Uniqueness of left Haar measure up to scale

First discharge the dependent-choice hypothesis of [F3] directly. For any set $A$, a prescribed $a\in A$, and a relation $R\subseteq A\times A$ with every successor set $R[x]=\{y\in A:xRy\}$ nonempty, [A1] chooses one member of each set in $\{R[x]:x\in A\}$. Composing this choice function with $x\mapsto R[x]$ gives $f:A\to A$ with $xRf(x)$. By [F4], $g(0)=a$ and $g(n+1)=f(g(n))$ define an infinite $R$-chain. This proves exactly the required dependent-choice principle, including the prescribed starting point. Now let $I(f)=\int fd\mu$ and $J(f)=\int fd\nu$. Compact finiteness makes these real-valued on real $C_c$. They are positive and invariant. They are nonzero: for each measure, nonzeroness and outer regularity give an open set of positive mass, open inner regularity gives a compact subset of positive mass, and [F3] gives a cutoff dominating its indicator. Use one such nonzero cutoff as $f_0$; strict positivity in [F1] gives $I(f_0),J(f_0)>0$.

Fix $f\ge 0$ and $ϵ>0$. Choose the same symmetric $u$ so the [F1] bounds for $f$ and $f_0$ are both at most $ϵ$, using the intersection of their identity neighbourhoods. Divide by $I(u)>0$ and set $r=J(u)\mathrm{/}I(u)$. Then $\mathrm{\mid }rI(f)-J(f)\mathrm{\mid }\le ϵ$ and $\mathrm{\mid }rI(f_0)-J(f_0)\mathrm{\mid }\le ϵ$. Multiplying by $I(f_0)$ and $I(f)$ respectively and subtracting cancels $r$, giving $\mathrm{\mid }I(f_0)J(f)-I(f)J(f_0)\mathrm{\mid }\le ϵ(I(f_0)+I(f))$. Letting $ϵ\downarrow 0$ proves $J(f)=cI(f)$ with $c=J(f_0)\mathrm{/}I(f_0)>0$.

For real $h=h^{+}-h^{-}$, linearity extends $J=cI$ from the positive cone to all $C_c(G;\mathbb{R})$. The measure $c\mu$ is Radon because scaling by a finite positive scalar preserves each regularity equality and compact finiteness. Therefore [F2] gives $\nu =c\mu$ on all Borel sets, including those of infinite measure.