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!
