Bounded finely additive signed measure is a signed measure. I am reading Royden Real Analysis Forth edition chapter 19.3 named: "The Kantrovitch Representation Theorem for the dual of $L^\infty$" He gives defenition of bounded finitely additive signed measure $v$ as: 
set function on measurable space which is finitely additive and has finite total variation of $v$ over whole space $X$.
I am wondering why does this imply that this is in fact a signed measure. So far, I've derived that $v$ of empty set is zero, and some other facts but I can't show countable additivity.
Is this even true? 
Thank you!
 A: In general it is false, that a finitely additive measure of bounded variation is countably additive. 
For this consider the measurable space $(X,\Sigma) = (\mathbb{N},2^{\mathbb{N}})$. 
Let $\phi$ be a Banach limit on $\ell^\infty = L^\infty(\mathbb{N},2^{\mathbb{N}},\#)$ with $\#$ counting measure.
Then let 
$$ \mu(A) = \phi(\mathbb{1}_A) \quad \text{for} \quad  A \in 2^{\mathbb{N}}\,,$$
where $1_A$ is the characteristic (indicator) function of $A$.
Then $\mu$ is obviously non-negative, finitely additive and of bounded variation, since if $A = \bigsqcup_{k=1}^\infty A_k$ is a partition, then 
$$
\sum_{k=1}^\infty |\mu(A_k)| = \sum_{k=1}^\infty \mu(A_k) = \lim_{n \to \infty} \phi(1_{\bigsqcup_{k=1}^n A_k}) \leq ||\phi|| \lim_{n \to \infty} ||1_{\bigsqcup_{k=1}^n A_k}||_{\ell^\infty} \leq ||\phi|| < \infty\,. 
$$
But $\mu$ is not countably additive, since $\mu(\{n\}) = 0$ for all $n \in \mathbb{N}$, but $\mu(\mathbb{N}) = 1$ by the definition of a Banach limit.
Note that the total variation of a finitely-additive measure is not necessarily countably additive. 
A: Let $\nu$ be the signed finitely additive set function over $\mathfrak M$ and $\|\nu\|$ its total variation. Then $\|\nu\|$ is a bounded positive measure, and we have that
$\lvert\nu(A)\rvert\le\|\nu\|(A)$, for all $A\in\mathfrak{M}$.
In order to show that $\nu$ is a signed measure, it suffices to show that
$$
\nu(A)=\sum_{n\in\mathbb N}\nu(A_n), 
$$
whenever $\{A_n\}_{n\in\mathbb N}$ is a sequence of disjoint measurable sets with union $A$. 
Let $S_n=A_1\cup\cdots\cup A_n$, $T_n=A\setminus S_n$. Then
$$
\nu(A)-\nu(T_n)=\nu(S_n)=\nu(A_1)+\cdots+\nu(A_n).
$$
As $\|\nu(T_n)\|\to 0$, then $\nu(T_n)\to 0$.
Next observe that
$$
\Big|\,\nu(A)-\sum_{n\in\mathbb N}\nu(A_n)\Big|=\Big|\,\nu(S_n)+\nu(T_n)-\sum_{k=1}^n\nu(A_k)-\sum_{k>n}\nu(A_k)\Big| \\ =\Big|\,\nu(T_n)-\sum_{k>n}\nu(A_k)\Big|\le \lvert \nu(T_n)\rvert+\sum_{k>n}\lvert\nu(A_k)\rvert
\le \| \nu(T_n)\|+\sum_{k>n}\|\nu\|(A_k) \\=\| \nu(T_n)\|+\|\nu\|\Big(\bigcup_{k>n}A_k\Big)=2\|\nu\|(T_n)\to 0.
$$
