Proof of Existence of the Total Variation of a Finite Signed Measure

Throughout, $\mathcal{F}$ is closed under the finite set operations used below, being a $\sigma$-algebra, and $S_{\nu}(A)$ is a set of real numbers, the image of a set under a function given by an expression (routine facts of the setting).

Step 1: a Hahn set exists. By the Hahn decomposition theorem (which uses the axiom of dependent choice), $\nu$ has a Hahn set. Let $P$ be any Hahn set for $\nu$, and let $\nu^{+}$ and $\nu^{-}$ be the Jordan decomposition formed with it: finite measures on $(X,\mathcal{F})$ with $\nu=\nu^{+}-\nu^{-}$ setwise. Their values are real numbers $\ge0$.

Step 2: an upper bound. Let $B\in\mathcal{F}$ with $B\subseteq A$. Then $A\setminus B\in\mathcal{F}$, and both $B$ and $A\setminus B$ are contained in $A$. Since $\nu^{-}(B)\ge0$ and $\nu^{+}(A\setminus B)\ge0$, and measures are monotone,

$$\nu(B)-\nu(A\setminus B)=\nu^{+}(B)-\nu^{-}(B)-\nu^{+}(A\setminus B)+\nu^{-}(A\setminus B)\le\nu^{+}(B)+\nu^{-}(A\setminus B)\le\nu^{+}(A)+\nu^{-}(A).$$

So $\nu^{+}(A)+\nu^{-}(A)$ is an upper bound of $S_{\nu}(A)$.

Step 3: the bound is attained. The set $B_{0}=A\cap P$ lies in $\mathcal{F}$ and is contained in $A$, and $A\setminus B_{0}=A\setminus P$. Hence

$$\nu(A\cap P)-\nu(A\setminus P)=\nu(B_{0})-\nu(A\setminus B_{0})\in S_{\nu}(A),$$

and by the defining formulas of the Jordan decomposition this number equals $\nu^{+}(A)+\nu^{-}(A)$. By Step 2 it is therefore the greatest element of $S_{\nu}(A)$; as $P$ was an arbitrary Hahn set, this proves (i).