I am self-studying signed measure, and I come across the following construction:
Let $\mu$ be a signed measure on the measurable space $(X,\mathscr{A})$, and let $A$ be a subset of $X$ that belongs to $\mathscr{A}$ and satisfies $-\infty<\mu(A)<0$. Let \begin{align} \delta_1 = \sup\{\mu(E):E\in\mathscr{A}\ \text{and}\ E\subseteq A\},\tag1 \end{align} and choose an $\mathscr{A}$-measurable subset $A_1$ of $A$ that satisfies \begin{align*} \mu(A_1)\geq\min\left\{\frac{1}{2}\delta_1,1\right\}. \end{align*} Then $\delta_1$ and $\mu(A_1)$ are nonnegative (note that (1) implies that $\delta_1\geq\mu(\emptyset)=0$). We proceed by induction, constructing sequences $\{\delta_n\}$ and $\{A_n\}$ by letting \begin{align*} \delta_n = \sup\left\{\mu(E):E\in\mathscr{A}\ \text{and}\ E\subseteq\left(A-\bigcup_{i=1}^{n-1}A_1\right)\right\}, \end{align*} and then choosing an $\mathscr{A}$-measurable subset $A_n$ of $A-\bigcup_{i=1}^{n-1}A_i$ that satisfies \begin{align*} \mu(A_n)\geq\min\left\{\frac{1}{2}\delta_n,1\right\}. \end{align*} Now define $A_{\infty}$ and $B$ by $A_{\infty}=\bigcup_{n=1}^{\infty}A_n$ and $B=A-A_{\infty}$.
The finiteness of $\mu(A)$ implies the finiteness of $\mu(A_{\infty})$, and so, since $\mu(A_{\infty})=\sum_n\mu(A_n)$, implies that $\lim_{n\to\infty}\mu(A_n)=0$. Consequently $\lim_{n\to\infty}\delta_n=0$.
I have such a difficult time understanding why $\lim_{n\to\infty}\delta_n=0$ would follow from the construction. Intuitively, I feel that what this procedure did is that it removes every subset of $A$ with nonnegative signed measure from $A$. Then what remains ($B$) must be of nonpositive signed measure. But I really have trouble proving $\lim_{n\to\infty}\delta_n=0$ formally; that is, proving $$ \lim_{n\to\infty}\delta_n = \lim_{n\to\infty}\sup\left\{\mu(E):E\in\mathscr{A}\ \text{and}\ E\subseteq\left(A-\bigcup_{i=1}^{n-1}A_1\right)\right\}=0. $$
Could someone please help me out? Thank you very much in advance!
Update$\quad$ The $\delta_n$'s cannot all be bigger than or equal to 2. There are two ways to see this: First, $\mu(A_n)\geq\min\left\{\frac{1}{2}\delta_n,1\right\}$ and $\lim_{n\to\infty}\mu(A_n)=0$. This make sure that $\delta_n$'s would eventually be smaller than 2. Second, from the construction, if all $\delta_n$'s are bigger than or equal to 2, then all the sets $E_n$'s must satisfy $E_n\subseteq A$ and $\mu(E_n)>1.9$. But this would imply that $\mu\left(\sum_{n=1}^{\infty}E_n\right)=+\infty$, contradicts $\bigcup_{n=1}^{\infty}E_n\subseteq A$.
