decomposition of finitely additive measure Let $(X,\mathcal{A},\mu)$ be a measure space. Let $\tau:\mathcal{A}\to\mathbb{R}$ be a function satisfying conditions:
$\sup\{|\tau(A)|;\;A\in\mathcal{A}\}<\infty$,
$\tau(A\cup B)=\tau(A)+\tau(B)$ if $A,B\in\mathcal{A},\;A\cap B=\emptyset,$
$\tau(A)=0$ if $A\in\mathcal{A}$ and $A$ if locally $\mu$-null.
This is a definition of finitely additive measure taken from Hewitt, Stromberg : Real and abstract analysis, Springer Verlag 1965, Definition 20.27.
My question: Let such measure $\tau$ be given. Do there exist nonnegative measures $\tau_1,\tau_2:\mathcal{A}\to[0,\infty)$ satisfying above conditions such that $\tau=\tau_1-\tau_2$ ? Thank in advance for any help.
 A: I think this result is known as Jordan decomposition.
Let $\mathcal{A}$ be a Boolean algebra and $\tau\colon\mathcal{A}\to\mathbb{R}$ be a finitely additive measure (i.e. a function such that $\tau(\emptyset)=0$ and $\tau(A\cup B)=\tau(A)+\tau(B)$ whenever $A,B\in\mathcal{A}$ are disjoint). Suppose that $\tau$ is bounded in the sense that $\sup\{\tau(A):\ A\in\mathcal{A}\}<+\infty$.
We define the functions $\tau^+\colon\mathcal{A}\to[0,+\infty)$ and $\tau^+\colon\mathcal{A}\to[0,+\infty)$ by setting: $\tau^+(A):=\sup\{\tau(B):\ B\subseteq A,\ B\in\mathcal{A}\}$ and $\tau^-(A):=\sup\{-\tau(B):\ B\subseteq A,\ B\in\mathcal{A}\}$ for every $A\in\mathcal{A}$. Please observe that, since $\tau$ is bounded, $\tau^+$ and $\tau^-$ must be bounded too.

Lemma 1:
  $\tau^+$ and $\tau^-$ are finitely additive measure.

proof: We prove that $\tau^+$ is a finitely additive measure (the proof for $\tau^-$ is similar). It is enough to take $A,B\in\mathcal{A}$ disjoint and prove that $\tau^+(C)=\tau^+(A)+\tau^-(B)$.
For all $C\in\mathcal{A}$ with $C\subseteq A\cup B$ we have:
$$\tau(C)=\tau(C\cap A)+\tau(C\cap B)\leq \tau^+(A)+\tau^+(B)$$
so $\sup\{\tau(C):C\in\mathcal{A},\ C\subseteq A\cup B\}\leq \tau^+(A)+\tau^+(B)$. 
To prove the reverse inequality we let $\epsilon>0$ and take $A_1\subseteq A$, $B_1\subseteq B$ in $\mathcal{A}$ such that $\tau(A_1)>\tau^+(A)-\epsilon/2$ and $\tau(B_1)>\tau^+(B)-\epsilon/2$ (such $A_1$ and $B_1$ exist because $\tau^+$ is bounded). Then, by setting $C:=A_1\cup B_1$, we have:
$$\tau^+(A\cup B)\geq\tau(C)=\tau(A_1)+\tau(B_1)>\tau^+(A)-\epsilon/2+\tau^+(B)-\epsilon/2=\tau^+(A)+\tau^+(B)-\epsilon.$$
But then, by the generality of $\epsilon$, we have $\tau^+(A\cup B)\geq\tau^+(A)+\tau^+(B)$ as claimed. We have proved that $\tau^+$ is a finitely additive measure.

Lemma 2:
  $\tau=\tau^+-\tau^-$.

proof:
 We show that the equality $\tau^-=\tau^+-\tau$ holds. For $A\in\mathcal{A}$ we have:
\begin{equation}
\begin{aligned}
\tau^+(A)-\tau(A) &=\sup\{\tau(B)-\tau(A):\ B\in\mathcal{A},\ B\subseteq A\}=\\
    &=\sup\{-\tau(A\setminus B):\ B\in\mathcal{A},\ B\subseteq A\}=\\
    &=\sup\{-\tau(C):\ C\in\mathcal{A},\ C\subseteq A\}=\\
    &=\tau^-(A).
\end{aligned}
\end{equation}
It is important to keep in mind that, differently from what happens in the case of $\sigma$-additive measures defined on $\sigma$-algebras, it is not true that $\tau^+$ can be written as the measure $A\mapsto\tau(A\cap P)$ for a specific $P\in\mathcal{A}$. This is because, in general, finitely additive measures do not admit a Hahn decomposition.
