Definition 6.1.1 (Haar Measure).label Let $G$ be a locally compact group and denote $C_{c}^{+}(G)\define\curl{f\in C_c(G;\com):f\geq 0,f\no\equiv 0}$ whose linear span is $C_{c}(G;\com)$. Let $\mu$ be a nonzero Radon measure on $G$ then using notation as in Definition 3.7.3 the following are equivalent:
- (1)
(Left Invariant) $(L_{x})_{*}\mu\equiv \mu$ on Borel sets of $G$, $\fall x\in G$
- (2)
(Right Invariant) $(R_{x})_{*}(\iota_{*}\mu)\equiv \iota_{*}\mu$ on Borel sets of $G$, $\fall x\in G$
- (3)
$\integral{}{}L_{y}fd\mu=\integral{}{}fd\mu$, $\fall f\in C_{c}^{+}(G),y\in G$
- (4)
$\integral{}{}R_{y}fd\mu=\integral{}{}fd\mu$, $\fall f\in C_{c}^{+}(G),y\in G$
If the above conditions hold, we say $\mu$ is a left (and $\iota_{*}\mu$ a right) Haar measure on $G$.
Every locally compact group $G$ possesses a left Haar measure $\mu$ which is unique up to rescaling. In addition $\mu$ is positive in the following sense:
For each $\emp\neq U\suf G$ open, $\mu(U)>0$.
For each $f\in C_{c}^{+}(G)$, $\integral{}{}fd\mu>0$.
For each left Haar measure $\lam$ the map
\begin{align*}L_{\lam}:C_{c}^{+}(G)\to (0,\infty)&&f\mto \frac{\integral{}{}fd\lam}{\integral{}{}fd\mu}\end{align*}is constant.
Proof. Since (1) and (2) are equivalent to
- (1)
$\mu(L_{x}\inv E)=\mu(x\inv E)=\mu(E)$ for each $x\in G,E\suf G$
- (2)
$\iota_{*}\mu(R_{x}\inv E)=\mu(x E\inv)=\mu(E\inv)$ for each $x\in G,E\suf G$
respectively and inversion is a homeomorphism (1)$\iff$(2). For any Radon measure $\mu$ we have for each $f\in C_{c}(G;\com)$
by approximating $f$ by simple functions hence (1) implies (3). Conversely suppose (3) holds then (3) holds for $f\in C_{c}(G;\com)$ and by Riesz Representation Theorem on LCH space (1) holds. Similarly one can show (2)$\iff$(4).
Fix $f_{0}\in C_{c}^{+}(G)$ then for each $\phi\in C_{c}^{+}(G)$, the functional
inherits the properties of the Outer Integral and Definition 6.1.2(7) gives
For each $V\in\cali{N}(1)$, define
Since $\cali{N}_{G}(1)$ is a filter, $\curl{F_V:V\in\cali{N}_G(1)}$ is a filter base on $\cali{I}=\curl{I_\phi:\phi\in C_c^+(G)}$ by Definition 3.4.1(F2) and Urysohn’s Lemma so let $\scr{F}$ be the filter generated by $\curl{F_V:V\in\cali{N}_G(1)}$. By Tychonoff’s theorem the product space
is compact. Each $I_{\phi}$ defines a point on $X$ by evaluation hence $\scr{F}$ induces a filter on $X$ which has a cluster point $I\in X$ by compactness. In particular $I\in \caps{}{F\in \scr{F}}\cl{F}=\caps{}{V\in\cali{N}_G(1)}\cl{F_V}$. By continuity of coordinate projections, each properties of the functionals $\curl{I_\phi}_{\phi\in C_c^+(G)}$ from Definition 6.1.2 determines a closed subset of $X$ which contains $F_{V}$ and hence $\cl{F_V}$. Let $\e>0$ and $f_{1},f_{2}\in C_{c}^{+}(G)$ then there exists $\phi\in C_{c}^{+}(G)$ such that $\abs{I_\phi(f_j)-I(f_j)}<\e$ for each $j=1,2$ and pick $V$ as in Lemma 6.1.3 then
Since $\e$ was arbitrary, $I$ is additive on $C_{c}^{+}(G)$. Let $0\neq f\in C_{c}(G)$ the $f=g-h$ for some $g,h\in C_{c}^{+}(G)$ so define
which is well-defined for if $g-h=g'-h'$ then $g+h'=g'+h$ and additivity gives
Existence of $\mu$ then follows from Riesz representation theorem and (3).
Suppose $\mu(U)=0$ for some open nonempty $U$ then $\mu(xU)=0$ for each $x\in G$. For any compact set $K$, there exists a finite subcover $\curl{x_i U}$ of $K$ so $\mu(K)=0$ since $\mu$ is a Radon measure hence $\mu(G)=0$ by inner regularity which contradicts $\mu$ is a Haar measure. Let $f\in C_{c}^{+}(G)$ and $U\define \curl{x\in G:f(x)>\frac{\normu{f}}{2}}$ then $\integral{}{}f d\mu>\frac{\normu{f}}{2}\mu(U)>0$.
For uniqueness, by (a) and Riesz Representation Theorem on LCH space it is equivalent to prove $L_{\lam}$ is constant. Let $f,g\in C_{c}^{+}(G)$ then uniform continuity there exists a symmetric neighborhood $V$ of $1$ such that $\abs{f(xy)-f(yx)},\abs{g(xy)-g(yx)}<\e$ for all $x$ whenever $y\in V$. By Definition 3.7.6 and Theorem 3.8.2 there exists $h\in C_{c}^{+}(G)$ with $h(x)=h(x\inv)$ and $\supp{h}\suf V$. By left invariance of $\lam$ and Fubini’s Theorem
Similarly by symmetry of $h$
Therefore,
for some constant $C>0$ depending only on $f,g,h,\lam,\mu$. In particular dividing by $\integral{}{}hd\mu \integral{}{}fd\mu$ and $\integral{}{}hd\mu \integral{}{}gd\mu$ respectively, there exists a constant depending only on $f,g,h,\lam,\mu$ such that
hence
since $\e$ was arbitrary we conclude $L_{\lam}$ is constant.$\square$
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