[proofplan] The proof applies the [Prékopa-Leindler inequality](/theorems/4118) to the indicator functions of $K$, $L$, and their Minkowski convex combination. The pointwise hypothesis is exactly the defining containment $(1-t)K+tL \supset \{(1-t)x+ty:x\in K,\ y\in L\}$. Integrating the three indicator functions converts the Prékopa-Leindler conclusion into the desired multiplicative volume inequality. The endpoint cases $t=0$ and $t=1$ are handled separately to avoid endpoint exponent conventions. [/proofplan] [step:Dispose of the endpoint cases $t=0$ and $t=1$] If $t=0$, then \begin{align*} (1-t)K+tL = K, \end{align*} so \begin{align*} \mathcal{L}^n((1-t)K+tL)=\mathcal{L}^n(K)=\mathcal{L}^n(K)^{1}\mathcal{L}^n(L)^0 \end{align*} with the usual convention that $a^0=1$ for $a>0$. If $\mathcal{L}^n(L)=0$, the desired inequality reads $\mathcal{L}^n(K)\geq \mathcal{L}^n(K)$ after interpreting the endpoint case directly from the formula before exponentiation. If $t=1$, then \begin{align*} (1-t)K+tL = L, \end{align*} and the same direct argument gives \begin{align*} \mathcal{L}^n((1-t)K+tL)=\mathcal{L}^n(L). \end{align*} Hence the asserted inequality holds for $t=0$ and $t=1$. For the rest of the proof assume $0<t<1$. [/step] [step:Define the indicator functions for Prékopa-Leindler] Since $K$ and $L$ are compact subsets of $\mathbb{R}^n$, they are Borel measurable. The map \begin{align*} A: K \times L &\to \mathbb{R}^n \\ (x,y) &\mapsto (1-t)x+ty \end{align*} is continuous, and $K\times L$ is compact, so $(1-t)K+tL=A(K\times L)$ is compact and therefore Borel measurable. Define [measurable functions](/page/Measurable%20Functions) \begin{align*} f: \mathbb{R}^n &\to [0,\infty) \\ x &\mapsto \mathbb{1}_K(x), \end{align*} \begin{align*} g: \mathbb{R}^n &\to [0,\infty) \\ y &\mapsto \mathbb{1}_L(y), \end{align*} and \begin{align*} h: \mathbb{R}^n &\to [0,\infty) \\ z &\mapsto \mathbb{1}_{(1-t)K+tL}(z). \end{align*} Because $K$, $L$, and $(1-t)K+tL$ are compact, these functions are integrable with respect to $\mathcal{L}^n$. [/step] [step:Verify the pointwise Prékopa-Leindler hypothesis] Let $x,y\in \mathbb{R}^n$. If $x\in K$ and $y\in L$, then by the definition of $(1-t)K+tL$, \begin{align*} (1-t)x+ty \in (1-t)K+tL. \end{align*} Hence \begin{align*} h((1-t)x+ty)=1=f(x)^{1-t}g(y)^t. \end{align*} If $x\notin K$ or $y\notin L$, then, since $0<t<1$, at least one of the factors $f(x)^{1-t}$ and $g(y)^t$ is $0$, so \begin{align*} f(x)^{1-t}g(y)^t=0 \leq h((1-t)x+ty). \end{align*} Thus, for every $x,y\in\mathbb{R}^n$, \begin{align*} h((1-t)x+ty)\geq f(x)^{1-t}g(y)^t. \end{align*} [guided] The only substantive hypothesis in the Prékopa-Leindler inequality is a pointwise comparison between $h$ evaluated at the convex combination $(1-t)x+ty$ and the weighted geometric mean of $f(x)$ and $g(y)$. Here the three functions are indicators, so the verification is a membership check. Take arbitrary $x,y\in\mathbb{R}^n$. If $x\in K$ and $y\in L$, then the definition of the Minkowski convex combination gives \begin{align*} (1-t)x+ty \in (1-t)K+tL. \end{align*} Therefore $h((1-t)x+ty)=1$. At the same time $f(x)=1$ and $g(y)=1$, so \begin{align*} f(x)^{1-t}g(y)^t=1^{1-t}1^t=1. \end{align*} If $x\notin K$ or $y\notin L$, then at least one of $f(x)$ and $g(y)$ is $0$. Because $0<t<1$, both exponents $1-t$ and $t$ are positive, and hence \begin{align*} f(x)^{1-t}g(y)^t=0. \end{align*} Since $h$ is an indicator function, it is nonnegative everywhere, so \begin{align*} 0 \leq h((1-t)x+ty). \end{align*} Combining the two cases proves that for all $x,y\in\mathbb{R}^n$, \begin{align*} h((1-t)x+ty)\geq f(x)^{1-t}g(y)^t. \end{align*} [/guided] [/step] [step:Apply Prékopa-Leindler and identify the three integrals] By the Prékopa-Leindler inequality (citing a result not yet in the wiki: Prékopa-Leindler inequality), applied to the nonnegative integrable measurable functions $f,g,h:\mathbb{R}^n\to[0,\infty)$ and the parameter $t\in(0,1)$, the verified pointwise hypothesis implies \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)^{1-t} \left(\int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y)\right)^t. \end{align*} By the definition of the indicator functions, \begin{align*} \int_{\mathbb{R}^n} h(z)\,d\mathcal{L}^n(z) &= \int_{\mathbb{R}^n} \mathbb{1}_{(1-t)K+tL}(z)\,d\mathcal{L}^n(z) = \mathcal{L}^n((1-t)K+tL),\\ \int_{\mathbb{R}^n} f(x)\,d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n} \mathbb{1}_{K}(x)\,d\mathcal{L}^n(x) = \mathcal{L}^n(K),\\ \int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y) &= \int_{\mathbb{R}^n} \mathbb{1}_{L}(y)\,d\mathcal{L}^n(y) = \mathcal{L}^n(L). \end{align*} Substituting these identities gives \begin{align*} \mathcal{L}^n((1-t)K+tL) \geq \mathcal{L}^n(K)^{1-t}\mathcal{L}^n(L)^t. \end{align*} [guided] We now use the Prékopa-Leindler inequality (citing a result not yet in the wiki: Prékopa-Leindler inequality). Its hypotheses require nonnegative measurable functions, integrability of the relevant functions, a parameter $t\in(0,1)$, and the pointwise inequality \begin{align*} h((1-t)x+ty)\geq f(x)^{1-t}g(y)^t \end{align*} for all $x,y\in\mathbb{R}^n$. We have already checked the pointwise inequality. The functions are nonnegative because they are indicator functions, measurable because their defining sets are Borel, and integrable because their defining sets are compact and hence have finite Lebesgue measure. Therefore Prékopa-Leindler 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)^{1-t} \left(\int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y)\right)^t. \end{align*} The reason indicators were chosen is that their integrals are exactly the volumes of the underlying sets. Indeed, \begin{align*} \int_{\mathbb{R}^n} h(z)\,d\mathcal{L}^n(z) &= \int_{\mathbb{R}^n} \mathbb{1}_{(1-t)K+tL}(z)\,d\mathcal{L}^n(z) = \mathcal{L}^n((1-t)K+tL),\\ \int_{\mathbb{R}^n} f(x)\,d\mathcal{L}^n(x) &= \int_{\mathbb{R}^n} \mathbb{1}_{K}(x)\,d\mathcal{L}^n(x) = \mathcal{L}^n(K),\\ \int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y) &= \int_{\mathbb{R}^n} \mathbb{1}_{L}(y)\,d\mathcal{L}^n(y) = \mathcal{L}^n(L). \end{align*} Replacing the three integrals by these measures yields \begin{align*} \mathcal{L}^n((1-t)K+tL) \geq \mathcal{L}^n(K)^{1-t}\mathcal{L}^n(L)^t. \end{align*} [/guided] [/step] [step:Relate the multiplicative inequality to volume-radius concavity for convex bodies] Assume in addition that $K$ and $L$ are convex bodies. The usual Brunn-Minkowski concavity inequality for volume radius states that \begin{align*} \mathcal{L}^n((1-t)K+tL)^{1/n} \geq (1-t)\mathcal{L}^n(K)^{1/n} + t\mathcal{L}^n(L)^{1/n}. \end{align*} By the weighted arithmetic-geometric mean inequality applied to the nonnegative numbers $\mathcal{L}^n(K)^{1/n}$ and $\mathcal{L}^n(L)^{1/n}$, this implies \begin{align*} \mathcal{L}^n((1-t)K+tL)^{1/n} \geq \mathcal{L}^n(K)^{(1-t)/n}\mathcal{L}^n(L)^{t/n}. \end{align*} Raising both sides to the power $n$ gives the multiplicative inequality. Thus, for convex bodies, the multiplicative form proved above is compatible with the standard Brunn-Minkowski concavity statement. [/step]