This post is dedicated to my number one follower for her fourteenth birthday,which I spoiled…

Let be a **compact abelian group.** By this we mean that is at once both an abelian group and a compact Hausdorff topological space, and that the group operations are continuous, meaning that is continuous on and is continuous as a map from to . It is known that there exists a regular Borel measure on , called the Haar measure, which is non-negative, satisfies , and is translation invariant:

for every Borel set . In fact, the Haar measure is known to exist in greater generality ( does not have to be commutative and if one allows to be infinite then can also be merely locally compact). The Haar measure is an indispensable tool in representation theory and in ergodic theory. In this post we will use the weak* compactness of the unit ball of the dual to give a slick proof of the existence of the Haar measure in the abelian compact case.

#### 1. The Kakutani–Markov fixed point theorem

The following theorem can be stated and proved in greater generality with more–or–less the same proof as we present.

**Theorem 1 (Kakutani–Markov fixed point theorem): ***Let be a Banach space, and let be a commuting family of weak- continuous linear maps on . Suppose that is a non-empty, weak*–compact and convex subset of such that for all . Then has a common fixed point in , i.e., there is an such that for all . *

**Proof: **Let us first prove the theorem in the case where . Choose some . Construct the averages

Since is convex and , we have that for all . Since is compact, the sequence contains a subnet that converges to some . To show that , we will prove that for all .

Fix . Then is continuous on in the weak* topology, hence bounded on the weak*–compact set , say for all . Now for any ,

Thus (recall that part of the definition of *subnet *is that goes to infinity with ). That completes the proof for the case where is a singleton.

Now let be arbitrary, and for every finite , denote . is evidently closed. We will show that the family has the finite intersection property. It will follow from compactness that there is some , which will be the sought after fixed point. Now, , so it follows that we only have to show that every is non-empty. This is done by induction on the cardinality of . If , then by the first half of the proof. Suppose that is not empty, and let . Then for every , we have for all

Therefore , so by the first half of the proof there is some fixed under . In other words, , so this set is not empty.

#### 2. The existence of Haar measure for abelian compact groups

We can now prove the existence of Haar measure for compact abelian groups.

**Theorem 2: ***Let be a compact Hausdorff abelian group. Then there exists a Haar measure for . That is, there is a regular Borel probability measure on that is translation invariant. *

**Proof:** For every , let be the translation operator given by . We will find a regular Borel probability measure on such that for all ,

(*)

The regularity of the measure together with Urysohn’s Lemma then implies that satisfies for all Borel and all (this might be trickier than it first seems).

Consider the family of operators on given by . Then by Exercise G in Notes 11 is weak* continuous for all . Moreover, for all and ,

Therefore is a commuting family. Now let be the subset of consisting of all probability measures. Then it is easy to see that is weak* closed and convex, and that leaves invariant. By the Kakutani–Markov Theorem, has a fixed point . By definition of , is a regular Borel probability measure on . By the fact that is a fixed point for we have , which is just another way of writing (*). That completes the proof.

**Exercise A:** It may seem as if the same argument would give the existence of a translation invariant regular Borel probability measure on a **locally** compact Hausdorff space.

- Prove why the theorem fails for non–compact spaces.
- What part of the argument breaks down?
- Make sure you know why that part
*doesn’t* break down in the compact case.