The following should be rather elementary, but since I never found a result of this kind in the literature and I never thought about this before, I wonder whether I'm missing something or not.
First of all, I claim the following:
Claim 1: Let
- $(E,\mathcal E,\mu)$ be a measure space;
- $\mathcal A\subseteq\mathcal E$ be disjoint;
- $B\in\mathcal E$ with $\biguplus\mathcal A\subseteq B$ and > $\mu(B)<\infty$.
Then $(\mu(A))_{A\in\mathcal A}$ is summable and $$\sum_{A\in\mathcal A}\mu(A)\le\mu(B)\tag1.$$
Proof: Since $\mathcal A$ is disjoint, $$\sum_{A\in\mathcal F}\mu(A)=\mu\left(\biguplus\mathcal F\right)\le\mu(B)<\infty\tag2$$ for all countable $\mathcal F\subseteq\mathcal A$ and hence, since $\mu$ is nonnegative, $(\mu(A))_{A\in\mathcal A}$ is summable with $$\sum_{A\in\mathcal A}\mu(A)=\sup_{\substack{\mathcal F\subseteq\mathcal A\\\mathcal F\text{ is finite}}}\sum_{A\in\mathcal F}\mu(A)\le\mu(B)\tag3.$$
Now I would like to use Claim 1 to conclude the following:
Claim 2: Let $E$ be a metric space, $\mu$ be a finite signed measure on $\mathcal B(E)$ and $x\in E$. Then, $$\forall\varepsilon>0:\exists\delta\in(0,\varepsilon]:|\mu|(\partial B_\delta(x))=0\tag4.$$
Proof: Let $\varepsilon>0$. Then, $$\biguplus_{\delta\in(0,\:\varepsilon)}\partial B_\delta(x)=B_\varepsilon(x)\setminus\{x\}\tag5$$ and hence, since $|\mu|$ is finite, $(\mu(\partial B_\delta(x)))_{\delta\in(0,\:\varepsilon)}$ is summable by Claim 1. Thus, $$\left\{\delta\in(0,\varepsilon):\mu(\partial B_\delta(x))\ne0\right\}\tag6$$ is countable. This immediately yields the claim.