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PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedaudited 2026-09-22
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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:GC and every hG, Gf(hx)dμ(x)=Gf(xh)dμ(x)=Gf(hxh1)dμ(x)=Gf(x1)dμ(x)=Gf(x)dμ(x).

Facts & Assumptions

Given: Assume the Axiom of Choice, a compact Lie group G with normalized Haar measure μ, an integrable f:GC, and hG.

[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(hE)=μ(E), μ(Eh)=μ(E) and μ(E1)=μ(E) for every Borel EG (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 (Ref)±,(Imf)± are integrable, and its integral is the corresponding signed combination of their integrals (Integrable real and complex functions, and their integrals).

Proof

technique · direct
1.1

Let EG be Borel. Since x1E(hx) is the indicator of h1E, the left invariance in [L1] gives G1E(hx)dμ(x)=μ(h1E)=μ(E); since x1E(xh) is the indicator of Eh1, the right invariance gives G1E(xh)dμ(x)=μ(Eh1)=μ(E); and since x1E(x1) is the indicator of E1, inversion invariance gives G1E(x1)dμ(x)=μ(E1)=μ(E).

L1algebra
2.1

Let g0 be Borel. By [L2] choose an increasing sequence of nonnegative simple Borel functions sng. Each sn 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 sn(hx)dμ=sn(xh)dμ=sn(x1)dμ=sndμ; the same sequences sn(h), sn(h), sn(1) increase to g(h), g(h), g(1), so two applications of monotone convergence in [L2] identify all four nonnegative integrals.

L2step 1.1
3.1

For the conjugation identity write ϕ(x)=hxh1. Applying step 2.1 to the nonnegative Borel function xg(hx) in place of g and then the left-translation identity gives g(hxh1)dμ(x)=g(hx)dμ(x)=g(x)dμ(x) for nonnegative Borel g.

step 2.1
4.1

Now let f be integrable complex-valued. The functions f(h),f(h),f(ϕ()),f(1) are Borel because xhx, xxh, 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 fdμ. 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].

A1L3step 2.1step 3.1

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