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