# 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

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Do you know about the Hahn decomposition? – Nate Eldredge Aug 1 '12 at 18:26
Yes, I know (for real measures). – Richard Aug 1 '12 at 18:37

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.