Compactness, finite Haar volume and invariant vectors in the regular representation

The group $G$ is nonempty and open in itself, so [F1] gives $\mu (G)>0$. If $G$ is compact, [F1] also gives $\mu (G)<\mathrm{\infty }$.

Suppose $0\mathrm{\ne }f\in L^2(G)$ is invariant and put $a=\mathrm{\parallel }f{\mathrm{\parallel }}_2>0$. By [F4] choose $h\in C_c(G)$ with $\mathrm{\parallel }f-h{\mathrm{\parallel }}_2<a\mathrm{/}2$. Then $\mathrm{\parallel }h{\mathrm{\parallel }}_2>a\mathrm{/}2$, so $h\mathrm{\ne }0$; let $K=\mathrm{supp}h$, a nonempty compact Borel set by [F6]. If ${\int }_K\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu =0$, then [F7] gives $f=0$ almost everywhere on $K$ and $h=0$ off $K$, whence $\mathrm{\parallel }f-h{\mathrm{\parallel }}_2^2=\mathrm{\parallel }h{\mathrm{\parallel }}_2^2+{\int }_{G\setminus K}\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu \ge \mathrm{\parallel }h{\mathrm{\parallel }}_2^2$, contradicting $\mathrm{\parallel }f-h{\mathrm{\parallel }}_2<a\mathrm{/}2<\mathrm{\parallel }h{\mathrm{\parallel }}_2$. Therefore $c:={\int }_K\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu >0$.

The set $D:=K{K}^{-1}$ is compact: inversion maps $K$ continuously to a compact set by [F9], the finite product $K\times K^{-1}$ is compact by [F5], and the inclusion into $G\times G$ followed by multiplication maps it continuously onto $D$. It is closed and Borel by [F6].

Suppose $G$ is noncompact. Let $\mathcal{H}={\bigcup }_{n\in \mathbb{N}}{G}^n$ be the set of finite histories. For every $h=(g_1,\dots ,g_n)\in \mathcal{H}$, the set $C_h:=G\setminus {\bigcup }_{i=1}^n{g}_{i}D$ is nonempty, since the removed finite union of compact translates is compact by [F9, F10] and cannot equal $G$; set $C_{\mathrm{\varnothing }}=G$. AC chooses a selector $s(h)\in C_h$ for all histories. Define $T(h)$ by appending $s(h)$ to $h$; [F11] recursively produces histories $h_{n+1}=T(h_n)$ from $h_0=\mathrm{\varnothing }$, hence a sequence $g_1,g_2,\dots$ of appended entries. The translates $g_{n}K$ are pairwise disjoint: an intersection $g_{j}K\cap g_{i}K\mathrm{\ne }\mathrm{\varnothing }$ would imply $g_j\in g_{i}K{K}^{-1}=g_{i}D$, contrary to the choice of $g_j$.

The constant function has $\mathrm{\parallel }1{\mathrm{\parallel }}_2^2={\int }_{G}1d\mu =\mu (G)$, so $\mu (G)<\mathrm{\infty }$ exactly when $1\in L^2(G)$. By step 1.1 this class is nonzero; it is fixed by every ${\lambda }_G(g)$. Thus (ii) is equivalent to (iv), and (iv) implies (iii).

For each $n$, invariance of $f$ means $f(g_{n}x)=f(x)$ almost everywhere, since ${\lambda }_G(g_n^{-1})f=f$; [F2] therefore gives ${\int }_{g_{n}K}\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu ={\int }_K\mathrm{\mid }f(g_{n}x){\mathrm{\mid }}^{2}d\mu (x)=c$. The sets $g_{n}K$ are disjoint Borel sets by step 1.4 and [F6], so [F7] gives $a^2={\int }_G\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu \ge {\sum }_{n=1}^N{\int }_{g_{n}K}\mathrm{\mid }f{\mathrm{\mid }}^{2}d\mu =Nc$ for every $N$. Since $c>0$, [F8] supplies $N$ with $Nc>a^2$, a contradiction. Thus (iii) implies (i).

Step 1.1 gives (i)$\Rightarrow$(ii), step 2.1 gives (ii)$\iff$(iv)$\Rightarrow$(iii), and step 2.2 gives (iii)$\Rightarrow$(i); hence all four conditions are equivalent. When they hold, $0<\mu (G)<\mathrm{\infty }$ by step 1.1, so scaling $\mu$ by $1\mathrm{/}\mu (G)$ gives a left Haar probability measure.