Today I want to prove the Brunn-Minkowski inequality. Let $\mu$ be the lebesgue measure. For any two sets $A, B \subset \mathbb{R}^d$ define $ A+B=\{a+b:a\in A, b\in B\}$. The Brunn-Minkowski theorem then states that if $A$,$B$ and $A+B$ are all measurable then the following inequality holds $$\mu (A+B)^{1/d} \geq \mu (A)^{1/d} + \mu (B)^{1/d}$$

We will first prove this for the case that $A,B$ are cuboids(with axe parallel edges):

Let $A$ be a cube with side lengths $a_1,...,a_n$ and $B$ a cube with side length $b_1,...b_n$. Then $A+B$ is a cube with side lengths $a_1+b_1,...a_n+b_n$. Thus we have:

\begin{equation}

\begin{split}

& \frac{ \mu (A)^{1/n} + \mu (B)^{1/n}} {\mu (A+B)^{1/n}}

=\frac{(\prod_{i=1}^n a_i)^{1/n}+(\prod_{i=1}^n b_i)^{1/n}}{(\prod_{i=1}^n a_i+b_i)^{1/n}} \\

& = (\prod_{i=1}^n \frac{ a_i}{a_i+b_i})^{1/n}+(\prod_{i=1}^n \frac{ b_i}{ a_i+b_i})^{1/n} \\

& \leq \frac{1}{n} \sum_{i=1}^n \frac{ a_i}{a_i+b_i}+\frac{1}{n} \sum_{i=1}^n \frac{ b_i}{ a_i+b_i}=1

\end{split}

\end{equation}

Now comes the cool part. We proof the inequality in the case that $A,B$ are finite unions of cuboids with disjoint interior.For this we do induction over the the number of cubes in the union $A \cup B$. We already dealt with the case when $A$ and $B$ are cuboids. So we may assume that $A$ is the union of at least two cuboids. Furthermore we may assume without loss of generality that the hyperplane $\{x \in \mathbb{R}: x_1=0\}$ seperates two cubes in $A$.

Define the sets $$A^+=\{x\in A: x_1 \geq 0\}, A^-=\{x\in A: x_1 \leq 0\}, B^+=\{x\in B: x_1 \geq 0\}, B^-=\{x\in B: x_1 \leq 0\},$$

Because $B$ is bounded we have $\frac{\mu(B^++te_2}{\mu(B+te_2)}$ is equal to $1$ for $t$ sufficiently large and $0$ for $-t$ sufficiently large. Since the map $ t \mapsto \frac{\mu(B^++te_2}{\mu(B+te_2)}$ is continuous we conclude that there exists $t$ such that $\frac{\mu(B^++te_2)}{\mu(B+te_2)}=\frac{\mu(A^++te_2)}{\mu(A+te_2)}$. So we can assume that $\frac{\mu(B^+}{\mu(B)}=\frac{\mu(A^+)}{\mu(A)}$. Since $A^+$ is a strict subset of $A$ the sets $A^+ \cup B^+$ and $A^- \cup B^-$ are made up of fewer cuboids then $A \cup B$. We have

\begin{equation}

\begin{split}

& \mu(A+ B)\geq \mu(A^+ + B^+)+\mu(A^- + B^-) \\

& \geq (\mu(A^+)^{1/n} +\mu(B^+)^{1/n})^n+(\mu(A^-)^{1/n} +\mu(B^-)^{1/n})^n \\

&=\mu(A^+)(1+(\frac{\mu(B^+)}{\mu(A^+)})^{1/n})^n+\mu(A^-)(1+(\frac{\mu(B^-)}{\mu(A^-)})^{1/n})^n \\

&= (\mu(A^+)+\mu(A^-))(1+(\frac{\mu(B)}{\mu(A)})^{1/n})^n \\

&= (\mu(A)^{1/n}+\mu(B)^{1/n})^n

\end{split}

\end{equation}

So by induction we are done. Now we can approximate any open set by boxes. So the inequalities follows for open sets. Now the inequality easily follows for measurable sets, because any measurable set can be approximated by open sets.