What Haar’s Theorem Provides
For a locally compact Hausdorff group, translation symmetry is compatible with a measure that is finite on compact sets and regular on Borel sets. Haar’s theorem says more: such a nonzero left-invariant measure always exists, and any two choices differ only by a positive constant. The hypotheses matter. In particular, the zero measure is invariant and regular, but it is not a Haar measure under the definition below.
Let \(G\) be a locally compact Hausdorff topological group, with Borel sigma-algebra \(\mathcal{B}_G\). For \(g\in G\), left translation is the map \(L_g:G\to G\) given by \(L_g(x)=gx\). A Borel measure \(\mu\) on \((G,\mathcal{B}_G)\) is locally finite if every compact subset of \(G\) has finite measure. It is regular here if it is outer regular on Borel sets and inner regular on open sets: for \(E\in\mathcal{B}_G\) and open \(U\subseteq G\),
Local finiteness is not the same as finiteness: it requires finite measure on compact sets, not necessarily on all of \(G\). Nor does the theorem assume that \(\mu(G)\) is finite or that \(\mu\) is sigma-finite.
Left invariance is an equality of measures of sets; it does not assert that the group is commutative. In particular, \(gE=\{gx:x\in E\}\), with the translating element on the left. The theorem below concerns left Haar measures only.
Haar’s Existence and Uniqueness Theorem
This is the standard Haar existence and uniqueness theorem. Its existence proof is a general construction for locally compact groups, beyond the scope of this tutorial; the uniqueness proof by translation averaging is treated separately. The statement makes explicit that the measures are nonzero, regular, locally finite Borel measures. Omitting “nonzero” from the uniqueness clause would make it false, since the zero measure also satisfies regularity, local finiteness, and left invariance.
One useful consequence is that every nonempty open set has positive Haar measure. This conclusion uses nonzeroness and regularity, not just invariance.
Proof. Suppose first, toward a contradiction, that every open set has measure zero. For every Borel set \(E\), outer regularity then gives \(\mu(E)=\inf_{V\supseteq E}\mu(V)=0\), contradicting that \(\mu\) is nonzero. Thus some open set \(O\) has positive measure. By inner regularity on open sets, there is a compact \(K\subseteq O\) with \(\mu(K)>0\). Now fix any nonempty open \(U\). The sets \(gU\), for \(g\in G\), cover \(G\): if \(u\in U\), then \(x\in (xu^{-1})U\) for every \(x\in G\). In particular, they cover \(K\). Compactness gives a finite subcover \(K\subseteq\bigcup_{j=1}^r g_jU\). If \(\mu(U)=0\), left invariance gives \(\mu(g_jU)=0\) for each \(j\), and finite subadditivity gives \(\mu(K)=0\), a contradiction. Hence \(\mu(U)>0\). \(\square\)
Worked Applications
Worked Example: Fixing the Scale on Euclidean Space
Take the additive group \(G=\mathbb{R}^2\) with its usual topology. It is locally compact and Hausdorff. Lebesgue measure \(m\) on \((\mathbb{R}^2,\mathcal{B}_{\mathbb{R}^2})\) is regular, locally finite, and translation-invariant, so it is a left Haar measure. By the uniqueness part of Haar’s theorem, any nonzero regular, locally finite, translation-invariant Borel measure \(\nu\) on this group has the form \(\nu=c\,m\) for some \(c>0\).
The constant is determined by a single set. Put \(Q=[0,1)^2\) and \(c=\nu(Q)\), since \(m(Q)=1\). This value is finite: \(Q\) lies in the compact square \([0,1]^2\), which has finite \(\nu\)-measure. It is positive because \(Q\) contains the nonempty open square \((0,1)^2\), and the proposition gives \(\nu((0,1)^2)>0\). Thus the scale is indeed a positive finite number.
For \(R=[0,2)\times[0,3)\), Lebesgue measure gives \(m(R)=2\cdot3=6\). Therefore \(\nu(R)=6c\). Check: \(R\) is the disjoint union of six translates of \(Q\), each with \(\nu\)-measure \(c\); finite additivity gives \(\nu(R)=c+c+c+c+c+c=6c\), as required.
Worked Example: Normalizing Haar Measure on the Circle
Let \(G=\mathbb{T}=\mathbb{R}/\mathbb{Z}\), with addition modulo \(1\). This is a compact Hausdorff group. Its normalized arc-length measure \(\lambda\) is the pushforward of Lebesgue measure on \([0,1)\) under \(t\mapsto[t]\). It has total mass \(1\), and it is invariant under addition in \(\mathbb{T}\): translating an arc-length parametrization by a fixed \(s\in[0,1)\) amounts to rotating the unit interval modulo \(1\). The interval splits at \(1-s\); ordinary translation invariance of Lebesgue measure on each of the two resulting intervals shows that the measure of any Borel set is unchanged by this rotation. Thus \(\lambda\) is a left Haar measure.
Consider the arc \(A=\{[t]:t\in[3/4,1)\cup[0,1/4)\}\). The two intervals are disjoint and have lengths \(1/4\) and \(1/4\), so \(\lambda(A)=1/4+1/4=1/2\). If \(\nu\) is any nonzero regular, locally finite, left-invariant Borel measure on \(\mathbb{T}\), uniqueness gives \(\nu=c\lambda\) with \(c>0\), and hence \(\nu(A)=c/2\). Check: the complementary arc, apart from its two endpoints, also has length \(1/2\); endpoints have \(\lambda\)-measure zero, so \(\lambda(\mathbb{T})=1/2+1/2=1\).
Worked Example: Counting Measure on a Discrete Group
Let \(G=\mathbb{Z}\) with the discrete topology. Every subset is Borel, compact subsets are finite, and counting measure \(\#\) is locally finite. It is regular: for any subset \(E\subseteq\mathbb{Z}\), it is open, so outer regularity follows by taking \(E\) itself; its measure is the supremum of the measures of its finite subsets, which are compact, giving inner regularity. Translation by any integer is a bijection, so it preserves the number of elements of every set. Consequently, counting measure is a left Haar measure.
By Haar uniqueness, any nonzero regular, locally finite, translation-invariant Borel measure \(\nu\) on \(\mathbb{Z}\) equals \(c\,\#\) for some \(c>0\). For \(F=\{-2,0,3,7\}\), there are four distinct elements, so \(\#(F)=4\) and \(\nu(F)=4c\). Check: \(F\) is the disjoint union of four singleton sets, and translation invariance gives each singleton measure \(\nu(\{0\})=c\); finite additivity again gives \(c+c+c+c=4c\).
Compact Groups and Normalization
On a compact group, local finiteness implies finite total mass because the entire group is compact. The proposition ensures that this total mass is positive: the group itself is a nonempty open set. It is therefore possible to rescale any Haar measure to have total mass \(1\). The resulting measure is often called normalized Haar measure.
Proof. Since \(G\) is compact, local finiteness gives \(\mu(G)<\infty\); since \(G\) is nonempty and open, the preceding proposition gives \(\mu(G)>0\). Thus division by \(\mu(G)\) is defined, and \(\bar\mu(G)=1\). Scaling preserves regularity and left invariance, so \(\bar\mu\) is a left-invariant regular Borel probability measure. If \(\rho\) is any other left-invariant regular Borel probability measure, it is nonzero and locally finite because \(\rho(G)=1\). Haar uniqueness gives \(\rho=c\mu\) for some \(c>0\). Evaluating on \(G\) yields \(1=\rho(G)=c\mu(G)\), so \(c=1/\mu(G)\) and \(\rho=\bar\mu\). \(\square\)
This proposition explains the distinction between a Haar measure and a normalized Haar probability: on a compact group, normalization fixes the otherwise arbitrary positive scale. On a noncompact group such as \(\mathbb{R}^2\), translation-invariant Lebesgue measure has infinite total mass, so it cannot be rescaled to a probability measure while retaining the same measure class and invariance.
A Common Uniqueness Pitfall
A frequent error is to state that every regular, locally finite, invariant Borel measure is a positive multiple of Haar measure, without excluding the zero measure. For example, on \(\mathbb{R}^2\), the zero measure is regular, locally finite, and translation-invariant. The incorrect conclusion would require \(0=c\,m\) for some \(c>0\), which is impossible because \(m([0,1)^2)=1\) and therefore \((c\,m)([0,1)^2)=c>0\). The correction is to require the competing measure to be nonzero. The same qualification is necessary on \(\mathbb{T}\): the zero measure is not a positive scalar multiple of normalized Haar measure.
The theorem also does not identify an invariant measure from invariance alone. Regularity and local finiteness are part of the hypotheses, and the measure must be a Borel measure. When applying the theorem, check that the group is locally compact and Hausdorff, that the measure is nonzero, and that its invariance is under the specified left translations. These checks distinguish Haar’s theorem from a claim about arbitrary set functions or arbitrary invariant measures.
Check Your Understanding
For each question, keep the distinction between a Haar measure and its normalization explicit.
- State the regularity, local finiteness, nonzeroness, and invariance conditions in the definition of a left Haar measure.
- Why must the uniqueness statement exclude the zero measure? Give a concrete Borel set that demonstrates the problem on \(\mathbb{R}^2\).
- If \(\nu=c\,m\) on \(\mathbb{R}^2\) and \(\nu([0,1)^2)=5\), what is \(\nu([0,2)\times[0,3))\)?
- Explain why a Haar measure on a compact group has finite, positive total mass and how to normalize it.
- For counting measure on \(\mathbb{Z}\), compute the measure of \(\{-2,0,3,7\}\), and state what uniqueness says about any nonzero regular, locally finite, translation-invariant Borel measure on \(\mathbb{Z}\).