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$\mu$ and $\nu$ are complex measure, and $|\mu|$ is the total variation, that is,

$$|\mu|(E):=\sup\left\{\sum_{i=1}^\infty|\mu(E_i)|, \{E_i\}_{i=1}^{+\infty}\mbox{ is partition of }E\right\}.$$

Is this always true ?


It seems directly use $|A+B|\leqslant|A|+|B|$to the definition of $|\mu+\nu|$. But...this is too easy, I think I make a mistake or miss something .

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Your inequality is correct. For any partition $\{E_i\}$ of $E$, we have

$$ \sum_{i=1}^\infty\vert(\mu+\nu)(E_i)\vert=\sum_{i=1}^\infty\vert\mu(E_i)+\nu(E_i)\vert\leq\sum_{i=1}^\infty\vert\mu(E_i)\vert+\sum_{i=1}^\infty\vert\nu(E_i)\vert $$

The inequality is justified since $\vert\mu\vert(E)<\infty$ and $\vert\nu\vert(E)<\infty$ for all measurable $E$. Thus taking the supremum on both sides we have:

$$\vert \mu+\nu\vert(E)=\sup\sum_{i=1}^\infty\vert\mu(E_i)+\nu(E_i)\vert\leq\sup\left(\sum_{i=1}^\infty\vert\mu(E_i)\vert+\sum_{i=1}^\infty\vert\nu(E_i)\vert\right)\\ \leq\sup\sum_{i=1}^\infty\vert\mu(E_i)\vert+\sup\sum_{i=1}^\infty\vert\nu(E_i)\vert=\vert\mu\vert(E)+\vert\nu\vert(E) $$

Hope this helps.

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There are indices $i$ missing, right? Moreover, the first inequality is wrong and only valid up to (arbitrary small) $\epsilon$. Then it should go through... – Dirk Nov 20 '12 at 9:54
Corrected. The first inequality wasn't even necessary, thanks for pointing it out. – icurays1 Nov 20 '12 at 14:13

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