Theorem 1

On every locally compact group, there exists a left (resp. right) Haar measure, and, up to a constant factor, there exists only one.

A) Existence. — Set $\mathscr{K}(G) = \mathscr{K}$, $\mathscr{K}_+(G) = \mathscr{K}_+$, $\mathscr{K}_+^* = \mathscr{K}_+ - \{0\}$. If C is a compact subset of G, we denote by $\mathscr{K}_+^*(C)$ the set of $f \in \mathscr{K}_+^*$ with support in C. For $f \in \mathscr{K}$ and $g \in \mathscr{K}_+^*$, there exist numbers $c_1, \ldots, c_n \geq 0$ and elements $s_1, \ldots, s_n$ of G such that $f \leq \sum_{i=1}^n c_i \gamma(s_i)g$: for, there exists a nonempty open set U in G such that $\inf_{s \in U} g(s) > 0$, and the support of $f$ can be covered by a finite number of left-translates of U. Let $(f : g)$ be the infimum of the numbers $\sum_{i=1}^n c_i$ for all systems $(c_1, \ldots, c_n, s_1, \ldots, s_n)$ of numbers $\geq 0$ and elements of G such that $f \leq \sum_{i=1}^n c_i \gamma(s_i)g$. Then:

(i) $(\gamma(s)f : g) = (f : g)$ for $f \in \mathscr{K}$, $g \in \mathscr{K}_+^*$, $s \in G$; (ii) $(\lambda f : g) = \lambda(f : g)$ for $f \in \mathscr{K}$, $g \in \mathscr{K}_+^*$, $\lambda \geq 0$; (iii) $((f + f') : g) \leq (f : g) + (f' : g)$ for $f \in \mathscr{K}$, $f' \in \mathscr{K}$, $g \in \mathscr{K}_+^*$; (iv) $(f : g) \geq (\sup f)/(\sup g)$ for $f \in \mathscr{K}$, $g \in \mathscr{K}_+^*$; (v) $(f : h) \leq (f : g)(g : h)$ for $f \in \mathscr{K}$, $g \in \mathscr{K}_+^*$, $h \in \mathscr{K}_+^*$; (vi) $0 < \frac{1}{(f_0 : f)} \leq \frac{(f : g)}{(f_0 : g)} \leq (f : f_0)$ for $f, f_0, g$ in $\mathscr{K}_+^*$; (vii) let $f, f', h$ be in $\mathscr{K}_+$ with $h(s) \geq 1$ on the support of $f + f'$, and let $\varepsilon > 0$; there exists a compact neighborhood V of e such that, for every $g \in \mathscr{K}_+^*(V)$,

The properties (i), (ii), (iii) are obvious. Let $f \in \mathscr{K}$, $g \in \mathscr{K}_+^*$; if $f \leq \sum_{i=1}^n c_i \gamma(s_i)g$ with the $c_i \geq 0$, then $\sup f \leq \sum_{i=1}^n c_i g(s_i^{-1}s)$ for some s \in G, therefore $\sup f \leq (\sum_{i=1}^n c_i) \sup g$, whence (iv). Let us now prove (v); let $f \in \mathscr{K}$, $g, h$ in $\mathscr{K}_+^*$; if $f \leq \sum_{i=1}^n c_i \gamma(s_i)g$ and $g \leq \sum_{j=1}^p d_j \gamma(t_j)h$ ($c_i \geq 0, d_j \geq 0, s_i, t_j$ in $G$), then $f \leq \sum_{i,j} c_i d_j \gamma(s_i t_j)h$, therefore $(f : h) \leq \sum_{i,j} c_i d_j = (\sum_i c_i)(\sum_j d_j)$; thus $(f : h) \leq (f : g)(g : h)$. Applying (v) to $f_0, f, g$ on the one hand and to $f, f_0, g$ on the other, one obtains (vi). Finally, let $f, f', h$ be in $\mathscr{K}_+$ with $h(s) \geq 1$ on the support of $f + f'$, and let $\varepsilon > 0$. Set $F = f + f' + \frac{1}{2} \varepsilon h$; the functions $\varphi, \varphi'$, that coincide respectively with $f/F$ and $f'/F$ on the support of $f + f'$ and are zero outside it, belong to $\mathscr{K}_+$; for every $\eta > 0$, there exists a compact neighborhood $V$ of $e$ such that $|\varphi(s) - \varphi(t)| \leq \eta$ and $|\varphi'(s) - \varphi'(t)| \leq \eta$ for $s^{-1} t \in V$. Then let $g \in \mathscr{K}_+^*(V)$; for every $s \in G$,

for, this is obvious at the points where $\gamma(s)g$ is zero, therefore outside of $sV$; and in $sV$, $\varphi \leq \varphi(s) + \eta$; similarly, $\varphi' \cdot \gamma(s)g \leq (\varphi'(s) + \eta) \cdot \gamma(s)g$. This stated, let $c_1, \ldots, c_n$ be numbers $\geq 0$ and $s_1, \ldots, s_n$ elements of $G$ such that $F \leq \sum_{i=1}^n c_i \gamma(s_i)g$; then

and similarly for $f'$; consequently

since $\varphi + \varphi' \leq 1$. Applying the definition of $F$, then (ii), (iii) and (v), one infers that

and, if $\eta$ has been chosen so that $\eta[2((f + f') : h) + \varepsilon] \leq \frac{1}{2}\varepsilon$, one obtains (vii).

As V runs over the set of compact neighborhoods of e, the $\mathscr{K}_+^*(V)$ form a base of a filter $\mathscr{B}$ on $\mathscr{K}_+^*$. Let $\mathscr{F}$ be an ultrafilter on $\mathscr{K}_+^*$ finer than $\mathscr{B}$. On the other hand, let us fix $f_0 \in \mathscr{K}_+^*$ and let us set, for $f \in \mathscr{K}_+^*$ and $g \in \mathscr{K}_+^*$,

By (vi), $\lim_{g,\mathscr{F}} I_g(f) = I(f)$ exists in the compact space $[1/(f_0 : f), (f : f_0)]$.

By (iii), $I(f + f') \leq I(f) + I(f')$. By (vii), $I(f) + I(f') \leq I(f + f') + \varepsilon I(h)$ for all $\varepsilon > 0$ provided $h$ is $\geq 1$ on the support of $f + f'$; it follows that $I(f + f') = I(f) + I(f')$. By Ch. II, §2, No. 1, Prop. 2, I is extendible to a linear form on $\mathscr{K}$; this linear form is a nonzero positive measure on G, left-invariant by (i); this is the sought-for left Haar measure. Passing to the opposite group, one deduces from this the existence of a right Haar measure.

B) Uniqueness. — Let $\mu$ be a left Haar measure, $\nu$ a right Haar measure. Then $\check{\nu}$ is a left Haar measure. We are going to show that $\mu$ and $\check{\nu}$ are proportional. This will prove that any two left Haar measures are indeed proportional.

Let $f \in \mathscr{K}$ be such that $\mu(f) \neq 0$. By Lemma 1, the function $D_f$ defined on G by the formula

is continuous on G. Let $g \in \mathscr{K}$. The function $(s, t) \mapsto f(s)g(ts)$ is continuous with compact support in $G \times G$. By Ch. III, §4, No. 1, Th. 2,

whence

This proves, first, that $D_f$ does not depend on $f$. For, if $f' \in \mathscr{K}$ is such that $\mu(f') \neq 0$, then $D_f \cdot \mu = D_{f'} \cdot \mu$, therefore $D_f = D_{f'}$ locally almost everywhere for $\mu$, hence everywhere, since $D_f$ and $D_{f'}$ are continuous and the support of $\mu$ is $G$. We may therefore write $D_f = D$. The formula (16) gives

The formula (19) may be extended by linearity to the functions $f \in \mathscr{K}$ such that $\mu(f) = 0$. We have $D(e) \neq 0$ since $\check{\nu} \neq 0$. This indeed establishes the proportionality of $\mu$ and $\check{\nu}$.