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 $\mu = \lambda \nu$.$\square$

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