Locally Compact Groups and Radon Measures

Topological and Measure-Theoretic Setting

Haar measure is a measure adapted to the topology of a group. The topology supplies compact sets and open neighborhoods; the measure must interact well with both. This makes local compactness, the Hausdorff property, regularity, and finiteness on compact sets central assumptions. Here we define these terms and verify them in elementary examples. We do not establish the existence or uniqueness of an invariant measure.

Definition: A topological group $G$ is a group equipped with a topology for which the multiplication map $(x,y)\mapsto xy$ and the inversion map $x\mapsto x^{-1}$ are continuous. It is Hausdorff if any two distinct points have disjoint open neighborhoods. It is locally compact if every point has a compact neighborhood: a compact set containing an open neighborhood of that point. A locally compact Hausdorff group is abbreviated LCH group.

A neighborhood need not itself be open. For example, in the additive group $\mathbb{R}$ with its usual topology, $[-1,1]$ is a compact neighborhood of $0$, since it contains the open interval $(-1,1)$. The interval $[-1,1]$ is not open. The group operations on $\mathbb{R}$ are continuous, and $\mathbb{R}$ is Hausdorff, so it is an LCH group. A discrete group is also an LCH group: every singleton is open and compact.

Compactness is useful because it provides finite subcovers, while the Hausdorff property ensures compact sets are closed. In an LCH space, these properties together also let us find open sets whose closures are compact.

Proposition: In a locally compact Hausdorff space $G$, every point $x$ has an open neighborhood $U$ whose closure $\overline U$ is compact.

Proof. Choose a compact neighborhood $C$ of $x$, and choose an open set $W$ in $G$ with $x\in W\subseteq C$. The compact set $C$ is closed because $G$ is Hausdorff. The subspace $C$ is compact Hausdorff. We first use the following consequence of compactness and the Hausdorff property: if $O$ is open in $C$ and $x\in O$, there is a relatively open set $V$ in $C$ such that $x\in V$ and $\overline V^{\,C}\subseteq O$. To verify it, the complement $F=C\setminus O$ is compact. For each $y\in F$, choose disjoint relatively open neighborhoods $U_y$ of $x$ and $V_y$ of $y$. Finitely many of the $V_y$ cover $F$. The intersection of the corresponding finitely many $U_y$ is a neighborhood of $x$ whose closure misses each of those $V_y$, and hence misses $F$. Apply this fact to the relatively open set $O=W$ in $C$, obtaining $V$ with $x\in V$ and $\overline V^{\,C}\subseteq W$. Write $V=C\cap V_0$ for an open set $V_0$ in $G$, and set $U=V_0\cap W$. Then $x\in U$, $U\subseteq C$, and, since $C$ is closed, $\overline U\subseteq C$. Moreover, $U\subseteq V$, so $\overline U\subseteq\overline V^{\,C}\subseteq W$. Thus $\overline U$ is a closed subset of the compact set $C$, and is compact. $\square$

Regular Borel Measures and Radon Measures

For a topological space $G$, its Borel sigma-algebra $\mathcal{B}_G$ is the sigma-algebra generated by its open sets. A Borel measure is a measure on $\mathcal{B}_G$. Regularity means that Borel sets can be approximated in measure from outside by open sets and from inside by compact sets.

Definition: A Borel measure $\mu$ on a Hausdorff space $G$ is regular if, for every $E\in\mathcal{B}_G$, $$ \mu(E)=\inf\{\mu(U): E\subseteq U,\ U\text{ open}\} =\sup\{\mu(K): K\subseteq E,\ K\text{ compact}\}. $$ It is finite on compact sets if $\mu(K)<\infty$ for every compact $K\subseteq G$. In this tutorial, a Radon measure means a regular Borel measure finite on compact sets.

The infimum and supremum in the definition may take the value $+\infty$. The empty set is compact, so the inner approximation is also meaningful when $E$ contains no nonempty compact subset. Some conventions define regularity using inner approximation only for open sets. The definition above states explicitly the Borel-set approximation used here.

Finiteness on compact sets is stronger than simply requiring the measure to be finite on some sets, but it need not make the measure finite on all of $G$. If $\mu$ is finite on compact sets and $G$ is LCH, the proposition gives, around each point, an open $U$ with compact closure. Then $$ \mu(U)\leq\mu(\overline U)<\infty. $$ Thus the measure is locally finite: every point has an open neighborhood of finite measure. The measure of the whole group may nevertheless be infinite.

Proposition: If $\mu$ is a nonzero Radon measure on an LCH space $G$, then some compact set has strictly positive measure.

Proof. Since $\mu$ is nonzero, there is a Borel set $E$ with $\mu(E)>0$. By inner regularity, $\mu(E)=\sup\{\mu(K):K\subseteq E,\ K\text{ compact}\}$. If every such compact set had measure zero, the supremum would be zero, contradicting $\mu(E)>0$. Therefore some compact $K\subseteq E$ satisfies $\mu(K)>0$. $\square$

Worked Examples

Worked Example: A Dirac Measure on the Real Line

Let $G=\mathbb{R}$ with its usual topology and let $\delta_0$ be the Dirac measure at $0$, defined on Borel sets by $\delta_0(E)=1$ if $0\in E$ and $\delta_0(E)=0$ if $0\notin E$. Verify that it is a Radon measure.

First, $\delta_0$ is a measure: for pairwise disjoint Borel sets, at most one set contains $0$, so the measure of their union equals the sum of their measures. For every compact $K$, $\delta_0(K)$ is either $0$ or $1$, hence is finite.

For regularity, consider any Borel set $E$. If $0\in E$, the compact set $\{0\}\subseteq E$ has measure $1$; no compact subset of $E$ has measure greater than $1$. The open set $\mathbb{R}$ contains $E$ and has measure $1$; every open superset of $E$ contains $0$ and therefore also has measure $1$. Both the inner supremum and outer infimum are $1=\delta_0(E)$. If $0\notin E$, the empty compact set has measure $0$, and the open set $\mathbb{R}\setminus\{0\}$ contains $E$ and has measure $0$. Since measures are nonnegative, the inner supremum and outer infimum are both $0=\delta_0(E)$.

Check: for $E=[-1,1]$, the set contains $0$, so $\delta_0(E)=1$. The compact subset $\{0\}$ gives inner measure $1$, and the open superset $\mathbb{R}$ gives outer measure $1$. Thus both approximations recover the same value.

Worked Example: Counting Measure on a Discrete Group

Let $G=\mathbb{Z}$ with the discrete topology and let $\mu$ be counting measure, so $\mu(E)$ is the number of elements of $E$ when $E$ is finite and is $+\infty$ when $E$ is infinite. Verify the Radon conditions and compute the measure of $E=\{-2,0,3\}$.

Every subset of a discrete space is open, so every subset of $\mathbb{Z}$ is Borel. A compact subset of a discrete space must be finite: its cover by the open singletons has a finite subcover only if the set is finite. Conversely, every finite set is compact. Hence every compact set has finite counting measure.

For any $E\subseteq\mathbb{Z}$, outer regularity holds by taking the open superset $U=E$; no superset can have measure less than $\mu(E)$. For inner regularity, every finite subset $F\subseteq E$ is compact. If $E$ has $N$ elements, where $N$ is finite, then $E$ itself is compact and the inner supremum is $N=\mu(E)$. If $E$ is infinite, it contains finite subsets of arbitrarily large size, so the inner supremum is $+\infty=\mu(E)$. Thus counting measure is Radon.

The set $E=\{-2,0,3\}$ has exactly three elements, so $\mu(E)=3$. As an independent check, its three singleton subsets are disjoint and each has measure $1$, giving $\mu(E)=1+1+1=3$.

Worked Example: A Finite Atomic Measure

On $G=\mathbb{R}$, define $\mu=2\delta_0+3\delta_1$, meaning $\mu(E)=2\delta_0(E)+3\delta_1(E)$ for each Borel set $E$. Verify regularity and calculate the measure of $A=[0,\tfrac12]$.

For any Borel set $E$, $\mu(E)$ is the sum of the weights of the points $0$ and $1$ that belong to $E$. This defines a measure because each Dirac measure is countably additive and a finite nonnegative linear combination of measures is a measure. For every compact set, its measure is at most $2+3=5$, so it is finite.

To check regularity, if $E$ contains neither atom, the open set $\mathbb{R}\setminus\{0,1\}$ contains $E$ and has measure $0$, and the empty compact set has measure $0$. If $E$ contains at least one atom, let $K$ be the finite set of atoms in $E$. This set is compact and $\mu(K)=\mu(E)$. The open set $\mathbb{R}$ contains $E$ and has measure $5$; more precisely, $\mathbb{R}\setminus F$, where $F$ is the finite set of atoms not in $E$, is open, contains $E$, and has measure $\mu(E)$. No open superset of $E$ can omit any atom in $E$, so its measure is at least $\mu(E)$. These inner and outer approximations give regularity in both cases.

The set $A=[0,\tfrac12]$ contains $0$ and does not contain $1$, since $1>\tfrac12$. Therefore $\mu(A)=2\cdot1+3\cdot0=2$. Directly, $A$ contains the atom of weight $2$ and excludes the atom of weight $3$, giving the same value.

Why These Conditions Matter for Haar Measure

Haar measure is sought on a topological group as a nonzero regular Borel measure that is finite on compact sets and invariant under left translation. The invariance condition concerns how the measure responds to the group operation; the present definitions identify the space and measure class in which that question is posed. Local compactness provides relatively compact open neighborhoods, so compact finiteness implies local finiteness. Regularity makes the measure of a Borel set accessible through open supersets and compact subsets. Nonzero excludes the zero measure, which would satisfy an invariance equation without giving a meaningful measure of size.

Two distinctions are worth keeping explicit. First, locally finite does not mean finite: counting measure on the discrete group $\mathbb{Z}$ is finite on every compact set but has infinite measure on $\mathbb{Z}$. Second, a Borel measure need not be Radon merely because it is defined on Borel sets; it must also have the stated regularity and compact-finiteness properties.

Common mistake. Counting measure on $\mathbb{R}$ is sometimes mistakenly described as finite on compact sets because each point has measure $1$. But the compact interval $[0,1]$ contains infinitely many points, so its counting measure is $+\infty$. Counting measure on $\mathbb{R}$ therefore fails compact finiteness. By contrast, on a discrete group compact sets are finite, which is why counting measure there is finite on compact sets.

These assumptions provide a workable framework, not a guarantee that every measure one can write down has the desired invariance. The zero measure is regular and finite on compact sets but is excluded by nontriviality. The Dirac and atomic measures above are nonzero Radon measures, but their definitions alone do not establish translation invariance. The existence and uniqueness questions for invariant measures require additional arguments.

Check Your Understanding

Use the definitions to check each statement, paying attention to whether a neighborhood must be open and whether finiteness is required on all of $G$.

Explain why $[-1,1]$ is a compact neighborhood of $0$ in $\mathbb{R}$, even though it is not open.
Show that every compact subset of a discrete space is finite, using the open cover by singletons.
For a Dirac measure $\delta_x$, determine the inner and outer regular approximations for a Borel set that contains $x$ and one that does not.
Why does compact finiteness imply local finiteness on an LCH space?
Give two distinct reasons why the zero measure is not a Haar measure.

Locally Compact Groups and Radon Measures

What You'll Learn