If $\Phi$ is an alternate sign measure then exists $c \in \mathbb{R}$ such that $|\Phi(A)| \leq c$ Definition. Alternate sign measure is a function $\sigma$-aditive $\Phi$ of sets defined in $\sigma$-ring of subsets in a given $X$. 
ProblemIf $\Phi$ is an alternate sign measure then exists $c \in \mathbb{R}$ such that $|\Phi(A)| \leq c$
My attempt:
If $E \in \sigma$-ring is unit, i.e, $A \subset E$ $\forall A \in \sigma$-ring
then I define $\Phi(E)=c$.So $\Phi(A)\leq \Phi(E)=c$. But I have problem, because if $c < 0$ the $c \leq \Phi(A)$ and it isn't what I want, besides I suppose that exists unit and I don't know what I should do if it doesn't exists. 
Another, let $A_k \in \sigma$-ring such that $A=\cup_k A_k$, then 
$\Phi(A)=\Sigma_k \Phi(A_k)$, so $|\Phi(A)|=|\Sigma_k \Phi(A_k)| \leq \Sigma_k |\Phi(A_k)|$ then if I could conclude that $\Sigma_k |\Phi(A_k)|$ converges we finish, but I don't see how.
Could you guide me? please. 
I'm using this book in page 392 is where You can find the definition.
 A: Given the space $X$, let $\Sigma$ be a $\sigma$-ring of subsets of $X$ and let 
$\Phi$ be a $\sigma$-aditive function from $\Sigma$ to $\mathbb{R}$. 
You want to prove that exists $c \in \mathbb{R}$ such that $|\Phi(A)| \leq c$, for all $A \in \Sigma$.
Proof: 
First, suppose that for any $n \in \mathbb{N}$ there is $A_n\in \Sigma$, such that $\Phi(A_n) > n$. For all $n \in \mathbb{N}$ let $A_n^+$ be the positive part of $A_n$ (using Hahn decomposition). We have, $\Phi(A_n^+) \geqslant\Phi(A_n) > n$.
Let $D = \bigcup_{n \in \mathbb{N}}A_n^+$. Then $D\in \Sigma$ and for all  $n \in \mathbb{N}$, we have
$$ \Phi(D) =\Phi \left(\bigcup_{n \in \mathbb{N}}A_n^+ \right) \geqslant \Phi(A_n^+)\geqslant \Phi(A_n)> n$$
Contradiction, because $\Phi(D) \in \mathbb{R}$.
So there is $n_0 \in \mathbb{N}$, such that for all $A \in \Sigma, \Phi(A) \leqslant n_0$.
Second, (it is similar to  the previous argument) suppose that for any $n \in \mathbb{N}$ there is $A_n\in \Sigma$, such that $\Phi(A_n) < -n$. For all $n \in \mathbb{N}$ let $A_n^-$ be the negative part of $A_n$ (using Hahn decomposition). We have, $\Phi(A_n^-) \leqslant\Phi(A_n) < -n$.
Let $E = \bigcup_{n \in \mathbb{N}}A_n^-$. Then $E\in \Sigma$ and for all  $n \in \mathbb{N}$, we have
$$ \Phi(E) =\Phi \left(\bigcup_{n \in \mathbb{N}}A_n^- \right) \leqslant \Phi(A_n^-)\leqslant \Phi(A_n)<-n$$
Contradiction, because $\Phi(E) \in \mathbb{R}$.
So there is $n_1 \in \mathbb{N}$, such that for all $A \in \Sigma, \Phi(A) \geqslant -n_1$.
Third, let $c$ be any number greater or equal to both $n_0$ and $n_1$, then we have,for all $A \in \Sigma$,
$$ -c\leqslant -n_1 \leqslant \Phi(A) \leqslant n_0 \leqslant c $$
so 
$$ |\Phi(A)| \leqslant c $$
Remark: What Kolmogorov calls "alternate sign measure" or "charge" in his book (page 392) is today more commonly called "finite signed measure defined on a $\sigma$-ring".
