I was reading the book of Bogachev "Measure Theory. Vol 1" the following theorem and important remark to this theorem:
I understood the proof of this theorem completely but I am not able to understand the meaning of this example.
1) What does mean the term "has no signed countably.."?
2) Can anyone explain why it has no signed extension to $\sigma(\mathcal{A})$ and how it follows from (iii)?
3) What is the essence of this example?
Would be very grateful for detailed explanation!
Suppose that there exists such a signed measure $\nu$. Then $\nu$ is nonnegative on $\mathcal{A}$, i.e. $\nu^{+}\ge\nu^{-}$ on $\mathcal{A}$ and, therefore, $\nu^{+}\ge\nu^{-}$ on $\sigma(\mathcal{A})$ (e.g., by Theorem 1.9.3, where you take $\mathcal{E}=\{A\in\sigma(\mathcal{A}):\nu^{+}(A)\ge \nu^{-}(A)\}$). This implies that $\nu$ is nonnegative on $\sigma(\mathcal{A})$ and must coincide with $\mu^{*}$ by uniqueness.
On the other hand, the example with $X=\{0,1\}$ and $\mathcal{A}=\{\emptyset,X\}$ shows that signed extensions to $\mathcal{A}_{\mu}$ do exist.