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

then by (5) of Proposition 27.2.4, for each $f \in C_{c}^{+}(G)$ with $f \ne 0$,

Thus $\mathcal{I}(f) = \bracs{I_g(f)|g \in C_c^+(G) \setminus \bracs{0}}$ is precompact 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, there exists $\bigcap_{V \in \cn_G(1)}\ol{E_V}\ne \emptyset$.

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

Let $f, f' \in C_{c}^{+}(G)$ and $\eps > 0$. By Lemma 27.2.5, 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 $f, f' \in C_{c}^{+}(G)$, $I(f + g) \ge I(f) + I(g)$.

Using Lemma 16.1.13, $I$ 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 $C_{c}(\mu; \real)$ in $L^{1}(\mu; \real)$ and (LH), $\mu$ is a left Haar measure.$\square$