It is possible for to be Lebesgue measurable and to not be; a counter example can be found in "Measure zero sets with non-measurable sum." On the other hand, if are Borel measurable, then is the continuous image of the Borel set , so analytic and thus measurable. See the discussion in Gardner's survey for more on this, as well as ways to avoid measurability hypothesis.
We note that in the case that A and B are compact, so is A + B, being the image of the compact set under the continuous addition map : , so the measurability conditions are easy to verify.
We give a well-known argument that follows a general recipe of arguments in measure theory; namely, it establishes a simple case by direct analysis, uses induction to establish a finitary extension of that special case, and then uses general machinery to obtain the general case as a limit. A discussion of this history of this proof can be found in Theorem 4.1 in Gardner's survey on Brunn–Minkowski.
We prove the version of the Brunn–Minkowski theorem that only requires to be measurable and non-empty.
The case that A and B are axis aligned boxes:
By translation invariance of volumes, it suffices to take . Then . In this special case, the Brunn–Minkowski inequality asserts that . After dividing both sides by , this follows from the AM–GM inequality: .
The case where A and B are both disjoint unions of finitely many such boxes:
We will use induction on the total number of boxes, where the previous calculation establishes the base case of two boxes. First, we observe that there is an axis aligned hyperplane H that such that each side of H contains an entire box of A. To see this, it suffices to reduce to the case where A consists of two boxes, and then calculate that the negation of this statement implies that the two boxes have a point in common.
For a body X, we let denote the intersections of X with the "right" and "left" halfspaces defined by H. Noting again that the statement of Brunn–Minkowski is translation invariant, we then translate B so that ; such a translation exists by the intermediate value theorem because is a continuous function, if v is perpendicular to H has limiting values 0 and as , so takes on at some point.
We now have the pieces in place to complete the induction step. First, observe that and are disjoint subsets of , and so Now, both have one fewer box than A, while each have at most as many boxes as B. Thus, we can apply the induction hypothesis: and .
Elementary algebra shows that if , then also , so we can calculate:
The case that A and B are bounded open sets:
In this setting, both bodies can be approximated arbitrarily well by unions of disjoint axis aligned rectangles contained in their interior; this follows from general facts about the Lebesgue measure of open sets. That is, we have a sequence of bodies , which are disjoint unions of finitely many axis aligned rectangles, where , and likewise . Then we have that , so . The right hand side converges to as , establishing this special case.
The case that A and B are compact sets:
For a compact body X, define to be the -thickening of X. Here each is the open ball of radius , so that is a bounded, open set. We note that , so that if X is compact, then . By using associativity and commutativity of Minkowski sum, along with the previous case, we can calculate that . Sending to 0 establishes the result.
The case of bounded measurable sets:
Recall that by the regularity theorem for Lebesgue measure for any bounded measurable set X, and for any , there is a compact set with . Thus, for all k, using the case of Brunn–Minkowski shown for compact sets. Sending establishes the result.
The case of measurable sets:
We let , and again argue using the previous case that , hence the result follows by sending k to infinity.
Proof as a corollary of the Prékopa–Leindler inequality
We give a proof of the Brunn–Minkowski inequality as a corollary to the Prékopa–Leindler inequality, a functional version of the BM inequality. We will first prove PL, and then show that PL implies a multiplicative version of BM, then show that multiplicative BM implies additive BM. The argument here is simpler than the proof via cuboids, in particular, we only need to prove the BM inequality in one dimensions. This happens because the more general statement of the PL-inequality than the BM-inequality allows for an induction argument.
The multiplicative form of the BM inequality
First, we note that the Brunn–Minkowski inequality implies a multiplicative version, using the inequality , which holds for . In particular, . The Prékopa–Leindler inequality is a functional generalization of this version of Brunn–Minkowski.
We will need the one dimensional version of BM, namely that if are measurable, then . First, assuming that are bounded, we shift so that . Thus, , whence by almost disjointedness we have that . We then pass to the unbounded case by filtering with the intervals
We first show the case of the PL inequality. Let , and note that . Thus, by the one-dimensional version of Brunn–Minkowski, we have that . We recall that if is non-negative, then Fubini's theorem implies . Then, we have that , where in the last step we use the weighted AM–GM inequality, which asserts that for .
Now we prove the case. For , we pick and set . For any c, we define , that is, defining a new function on n-1 variables by setting the last variable to be . Applying the hypothesis and doing nothing but formal manipulation of the definitions, we have that .
Thus, by the inductive case applied to the functions , we obtain . We define and similarly. In this notation, the previous calculation can be rewritten as: . Since we have proven this for any fixed , this means that the function satisfy the hypothesis for the one dimensional version of the PL theorem. Thus, we have that , implying the claim by Fubini's theorem. QED
PL implies multiplicative BM
The multiplicative version of Brunn–Minkowski follows from the PL inequality, by taking .
Multiplicative BM implies Additive BM
We now explain how to derive the BM-inequality from the PL-inequality. First, by using the indicator functions for Prékopa–Leindler inequality quickly gives the multiplicative version of Brunn–Minkowski: . We now show how the multiplicative BM-inequality implies the usual, additive version.
We assume that both A,B have positive volume, as otherwise the inequality is trivial, and normalize them to have volume 1 by setting . We define ; note that . With these definitions, and using that , we calculate using the multiplicative Brunn–Minkowski inequality that:
The additive form of Brunn–Minkowski now follows by pulling the scaling out of the leftmost volume calculation and rearranging.
The Brunn–Minkowski inequality gives much insight into the geometry of high dimensional convex bodies. In this section we sketch a few of those insights.
Concavity of the radius function (Brunn's theorem)
Consider a convex body . Let be vertical slices of K. Define to be the radius function; if the slices of K are discs, then r(x) gives the radius of the disc K(x), up to a constant. For more general bodies this radius function does not appear to have a completely clear geometric interpretation beyond being the radius of the disc obtained by packing the volume of the slice as close to the origin as possible; in the case when K(x) is not a disc, the example of a hypercube shows that the average distance to the center of mass can be much larger than r(x). We note that sometimes in the context of a convex geometry, the radius function has a different meaning, here we follow the terminology of this lecture.
By convexity of K, we have that . Applying the Brunn–Minkowski inequality gives , provided . This shows that the radius function is concave on its support, matching the intuition that a convex body does not dip into itself along any direction. This result is sometimes known as Brunn's theorem.
Brunn–Minkowski symmetrization of a convex body
Again consider a convex body . Fix some line and for each let denote the affine hyperplane orthogonal to that passes through . Define, ; as discussed in the previous section, this function is concave. Now, let . That is, is obtained from by replacing each slice with a disc of the same -dimensional volume centered inside of . The concavity of the radius function defined in the previous section implies that that is convex. This construction is called the Brunn–Minkowski symmetrization.
Theorem (Grunbaum's theorem): Consider a convex body . Let be any half-space containing the center of mass of ; that is, the expected location of a uniform point sampled from Then .
Grunbaum's theorem can be proven using Brunn–Minkowski inequality, specifically the convexity of the Brunn–Minkowski symmetrization. See these lecture notes for a proof sketch.
Grunbaum's inequality has the following fair cake cutting interpretation. Suppose two players are playing a game of cutting up an dimensional, convex cake. Player 1 chooses a point in the cake, and player two chooses a hyperplane to cut the cake along. Player 1 then receives the cut of the cake containing his point. Grunbaum's theorem implies that if player 1 chooses the center of mass, then the worst that an adversarial player 2 can do is give him a piece of cake with volume at least a fraction of the total. In dimensions 2 and 3, the most common dimensions for cakes, the bounds given by the theorem are approximately respectively. Note, however, that in dimensions, calculating the centroid is hard, limiting the usefulness of this cake cutting strategy for higher dimensional, but computationally bounded creatures.
Applications of Grunbaum's theorem also appear in convex optimization, specifically in analyzing the converge of the center of gravity method. See theorem 2.1 in these notes.
Let denote the unit ball. For a convex body, K, let define its surface area. This agrees with the usual meaning of surface area by the Minkowski-Steiner formula. Consider the function . The isoperimetric inequality states that this is maximized on Euclidean balls.
Proof of isoperimetric inequality via Brunn–Minkowski
First, observe that Brunn–Minkowski implies where in the last inequality we used that for . We use this calculation to lower bound the surface area of via Next, we use the fact that , which follows from the Minkowski-Steiner formula, to calculate Rearranging this yields the isoperimetric inequality:
Applications to inequalities between mixed volumes
The Brunn–Minkowski inequality can be used to deduce the following inequality , where the term is a mixed-volume. Equality holds iff K,L are homothetic. (See theorem 3.4.3 in Hug and Weil's course on convex geometry.)
We recall the following facts about mixed volumes : , so that in particular if , then .
Let . Brunn's theorem implies that this is concave for . Thus, , where denotes the right derivative. We also have that . From this we get , where we applied BM in the last inequality.
Concentration of measure on the sphere and other strictly convex surfaces
The proof of the Brunn–Minkowski theorem establishes that the function
is concave in the sense that, for every pair of nonempty compact subsets A and B of Rn and every 0 ≤ t ≤ 1,
For convex sets A and B of positive measure, the inequality in the theorem is strict
for 0 < t < 1 unless A and B are positive homothetic, i.e. are equal up to translation and dilation by a positive factor.
It is instructive to consider the case where an square in the plane, and a ball of radius . In this case, is a rounded square, and its volume can be accounted for as the four rounded quarter circles of radius , the four rectangles of dimensions along the sides, and the original square. Thus, .
This example also hints at the theory of mixed-volumes, since the terms that appear in the expansion of the volume of correspond to the differently dimensional pieces of A. In particular, if we rewrite Brunn–Minkowski as , we see that we can think of the cross terms of the binomial expansion of the latter as accounting, in some fashion, for the mixed volume representation of . This same phenomenon can also be seen for the sum of an n-dimensional box and a ball of radius , where the cross terms in , up to constants, account for the mixed volumes. This is made precise for the first mixed volume in the section above on the applications to mixed volumes.
The left-hand side of the BM inequality can in general be much larger than the right side. For instance, we can take X to be the x-axis, and Y the y-axis inside the plane; then each has measure zero but the sum has infinite measure. Another example is given by the Cantor set. If denotes the middle third Cantor set, then it is an exercise in analysis to show that .
The Brunn–Minkowski inequality continues to be relevant to modern geometry and algebra. For instance, there are connections to algebraic geometry, and combinatorial versions about counting sets of points inside the integer lattice.