}|. \end{align*} Substituting back into the chain of inequalities: \begin{align*} |K| &\leq \prod_{A \in \mathcal{A}_-} |K_A|^{1/k} \cdot \prod_{j=1}^k |K_{B_j \cup \{n\}}|^{1/k} \\ &= \prod_{A \in \mathcal{A}_-} |K_A|^{1/k} \cdot \prod_{\substack{A \in \mathcal{A} \\ n \in A}} |K_A|^{1/k} \\ &= \prod_{A \in \mathcal{A}} |K_A|^{1/k}. \end{align*} The second equality holds because the terms with $B_j \cup \{n\}$ range over exactly the original sets $A \in \mathcal{A}$ that contain $n$, and the $\mathcal{A}_-$ terms are the sets $A \in \mathcal{A}$ not containing $n$. Together, they account for all $r$ members of $\mathcal{A}$. Raising both sides to the $k$-th power: \begin{align*} |K|^k \leq \prod_{A \in \mathcal{A}} |K_A|. \end{align*} [guided] The final reassembly is a bookkeeping exercise. The product $\prod_{A \in \mathcal{A}} |K_A|^{1/k}$ splits into two sub-products: - Over $A \in \mathcal{A}_-$ (sets not containing $n$): these were extracted as constants. - Over $A \in \mathcal{A}$ with $n \in A$: for each such $A$, Fubini's theorem identifies $\int |K(x_n)_{A \setminus \{n\}}| \, d\mathcal{L}^1(x_n)$ with $|K_A|$. Together, we have $|K| \leq \prod_{A \in \mathcal{A}} |K_A|^{1/k}$, which upon raising to the $k$-th power gives the desired $|K|^k \leq \prod_{A \in \mathcal{A}} |K_A|$. This completes the induction. [/guided] [/step]