6.1 Haar Measure

Proof. Since (1) and (2) are equivalent to

1. (1)

   $\mu(L_{x}\inv E)=\mu(x\inv E)=\mu(E)$ for each $x\in G,E\suf G$

2. (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.$\square$

Proof.

1. (1)

   Since $\emp\neq U_{\al}\define \curl{x\in G:\phi(x)>\al\supp{\phi}}\suf G$ is open, $\curl{xU_\al:x\in G}$ is an open cover of $G$ by compactness of $\supp{f}$ there exists $x_{1},...,x_{N_\al}$ such that $\supp{f}\suf \cups{N_\al}{j=1}x_{j}U_{\al}$. For each $x\in \supp{f}$ there exists $j$ such that $x\in x_{j}U$ hence $\phi(x_{j}\inv x)>\al\normu{\phi}$. Thus

   \begin{align*}\sums{N_\al }{j=1}\frac{\normu{f}}{\al\normu{\phi}}L_{x_j}\phi(x)=\sums{N_\al }{j=1}\frac{\normu{f}}{\al\normu{\phi}}\phi(x_{j}\inv x)> \normu{f}\geq f(x)\end{align*}

   
   
2. (2)

   It suffices to prove $(f:\phi)\leq (L_{y}f:\phi)$ since $x\mto L_{x}$ is a group homomorphism. Let $\e>0$ then $L_{y}f\leq \sums{n }{j=1}c_{j}L_{x_j}\phi$ for $c_{1},...,c_{n}$ satisfying $\sums{n }{j=1}c_{j}\leq (L_{y}f:\phi)+\e$ hence $f\leq \sums{n }{j=1}c_{j}L_{y\inv x_j}\phi$.

3. (3)

   The sum of two coverings is a covering of the sum

4. (4)

   Covering respects scaling up to the scaling constant

5. (5)

   By transitivity of $\leq$

6. (6)

   For any covering we have

   \begin{align*}f(x)\leq \sums{}{j}c_{j}L_{x_j}\phi(x)\leq \sums{}{j }c_{j}\normu{\phi}\end{align*}

   
   

   for each $x\in G$ hence taking supremum over $x$ gives the desired result.

7. (7)

   If $f\leq \sums{}{j }c_{j}L_{x_j}\psi$ and $\psi\leq \sums{}{k }b_{k} L_{y_k}\phi$ then $f\leq \sums{}{j,k }c_{j}b_{k}L_{x_jy_k}\phi$.

$\square$

Proof. $\square$