Haar Measure on Euclidean Spaces and Tori

Euclidean Space: Translation Invariance and Haar Measure

On the additive group \(\mathbb{R}^n\), translation invariance means that shifting a set does not change its measure. The familiar volume measure has this property, and it is the natural Haar measure for Euclidean space. On the quotient group \(\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n\), the same idea leads to a finite, normalized measure: integrate over one unit cube, which represents each point of the torus exactly once apart from boundary identifications.

Throughout, \(n\ge1\). The Lebesgue measure space on \(\mathbb{R}^n\) is \((\mathbb{R}^n,\mathcal{L},m)\); \(m\) is finite on bounded sets, sigma-finite, and not finite on all of \(\mathbb{R}^n\). A Haar measure on a locally compact group is a nonzero, translation-invariant, locally finite regular Borel measure. Here locally finite means finite on compact sets, and regular means that Borel sets can be approximated from inside by compact sets and from outside by open sets. Lebesgue measure restricted to the Borel sigma-algebra \(\mathcal{B}_{\mathbb{R}^n}\) is regular and locally finite.

Proof. Lebesgue outer measure \(m^*\) is defined using countable covers by boxes, with the sum of their volumes. Translating every box in a cover by \(a\) preserves its volume. Thus \(m^*(E+a)\le m^*(E)\); applying the same argument to \(E+a\) and the translation \(-a\) gives \(m^*(E)\le m^*(E+a)\). Hence the two outer measures are equal. The Carathéodory measurability criterion is also preserved by translation: for any \(A\subseteq\mathbb{R}^n\), the equality \(m^*(A)=m^*(A\cap E)+m^*(A\setminus E)\) is equivalent, after translating all three sets, to the same equality with \(A+a\) and \(E+a\). Therefore \(E\) is Lebesgue measurable if and only if \(E+a\) is, and their measures agree. Lebesgue measure is nonzero, finite on compact sets, and regular on Borel sets, so its Borel restriction satisfies the stated Haar-measure conditions. \(\square\)

This identifies a Haar measure; no uniqueness assertion is needed here. In particular, the measure is not a probability measure on \(\mathbb{R}^n\), since \(m(\mathbb{R}^n)=\infty\). The distinction between finite and locally finite matters: every compact set has finite measure, but the whole noncompact group has infinite measure.

One Fundamental Domain for the Torus

The \(n\)-torus is the quotient group \(\mathbb{T}^n=\mathbb{R}^n/\mathbb{Z}^n\), with addition inherited from \(\mathbb{R}^n\). Write \(q:\mathbb{R}^n\to\mathbb{T}^n\) for the quotient map. The half-open cube \(F=[0,1)^n\) is a fundamental domain: every coset has exactly one representative in \(F\). The quotient topology makes \(\mathbb{T}^n\) a compact group. For a Borel set \(A\subseteq\mathbb{T}^n\), define its measure by measuring the representatives in \(F\).

The integral formula follows first for indicators from the definition of \(\nu\), then for nonnegative simple functions by finite additivity of the integral, and then for nonnegative Borel functions by the definition of the Lebesgue integral (or monotone approximation by simple functions). For real- or complex-valued Borel \(g\) with \(\int_{\mathbb{T}^n}|g|\,d\nu<\infty\), applying the formula to the positive and negative parts, and to the real and imaginary parts when needed, gives the same formula for \(g\). The measure space \((\mathbb{T}^n,\mathcal{B}_{\mathbb{T}^n},\nu)\) is finite, with \(\nu(\mathbb{T}^n)=m(F)=1\).

We will use the following periodic integration fact. A function \(f:\mathbb{R}^n\to\mathbb{C}\) is \(\mathbb{Z}^n\)-periodic if \(f(x+k)=f(x)\) for every \(x\in\mathbb{R}^n\) and \(k\in\mathbb{Z}^n\). If it is measurable and \(\int_F|f(x)|\,dx<\infty\), periodicity implies it is integrable on every bounded rectangle: such a rectangle is covered by finitely many integer translates of \(F\), and the integral of \(|f|\) on each translate is finite. The integral over any translated unit cube is then the same as over \(F\).

Proof. First consider a measurable, integrable, \(1\)-periodic function \(u:\mathbb{R}\to\mathbb{C}\). Write \(a=k+r\), where \(k\in\mathbb{Z}\) and \(0\le r<1\). Integer periodicity and additivity of the integral give \(\int_a^{a+1}u=\int_r^{r+1}u=\int_r^1u+\int_1^{1+r}u\). In the second integral, substitute \(s=x-1\); periodicity gives \(\int_1^{1+r}u(x)\,dx=\int_0^r u(s)\,ds\). Thus \(\int_a^{a+1}u=\int_r^1u+\int_0^r u=\int_0^1u\).

For \(n\) variables, shift one coordinate at a time. Fix a coordinate \(j\) and hold the other \(n-1\) coordinates in their current intervals. The relevant bounded rectangle has finite integral of \(|f|\), as noted above. Fubini's theorem therefore gives integrable one-variable sections for almost every choice of the other coordinates. For each such choice, the section is \(1\)-periodic, so the one-variable identity just proved equates its integral over the old unit interval and the shifted unit interval. Integrating that equality over the other coordinates gives equality of the \(n\)-dimensional integrals. The exceptional sections form a null set and do not affect those integrals. Repeating this argument for each coordinate, with Fubini applied on each bounded intermediate rectangle, changes \(F\) to \(a+F\) without changing the integral. This proves the lemma. \(\square\)

The use of Fubini here is essential: integrability on a bounded rectangle ensures section integrability only for almost every choice of the other coordinates, not for every choice. The exceptional sections are harmless because their parameter set has measure zero.

Proof. Let \(t=q(a)\in\mathbb{T}^n\) and \(A\in\mathcal{B}_{\mathbb{T}^n}\). The function \(f(x)=\chi_A(q(x))\) is measurable, bounded, and \(\mathbb{Z}^n\)-periodic. The quotient identity \(q(x+a)=q(x)+t\) gives \(\chi_{A+t}(q(x))=f(x-a)\). Therefore, using the defining integral formula and changing variables \(y=x-a\), \(\nu(A+t)=\int_F f(x-a)\,dx=\int_{F-a}f(y)\,dy=\int_F f(y)\,dy=\nu(A)\). The third equality is the periodic integration lemma, applicable since \(f\) is bounded and hence integrable on \(F\). Also \(\nu(\mathbb{T}^n)=m(F)=1\), so \(\nu\) is a translation-invariant probability measure. It is finite and Borel on the compact group, and therefore is a normalized Haar measure. \(\square\)

Periodic Integrals in Practice

Normalization and a Common Pitfall

On \(\mathbb{R}^n\), Lebesgue measure is an infinite, locally finite Haar measure. On \(\mathbb{T}^n\), the unit-cube construction gives a finite Haar measure with total mass \(1\). These are related by the quotient and the choice of a fundamental domain, but their total masses are different because the groups have different compactness properties.

A common error is to use a cube that covers the torus more than once and still call its pushforward normalized Haar measure. For example, in \(\mathbb{T}^1\), pushing Lebesgue measure on \([0,2)\) forward gives total mass \(m([0,2))=2\), not \(1\). It is translation invariant, but it is not normalized. The correction is to use the fundamental interval \([0,1)\), whose measure is \(1\); alternatively, the pushforward from \([0,2)\) must be multiplied by \(1/2\). More generally, a fundamental domain is chosen to represent each coset once up to boundary sets of measure zero, and normalization is checked by computing the measure of that domain.