Jtdylktuy

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. $nu^{+}genu^{-}$ on $mathcal{A}$ and, therefore, $nu^{+}genu^{-}$ on $sigma(mathcal{A})$ (e.g. by Theorem 1.9.3, where you take $mathcal{E}={Ainsigma(mathcal{A}):nu^{+}(A)ge nu^{-}(A)}$). 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.