Kemperman's theorem

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:

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

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$$A_1A_2\dots A_n$$ is also a compact subset and:

$$\! \mu(A_1A_2 \dots A_n) \ge \min \{ \mu(A_1) + \mu(A_2) + \dots + \mu(A_n), 1 \}$$

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.

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.