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Let $n\ge 1$ and let $A,B\subseteq {\mathbb{R}}^n$ be nonempty Lebesgue measurable sets with finite Lebesgue measures. Assume that $A+B$ is also a Lebesgue measurable set; its measure may be infinite. ThenHere ${\lambda }_n$ denotes $n$-dimensional Lebesgue measure in Euclidean space. Nonempty compact sets and nonempty open sets of finite measure satisfy the measurability assumptions. Measurability of arbitrary summands alone does not guarantee measurability of their Minkowski sum. The theorem does not require convexity. See R. J. Gardner, The Brunn-Minkowski Inequality (2002), Theorem 4.1 and the following remarks.

The $n$th root converts a volume into a quantity that scales like a length: uniform dilation gives ${\lambda }_n(tA{)}^{1/n}=t{\lambda }_n(A{)}^{1/n}$ for $t>0$. In the Euclidean plane the theorem compares square roots of areas, not the areas themselves. For the two rectangles above it reads $3\ge \sqrt{2}+\sqrt{2}=2\sqrt{2}$. Equivalently their sum has area $9\ge 8$. For the two disks, it reads $3\sqrt{\pi }=\sqrt{\pi }+2\sqrt{\pi }$, an equality. These are exact geometric examples, not measured data.

Applying the theorem to the scaled sets gives, whenever that sum is a Lebesgue measurable set,For convex bodies, this means the root-volume of the interpolated shape lies above the straight line between the input root-volumes. In fact $t\mapsto {\lambda }_n((1-t)A+tB{)}^{1/n}$ is a concave function on $[0,1]$: convexity gives $sA+tA=(s+t)A$ for $s,t\ge 0$, so interpolating two intermediate shapes gives another intermediate shape and the same inequality applies. The beauty is that no particular boundary shape appears in the bound: possible pointwise displacements force a minimum amount of space.

For convex bodies $A,B$, equality holds exactly when they are homothetic convex bodies, meaning $B=v+sA$ for some vector $v$ and $s>0$. This equality qualification is part of the Brunn-Minkowski theorem for convex bodies. ^[ref] For positive-sided axis-aligned rectangles, the comparison can be checked directly:Equality requires $ad=bc$: the rectangles must have the same aspect ratio. Our $2\times 1$ and $1\times 2$ rectangles do not. All positive-radius disks are homothetic convex bodies, explaining their equality.

Do not apply that equality criterion without its hypotheses. If $A=\{a\}$, its Lebesgue measure is zero and $A+B$ is a translation of $B$, so equality holds even for nonconvex $B$. Also, changing a summand on a null set can change the measure of its Minkowski sum: in one dimension, $A=[0,1]\cup \{3\}$ and $B=[0,1]$ both have measure $1$, but $A+B=[0,2]\cup [3,4]$ has measure $3$. Removing the single extra point restores equality with sum measure $2$. Thus merely being equal almost everywhere to homothetic convex bodies is not a sufficient equality test for arbitrary sets.

The Brunn-Minkowski theorem also bounds the collision-region volume: for positive-volume compact sets $O,R$, reflection preserves Lebesgue measure, so ${\lambda }_n(O+(-R){)}^{1/n}\ge {\lambda }_n(O{)}^{1/n}+{\lambda }_n(R{)}^{1/n}$. In the toy collision example this gives area at least $(2+\sqrt{2}{)}^2=6+4\sqrt{2}\approx 11.657$, compared with the exact area $12$. A lower bound alone does not identify safe reference positions; the actual Minkowski sum supplies those.

A convex body in ${\mathbb{R}}^n$ is a compact convex set with nonempty interior in the ambient Euclidean space. It consequently has positive finite Lebesgue measure. A filled rectangle with positive side lengths and a closed positive-radius Euclidean disk are planar convex bodies. A line segment in ${\mathbb{R}}^2$ has empty ambient interior and is not a convex body under this convention. This is the convention used in the Brunn-Minkowski theorem. ^[ref]

Two convex bodies are homothetic convex bodies if one is a translation of a positive uniform dilation of the other. Their lengths scale by $s$ and their volumes by $s^n$. Closed disks of positive radii are homothetic convex bodies; axis-aligned positive-sided rectangles are homothetic convex bodies exactly when their aspect ratios agree. The Brunn-Minkowski theorem characterizes this relation by equality of root-volumes in Minkowski addition.

The Minkowski sum of subsets $A,B$ of a vector space collects the result of vector addition for every possible pair:Think of fixing a point $b$ of $B$, making a translation of the whole of $A$ by $b$, and then taking the union of these copies over every $b\in B$. The points of $B$ specify displacement vectors; they need not be the centre of a shape. Minkowski addition applies to nonconvex sets too. It is different from set union: for $A=B=\{0,1\}$ on the real line, $A\cup B=\{0,1\}$ but $A+B=\{0,1,2\}$.

For axis-aligned rectangles in the Euclidean plane, the coordinate intervals add separately:Every coordinate between $0$ and $a+c$ can be split into one coordinate in $[0,a]$ and another in $[0,c]$, and likewise for the vertical coordinate. Thus a $2\times 1$ rectangle plus a $1\times 2$ rectangle gives a $3\times 3$ square of area $9$, although the two input areas total only $4$. The Minkowski sum adds possible displacements, so its area need not be the sum of input areas.

For closed Euclidean disks $D_r=\{x\in {\mathbb{R}}^2:\Vert x\Vert \le r\}$ centred at the origin, the triangle inequality gives $D_r+D_s\subseteq D_{r+s}$. Conversely, if $r+s>0$, split any $x\in D_{r+s}$ as $x=rx/(r+s)+sx/(r+s)$. Therefore $D_r+D_s=D_{r+s}$, including the case $r=s=0$. Disks of radii $1$ and $2$ give a disk of radius $3$, with areas $\pi ,4\pi ,9\pi$. For disks with centres $u,v$, the resulting centre is $u+v$. These two examples lead to the root-of-area comparison in the Brunn-Minkowski theorem.

Minkowski addition is commutative and associative because vector addition is. Its identity is the singleton $\{0\}$, and $A+\varnothing =\varnothing$. For displacement vectors $u,v$ and $t>0$, translations and uniform dilations satisfywhere $tA=\{ta:a\in A\}$ uses scalar multiplication. A Minkowski sum of convex sets is convex: a line segment between $a_1+b_1$ and $a_2+b_2$ is obtained by adding the corresponding points of the line segments in $A$ and $B$. Closure requires care; see sum of two closed sets need not be closed and sum of a compact set and a closed set.