The modular function is a continuous homomorphism

Discharge the choice hypothesis of [F5] from AC. Given a set $A$, a point $a\in A$ and a serial relation $R$ on $A$ (every $R[x]$ nonempty), [A1] chooses one member of each set in the family $\{R[x]:x\in A\}$, and composing this choice function with $x\mapsto R[x]$ gives $f:A\to A$ with $xRf(x)$. By [F6] the recursion $g(0)=a$, $g(n+1)=f(g(n))$ produces an infinite $R$-chain from $a$. So Dependent Choice is available.

Multiplicativity. Fix $g,h\in G$ and a Borel set $E$ with $0<\mu (E)<\mathrm{\infty }$, which exists by inner regularity and finiteness on compact sets in [F4]. Using the Borel-level identity of [F1] twice gives $\mu (E(gh{)}^{-1})=\mu (E{h}^{-1}{g}^{-1})=c(g)\mu (E{h}^{-1})=c(g)c(h)\mu (E)$, while the defining property of $c$ gives $\mu (E(gh{)}^{-1})=c(gh)\mu (E)$; the scalar $c(gh)-c(g)c(h)$ therefore annihilates the nonzero measure $\mu$, so $c(gh)=c(g)c(h)$, and replacing $g,h$ by their inverses and using ${\mathrm{\Delta }}_G(x)=c(x^{-1})$ from [F2] gives ${\mathrm{\Delta }}_G(gh)={\mathrm{\Delta }}_G(g){\mathrm{\Delta }}_G(h)$.

There is $f_0\in C_c(G)$ with $f_0\ge 0$ and ${\int }_G{f}_{0}d\mu >0$. Since $\mu$ is nonzero and outer regular on Borel sets by [F4], some open $U$ has $\mu (U)>0$; inner regularity of $\mu$ on the open set $U$ gives a compact $K\subseteq U$ with $\mu (K)>0$. By [F5], applied under the Dependent Choice just derived, choose $f_0\in C_c(G)$ with ${\mathbf{1}}_K\le f_0\le {\mathbf{1}}_U$; then $f_0\ge 0$ and $\int f_{0}d\mu \ge \mu (K)>0$.

Define $\phi (a):={\int }_G{f}_0(xa)d\mu (x)$ for $a\in G$ for the function $f_0$ of step 2.1. By [F1] applied to $f_0$ we have $\phi (a)=c(a)\int f_{0}d\mu ={\mathrm{\Delta }}_G(a^{-1})\phi (e)$ for every $a$, and $\phi (e)=\int f_{0}d\mu >0$ by step 2.1; hence ${\mathrm{\Delta }}_G(a)=\phi (a^{-1})\mathrm{/}\phi (e)$ for every $a$, so continuity of ${\mathrm{\Delta }}_G$ will follow from continuity of $\phi$ and of inversion.

The function $\phi$ is continuous. Fix $a_0\in G$. By [F3] there are a neighbourhood $N$ of $a_0$ and a compact $C\subseteq G$ with $\mathrm{supp}{R}_a{f}_0\subseteq C$ for every $a\in N$ and with $\mathrm{\parallel }{R}_a{f}_0-R_{a_0}{f}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\to 0$ as $a\to a_0$; for $a\in N$ the difference $R_a{f}_0-R_{a_0}{f}_0$ is supported in $C$, whence $\mathrm{\mid }\phi (a)-\phi (a_0)\mathrm{\mid }\le \mathrm{\parallel }{R}_a{f}_0-R_{a_0}{f}_0{\mathrm{\parallel }}_{\mathrm{\infty }}\mu (C)$ with $\mu (C)<\mathrm{\infty }$ by [F4], so $\phi (a)\to \phi (a_0)$.

${\mathrm{\Delta }}_G$ is continuous: by step 3.1 it is the composition of the continuous maps $a\mapsto a^{-1}$ (inversion in a topological group), the map $\phi$ that is continuous by step 3.2, and division by the positive constant $\phi (e)>0$.

Finally ${\mathrm{\Delta }}_G(e)=c(e)=1$, because $\mu (E{e}^{-1})=\mu (E)$ forces $c(e)=1$ by the same uniqueness of the scalar as in [F1]; and ${\mathrm{\Delta }}_G$ takes values in ${\mathbb{R}}_{>0}$ by definition. With the multiplicativity of step 1.2 this exhibits ${\mathrm{\Delta }}_G$ as a continuous homomorphism from $G$ to the multiplicative group ${\mathbb{R}}_{>0}$. ∎