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/Part 4: Measure Theory and Integration/Chapter 28: Integration on Locally Compact Groups

28.2 Haar Measures

Definition 28.2.1 (Haar Measure).label Let $G$ be a locally compact group and $\mu: \cb_{G} \to [0, \infty]$ be a non-zero Radon measure, then $\mu$ is a left Haar measure if

  1. (LH)

    For each $g \in G$ and $A \in \cb_{G}$, $\mu(gA) = \mu(A)$.

and a right Haar measure if

  1. (RH)

    For each $g \in G$ and $A \in \cb_{G}$, $\mu(Ag) = \mu(A)$.

Lemma 28.2.2.label Let $G$ be a locally compact group, then there exists an open and closed subgroup $H$ that is $\sigma$-compact.

Proof. Let $K \in \cn_{G}(1)$ and $K^{(1)}= K$. For each $n \in \natp$, let $K^{(n+1)}= KK^{(n)}$, then $K^{(n+1)}$ is compact by Proposition 5.16.3 with $K^{(n+1)}\in \cn_{G}(K^{(n)})$. Let $H = \bigcup_{n \in \natp}K^{(n)}$, then $H$ is a subgroup of $G$, which is open by Lemma 5.4.3. Since $H$ admits an exhaustion by compact sets, it is $\sigma$-compact.

Finally, since

\[G \setminus H = G \setminus \bigcup_{n \in \natp}K^{(n)}= \bigcap_{n \in \natp}G \setminus K^{(n)}\]

and $K^{(n)}$ is closed for each $n \in \natp$ by Proposition 5.16.4, $G \setminus H$ is closed, and hence $H$ is open.$\square$

Definition 28.2.3 (Covering Ratio).label Let $G$ be a locally compact group and $f, g \in C_{c}^{+}(G)$, then

\[(f: g) = \inf\bracs{\sum_{j = 1}^n c_j \bigg | \seqf{c_j} \subset [0, \infty), \seqf{x_j} \subset G, f \le \sum_{j = 1}^n c_j L_{x_j}g}\]

is the covering ratio of $f$ by $g$.

Proposition 28.2.4.label Let $G$ be a locally compact group and $f, h, g \in C_{c}^{+}(G)$, then:

  1. (1)

    If $g \ne 0$, then $(f: g) < \infty$.

  2. (2)

    $(f + h: g) \le (f: g) + (h: g)$.

  3. (3)

    For each $\lambda \ge 0$, $(\lambda f: g) = \lambda(f: g)$.

  4. (4)

    If $f \le h$, then $(f: g) \le (h: g)$.

  5. (5)

    $(f: g) \le (f: h)(h: g)$.

  6. (6)

    $(f: g) \ge \norm{f}_{u}/\norm{g}_{h}$.

  7. (7)

    For each $x \in G$, $(L_{x}f: g) = (f: g)$.

Lemma 28.2.5.label Let $G$ be a locally compact group, $f, f'\in C_{c}^{+}(G)$, and $\eps > 0$, then there exists $V \in \cn_{G}(1)$ such that for any $g \in C_{c}^{+}(V)$ with $g \ne 0$,

\[(f: g) + (f': g) \le (f + f': g) + \eps\]

Proof, [Lemma 2.18, Fol16]. By Urysohn’s Lemma, there exists $\eta \in C_{c}^{+}(G; [0, 1])$ such that $\eta|_{\supp{f} \cup \supp{f'}}= 1$.

Let $\delta > 0$, and define

\[H = f + f' + \delta \eta \quad h = \frac{f}{H}\quad h' = \frac{f'}{H}\]

By Proposition 28.1.2, there exists $V \in \cn_{G}(1)$ such that for any $x, y \in G$ with $x^{-1}y \in V$,

\[|h(x) - h(y)|, |h'(x) - h'(y)| < \delta\]

Let $g \in C_{c}^{+}(V)$, $\seqf{c_j}\subset [0, \infty)$, and $\seqf{x_j}\subset G$ such that $H \le \sum_{j = 1}^{n} c_{j} L_{x_j}\phi$, then for each $x \in G$,

\begin{align*}f(x)&= H(x)h(x) \le \sum_{j = 1}^{n} c_{j} L_{x_j}g(x)h(x) = \sum_{j = 1}^{n} c_{j}g(x_{j}^{-1}x)h(x) \\&\le \sum_{j = 1}^{n} c_{j}[h(x_{j}) + \delta] \cdot L_{x_j}g(x)\end{align*}

Likewise,

\[f'(x) \le \sum_{j = 1}^{n} c_{j}[h'(x_{j}) + \delta] \cdot L_{x_j}g(x)\]

As $h + h' \le 1$,

\[(f: g) + (f': g) \le \sum_{j = 1}^{n} c_{j}[h(x_{j}) + h'(x_{j}) + 2\delta]\]

Since the above holds for all such $\seqf{c_j}\subset [0, \infty)$ and $\seqf{x_j}\subset G$,

\begin{align*}(f: g) + (f': g)&\le (1 + 2\delta)(H: g) \\&\le (1 + 2\delta)[(f + f': g) + \delta(\eta: g)]\end{align*}

$\square$

Theorem 28.2.6 (Haar).label Let $G$ be a locally compact group, then:

  1. (1)

    There exists a left/right Haar measure on $G$.

  2. (2)

    For any two left/right Haar measures $\mu$ and $\nu$ on $G$, there exists $\lambda > 0$ such that $\mu = \lambda \nu$.

Proof, [Theorem 2.10, 2.20, Fol16]. (1): Fix $h \in C_{c}^{+}(G)$ with $h \ne 0$. For each $g \in C_{c}^{+}(G)$ with $g \ne 0$, let

\[I_{g}: C_{c}^{+}(G) \to [0, \infty) \quad f \mapsto \frac{(f: g)}{(h: g)}\]

then by (5) of Proposition 28.2.4, for each $f \in C_{c}^{+}(G)$ with $f \ne 0$,

\[\frac{1}{(h: f)}= \frac{(f: g)}{(h: f)(f: g)}\le I_{g}(f) \le \frac{(f: h)(h: g)}{(h: g)}= (f: h)\]

Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is relatively compact for each $f \in C_{c}^{+}(G)$.

For each $V \in \cn_{G}(1)$, let $E_{V} = \bracs{I_g|g \in C_c^+(V) \setminus \bracs{0}}$, then $\fF = \bracs{E_V|V \in \cn_G(1)}$ is a filter on the product space $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$. By Tychonoff’s Theorem, $\prod_{f \in C_c^+(G)}\ol{\mathcal{I}(f)}$ is compact, and $\bigcap_{V \in \cn_G(1)}\ol{E_V}\ne \emptyset$.

Let $I \in \bigcap_{V \in \cn_G(1)}\ol{E_V}$, then by continuity,

  1. (i)

    For each $f \in C_{c}^{+}(G) \setminus \bracs{0}$, $I(f) \in [(h: f)^{-1}, (f: h)]$.

  2. (ii)

    For every $\lambda \ge 0$ and $f \in C_{c}^{+}(G)$, $I(\lambda f) = \lambda I(f)$.

  3. (iii)

    For any $x \in G$, $I(L_{x}f) = I(f)$.

  4. (iv)

    For each $f, f' \in C_{c}^{+}(G)$, $I(f + f') \le I(f) + I(f')$.

Let $f, f' \in C_{c}^{+}(G)$ and $\eps > 0$. By Lemma 28.2.5, there exists $V \in \cn_{G}(1)$ such that for each $g \in E_{V}$,

  1. (a)

    $|I_{g}(f) - I(f)|, |I_{g}(f') - I(f')| < \eps$.

  2. (b)

    $I_{g}(f) + I_{g}(f') \le I_{g}(f + f') + \eps$.

In which case, $I(f) + I(f') \le I(f + f') + 3\eps$. Since this holds for all $\eps > 0$,

  1. (v)

    For each $f, f' \in C_{c}^{+}(G)$, $I(f + f') \ge I(f) + I(f')$.

Using Lemma 16.3.7, $I$ extends to a positive linear functional on $C_{c}(G; \real)$, with $I(f) > 0$ for all $f \in C_{c}^{+}(G) \setminus \bracs{0}$, and

  1. (LH)

    For each $f \in C_{c}(G)$ and $x \in G$, $I(L_{x}f) = I(f)$.

By (i) and the Riesz Representation Theorem, there exists a unique non-zero Radon measure $\mu: \cb_{G} \to [0, \infty]$ such that for each $f \in C_{c}^{+}(G)$, $I(f) = \int_{G} f d\mu$. Finally, by density of $C_{c}(G; \real)$ in $L^{1}(\mu; \real)$ and (LH), $\mu$ is a left Haar measure.

(2): Let $f, g \in C_{c}^{+}(G) \setminus \bracs{0}$ and $\eps > 0$, then by Proposition 28.1.2, there exists a symmetric neighbourhood $V \in \cn_{G}(1)$ such that for any $x \in G$ and $y \in V$,

\[|f(xy) - f(yx)|, |g(xy) - g(yx)| < \eps\]

By Urysohn’s lemma, there exists $h \in C_{c}^{+}(V) \setminus \bracs{0}$ such that $h(x) = h(x^{-1})$ for all $x \in V$. Since $f$ and $h$ are both compactly supported and $\mu, \nu$ are locally finite, by Tonelli’s Theorem,

\begin{align*}\paren{\int f d\mu}\paren{\int h d\nu}&= \iint f(x)h(y) \mu(dx)\nu(dy) \\&= \iint f(yx)h(y) \nu(dx)\mu(dy) \\\end{align*}

Similarly, by symmetry of $h$,

\begin{align*}\paren{\int h d\mu}\paren{\int f d\nu}&= \iint h(x)f(y) \mu(dx)\nu(dy) \\&= \iint h(y^{-1}x)f(y) \mu(dx)\nu(dy) \\&= \iint h(x^{-1}y)f(y) \nu(dy)\mu(dx) \\&= \iint h(y)f(xy) \nu(dy)\mu(dx) \\&= \iint h(y)f(xy) \mu(dx)\nu(dy)\end{align*}

Thus there exists $C > 0$ such that

\begin{align*}&\abs{\paren{\int f d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int f d\nu}}\\&\le \abs{\iint h(y)[f(xy) - f(yx)]\mu(dx)\nu(dy)}\le C\eps\end{align*}

and

\[\abs{\paren{\int g d\mu}\paren{\int h d\nu} - \paren{\int h d\mu}\paren{\int g d\nu}}\le C\eps\]

so there exists $C' > 0$ such that

\[\abs{\frac{\int h d\nu}{\int h d\mu} - \frac{\int g d\nu}{\int g d\mu}}\le C'\eps\]

and

\[\abs{\frac{\int f d\nu}{\int f d\mu} + \frac{\int h d\nu}{\int h d\mu}}\le C'\eps\]

Therefore

\[\abs{\frac{\int f d\nu}{\int f d\mu} - \frac{\int g d\nu}{\int g d\mu}}\le 2C' \eps\]

As the above holds for all $\eps > 0$, $\int f d\nu/\int f d\mu = \int g d\nu/\int g d\mu$. By uniqueness from the Riesz Representation Theorem, there exists $\lambda > 0$ such that $\mu = \lambda \nu$.$\square$

Proposition 28.2.7.label Let $G$ be a locally compact group and $\mu: \cb_{G} \to [0, \infty]$ be a left/right Haar measure, then:

  1. (1)

    For each $U \subset G$ open with $U \ne \emptyset$, $\mu(U) > 0$.

  2. (2)

    For each $f \in C_{c}^{+}(G) \setminus \bracs{0}$, $\int \phi d\mu > 0$.

Proof, for left Haar measures. Since $\mu \ne 0$, there exists $g \in C_{c}^{+}(G) \setminus \bracs{0}$ such that $\int g d\mu > 0$. By compactness of $\supp{g}$, there exists $\seqf{x_j}\subset G$ such that:

\[\supp{g}\subset \bigcup_{j = 1}^{n} x_{j}^{-1}U \quad g \le \sum_{j = 1}^{n} L_{x_j}f\]

Thus $0 < \int g d\mu \le n\mu(U)$ and $0 < \int g d\mu \le n \int f d\mu$.$\square$

Proposition 28.2.8.label Let $G$ be a locally compact group, $\mu: \cb_{G} \to [0, \infty]$ be a left Haar measure, $p \in [1, \infty)$, and $E$ be a normed vector space over $K \in \RC$, then the mapping

\[G \times L^{p}(\mu; E) \quad (x, f) \mapsto L_{x}f\]

is jointly continuous. Similarly, if $\nu: \cb_{G} \to [0, \infty]$ is a right Haar measure, then

\[G \times L^{p}(\mu; E) \quad (x, f) \mapsto R_{x}f\]

is also jointly continuous.

Proof of the left case. Let $\eps > 0$, $x, y \in G$, and $f, g \in L^{p}(\mu; E)$, then

\begin{align*}\norm{L_xf - L_yg}_{L^p(\mu; E)}&\le \norm{L_xf - L_yf}_{L^p(\mu; E)}+ \norm{L_yf - L_y g}_{L^p(\mu; E)}\\&= \norm{L_xf - L_yf}_{L^p(\mu; E)}+ \norm{f - g}_{L^p(\mu; E)}\end{align*}

By Proposition 24.1.7, there exists $\phi \in C_{c}(G)$ such that $\norm{\phi - f}_{L^p(\mu; E)}< \eps$. In which case,

\begin{align*}\norm{L_xf - L_yf}_{L^p(\mu; E)}&\le \norm{L_xf - L_x \phi}_{L^p(\mu; E)}+ \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)}\\&+ \norm{L_yf - L_y \phi}_{L^p(\mu; E)}\\&= 2\norm{f - \phi}_{L^p(\mu; E)}+ \norm{L_x\phi - L_y\phi}_{L^p(\mu; E)}\\&\le 2\eps + \normn{L_{x^{-1}y}\phi - \phi}_{u}\mu\bracs{\phi \ne 0}\end{align*}

By Proposition 28.1.2, there exists $V \in \cn_{G}(1)$ such that if $x^{-1}y \in V$, then $\norm{L_{x^{-1}y}\phi - \phi}_{u} < \eps/\mu\bracs{\phi \ne 0}$. Thus if $x^{-1}y \in V$, then

\[\norm{L_xf - L_yf}_{L^p(\mu; E)}\le 3\eps + \norm{f - g}_{L^p(\mu; E)}\]

$\square$

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