Haar Measure on Compact Groups

Why Compactness Makes Normalization Possible

A left Haar measure is invariant under left translation and is locally finite: every point has an open neighborhood of finite measure. On a compact group, local finiteness and compactness together force the measure of the entire group to be finite. Nonzeroness forces that total mass to be positive. These two facts make it possible to rescale Haar measure to a probability measure.

Throughout this tutorial, \(G\) is a compact Hausdorff topological group, with identity \(e\), and its measurable sets are the Borel sets \(\mathcal{B}_G\). A measure on these sets is called a left Haar measure here if it is a nonzero, left-invariant Radon measure. In particular, it is locally finite, and left invariance means \(\mu(gE)=\mu(E)\) for every \(g\in G\) and \(E\in\mathcal{B}_G\). We use the standard Haar existence theorem to take such a measure as given; its existence theorem is treated separately in the course.

Proof. For each \(x\in G\), local finiteness gives an open neighborhood \(U_x\) of \(x\) with \(\mu(U_x)<\infty\). These neighborhoods cover \(G\). Compactness gives a finite subcover \(U_{x_1},\ldots,U_{x_k}\). By finite subadditivity and monotonicity of measures,

Because \(\mu\) is nonzero, some Borel set \(E\) has \(\mu(E)>0\). Since \(E\subseteq G\), monotonicity gives \(\mu(G)\geq\mu(E)>0\). Thus division by \(\mu(G)\) is defined and finite. For pairwise disjoint Borel sets \(E_n\), countable additivity gives \(\lambda(\bigcup_n E_n)=\sum_n\lambda(E_n)\), since the common divisor \(\mu(G)\) is positive and finite. Also \(\lambda(G)=1\), and for \(g\in G\), \(\lambda(gE)=\mu(gE)/\mu(G)=\mu(E)/\mu(G)=\lambda(E)\). So \(\lambda\) is a left-invariant probability measure. \(\square\)

The proof uses two separate parts of the assumptions: local finiteness supplies the finite-measure neighborhoods, while compactness reduces their cover to finitely many sets. Nonzeroness is what ensures the total mass is not zero. Normalization changes the scale of a measure, not its left-translation invariance.

Finite Groups: Uniform Mass on Each Element

For a finite group, every subset is measurable. Suppose \(\nu\) is a finite left-invariant measure on all subsets of a finite group \(G\), and assume \(0<\nu(G)<\infty\). Left translations act transitively on the elements: for \(x,y\in G\), the translation by \(yx^{-1}\) sends \(x\) to \(y\). Thus all singleton sets have the same mass. Finite additivity then determines the mass of every subset.

Proof. Given \(x,y\in G\), set \(g=yx^{-1}\), so \(gx=y\). Left invariance gives \(\nu(\{y\})=\nu(g\{x\})=\nu(\{x\})\). Write their common mass as \(c\). Since \(G\) is a disjoint union of its \(|G|\) singletons, finite additivity gives \(\nu(G)=|G|c\). The assumptions \(0<\nu(G)<\infty\) imply \(c=\nu(G)/|G|\), which is positive and finite. For any \(E\subseteq G\), additivity over its finitely many singletons gives \(\nu(E)=|E|c\). Dividing by \(\nu(G)\) gives normalized mass \(|E|/|G|\), and hence mass \(1/|G|\) for each singleton. \(\square\)

Haar Probability on a Compact Group

For any compact group, the theorem lets us choose the scale of a left Haar measure so that the whole group has mass one. This does not assert that every invariant measure already has mass one, nor does the argument establish uniqueness among invariant probability measures. The normalized measure is a probability because it is countably additive, nonnegative, and assigns mass one to the whole space.

The Circle: Normalized Arc Length

Represent the circle group as \(\mathbb{T}=\mathbb{R}/\mathbb{Z}\), with addition modulo \(1\). Let \(q(t)\) denote the class of \(t\in\mathbb{R}\). Each point of \(\mathbb{T}\) has a unique representative in \([0,1)\). For a Borel set \(A\subseteq\mathbb{T}\), define \(\widetilde{A}=\{t\in\mathbb{R}:q(t)\in A\}\) and set \(\lambda(A)=m(\widetilde{A}\cap[0,1))\), where \(m\) is Lebesgue measure on \(\mathbb{R}\). The map \(q\) identifies the circle with \([0,1)\) as a measurable space, so this defines a Borel measure; in particular, \(\lambda(\mathbb{T})=m([0,1))=1\).

To verify invariance, fix \(\alpha\in[0,1)\) and let \(h(t)=\chi_{\widetilde{A}}(t)\). This is a measurable, \(1\)-periodic function taking values in \(\{0,1\}\). The measure of the translate \(q(\alpha)+A\) is \(\int_0^1 h(t-\alpha)\,dt\). By translation invariance of Lebesgue measure on \(\mathbb{R}\), this equals \(\int_{-\alpha}^{1-\alpha}h(s)\,ds\). Periodicity lets us move this interval by \(1\), obtaining \(\int_{1-\alpha}^{2-\alpha}h(s)\,ds\). Splitting at \(1\), and using \(h(s)=h(s-1)\) on \([1,2-\alpha]\), gives

Thus \(\lambda(q(\alpha)+A)=\lambda(A)\). Since every circle element has a representative in \([0,1)\), this proves left invariance for every translation. The measure is also finite and positive on the whole group, so it is a normalized Haar measure on the circle.

Open Sets and Points in Compact Groups

Although the whole group has positive mass, the measure of an individual point depends on whether the group is finite or infinite. A useful general fact is that every nonempty open set has positive Haar measure. In an infinite compact group, by contrast, individual points have measure zero.

Proof. Choose \(u\in U\). For every \(x\in G\), \(x\in(xu^{-1})U\), so the family of translates \(\{gU:g\in G\}\) covers \(G\). Each translate is open. Compactness provides a finite subcover \(g_1U,\ldots,g_rU\). If \(\lambda(U)=0\), invariance gives \(\lambda(g_jU)=0\) for each \(j\), and subadditivity then gives \(\lambda(G)\leq\sum_{j=1}^r\lambda(g_jU)=0\), contradicting \(\lambda(G)=1\). Hence \(\lambda(U)>0\).

For the second claim, left translation sends any singleton to any other singleton, so every singleton has the same mass \(c\). Since \(\lambda\) is a probability, \(c\) is finite. If \(G\) is infinite, then for every positive integer \(n\) it contains \(n\) distinct points. Their singletons are disjoint, so finite additivity and monotonicity give \(nc\leq1\). If \(c>0\), choose an integer \(n>1/c\); then \(nc>1\), a contradiction. Therefore \(c=0\), as claimed. \(\square\)

On the circle, this distinguishes open arcs from individual points: an arc of positive length has positive measure, while each singleton has measure zero. In a finite group with normalized counting measure, every singleton instead has mass \(1/|G|\). Neither conclusion is a contradiction: the group and the measure differ.

A Common Normalization Pitfall

A tempting but invalid argument is: “A measure is invariant, so divide every set's mass by the total mass.” This division is justified only when the total mass is finite and strictly positive. The infinite-valued measure in the finite-group example is left-invariant and nonzero, yet its total mass is \(+\infty\); it cannot be normalized by division. For compact groups, the Haar hypotheses repair both issues: nonzeroness gives positive total mass, and local finiteness together with compactness gives finite total mass.