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Normalized Haar probability on a compact group

Statement

Assume AC. Every compact Hausdorff group has a unique left Haar probability measure. It is right invariant and inversion invariant as well.

Facts & Assumptions

Given: A compact Hausdorff group and AC.

[F1]

A left Haar measure exists under AC. (Existence of left and right Haar measures)

[F2]

The whole compact nonempty group has positive finite Haar mass. (Haar measure is positive on nonempty open sets and finite on compact sets)

[F3]

Left Haar measures differ by a positive scalar. (Uniqueness of left Haar measure up to scale)

[A1]

AC is assumed in the choice-function form stated in the cited definition. (The Axiom of Choice)

Proof

technique · direct

1.1

Under the assumed AC, take a left Haar measure $λ$ . Since $G$ is open, nonempty and compact, $0 < λ (G) < \infty$ . Put $μ (E) = λ (E) / λ (G)$ ; it is left Haar and $μ (G) = 1$ . Any other left Haar probability is $c μ$ , and evaluation on $G$ forces $c = 1$ .

F1F2F3A1

2.1

For fixed $a$ , $ν (E) = μ (E a)$ is Radon by the right-translation homeomorphism and is a probability. Left and right translations commute, so $ν (b E) = μ (b E a) = μ (E a) = ν (E)$ . Uniqueness gives $ν = μ$ , proving right invariance. Now $η (E) = μ (E^{- 1})$ is a Radon probability and $η (a E) = μ (E^{- 1} a^{- 1}) = μ (E^{- 1})$ by right invariance; uniqueness again gives $η = μ$ .

F1F3step 1.1∎

Sources

Knapp, Advanced Real Analysis, VI §2, pp.225–230, Lemmas 6.9–6.13. Local argument and conventions as displayed above.

Depends on

Used by

Haar measure on an infinite product of compact groups Example
Normalized Haar measure on a finite group Example

Dependency tree · two levels

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Sources

Knapp, Advanced Real Analysis, VI §2, pp.225–230, Lemmas 6.9–6.13 (standard reference, not scraped)