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Normalized Haar measure on a torus

Example

Assume the Axiom of Choice and let r0 be an integer. Put Tr=(R/Z)r and let q:RrTr be the quotient map. For nonnegative Borel f, or Haar-integrable complex Borel f, normalized Haar integration is Trf(t)dt=[0,1)rf(q(x))dx. For r=0 the cube and torus are singletons, and the right side uses mass one on the singleton (the empty product convention).

Facts & Assumptions

Given: A nonnegative integer r, Tr, q, and Q=[0,1)r.

[A1]

AC is assumed (The Axiom of Choice); it covers the countable-choice measure suppliers and Haar existence and uniqueness.

[L1]

A compact Hausdorff group has a unique left Haar probability measure, which is also right and inversion invariant (Normalized Haar probability on a compact group). This is the normalized measure on a compact Lie group (Normalized Haar measure on a compact Lie group). A left Haar measure is a nonzero left-invariant Borel measure, compact finite, outer regular on Borel sets and inner regular on open sets (Left Haar integral and left Haar measure).

[L3]

For r1, Lebesgue measure is Radon and compact-inner regular on every Borel set (Lebesgue measure is a Radon measure on R^n).

Verification

technique · direct
1.1

If r=0, both spaces are singletons, their probability measure is Dirac, all translations and inversion are identity, and the asserted integral is evaluation at the point. Now suppose r1. Every coset modulo Zr has a unique representative in Q, by subtracting the coordinatewise integer floors. Define ν(B)=λr(q1(B)Q) for Borel BTr. The preimage is Borel and countable disjoint unions pull back to disjoint unions, so this is a Borel measure; [L2] gives ν(Tr)=λr(Q)=1.

A1L2L3algebra
2.1

Let aRr and E=q1(B)Q. Partition E into the Borel sets Em={xE:x+amQ}, mZr. Only finitely many are nonempty, since xQ and a is fixed. Unique representatives imply that the translates Em+am are pairwise disjoint and their union is exactly q1(q(a)+B)Q: surjectivity follows by subtracting q(a), and injectivity follows because two points of Q differing by an integer vector coincide. Translation invariance and finite additivity give ν(q(a)+B)=mλr(Em+am)=mλr(Em)=ν(B). Half-open faces cause no overlap or omitted boundary points.

L2step 1.1algebra
2.2

To prove regularity, let B be Borel and E=q1(B)Q. By [L3], for each ε>0 there is compact CE with λr(EC)<ε. Then q(C) is compact in B, and ν(Bq(C))λr(EC)<ε. Applying this to Bc gives a compact DBc with ν(BcD)<ε; the open set TrD contains B and its excess over B has measure less than ε. Thus ν is both inner and outer regular.

L3step 1.1algebra
3.1

The probability measure ν is compact finite and nonzero by step 1.1, left invariant by step 2.1, and regular by step 2.2. It is therefore a left Haar probability in the exact sense of [L1]. Apply the unique-left-Haar-probability clause of [L1]; it equals normalized Haar measure and in particular is also inversion invariant. We do not invoke uniqueness restricted to measures already known to be inversion invariant.

A1L1step 1.1step 2.1step 2.2
4.1

The integral formula holds for indicators of Borel sets by the definition of ν in step 1.1 and its identification in step 3.1. Finite linearity gives it for nonnegative simple functions. Increasing simple approximations, or equivalently the defining supremum for the nonnegative integral, give it for nonnegative Borel f. Applying this to the positive and negative parts of the real and imaginary parts proves the formula for integrable complex f; applying it first to f verifies integrability on the cube. The singleton case was established separately in step 1.1.

step 1.1step 3.1algebra

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