[proofplan] We first dispatch the case $n=1$ by the classical one-dimensional Prékopa-Leindler theorem of Henstock and Macbeath for extended non-negative [measurable functions](/page/Measurable%20Functions) on $\mathbb R$. For $n\geq 2$, we prove the inequality by converting functions into logarithmic superlevel sets and applying the Brunn-Minkowski set inequality, in its outer-measure form, to those sets. After treating the zero and infinite cases, we normalise $f$ and $g$ to have integral $1$. The hypothesis implies that every logarithmic superlevel set of $h$ contains the Minkowski interpolation of suitable logarithmic superlevel sets of $f$ and $g$, and Brunn-Minkowski gives the corresponding lower bound for its measure. Integrating this lower bound over all real logarithmic levels, and using the same cited one-dimensional theorem for the level-parameter functions, gives the desired estimate by the [layer-cake formula](/theorems/2956). [/proofplan] [step:Reduce to the case where $f$ and $g$ have finite positive integral] Define the extended non-negative numbers \begin{align*} A := \int_{\mathbb R^n} f\,d\mathcal L^n, \qquad B := \int_{\mathbb R^n} g\,d\mathcal L^n, \qquad H := \int_{\mathbb R^n} h\,d\mathcal L^n. \end{align*} If $n=1$, the conclusion is exactly the cited one-dimensional Prékopa-Leindler theorem for extended non-negative Lebesgue measurable functions on $\mathbb R$, so assume $n\geq 2$ for the remaining steps. If $A = 0$ or $B = 0$, then $A^{1-t}B^t = 0$ and the conclusion follows from $H \geq 0$. For each $m \in \mathbb N$, let $B(0,m) := \{x \in \mathbb R^n : |x| < m\}$ denote the open Euclidean ball of radius $m$ centred at $0$. If $A = \infty$ or $B = \infty$, define measurable maps \begin{align*} f_m:\mathbb R^n&\to[0,\infty], & f_m(x)&:=\min\{f(x),m\}\mathbb 1_{B(0,m)}(x),\\ g_m:\mathbb R^n&\to[0,\infty], & g_m(x)&:=\min\{g(x),m\}\mathbb 1_{B(0,m)}(x). \end{align*} These measurable functions have finite integrals, satisfy $0 \leq f_m \leq f$ and $0 \leq g_m \leq g$, and obey $f_m \uparrow f$ and $g_m \uparrow g$ pointwise as $m\to\infty$. Since the case $A=0$ or $B=0$ has already been handled, the [monotone convergence theorem](/theorems/509) implies that, for all sufficiently large $m$, both integrals $\int f_m\,d\mathcal L^n$ and $\int g_m\,d\mathcal L^n$ are positive. For those $m$, the original hypothesis gives \begin{align*} h((1-t)x+ty) \geq f_m(x)^{1-t}g_m(y)^t \end{align*} for all $x,y \in \mathbb R^n$. Hence the finite case gives \begin{align*} H \geq \left(\int_{\mathbb R^n} f_m\,d\mathcal L^n\right)^{1-t} \left(\int_{\mathbb R^n} g_m\,d\mathcal L^n\right)^t. \end{align*} By monotone convergence, the right-hand side tends to $A^{1-t}B^t$, interpreted in $[0,\infty]$. Thus the desired inequality follows once the finite positive case is proved. [/step] [step:Normalise the finite positive case] Assume now that $0 < A < \infty$ and $0 < B < \infty$. Using the conventions for products in $[0,\infty]$, define measurable maps \begin{align*} F: \mathbb R^n &\to [0,\infty], & F(x) &:= A^{-1}f(x), \\ G: \mathbb R^n &\to [0,\infty], & G(y) &:= B^{-1}g(y), \\ K: \mathbb R^n &\to [0,\infty], & K(z) &:= A^{-(1-t)}B^{-t}h(z). \end{align*} Then \begin{align*} \int_{\mathbb R^n} F\,d\mathcal L^n = 1, \qquad \int_{\mathbb R^n} G\,d\mathcal L^n = 1, \end{align*} and the hypothesis becomes \begin{align*} K((1-t)x+ty) \geq F(x)^{1-t}G(y)^t \end{align*} for all $x,y \in \mathbb R^n$. It is therefore enough to prove \begin{align*} \int_{\mathbb R^n} K\,d\mathcal L^n \geq 1, \end{align*} because multiplying by $A^{1-t}B^t$ gives the stated inequality. [/step] [step:Compare the logarithmic superlevel sets by the functional hypothesis] For each $r \in \mathbb R$, define the measurable logarithmic superlevel sets \begin{align*} E_r := \{x \in \mathbb R^n : F(x) > e^{-r}\}, \qquad D_r := \{y \in \mathbb R^n : G(y) > e^{-r}\}, \qquad K_r := \{z \in \mathbb R^n : K(z) > e^{-r}\}. \end{align*} Let $a,b \in \mathbb R$. If $x \in E_a$ and $y \in D_b$, then $F(x) > e^{-a}$ and $G(y) > e^{-b}$. Hence \begin{align*} K((1-t)x+ty) \geq F(x)^{1-t}G(y)^t > e^{-((1-t)a+tb)}. \end{align*} Thus \begin{align*} (1-t)E_a + tD_b \subset K_{(1-t)a+tb}, \end{align*} where $(1-t)E_a+tD_b := \{(1-t)x+ty : x \in E_a,\ y \in D_b\}$. [/step] [step:Apply Brunn-Minkowski in outer-measure form to bound the level-set measures] We use the following outer-measure form of Brunn-Minkowski: if $U,V\subset\mathbb R^n$ are nonempty arbitrary sets and $\lambda\in(0,1)$, then \begin{align*} \mathcal L^{n,*}((1-\lambda)U+\lambda V)^{1/n} \geq (1-\lambda)\mathcal L^{n,*}(U)^{1/n} +\lambda\mathcal L^{n,*}(V)^{1/n}, \end{align*} where $\mathcal L^{n,*}$ denotes $n$-dimensional Lebesgue outer measure, with the usual convention that the right side is $+\infty$ if one of the outer measures is infinite. It is also valid for infinite outer measure by the stated extended-real convention. For measurable sets, $\mathcal L^{n,*}$ agrees with $\mathcal L^n$. For $r \in \mathbb R$ and $a,b \in \mathbb R$ with $(1-t)a+tb=r$, the inclusion from the previous step and this outer-measure [Brunn-Minkowski inequality](/theorems/4080) give \begin{align*} \mathcal L^n(K_r)^{1/n} &\geq \mathcal L^{n,*}((1-t)E_a+tD_b)^{1/n} \\ &\geq (1-t)\mathcal L^n(E_a)^{1/n}+t\mathcal L^n(D_b)^{1/n}, \end{align*} whenever $E_a$ and $D_b$ are both nonempty. If either $E_a$ or $D_b$ is empty, then $\mathcal L^n(E_a)^{1-t}\mathcal L^n(D_b)^t=0$, so the geometric-mean estimate below is immediate. In the nonempty case, the [first inequality](/theorems/2897) is valid because $K_r$ is Lebesgue measurable and contains $(1-t)E_a+tD_b$. In particular, using the arithmetic-geometric mean inequality with weights $1-t$ and $t$, we obtain \begin{align*} \mathcal L^n(K_r) \geq \mathcal L^n(E_a)^{1-t}\mathcal L^n(D_b)^t. \end{align*} Taking the supremum over all $a,b \in \mathbb R$ satisfying $(1-t)a+tb=r$ yields \begin{align*} \mathcal L^n(K_r) \geq \sup_{(1-t)a+tb=r}\mathcal L^n(E_a)^{1-t}\mathcal L^n(D_b)^t. \end{align*} [/step] [step:Integrate the logarithmic level-set estimate by the layer-cake formula] Let $\mathcal L^1$ denote one-dimensional Lebesgue measure on $\mathbb R$. The layer-cake formula for non-negative measurable functions gives \begin{align*} \int_{\mathbb R^n} K\,d\mathcal L^n = \int_0^\infty \mathcal L^n(\{z \in \mathbb R^n : K(z)>s\})\,d\mathcal L^1(s). \end{align*} Using the substitution $s=e^{-r}$, the interval $s \in (0,\infty)$ corresponds exactly to $r \in \mathbb R$, and reversing the limits gives $d\mathcal L^1(s)=e^{-r}\,d\mathcal L^1(r)$ in the transformed integral. Therefore \begin{align*} \int_{\mathbb R^n} K\,d\mathcal L^n = \int_{-\infty}^{\infty} e^{-r}\mathcal L^n(K_r)\,d\mathcal L^1(r). \end{align*} Likewise, \begin{align*} 1 = \int_{\mathbb R^n} F\,d\mathcal L^n = \int_{-\infty}^{\infty} e^{-a}\mathcal L^n(E_a)\,d\mathcal L^1(a), \end{align*} and \begin{align*} 1 = \int_{\mathbb R^n} G\,d\mathcal L^n = \int_{-\infty}^{\infty} e^{-b}\mathcal L^n(D_b)\,d\mathcal L^1(b). \end{align*} Define the measurable functions \begin{align*} \phi: \mathbb R &\to [0,\infty), & \phi(a) &:= e^{-a}\mathcal L^n(E_a), \\ \psi: \mathbb R &\to [0,\infty), & \psi(b) &:= e^{-b}\mathcal L^n(D_b), \\ \omega: \mathbb R &\to [0,\infty], & \omega(r) &:= e^{-r}\mathcal L^n(K_r). \end{align*} The functions $a \mapsto \mathcal L^n(E_a)$ and $b \mapsto \mathcal L^n(D_b)$ are monotone non-decreasing, hence Borel measurable; moreover $\mathcal L^n(E_a) \leq e^a$ and $\mathcal L^n(D_b) \leq e^b$ by Markov's inequality and the identities $\int_{\mathbb R^n}F\,d\mathcal L^n=\int_{\mathbb R^n}G\,d\mathcal L^n=1$. The function $r \mapsto \mathcal L^n(K_r)$ is also monotone non-decreasing, hence $\omega$ is measurable as an extended non-negative function. For all $a,b \in \mathbb R$ with $r=(1-t)a+tb$, the preceding level-set estimate gives \begin{align*} \omega((1-t)a+tb) &= e^{-((1-t)a+tb)}\mathcal L^n(K_{(1-t)a+tb}) \\ &\geq \left(e^{-a}\mathcal L^n(E_a)\right)^{1-t} \left(e^{-b}\mathcal L^n(D_b)\right)^t \\ &= \phi(a)^{1-t}\psi(b)^t. \end{align*} The one-dimensional Prékopa-Leindler theorem used here says: if $\phi,\psi,\omega:\mathbb R\to[0,\infty]$ are Lebesgue measurable and satisfy \begin{align*} \omega((1-t)a+tb)\geq \phi(a)^{1-t}\psi(b)^t \end{align*} for every $a,b\in\mathbb R$, then \begin{align*} \int_{\mathbb R}\omega\,d\mathcal L^1 \geq \left(\int_{\mathbb R}\phi\,d\mathcal L^1\right)^{1-t} \left(\int_{\mathbb R}\psi\,d\mathcal L^1\right)^t, \end{align*} with all integrals interpreted in $[0,\infty]$. The three functions just defined are measurable and satisfy this pointwise hypothesis, so the one-dimensional theorem gives \begin{align*} \int_{-\infty}^{\infty} \omega(r)\,d\mathcal L^1(r) \geq \left(\int_{-\infty}^{\infty} \phi(a)\,d\mathcal L^1(a)\right)^{1-t} \left(\int_{-\infty}^{\infty} \psi(b)\,d\mathcal L^1(b)\right)^t =1. \end{align*} Therefore $\int_{\mathbb R^n} K\,d\mathcal L^n \geq 1$, which proves the normalised case. [/step] [step:Undo the normalisation] From the normalised case applied to $F,G,K$, we have \begin{align*} A^{-(1-t)}B^{-t}\int_{\mathbb R^n} h\,d\mathcal L^n = \int_{\mathbb R^n} K\,d\mathcal L^n \geq 1. \end{align*} Multiplying by $A^{1-t}B^t$ gives \begin{align*} \int_{\mathbb R^n} h\,d\mathcal L^n \geq \left(\int_{\mathbb R^n} f\,d\mathcal L^n\right)^{1-t} \left(\int_{\mathbb R^n} g\,d\mathcal L^n\right)^t. \end{align*} Together with the zero and infinite cases handled at the start, this proves the theorem. [/step]