Jordan-Hahn decomposition in Robert Ash's book I was reading through the book "Real Analysis and Probability" by Robert Ash, and got really confused by the proof given to the Jordan-Hahn decomposition. The theorem states the following.
Let $\lambda$ be a countably additive extended real valued function on the $\sigma$ field F, then defining:
$\lambda ^+(A)= \sup\{\lambda(B): B \in F , B\subset A\}$
$\lambda ^-(A)= -\inf\{\lambda(B): B \in F , B\subset A\}$
Then $\lambda^+, \lambda^-$ are measures on F and $\lambda=\lambda^+-\lambda^-$
In the proof he says we can assume that $\lambda$ doesn't assume the value $- \infty$, by using the following reasoning. If $- \infty$ belongs to the range of $\lambda$ then, $\infty$ does not, by the definition of countable additive function.Thus $-\lambda$ never takes the value $- \infty$, but I don't understand how this last sentence justifies that we can assume such fact.

Suppose that $\lambda (A)=\infty $ and for all measurable $B\subset C=\left(\bigcup \{X:X\in F\}\right)\setminus A$ we have $\lambda (B)-$ is finite or $\lambda (B) =\infty.$ Then for every $D\in F$ we have $\lambda (D) =\lambda ((D\cap A)\cup (D\cap C))=\lambda (D\cap A) +\lambda (D\cap C)$ is finite or equal $\infty.$
Now assume that $\lambda (T) =\infty$ and $\lambda (S)=-\infty $ for some $T, S\in F$ then by the above there exists $S_1 \in F$ such that $T\cap S_1 =\emptyset $ and $\lambda (S_1 ) =-\infty.$ But this is imposible since $T\cup S_1 \in F$ and $$\lambda (T\cup S_1 ) =\lambda (T) +\lambda (S_1 ) =\infty -\infty$$ is undefined.
