A question on total variation of signed measures This is a variant on a problem from Royden's Real Analysis:
Let $\nu$ be a signed measure on a measurable space $\left\langle X,\mathcal{M}\right\rangle$, and let $$\Sigma:=\left\{{\left.\sum_{k=1}^{n}{\left|\nu{\left(E_k\right)}\right|}\ \right|}\ E_k\in\mathcal{M}\hspace{4pt}\forall k=1,2,\dots,n;n\in\mathbb{N}\right\}.$$  We define the total variation of $\nu$, denoted by $|\nu|$, as follows:  given the Jordan decomposition $\nu=\nu^+-\nu^-$, $|\nu|:=\nu^++\nu^-$.  I’m trying to prove that $|\nu|(X)=\sup\left(\Sigma\right)$.  I have $\mathrm{LHS}\leq\mathrm{RHS}$:  using a Hahn decomposition $X=A\sqcup B$ with $A\hspace{4pt}\nu$-positive and $B\hspace{4pt}\nu$-negative, I’ve proven that $|\nu|(X)=|\nu(A)|+|\nu(B)|\in\Sigma$, and hence $|\nu|(X)\leq\sup\left(\Sigma\right)$.  I’m not sure quite how to prove the opposite inequality.  I’ve tried to show that $|\nu|(X)$ majorizes $\Sigma$, but not gotten anywhere with it.  Any suggestions would be appreciated.
 A: I've figured out the opposite inequality:
let $\left\{E_k\right\}_{k=1}^{n}$ be an arbitrary sequence of measurable sets, and assume without loss of generality that $E_i\cap E_j=\emptyset\hspace{4pt}\forall i\ne j$.  Let $X=A\sqcup B$ be a Hahn decomposition of $X$ as above, with $A$ $\nu$-positive and $B$ $\nu$-negative.  Then $\forall k=1,2,\dots,n$, we have that $ E_k=\left(E_k\cap A\right)\sqcup\left(E_k\cap B\right)$.  Hence $\sum_{k=1}^{n}{\left|\nu\left(E_k\right)\right|}=\sum_{k=1}^{n}{\left|\nu\left(E_k\cap A\right)+\nu\left(E_k\cap B\right)\right|}\leq\sum_{k=1}^{n}{\left|\nu\left(E_k\cap A\right)\right|}+\sum_{k=1}^{n}{\left|\nu\left(E_k\cap B\right)\right|}=\sum_{k=1}^{n}{\nu^+\left(E_k\cap A\right)}+\sum_{k=1}^{n}{\nu^-\left(E_k\cap B\right)}$.  Now $\bigcup_{k=1}^{n}{\left(E_k\cap A\right)}\subseteq A$ and $\nu^+\left[\bigcup_{k=1}^{n}{\left(E_k\cap A\right)}\right]=\sum_{k=1}^{n}{\nu^+\left(E_k\cap A\right)}$, so by monotonicity, $\nu^+\left[\bigcup_{k=1}^{n}{\left(E_k\cap A\right)}\right]=\sum_{k=1}^{n}{\nu^+\left(E_k\cap A\right)}\leq\nu^+(A)$.  Similarly, $\nu^-\left[\bigcup_{k=1}^{n}{\left(E_k\cap B\right)}\right]=\sum_{k=1}^{n}{\nu^-\left(E_k\cap B\right)}\leq\nu^-(B)$.  Hence we have $\sum_{k=1}^{n}{\left|\nu\left(E_k\right)\right|}\leq\nu^+(A)+\nu^-(B)$.  But $\nu^+(A)+\nu^-(B)=\left|\nu\right|(X)$, so we have that $\sum_{k=1}^{n}{\left|\nu\left(E_k\right)\right|}\leq\left|\nu\right|(X)$.  Hence $\left|\nu\right|(X)$ is an upper bound for $S$.  $\therefore\left|\nu\right|(X)\geq\sup(S)$, as the supremum is the least upper bound.  $\therefore\left|\nu\right|(X)=\sup(S)$.  $\blacksquare$
