Kemperman's theorem

Statement

For two subsets

Suppose $G$ is a compact connected topological group with a Haar measure $\mu$ of total volume 1 (so that the Haar measure is a probability measure). Suppose $A,B$ are compact subsets of $G$. Then, the product of subsets$AB$ is also a compact subset and:

$\!\mu (AB)\geq \min\{\mu (A)+\mu (B),1\}$

For finitely many subsets

Suppose $G$ is a compact connected topological group with a Haar measure $\mu$ of total volume 1 (so that the Haar measure is a probability measure). Suppose $A_{1},A_{2},\dots ,A_{n}$ are compact subsets of $G$. Then, the product of subsets$A_{1}A_{2}\dots A_{n}$ is also a compact subset and:

$\!\mu (A_{1}A_{2}\dots A_{n})\geq \min\{\mu (A_{1})+\mu (A_{2})+\dots +\mu (A_{n}),1\}$

Note for abelian groups

When, $G$ satisfies the additional condition of being an abelian group, and the group operations are denoted additively, the product of subsets is termed the Minkowski sum. Thus, this provides a lower bound on the size of the Minkowski sum.

Caveats

Importance of connectedness

Connectedness is important because if $G$ has a proper open subgroup, then by compactness it must have finite index, and then the measure of the subgroup is the reciprocal of that index. Taking $A,B$ to both be that subgroup violates the theorem.

Sharpness of estimate

We can take $G$ to be the circle group and $A,B$ to be arcs starting at the identity element to show that the estimates cannot be improved.

Related facts

• Product of subsets whose total size exceeds size of group equals whole group
• Cauchy-Davenport theorem: A similar fact for a group of prime order.

External links

• Blog post by Terence Tao on Kemperman's theorem