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 inGardner's survey on Brunn–Minkowski.

We prove the version of the Brunn–Minkowski theorem that only requires$A,B,A+B$ to be measurable and non-empty.

• The case thatA andB are axis aligned boxes:

By translation invariance of volumes, it suffices to take$A=\prod _{i=1}^n[0,a_i],B=\prod _{i=1}^n[0,b_i]$. Then$A+B=\prod _{i=1}^n[0,a_i+b_i]$. In this special case, the Brunn–Minkowski inequality asserts that$\prod (a_i+b_i{)}^{1/n}\ge \prod a_i^{1/n}+\prod b_i^{1/n}$. After dividing both sides by$\prod (a_i+b_i{)}^{1/n}$ , this follows from theAM–GM inequality:$(\prod \frac{a_i}{a_i+b_i}{)}^{1/n}+(\prod \frac{b_i}{a_i+b_i}{)}^{1/n}\le \sum \frac{1}{n}\frac{a_i+b_i}{a_i+b_i}=1$.

• The case whereA andB 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 hyperplaneH that such that each side ofH contains an entire box ofA. To see this, it suffices to reduce to the case whereA 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$X^{-},X^{+}$ denote the intersections ofX 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$\frac{\mu (A^{+})}{\mu (B^{+})}=\frac{\mu (A^{-})}{\mu (B^{-})}$; such a translation exists by the intermediate value theorem because$t\to \mu ((B+tv{)}^{+})$ is a continuous function, ifv is perpendicular toH$\frac{\mu ((B+tv{)}^{+})}{\mu ((B+tv{)}^{-})}$ has limiting values0 and$\mathrm{\infty }$ as${\displaystyle t\to -\mathrm{\infty },t\to \mathrm{\infty }}$, so takes on$\frac{\mu (A^{+})}{\mu (A^{-})}$ at some point.

We now have the pieces in place to complete the induction step. First, observe that$A^{+}+B^{+}$ and${\displaystyle A^{-}+B^{-}}$ are disjoint subsets of$A+B$, and so$\mu (A+B)\ge \mu (A^{+}+B^{+})+\mu (A^{-}+B^{-}).$ Now,$A^{+},A^{-}$ both have one fewer box thanA, while$B^{+},B^{-}$ each have at most as many boxes asB. Thus, we can apply the induction hypothesis:$\mu (A^{+}+B^{+})\ge (\mu (A^{+}{)}^{1/n}+\mu (B^{+}{)}^{1/n}{)}^n$ and$\mu (A^{-}+B^{-})\ge (\mu (A^{-}{)}^{1/n}+\mu (B^{-}{)}^{1/n}{)}^n$.

Elementary algebra shows that if$\frac{\mu (A^{+})}{\mu (B^{+})}=\frac{\mu (A^{-})}{\mu (B^{-})}$, then also$\frac{\mu (A^{+})}{\mu (B^{+})}=\frac{\mu (A^{-})}{\mu (B^{-})}=\frac{\mu (A)}{\mu (B)}$, so we can calculate:

${\displaystyle \begin{array}{r}\mu (A+B)\ge \mu (A^{+}+B^{+})+\mu (A^{-}+B^{-})\ge (\mu (A^{+}{)}^{1/n}+\mu (B^{+}{)}^{1/n}{)}^n+(\mu (A^{-}{)}^{1/n}+\mu (B^{-}{)}^{1/n}{)}^n\\ =\mu (B^{+})(1+\frac{\mu (A^{+}{)}^{1/n}}{\mu (B^{+}{)}^{1/n}}{)}^n+\mu (B^{-})(1+\frac{\mu (A^{-}{)}^{1/n}}{\mu (B^{-}{)}^{1/n}}{)}^n=(1+\frac{\mu (A{)}^{1/n}}{\mu (B{)}^{1/n}}{)}^n(\mu (B^{+})+\mu (B^{-}))=(\mu (B{)}^{1/n}+\mu (A{)}^{1/n}{)}^n\end{array}}$

• The case thatA andB 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$A_k\subseteq A$, which are disjoint unions of finitely many axis aligned rectangles, where$\mu (A\setminus A_k)\le 1/k$, and likewise$B_k\subseteq B$. Then we have that$A+B\supseteq A_k+B_k$, so$\mu (A+B{)}^{1/n}\ge \mu (A_k+B_k{)}^{1/n}\ge \mu (A_k{)}^{1/n}+\mu (B_k{)}^{1/n}$. The right hand side converges to$\mu (A{)}^{1/n}+\mu (B{)}^{1/n}$ as$k\to \mathrm{\infty }$, establishing this special case.

• The case thatA andB are compact sets:

For a compact bodyX, define$X_{ϵ}=X+B(0,ϵ)$ to be the$ϵ$-thickening ofX. Here each$B(0,ϵ)$ is the open ball of radius$ϵ$, so that$X_{ϵ}$ is a bounded, open set.$\bigcap _{ϵ>0}{X}_{ϵ}=\text{cl}(X)$, so that ifX is compact, then$\underset{ϵ\to 0}{lim}\mu (X_{ϵ})=\mu (X)$. By using associativity and commutativity of Minkowski sum, along with the previous case, we can calculate that$\mu ((A+B{)}_{2ϵ}{)}^{1/n}=\mu (A_{ϵ}+B_{ϵ}{)}^{1/n}\ge \mu (A_{ϵ}{)}^{1/n}+\mu (B_{ϵ}{)}^{1/n}$. Sending$ϵ$ to0 establishes the result.

• The case of bounded measurable sets:

Recall that by theregularity theorem for Lebesgue measure for any bounded measurable setX, and for any$k>\ge$, there is a compact set$X_k\subseteq X$ with. Thus,$\mu (A+B)\ge \mu (A_k+B_k)\ge (\mu (A_k{)}^{1/n}+\mu (B_k{)}^{1/n}{)}^n$ for allk, using the case of Brunn–Minkowski shown for compact sets. Sending$k\to \mathrm{\infty }$ establishes the result.

• The case of measurable sets:

We let$A_k=[-k,k{]}^n\cap A,B_k=[-k,k{]}^n\cap B$, and again argue using the previous case that$\mu (A+B)\ge \mu (A_k+B_k)\ge (\mu (A_k{)}^{1/n}+\mu (B_k{)}^{1/n}{)}^n$, hence the result follows by sending k to infinity.

We give a proof of the Brunn–Minkowski inequality as a corollary to thePrékopa–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, the Brunn–Minkowski inequality implies a multiplicative version, using the inequality$\lambda x+(1-\lambda )y\ge x^{\lambda }{y}^{\lambda }$, which holds for$x,y\ge 0,\lambda \in [0,1]$. In particular,$\mu (\lambda A+(1-\lambda )B)\ge (\lambda \mu (A{)}^{1/n}+(1-\lambda )\mu (B{)}^{1/n}{)}^n\ge \mu (A{)}^{\lambda }\mu (B{)}^{1-\lambda }$. The Prékopa–Leindler inequality is a functional generalization of this version of Brunn–Minkowski.

• Prékopa–Leindler inequality

Theorem (Prékopa–Leindler inequality): Fix$\lambda \in (0,1)$. Let$f,g,h:{\mathbb{R}}^n\to {\mathbb{R}}_{+}$ be non-negative, measurable functions satisfying$h(\lambda x+(1-\lambda )y)\ge f(x{)}^{\lambda }g(y{)}^{1-\lambda }$ for all$x,y\in {\mathbb{R}}^n$. Then${\int }_{{\mathbb{R}}^n}h(x)dx\ge ({\int }_{{\mathbb{R}}^n}f(x)dx{)}^{\lambda }({\int }_{{\mathbb{R}}^n}g(x)dx{)}^{1-\lambda }$.

Proof (Mostly followingthis lecture):

We will need the one dimensional version of BM, namely that if$A,B,A+B\subseteq \mathbb{R}$ are measurable, then$\mu (A+B)\ge \mu (A)+\mu (B)$. First, assuming that$A,B$ are bounded, we shift$A,B$ so that$A\cap B=\{0\}$. Thus,$A+B\supset A\cup B$, whence by almost disjointedness we have that$\mu (A+B)\ge \mu (A)+\mu (B)$. We then pass to the unbounded case by filtering with the intervals$[-k,k].$

We first show the$n=1$ case of the PL inequality. Let$L_h(t)=\{x:h(x)\ge t\}$.$L_h(t)\supseteq \lambda L_f(t)+(1-\lambda )L_g(t)$. Thus, by the one-dimensional version of Brunn–Minkowski, we have that$\mu (L_h(t))\ge \mu (\lambda L_f(t)+(1-\lambda )L_g(t))\ge \lambda \mu (L_f(t))+(1-\lambda )\mu (L_g(t))$. We recall that if$f(x)$ is non-negative, then Fubini's theorem implies${\int }_{\mathbb{R}}h(x)dx={\int }_{t\ge 0}\mu (L_h(t))dt$. Then, we have that${\int }_{\mathbb{R}}h(x)dx={\int }_{t\ge 0}\mu (L_h(t))dt\ge \lambda {\int }_{t\ge 0}\mu (L_f(t))+(1-\lambda ){\int }_{t\ge 0}\mu (L_g(t))=\lambda {\int }_{\mathbb{R}}f(x)dx+(1-\lambda ){\int }_{\mathbb{R}}g(x)dx\ge ({\int }_{\mathbb{R}}f(x)dx{)}^{\lambda }({\int }_{\mathbb{R}}g(x)dx{)}^{1-\lambda }$, where in the last step we use theweighted AM–GM inequality, which asserts that$\lambda x+(1-\lambda )y\ge x^{\lambda }{y}^{1-\lambda }$ for$\lambda \in (0,1),x,y\ge 0$.

Now we prove the$n>1$ case. For$x,y\in {\mathbb{R}}^{n-1},\alpha ,\beta \in \mathbb{R}$, we pick$\lambda \in [0,1]$ and set$\gamma =\lambda \alpha +(1-\lambda )\beta$. For any c, we define$h_c(x)=h(x,c)$, that is, defining a new function on n-1 variables by setting the last variable to be$c$. Applying the hypothesis and doing nothing but formal manipulation of the definitions, we have that$h_{\gamma }(\lambda x+(1-\lambda )y)=h(\lambda x+(1-\lambda )y,\lambda \alpha +(1-\lambda )\beta ))=h(\lambda (x,\alpha )+(1-\lambda )(y,\beta ))\ge f(x,\alpha {)}^{\lambda }g(y,\beta {)}^{1-\lambda }=f_{\alpha }(x{)}^{\lambda }{g}_{\beta }(y{)}^{1-\lambda }$.

Thus, by the inductive case applied to the functions$h_{\gamma },f_{\alpha },g_{\beta }$, we obtain${\int }_{{\mathbb{R}}^{n-1}}{h}_{\gamma }(z)dz\ge ({\int }_{{\mathbb{R}}^{n-1}}{f}_{\alpha }(z)dz{)}^{\lambda }({\int }_{{\mathbb{R}}^{n-1}}{g}_{\beta }(z)dz{)}^{1-\lambda }$. We define$H(\gamma ):={\int }_{{\mathbb{R}}^{n-1}}{h}_{\gamma }(z)dz$ and$F(\alpha ),G(\beta )$ similarly. In this notation, the previous calculation can be rewritten as:$H(\lambda \alpha +(1-\lambda )\beta )\ge F(\alpha {)}^{\lambda }G(\beta {)}^{1-\lambda }$. Since we have proven this for any fixed$\alpha ,\beta \in \mathbb{R}$, this means that the function$H,F,G$ satisfy the hypothesis for the one dimensional version of the PL theorem. Thus, we have that${\int }_{\mathbb{R}}H(\gamma )d\gamma \ge ({\int }_{\mathbb{R}}F(\alpha )d\alpha {)}^{\lambda }({\int }_{\mathbb{R}}F(\beta )d\beta {)}^{1-\lambda }$, 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$h=1_{\lambda A+(1-\lambda )B},f=1_A,g=1_B$.

• 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$A,B,\lambda A+(1-\lambda )B$ Prékopa–Leindler inequality quickly gives the multiplicative version of Brunn–Minkowski:$\mu (\lambda A+(1-\lambda )B)\ge \mu (A{)}^{\lambda }\mu (B{)}^{1-\lambda }$. We now show how the multiplicative BM-inequality implies the usual, additive version.

We assume that bothA,B have positive volume, as otherwise the inequality is trivial, and normalize them to have volume 1 by setting$A^{\prime }=\frac{A}{\mu (A{)}^{1/n}},B^{\prime }=\frac{B}{\mu (B{)}^{1/n}}$. We define${\lambda }^{\prime }=\frac{\lambda \mu (B{)}^{1/n}}{(1-\lambda )\mu (A{)}^{1/n}+\lambda \mu (B{)}^{1/n}}$;$1-{\lambda }^{\prime }=\frac{(1-\lambda )\mu (A{)}^{1/n}}{(1-\lambda )\mu (A{)}^{1/n}+\lambda \mu (B{)}^{1/n}}$. With these definitions, and using that$\mu (A^{\prime })=\mu (B^{\prime })=1$, we calculate using the multiplicative Brunn–Minkowski inequality that:

${\displaystyle \mu (\frac{(1-\lambda )A+\lambda B}{(1-\lambda )\mu (A{)}^{1/n}+\lambda \mu (B{)}^{1/n}})=\mu ((1-{\lambda }^{\prime })A^{\prime }+{\lambda }^{\prime }B)\ge \mu (A^{\prime }{)}^{1-{\lambda }^{\prime }}\mu (B^{\prime }{)}^{{\lambda }^{\prime }}=1.}$

The additive form of Brunn–Minkowski now follows by pulling the scaling out of the leftmost volume calculation and rearranging.

First, observe that Brunn–Minkowski implies$\mu (K+ϵB)\ge (\mu (K{)}^{1/n}+ϵV(B{)}^{1/n}{)}^n=\mu (K)(1+ϵ(\frac{\mu (B)}{\mu (K)}{)}^{1/n}{)}^n\ge \mu (K)(1+nϵ(\frac{\mu (B)}{\mu (K)}{)}^{1/n}),$ where in the last inequality we used that$(1+x{)}^n\ge 1+nx$ for$x\ge 0$. We use this calculation to lower bound the surface area of$K$ via$S(K)=\underset{ϵ\to 0}{lim}\frac{\mu (K+ϵB)-\mu (K)}{ϵ}\ge n\mu (K)(\frac{\mu (B)}{\mu (K)}{)}^{1/n}.$ Next, we use the fact that$S(B)=n\mu (B)$, which follows from theMinkowski-Steiner formula, to calculate$\frac{S(K)}{S(B)}=\frac{S(K)}{n\mu (B)}\ge \frac{\mu (K)(\frac{\mu (B)}{\mu (K)}{)}^{1/n}}{\mu (B)}=\mu (K{)}^{\frac{n-1}{n}}\mu (B{)}^{\frac{1-n}{n}}.$ Rearranging this yields the isoperimetric inequality:$\frac{\mu (B{)}^{1/n}}{S(B{)}^{1/(n-1)}}\ge \frac{\mu (K{)}^{1/n}}{S(K{)}^{1/(n-1)}}.$

We recall the following facts aboutmixed volumes :$\mu ({\lambda }_1{K}_1+{\lambda }_2{K}_2)=\sum _{j_1,\dots ,j_n=1}^{r}V(K_{j_1},\dots ,K_{j_n}){\lambda }_{j_1}\dots {\lambda }_{j_n}$, so that in particular if$g(t)=\mu (K+tL)=\mu (V)+nV(K,\dots ,K,L)t+\dots$, then$g^{\prime }(0)=nV(K,\dots ,K,L)$.

Let$f(t):=\mu (K+tL{)}^{1/n}$. Brunn's theorem implies that this is concave for$t\in [0,1]$. Thus,$f^{+}(0)\ge f(1)-f(0)=\mu (K+L{)}^{1/n}-V(K{)}^{1/n}$, where$f^{+}(0)$ denotes the right derivative. We also have that$f^{+}(0)=\frac{1}{n}\mu (K{)}^{\frac{n-1}{n}}nV(K,\dots ,K,L)$. From this we get$\mu (K{)}^{\frac{1-n}{n}}V(K,\dots ,K,L)\ge \mu (K+L{)}^{1/n}-V(K{)}^{1/n}\ge V(L{)}^{1/n}$, where we applied BM in the last inequality.

Proof: Let$\delta ={ϵ}^2/8$, and let$Y=S\setminus X_{ϵ}$. Then, for$x\in X,y\in Y$ one can show, using$\frac{1}{2}||x+y|{|}^2=||x|{|}^2+||y|{|}^2-\frac{1}{2}||x-y|{|}^2$ and$\sqrt{1-x}\le 1-x/2$ for$x\le 1$, that$||\frac{x+y}{2}||\le 1-\delta$. In particular,$\frac{x+y}{2}\in (1-\delta )B(0,1)$.

We let$\overline{X}=\text{Conv}(X,\{0\}),\overline{Y}=\text{Conv}(Y,\{0\})$, and aim to show that$\frac{\overline{X}+\overline{Y}}{2}\subseteq (1-\delta )B(0,1)$. Let$x\in X,y\in Y,\alpha ,\beta \in [0,1],\overline{x}=\alpha x,\overline{y}=\alpha y$. The argument below will be symmetric in$\overline{x},\overline{y}$, so we assume without loss of generality that$\alpha \ge \beta$ and set$\gamma =\beta /\alpha \le 1$. Then,

${\displaystyle \frac{\overline{x}+\overline{y}}{2}=\frac{\alpha x+\beta y}{2}=\alpha \frac{x+\gamma y}{2}=\alpha (\gamma \frac{x+y}{2}+(1-\gamma )\frac{x}{2})=\alpha \gamma \frac{x+y}{2}+\alpha (1-\gamma )\frac{x}{2}}$.

This implies that$\frac{\overline{x}+\overline{y}}{2}\in \alpha \gamma (1-\delta )B+\alpha (1-\gamma )(1-\delta )B=\alpha (1-\delta )B\subseteq (1-\delta )B$. (Using that for any convex body K and$\gamma \in [0,1]$,$\gamma K+(1-\gamma )K=K$.)

Thus, we know that$\frac{\overline{X}+\overline{Y}}{2}\subseteq (1-\delta )B(0,1)$, so$\mu (\frac{\overline{X}+\overline{Y}}{2})\le (1-\delta {)}^n\mu (B(0,1))$. We apply the multiplicative form of the Brunn–Minkowski inequality to lower bound the first term by$\sqrt{\mu (\overline{X})\mu (\overline{Y})}$, giving us$(1-\delta {)}^n\mu (B)\ge \mu (\overline{X}{)}^{1/2}\mu (\overline{Y}{)}^{1/2}$.

${\displaystyle 1-\frac{\nu (X_{ϵ})}{\nu (S)}=\frac{\nu (Y)}{\nu (S)}=\frac{\mu (\overline{Y})}{\mu (B)}\le (1-\delta {)}^{2n}\frac{\mu (B)}{\mu (\overline{X})}\le (1-\delta {)}^{2n}\frac{\nu (S)}{\nu (X)}\le e^{-2n\delta }\frac{\nu (S)}{\nu (X)}=e^{-n{ϵ}^2/4}\frac{\nu (S)}{\nu (X)}}$. QED