nLab
Haar measure

Idea

If $G$ is a topological group, a Haar measure is a translation-invariant measure on the Borel sets of $G$. The archetypal example of Haar measure is the Lebesgue measure on the (additive group underlying) cartesian space ${ℝ}^n$.

Definition

The proper generality in which to discuss Haar measure is where the topological group $G$ is assumed to be locally compact Hausdorff, and from here on we assume this. (For topological groups, the Hausdorff assumption is rather mild; it is equivalent to the $T_0$ separation condition. See the discussion at uniform space.)

Let $C_c(G)$ denote the vector space of continuous real-valued functionals with compact support on $G$. This is a locally convex topological vector space where the locally convex structure is specified by the family of seminorms

$${\rho }_K(f)=\underset{x\in K}{\sup }\mid f(x)\mid ,$$\rho_K(f) = \sup_{x \in K} |f(x)|,

$K$ ranging over compact subsets of $G$. Recall that a Radon measure on $G$ may be described as a continuous linear functional

$$\mu :C_c(G)\to ℝ$$\mu: C_c(G) \to \mathbb{R}

which is positive in the sense that $\mu (f)\ge 0$ whenever $f\ge 0$. This defines a measure $\hat{\mu }$ on the $\sigma$-algebra of Borel sets in the usual sense of measure theory, where

$$\hat{\mu }(B)=\sup \{\mu (f):\mathrm{supp}(f)=K\subseteq B,{\rho }_K(f)=1\}$$\hat{\mu}(B) = sup \{\mu(f): supp(f) = K \subseteq B, \rho_K(f) = 1\}

By abuse of notation, we generally conflate $\mu$ and $\hat{\mu }$.

A left Haar measure on $G$ is a nonzero Radon measure $\mu$ such that

$$\mu (gB)=\mu (B)$$\mu(g B) = \mu(B)

for all $g\in G$ and all Borel sets $B$.

Existence and Uniqueness

Any locally compact Hausdorff topological group $G$ admits a Haar mesaure that is unique up to scalar multiple. This result was first proven by Weil. A proof by be found in these online notes by Rubinstein-Salzedo.

Left and Right Haar Measures that differ

The left and the right Haar measure may or may not coincide, groups for which they coincide are called unimodular. Consider the matrix subgroup

$$G:=\{\left(\begin{array}{cc}y& x\\ 0& 1\end{array}\right)\mid x,y\in ℝ,y>0\}$$G := \left\{ \left.\, \begin{pmatrix} y & x \\ 0 & 1 \end{pmatrix}\,\right|\, x, y \in \mathbb{R}, y \gt 0 \right\}

The left and right invariant measures are, respectively,

$$d{\mu }_L=y^{-2}\mathrm{dx}\mathrm{dy},d{\mu }_R=y^{-1}\mathrm{dx}\mathrm{dy}$$d\mu_L = y^{-2} dx dy,\quad d\mu_R = y^{-1} dx dy

and so G is not unimodular.