Invariant Measure on a Discrete Group
On a discrete group, the measurable sets are simple to describe, but normalization still matters. Counting measure assigns mass one to each point; on a finite group it therefore has total mass equal to the group’s order. Dividing by that order gives a probability measure, also left invariant, but it is a different measure. On a countably infinite discrete group, counting measure remains locally finite even though its total mass is infinite.
Let \(G\) be a finite or countable group with the discrete topology, identity \(e\), and multiplication written multiplicatively. Every subset of \(G\) is Borel measurable: every subset is open, so the Borel sigma-algebra is the power set \(\mathcal{P}(G)\). A compact subset of \(G\) is finite. Indeed, the cover of a subset \(K\) by its open singletons has a finite subcover only when \(K\) is finite; conversely, every finite set is compact.
For a countable discrete group, the counting measure \(\#\) is defined by \(\#(E)=|E|\) if \(E\) is finite and \(\#(E)=\infty\) if \(E\) is infinite. It is left invariant because the map \(x\mapsto gx\) is a bijection from \(E\) to \(gE\). It is Radon: finite sets have finite measure, every set is open, and every subset \(E\) is the increasing union of its finite subsets, so its measure is the supremum of the measures of those subsets.
Why the Singleton Mass Determines the Measure
Proof. For any \(x,y\in G\), left multiplication by \(yx^{-1}\) sends \(x\) to \(y\). Invariance therefore gives \(\mu(\{y\})=\mu(\{x\})\). Set \(c=\mu(\{e\})\). The singleton \(\{e\}\) is compact, so \(c<\infty\). Every singleton has measure \(c\). If \(E\) has exactly \(n\) elements, finite additivity gives \(\mu(E)=nc\). If \(E\) is infinite, countability lets us list its distinct elements as \(x_1,x_2,\ldots\). The finite sets \(E_n=\{x_1,\ldots,x_n\}\) increase to \(E\). By continuity from below for measures, \(\mu(E)=\lim_{n\to\infty}\mu(E_n)=\lim_{n\to\infty}nc\), which is \(0\) if \(c=0\) and \(\infty\) if \(c>0\). If \(c=0\), this argument shows that every subset has measure zero, so \(\mu\) is the zero measure. Thus a nonzero \(\mu\) has \(c>0\), and the asserted formula follows. \(\square\)
The theorem classifies left-invariant Radon measures here without needing an existence or uniqueness theorem for general locally compact groups. The direct argument depends on discreteness: each point is a measurable compact singleton, and left multiplication can carry any singleton to any other.
Finite Groups: Counting and Probability Normalization
If \(G\) is finite, counting measure is a left Haar measure with total mass \(|G|\). A left-invariant probability measure must instead have total mass one. The classification shows exactly how to obtain it: give each of the \(|G|\) elements mass \(1/|G|\). This measure is called normalized counting measure. The word “normalized” is essential; its singleton masses differ from those of counting measure unless \(|G|=1\).
Worked Example: The Cyclic Group of Order Four
Write \(C_4=\{e,a,a^2,a^3\}\), with \(a^4=e\). Counting measure assigns mass one to every element, so its total mass is \(1+1+1+1=4\). Normalized counting measure \(\nu\) assigns mass \(1/4\) to every element. For \(E=\{e,a^2\}\), \(\nu(E)=1/4+1/4=1/2\).
Check invariance directly with \(g=a\). The translated set is \(aE=\{a,a^3\}\), so \(\nu(aE)=1/4+1/4=1/2=\nu(E)\). More generally, multiplication by any fixed \(g\in C_4\) permutes the four elements. It therefore preserves the number of elements in every subset and hence preserves both counting measure and its scalar multiple \(\nu\).
Check: \(\nu(C_4)=4\cdot(1/4)=1\), whereas \(\#(C_4)=4\). Thus both measures are left invariant, but only \(\nu\) is a probability measure.
Proof. By the theorem, a nonzero left-invariant measure has the form \(\mu(E)=c|E|\). If it is a probability measure, then \(1=\mu(G)=c|G|\), so \(c=1/|G|\). Conversely, this choice gives total mass one and is left invariant because left multiplication is a bijection. This also covers the trivial group: if \(|G|=1\), the formula assigns mass one to its sole element. \(\square\)
Worked Example: A Nonuniform-Looking Subset of \(S_3\)
Let \(G=S_3\), the group of permutations of three symbols, which has six elements. Its normalized counting measure assigns mass \(1/6\) to each permutation. Consider \(E=\{\mathrm{id},(12),(123)\}\). Then \(\nu(E)=3\cdot(1/6)=1/2\). For \(g=(13)\), left multiplication sends \(E\) to \(\{(13),(13)(12),(13)(123)\}\). These are three distinct permutations because left multiplication is a bijection, so \(\nu(gE)=3\cdot(1/6)=1/2\).
Check: There are six permutations, each of mass \(1/6\), so \(\nu(S_3)=6\cdot(1/6)=1\). The calculation uses left multiplication throughout; it does not assume that \(S_3\) is commutative.
Countably Infinite Groups: Locally Finite, Not a Probability
For a countably infinite discrete group, counting measure assigns mass one to every point and infinite mass to the whole group. This does not prevent it from being a Haar measure: Haar measure is required to be finite on compact sets, not finite on the entire space. Compact subsets of a discrete group are finite, so counting measure is finite on every compact set.
The classification also rules out a left-invariant probability measure on such a group. If the common singleton mass \(c\) is positive, the measure of the infinite group is infinite. If \(c=0\), countable additivity makes the measure of every subset zero, so the measure cannot be a probability.
Worked Example: Counting Measure on the Integers
Give \(\mathbb{Z}\), under addition, its discrete topology. Counting measure satisfies \(\#(E+k)=\#(E)\) for each \(k\in\mathbb{Z}\), because \(n\mapsto n+k\) is a bijection. For \(E=\{-1,0,2\}\), \(\#(E)=3\); translating by \(k=4\) gives \(E+4=\{3,4,6\}\), and \(\#(E+4)=3\).
For a nonnegative function \(f\) on \(\mathbb{Z}\), its integral with respect to counting measure is the sum \(\int f\,d\#=\sum_{n\in\mathbb{Z}}f(n)\), with \(+\infty\) allowed. For \(f=\chi_{\{-1,0,2\}}\), the sum has three terms equal to one and all other terms equal to zero, so \(\int f\,d\#=3\). Translation preserves this integral: for the translated indicator \(\chi_{E+4}\), the sum again has exactly three terms equal to one.
Check: Every finite subset of \(\mathbb{Z}\), including \(E\), is compact and has finite counting measure. But \(\#(\mathbb{Z})=\infty\), since the increasing sets \(\{-n,\ldots,n\}\) have \(2n+1\) elements and \(2n+1\to\infty\).
Worked Example: The Infinite Dihedral Group
Consider the group \(D_\infty=\langle r,s\mid s^2=e,\ srs=r^{-1}\rangle\), with the discrete topology. Its elements have the forms \(r^n\) and \(r^ns\), for \(n\in\mathbb{Z}\), so it is countable and infinite. Counting measure gives each element mass one. Take \(E=\{e,s,r\}\) and left-translate by \(s\). The translated elements are \(s e=s\), \(s s=e\), and \(s r=r^{-1}s\); hence \(sE=\{s,e,r^{-1}s\}\). The elements in each set are distinct, so \(\#(E)=3=\#(sE)\).
Check: The computation uses the group relation \(sr=r^{-1}s\), and \(r^{-1}s\) is different from \(e\) and \(s\). Thus the translated set has exactly three elements. This illustrates that the invariance argument applies even when the group law is noncommutative.
A Common Normalization Error
A frequent mistake is to call counting measure on a finite group a probability measure. For \(C_4\), the incorrect calculation is “each of the four elements has mass one, so the total mass is one.” The correction is to add the four singleton masses: \(1+1+1+1=4\). Counting measure is left invariant but is not a probability measure. To obtain a probability, divide each mass by four: \(1/4+1/4+1/4+1/4=1\).
More generally, left invariance alone does not select a normalization: every positive scalar multiple of counting measure is left invariant. On a finite group, requiring total mass one fixes the scalar. On a countably infinite discrete group, no positive scalar gives finite total mass, so no such normalization to a probability is possible.
| Group | Counting measure of a singleton | Total counting measure | Invariant probability? |
|---|---|---|---|
| Finite \(G\), with \(|G|=N\) | 1 | \(N\) | Yes: mass \(1/N\) per point |
| Countably infinite \(G\) | 1 | \(\infty\) | No |
Check Your Understanding
In each question, distinguish invariance from normalization and use the discrete topology assumptions explicitly.
- Prove directly that counting measure on a finite group is left invariant by showing that left multiplication is a bijection on every subset.
- A finite group has eight elements. What is the mass of each singleton under normalized counting measure, and what is the measure of a subset with three elements?
- Suppose \(\mu\) is a nonzero left-invariant Radon measure on a countably infinite discrete group and \(\mu(\{e\})=2\). Find \(\mu(E)\) for a finite set of seven elements and for the whole group.
- Explain why a measure assigning mass zero to every singleton on a countable group must assign mass zero to every subset.
- Why does infinite total mass not prevent counting measure on \(\mathbb{Z}\) from being locally finite?