[proofplan]
We prove the Brunn-[Minkowski inequality](/theorems/517) for compact convex sets by reducing it to the multiplicative form of the [Prékopa-Leindler inequality](/theorems/4118). The reduction uses indicator functions of the sets $K$, $L$, and a scaled Minkowski combination of them. After applying Prékopa-Leindler, we choose the scaling parameter so that the resulting inequality is exactly homogeneous in the two volumes. The zero-volume cases are handled separately by translation monotonicity of Lebesgue measure.
[/proofplan]
[step:Handle the case where one set has zero Lebesgue measure]
Set
\begin{align*}
a := \mathcal{L}^n(K)^{1/n}, \qquad b := \mathcal{L}^n(L)^{1/n}.
\end{align*}
Suppose first that $a = 0$. Since $L$ is non-empty, choose a point $x_0 \in K$. The translation map
\begin{align*}
T_{x_0}: \mathbb{R}^n &\to \mathbb{R}^n \\
y &\mapsto x_0 + y
\end{align*}
sends $L$ into $K+L$. Lebesgue measure is translation-invariant, so
\begin{align*}
\mathcal{L}^n(K+L) \geq \mathcal{L}^n(T_{x_0}(L)) = \mathcal{L}^n(L).
\end{align*}
Taking $n$-th roots gives
\begin{align*}
\mathcal{L}^n(K+L)^{1/n} \geq b = a+b.
\end{align*}
The case $b=0$ is identical after interchanging $K$ and $L$.
[/step]
[step:Introduce the normalized Minkowski interpolation]
Assume now that $a>0$ and $b>0$. Define
\begin{align*}
\lambda := \frac{a}{a+b} \in (0,1).
\end{align*}
Define the indicator functions
\begin{align*}
f: \mathbb{R}^n &\to [0,\infty) \\
x &\mapsto \mathbb{1}_K(x),
\end{align*}
and
\begin{align*}
g: \mathbb{R}^n &\to [0,\infty) \\
y &\mapsto \mathbb{1}_L(y).
\end{align*}
Since $K$ and $L$ are compact, both functions are Borel measurable and integrable with respect to $\mathcal{L}^n$, and
\begin{align*}
\int_{\mathbb{R}^n} f(x)\,d\mathcal{L}^n(x) = \mathcal{L}^n(K)=a^n,
\qquad
\int_{\mathbb{R}^n} g(y)\,d\mathcal{L}^n(y) = \mathcal{L}^n(L)=b^n.
\end{align*}
Define
\begin{align*}
h: \mathbb{R}^n &\to [0,\infty) \\
z &\mapsto \mathbb{1}_{\lambda K+(1-\lambda)L}(z),
\end{align*}
where
\begin{align*}
\lambda K+(1-\lambda)L
:=
\{\lambda x+(1-\lambda)y \in \mathbb{R}^n : x \in K,\ y \in L\}.
\end{align*}
Let
\begin{align*}
\Phi: K \times L &\to \mathbb{R}^n \\
(x,y) &\mapsto \lambda x+(1-\lambda)y
\end{align*}
denote the continuous affine map defining this Minkowski interpolation. Since $K \times L$ is compact and $\Phi$ is continuous, the set $\lambda K+(1-\lambda)L = \Phi(K \times L)$ is compact. Hence $h$ is Borel measurable and integrable with respect to $\mathcal{L}^n$.
[/step]
[step:Verify the pointwise hypothesis of Prékopa-Leindler]
Let $x,y \in \mathbb{R}^n$. If $f(x)g(y)=0$, then
\begin{align*}
h(\lambda x+(1-\lambda)y) \geq 0 = f(x)^\lambda g(y)^{1-\lambda}.
\end{align*}
If $f(x)g(y)=1$, then $x \in K$ and $y \in L$, hence
\begin{align*}
\lambda x+(1-\lambda)y \in \lambda K+(1-\lambda)L.
\end{align*}
Therefore
\begin{align*}
h(\lambda x+(1-\lambda)y)=1=f(x)^\lambda g(y)^{1-\lambda}.
\end{align*}
Thus, 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*}
[/step]
[step:Apply Prékopa-Leindler to the indicator functions]
By the [Prékopa-Leindler Inequality](/theorems/???) applied on $(\mathbb{R}^n,\mathcal{B}(\mathbb{R}^n),\mathcal{L}^n)$ with parameter $\lambda \in (0,1)$ to the non-negative [measurable functions](/page/Measurable%20Functions) $f$, $g$, and $h$ satisfying the pointwise hypothesis above, we obtain
\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*}
Substituting the definitions of $f$, $g$, and $h$ gives
\begin{align*}
\mathcal{L}^n(\lambda K+(1-\lambda)L)
\geq
(a^n)^\lambda (b^n)^{1-\lambda}
=
a^{n\lambda} b^{n(1-\lambda)}.
\end{align*}
Taking $n$-th roots,
\begin{align*}
\mathcal{L}^n(\lambda K+(1-\lambda)L)^{1/n}
\geq
a^\lambda b^{1-\lambda}.
\end{align*}
[/step]
[step:Undo the normalization to obtain the Brunn-Minkowski inequality]
By the definition of $\lambda$,
\begin{align*}
\lambda K+(1-\lambda)L
=
\frac{a}{a+b}K+\frac{b}{a+b}L
=
\frac{1}{a+b}(aK+bL).
\end{align*}
Lebesgue measure scales under dilation by the factor $c^n$ for $c>0$, so
\begin{align*}
\mathcal{L}^n(aK+bL)^{1/n}
=
(a+b)\,\mathcal{L}^n(\lambda K+(1-\lambda)L)^{1/n}.
\end{align*}
Using the previous estimate,
\begin{align*}
\mathcal{L}^n(aK+bL)^{1/n}
\geq
(a+b)a^\lambda b^{1-\lambda}.
\end{align*}
This inequality is not yet the desired one, because it concerns $aK+bL$ rather than $K+L$. Instead, apply the preceding Prékopa-Leindler argument directly with the parameter
\begin{align*}
\lambda=\frac{a}{a+b}
\end{align*}
to the rescaled compact convex sets
\begin{align*}
K_0 := \frac{1}{a}K,
\qquad
L_0 := \frac{1}{b}L.
\end{align*}
Then
\begin{align*}
\mathcal{L}^n(K_0)=1,
\qquad
\mathcal{L}^n(L_0)=1,
\end{align*}
and the previous estimate gives
\begin{align*}
\mathcal{L}^n(\lambda K_0+(1-\lambda)L_0)^{1/n} \geq 1.
\end{align*}
Since
\begin{align*}
\lambda K_0+(1-\lambda)L_0
=
\frac{1}{a+b}K+\frac{1}{a+b}L
=
\frac{1}{a+b}(K+L),
\end{align*}
the scaling law for $\mathcal{L}^n$ gives
\begin{align*}
\mathcal{L}^n(K+L)^{1/n}
=
(a+b)\,\mathcal{L}^n(\lambda K_0+(1-\lambda)L_0)^{1/n}
\geq
a+b.
\end{align*}
Recalling $a=\mathcal{L}^n(K)^{1/n}$ and $b=\mathcal{L}^n(L)^{1/n}$, this is precisely
\begin{align*}
\mathcal{L}^n(K+L)^{1/n}
\geq
\mathcal{L}^n(K)^{1/n}+\mathcal{L}^n(L)^{1/n}.
\end{align*}
[/step]
