[proofplan] We prove the inequality from the [Prékopa-Leindler inequality](/theorems/4118), which is the functional form of Brunn-Minkowski. First we record compactness and measurability of the Minkowski sum. The zero-measure cases follow from translating one set into the sum. In the positive-measure case, we choose the interpolation parameter dictated by the two volumes, apply Prékopa-Leindler to scaled copies of the indicator functions, and compute the resulting constant exactly. [/proofplan] [step:Verify compactness and measurability of the sets involved] Since $K$ and $L$ are compact subsets of $\mathbb{R}^n$, they are Borel measurable and therefore $\mathcal{L}^n$-measurable. Define the addition map \begin{align*} A: K \times L &\to \mathbb{R}^n \\ (x,y) &\mapsto x+y. \end{align*} The map $A$ is continuous, and $K \times L$ is compact in $\mathbb{R}^{2n}$. Hence \begin{align*} K+L = A(K \times L) \end{align*} is compact. In particular, $K+L$ is Borel measurable and $\mathcal{L}^n(K+L)$ is defined. [guided] Before applying any integral inequality, we need to know that all the sets whose volumes occur are measurable. Compact subsets of $\mathbb{R}^n$ are closed, hence Borel measurable, so $K$ and $L$ are $\mathcal{L}^n$-measurable. We also need measurability of the Minkowski sum. Define the addition map \begin{align*} A: K \times L &\to \mathbb{R}^n \\ (x,y) &\mapsto x+y. \end{align*} This is the restriction to $K \times L$ of the continuous map $\mathbb{R}^n \times \mathbb{R}^n \to \mathbb{R}^n$, $(x,y) \mapsto x+y$. Since $K$ and $L$ are compact, the product $K \times L$ is compact. The continuous image of a compact set is compact, so \begin{align*} K+L = A(K \times L) \end{align*} is compact. Therefore $K+L$ is Borel measurable, and its Lebesgue measure $\mathcal{L}^n(K+L)$ is well-defined. [/guided] [/step] [step:Handle the cases where one set has zero volume] Let \begin{align*} a := \mathcal{L}^n(K), \qquad b := \mathcal{L}^n(L). \end{align*} If $b=0$, choose $y_0 \in L$. The translation map $T_{y_0}: \mathbb{R}^n \to \mathbb{R}^n$ defined by $T_{y_0}(x)=x+y_0$ satisfies $T_{y_0}(K) \subset K+L$, and translation invariance of Lebesgue measure gives \begin{align*} \mathcal{L}^n(K+L) \geq \mathcal{L}^n(T_{y_0}(K)) = \mathcal{L}^n(K)=a. \end{align*} Thus \begin{align*} \mathcal{L}^n(K+L)^{1/n} \geq a^{1/n}=a^{1/n}+b^{1/n}. \end{align*} The case $a=0$ is identical, choosing $x_0 \in K$ and using $x_0+L \subset K+L$. Hence it remains to prove the result when $a>0$ and $b>0$. [/step] [step:Choose the volume-weighted interpolation parameter] Assume $a>0$ and $b>0$. Define \begin{align*} s := a^{1/n}, \qquad t := b^{1/n}, \qquad \lambda := \frac{s}{s+t}. \end{align*} Then $s,t>0$ and $\lambda \in (0,1)$. Define the scaled compact sets \begin{align*} K_\lambda := \lambda^{-1}K = \{\lambda^{-1}x : x \in K\}, \qquad L_\lambda := (1-\lambda)^{-1}L = \{(1-\lambda)^{-1}y : y \in L\}. \end{align*} By the scaling property of Lebesgue measure, \begin{align*} \mathcal{L}^n(K_\lambda)=\lambda^{-n}a, \qquad \mathcal{L}^n(L_\lambda)=(1-\lambda)^{-n}b. \end{align*} Moreover, \begin{align*} \lambda K_\lambda + (1-\lambda)L_\lambda = K+L. \end{align*} [guided] The choice of $\lambda$ is the main normalization. We want an interpolation inequality to produce exactly the expression $(a^{1/n}+b^{1/n})^n$, so we set \begin{align*} s := a^{1/n}, \qquad t := b^{1/n}, \qquad \lambda := \frac{s}{s+t}. \end{align*} Because $a>0$ and $b>0$, we have $s,t>0$, so $\lambda \in (0,1)$. We now define scaled versions of the two sets: \begin{align*} K_\lambda := \lambda^{-1}K = \{\lambda^{-1}x : x \in K\}, \qquad L_\lambda := (1-\lambda)^{-1}L = \{(1-\lambda)^{-1}y : y \in L\}. \end{align*} These are compact because scalar multiplication by a nonzero real number is a homeomorphism of $\mathbb{R}^n$. Lebesgue measure scales by the absolute value of the determinant of the dilation, so \begin{align*} \mathcal{L}^n(K_\lambda)=\lambda^{-n}a, \qquad \mathcal{L}^n(L_\lambda)=(1-\lambda)^{-n}b. \end{align*} Finally, the definition of these scaled sets gives \begin{align*} \lambda K_\lambda + (1-\lambda)L_\lambda = \{\lambda(\lambda^{-1}x)+(1-\lambda)((1-\lambda)^{-1}y):x\in K,\ y\in L\} =K+L. \end{align*} Thus proving a lower bound for the interpolated set $\lambda K_\lambda+(1-\lambda)L_\lambda$ is exactly the same as proving a lower bound for $K+L$. [/guided] [/step] [step:Apply Prékopa-Leindler to the scaled indicator functions] Define [measurable functions](/page/Measurable%20Functions) \begin{align*} f: \mathbb{R}^n &\to [0,\infty) \\ x &\mapsto \mathbb{1}_{K_\lambda}(x), \end{align*} \begin{align*} g: \mathbb{R}^n &\to [0,\infty) \\ y &\mapsto \mathbb{1}_{L_\lambda}(y), \end{align*} and \begin{align*} h: \mathbb{R}^n &\to [0,\infty) \\ z &\mapsto \mathbb{1}_{K+L}(z). \end{align*} For every $x,y \in \mathbb{R}^n$, if $f(x)g(y)=1$, then $x \in K_\lambda$ and $y \in L_\lambda$, hence \begin{align*} \lambda x+(1-\lambda)y \in \lambda K_\lambda+(1-\lambda)L_\lambda=K+L. \end{align*} Therefore \begin{align*} h(\lambda x+(1-\lambda)y) \geq f(x)^\lambda g(y)^{1-\lambda} \end{align*} for all $x,y \in \mathbb{R}^n$, with the convention that $0^\alpha=0$ for $\alpha>0$. By the Prékopa-Leindler inequality (citing a result not yet in the wiki: Prékopa-Leindler Inequality), applied on $\mathbb{R}^n$ with Lebesgue measure $\mathcal{L}^n$ and parameter $\lambda \in (0,1)$, \begin{align*} \int_{\mathbb{R}^n} h(z)\,d\mathcal{L}^n(z) \geq \left(\int_{\mathbb{R}^n} f(x)\,d\mathcal{L}^n(x)\right)^\lambda \left(\int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y)\right)^{1-\lambda}. \end{align*} Since $f,g,h$ are indicators of compact sets, these integrals are finite and equal to the corresponding Lebesgue measures. Thus \begin{align*} \mathcal{L}^n(K+L) \geq \mathcal{L}^n(K_\lambda)^\lambda \mathcal{L}^n(L_\lambda)^{1-\lambda}. \end{align*} [guided] We now convert the set inclusion into the functional hypothesis required by Prékopa-Leindler. Define \begin{align*} f: \mathbb{R}^n &\to [0,\infty) \\ x &\mapsto \mathbb{1}_{K_\lambda}(x), \end{align*} \begin{align*} g: \mathbb{R}^n &\to [0,\infty) \\ y &\mapsto \mathbb{1}_{L_\lambda}(y), \end{align*} and \begin{align*} h: \mathbb{R}^n &\to [0,\infty) \\ z &\mapsto \mathbb{1}_{K+L}(z). \end{align*} These functions are measurable because $K_\lambda$, $L_\lambda$, and $K+L$ are compact. The Prékopa-Leindler hypothesis asks us to compare $h(\lambda x+(1-\lambda)y)$ with $f(x)^\lambda g(y)^{1-\lambda}$. If either $x \notin K_\lambda$ or $y \notin L_\lambda$, then $f(x)^\lambda g(y)^{1-\lambda}=0$, so the desired inequality is automatic because $h$ is nonnegative. If $x \in K_\lambda$ and $y \in L_\lambda$, then \begin{align*} \lambda x+(1-\lambda)y \in \lambda K_\lambda+(1-\lambda)L_\lambda=K+L, \end{align*} so $h(\lambda x+(1-\lambda)y)=1$. Therefore, for all $x,y \in \mathbb{R}^n$, \begin{align*} h(\lambda x+(1-\lambda)y) \geq f(x)^\lambda g(y)^{1-\lambda}. \end{align*} We may now apply the Prékopa-Leindler inequality (citing a result not yet in the wiki: Prékopa-Leindler Inequality). Its hypotheses are satisfied: $f,g,h$ are nonnegative measurable functions on $\mathbb{R}^n$, the parameter satisfies $\lambda \in (0,1)$, and the pointwise interpolation inequality has just been verified. It gives \begin{align*} \int_{\mathbb{R}^n} h(z)\,d\mathcal{L}^n(z) \geq \left(\int_{\mathbb{R}^n} f(x)\,d\mathcal{L}^n(x)\right)^\lambda \left(\int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y)\right)^{1-\lambda}. \end{align*} Because $f,g,h$ are indicator functions of compact sets, their integrals are exactly the Lebesgue measures of those sets. Hence \begin{align*} \mathcal{L}^n(K+L) \geq \mathcal{L}^n(K_\lambda)^\lambda \mathcal{L}^n(L_\lambda)^{1-\lambda}. \end{align*} [/guided] [/step] [step:Compute the constant and take the $n$th root] Using the scaling identities from the previous steps, \begin{align*} \mathcal{L}^n(K+L) &\geq (\lambda^{-n}a)^\lambda ((1-\lambda)^{-n}b)^{1-\lambda}. \end{align*} Since $a=s^n$, $b=t^n$, $\lambda=s/(s+t)$, and $1-\lambda=t/(s+t)$, the right-hand side is \begin{align*} (\lambda^{-n}s^n)^\lambda ((1-\lambda)^{-n}t^n)^{1-\lambda} &= \left(\frac{s}{\lambda}\right)^{n\lambda} \left(\frac{t}{1-\lambda}\right)^{n(1-\lambda)} \\ &= (s+t)^{n\lambda}(s+t)^{n(1-\lambda)} \\ &= (s+t)^n. \end{align*} Therefore \begin{align*} \mathcal{L}^n(K+L) \geq (s+t)^n. \end{align*} Taking the $n$th root, which preserves the inequality because both sides are nonnegative, gives \begin{align*} \mathcal{L}^n(K+L)^{1/n} \geq s+t = \mathcal{L}^n(K)^{1/n}+\mathcal{L}^n(L)^{1/n}. \end{align*} This proves the Brunn-[Minkowski inequality](/theorems/517). Since every convex body in $\mathbb{R}^n$ is, by definition, a nonempty compact convex subset of $\mathbb{R}^n$, the stated particular case follows. [/step]