= Solution
First take nonempty bounded <open sets> $A,B\subset\mathbb R^n$. Their <Minkowski sum> is open and hence a <Lebesgue measurable set>. Apply the <Prékopa–Leindler inequality> to their <indicator functions> and the <indicator function> of $(1-\theta)A+\theta B$. This gives the multiplicative form
$$
\lambda_n((1-\theta)A+\theta B)\geq\lambda_n(A)^{1-\theta}\lambda_n(B)^\theta.
$$
To obtain the usual additive <Brunn–Minkowski inequality>, put $a=\lambda_n(A)^{1/n}$, $b=\lambda_n(B)^{1/n}$, $A_0=A/a$ and $B_0=B/b$. Both normalized <open sets> have <Lebesgue measure> one. For $\theta=b/(a+b)$,
$$
\frac{A+B}{a+b}=(1-\theta)A_0+\theta B_0.
$$
The multiplicative bound and the scaling rule for <Lebesgue measure> imply
$$
\boxed{\lambda_n(A+B)^{1/n}\geq\lambda_n(A)^{1/n}+\lambda_n(B)^{1/n}.}
$$
Replacing $A,B$ by $A\cap B(0,R),B\cap B(0,R)$ and using continuity from below extends this form of the <Brunn–Minkowski inequality> to all nonempty <open sets>, including those of infinite <Lebesgue measure>. This open-set form is sufficient here and avoids any measurability qualification for sums of arbitrary measurable sets.
Back to article page
