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Haar integration is translation and conjugation invariant

Statement

Assume the Axiom of Choice. Let $G$ be a compact Lie group with normalized Haar measure $μ$ , so that $μ$ is the unique regular Borel probability measure that is left, right and inversion invariant (Normalized Haar measure on a compact Lie group). Then for every integrable $f : G \to C$ and every $h \in G$ , $\int_{G} f (h x) d μ (x) = \int_{G} f (x h) d μ (x) = \int_{G} f (h x h^{- 1}) d μ (x) = \int_{G} f (x^{- 1}) d μ (x) = \int_{G} f (x) d μ (x) .$

Facts & Assumptions

Given: Assume the Axiom of Choice, a compact Lie group $G$ with normalized Haar measure $μ$ , an integrable $f : G \to C$ , and $h \in G$ .

[A1]

The Axiom of Choice is the choice principle of The Axiom of Choice; it is used exactly through [L1].

[L1]

Normalized Haar measure $μ$ is the unique regular Borel probability measure on $G$ that is invariant under left translations, right translations and inversion; in particular $μ (h E) = μ (E)$ , $μ (E h) = μ (E)$ and $μ (E^{- 1}) = μ (E)$ for every Borel $E \subseteq G$ (Normalized Haar measure on a compact Lie group).

[L2]

For nonnegative Borel $g$ the nonnegative integral is the supremum of the integrals of the simple Borel functions below it, the integral of a nonnegative simple function is its finite linear combination of measure values, and an increasing sequence of nonnegative Borel functions with pointwise limit $g$ has integrals converging to the integral of $g$ (The nonnegative integral agrees with the simple integral on simple functions, Every nonnegative measurable function is the increasing limit of simple measurable functions, Monotone convergence for the integral).

[L3]

A complex-valued function is integrable exactly when the four nonnegative functions $(Re f)_{\pm}, (Im f)_{\pm}$ are integrable, and its integral is the corresponding signed combination of their integrals (Integrable real and complex functions, and their integrals).

Proof

technique · direct

1.1L1algebra

Let $E \subseteq G$ be Borel. Since $x \mapsto 1_{E} (h x)$ is the indicator of $h^{- 1} E$ , the left invariance in [L1] gives $\int_{G} 1_{E} (h x) d μ (x) = μ (h^{- 1} E) = μ (E)$ ; since $x \mapsto 1_{E} (x h)$ is the indicator of $E h^{- 1}$ , the right invariance gives $\int_{G} 1_{E} (x h) d μ (x) = μ (E h^{- 1}) = μ (E)$ ; and since $x \mapsto 1_{E} (x^{- 1})$ is the indicator of $E^{- 1}$ , inversion invariance gives $\int_{G} 1_{E} (x^{- 1}) d μ (x) = μ (E^{- 1}) = μ (E)$ .

2.1L2step 1.1

Let $g \geq 0$ be Borel. By [L2] choose an increasing sequence of nonnegative simple Borel functions $s_{n} ↗ g$ . Each $s_{n}$ is a finite linear combination of Borel indicators, so the three translation/inversion identities of step 1.1 and linearity of the simple integral give $\int s_{n} (h x) d μ = \int s_{n} (x h) d μ = \int s_{n} (x^{- 1}) d μ = \int s_{n} d μ$ ; the same sequences $s_{n} (h \cdot)$ , $s_{n} (\cdot h)$ , $s_{n} (\cdot^{- 1})$ increase to $g (h \cdot)$ , $g (\cdot h)$ , $g (\cdot^{- 1})$ , so two applications of monotone convergence in [L2] identify all four nonnegative integrals.

3.1step 2.1

For the conjugation identity write $ϕ (x) = h x h^{- 1}$ . Applying step 2.1 to the nonnegative Borel function $x \mapsto g (h x)$ in place of $g$ and then the left-translation identity gives $\int g (h x h^{- 1}) d μ (x) = \int g (h x) d μ (x) = \int g (x) d μ (x)$ for nonnegative Borel $g$ .

4.1A1L3step 2.1step 3.1∎

Now let $f$ be integrable complex-valued. The functions $f (h \cdot), f (\cdot h), f (ϕ (\cdot)), f (\cdot^{- 1})$ are Borel because $x \mapsto h x$ , $x \mapsto x h$ , $ϕ$ and inversion are homeomorphisms of $G$ , and they are integrable because the identities of steps 2.1 and 3.1 applied to $∣ f ∣$ show that each of these four functions has the finite integral $\int ∣ f ∣ d μ$ . Splitting $f$ into its four nonnegative parts by [L3] and applying the corresponding identity of steps 2.1 and 3.1 to each part, then recombining, gives the displayed chain of equalities. The Axiom of Choice entered only through [L1].

Depends on

Dependency tree · 2 levels

Sources

Anthony W. Knapp, Lie Groups Beyond an Introduction, 2nd ed.