Lemma: Let $\rho$ be a signed measure that omits the value $+\infty$. If $\rho(G)>0$, then there exists a subset $E\subseteq G$ such that $E$ is totally positive and $\rho(E)>0$.
Proof: Let $E_1=G$. If $E_1$ is totally positive, then there is nothing to prove; so suppose that $E_1$ is NOT totally positive. Then $\exists F\subseteq E_1$ such that $\rho(F)>\rho(E_1)$. In particular, the set $J_1=\{n\in\mathbb{N}: \exists F\subseteq E_1(\rho(F)\geq\rho(E_1)+\frac{1}{n})\}$ is nonempty, thus has a smallest element $n_1$. Choose $E_2\subseteq E_1$ such that $\rho(E_2)\geq\rho(E_1)+\frac{1}{n_1}$. Now, if $E_2$ were totally positive, then the proof is complete. Otherwise, let $n_2$ be the smallest element of $J_2=\{n\in\mathbb{N}: \exists F\subseteq E_2(\rho(F)\geq\rho(E_2)+\frac{1}{n})\}$. Continuing by induction produces a decreasing sequence of measurable sets $\mathbf{(E_i)_{i=1}^{\infty}}$ such that $\mathbf{\rho(E_{i+1})\geq\rho(E_i)+\frac{1}{n_i}}$.
I am almost certain that Choice has been invoked somewhere in the definition of $(E_i)_{i=1}^{\infty}$, but I just can't put my finger on exactly where. Any help would be greatly appreciated.
Let $A_1=\{F\subseteq G: \exists n\in\mathbb{N}(\rho(F)\geq\rho(G)+\frac{1}{n})\}$, $A_2=\{F'\subseteq G: \exists F\in A_1[F'\subseteq F\land\exists n\in\mathbb{N}(\rho(F')\geq \rho(F)+\frac{1}{n})]\}$, etc. Then one can invoke the Dependent Choice to show the existence of a sequence $(A_i)_{i=1}^{\infty}$, where the total relation $R$ on $\mathscr{A}=\{A_i: i\in\mathbb{N}\}$ is $$A_i R A_j\ \text{if and only if}\ \forall F'\in A_{j}\exists F\in A_i[F'\subseteq F\land\exists n\in\mathbb{N}(\rho(F')\geq\rho(F)+\frac{1}{n})].$$ Further, one can invoke the existence of a Choice Function $c: \mathscr{A}\to\overset{\infty}{\underset{i=1}{\bigcup}}A_i$ to show the existence of a sequence $(E_i)_{i=1}^{\infty}$ such that $\forall i(E_i\in A_i)$. Indeed, the definition of $(E_i)_{i=1}^{\infty}$ in the proof of the lemma requires even stricter conditions, namely $\rho(E_{i+1})\geq\rho(E_i)+\frac{1}{n_i}$, where $n_i$ is the smallest element of each $J_i$, hence further argument is still needed.