Variation of sum of measure A variation of arbitrary complex measure $\nu$ on the measurable set $E$ is called the number $\|\nu\|(E)=\sup \sum_{n=1}^\infty |\nu (E_n)|$, where supremum is taken over all sequences $(E_n)$ such that $E_n$ are  measurable, pairwise disjoint and their union is $E$.
Let $\mu$ and $\lambda$ be a complex measures on the same sigma-algebra in $X$ and assume that these measures are concentrate in disjoint measurable subsets $A,B \subset X$. 
I wish to show that  then $\| \lambda+\mu \| (E)=\|\lambda\|(E)+\|\mu\|(E)$ for arbitrary measurable set $E$. 
I know how to do "$\leq$ "inequality but don't know how to prove "$\geq$".
Thanks
 A: Fix $\varepsilon>0$, and $\{E_j\},\{F_j\}$ sequences of pairwise disjoint measurable sets such that 
$$\lVert\mu\rVert(E)\leq \sum_{j=0}^{+\infty}|\mu(E_j)|+\varepsilon,\quad \lVert\nu\rVert(E)\leq \sum_{j=0}^{+\infty}|\nu(F_j)|+\varepsilon.$$
Define for $(i,j,k)\in\Bbb N^2\times \{0,1\}$:
$$S_{i,j,0}=E_i\cap F_j\cap A,\quad S_{i,j,1}=E_i\cap F_j\cap B.$$
This gives pairwise disjoint sets. Using $\sigma$-additivity and the property of concentration, we get 
\begin{align}
\lVert \mu+\nu\rVert(E)&\geq \sum_{(i,j,k)\in \Bbb N^2\times \{0,1\}}|(\mu+\nu)(S_{i,j,k})|\\
&=\sum_{(i,j)\in\Bbb N}|\mu(E_i\cap F_j)|+\sum_{(i,j)\in\Bbb N}|\nu(E_i\cap F_j)|\\
&=\sum_{i\in\Bbb N}\sum_{j\in\Bbb N}|\mu(E_i\cap F_j)|+\sum_{j\in\Bbb N}\sum_{i\in\Bbb N}|\nu(E_i\cap F_j)|\\
&\geq \sum_{i\in\Bbb N}\left|\sum_{j\in\Bbb N}\mu(E_i\cap F_j)\right|+
\sum_{j\in\Bbb N}\left|\sum_{i\in\Bbb N}\nu(E_i\cap F_j)\right|\\
&=\sum_{i\in\Bbb N}|\mu(E_i)|+\sum_{j\in\Bbb N}|\nu(F_j)|\\
&\geq \lVert\mu\rVert(E)+\lVert\nu\rVert(E)-2\varepsilon.
\end{align}
This gives the result since $\varepsilon$ is arbitrary.
