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Assume $\nu$ is a complex measure, E is an arbitrary measurable set. How to prove $\sup\{\sum\limits_{i=1}^\infty |\nu(E_i)||E_i$ is a disjoint sequence of measurable sets such that $E=\bigcup\limits_{i=1}^\infty E_i\}\leq\sup\{|\int_E fd\nu|||f|\leq 1\}$? Thanks.

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You should invest a bit more work in your questions: What is the definition of a complex measure you're working with? (There are various ways to define them and the answer will depend quite drastically on the definition). What do you know about complex measures? What are your thoughts about this problem? Lacking this information it is quite hard to give a reasonable answer to this and your previous question without investing a serious amount of effort. Thus I'm voting to close this question. –  t.b. Mar 20 '11 at 19:36
    
Is it only me, who is failing to see, why the answer to this question depends on a specific definition of complex measure? –  Alexander Thumm Mar 21 '11 at 6:59
    
@Theo Buehler. As for your first question, I don't know there are various definition of complex measure. My definition of complex measure is natural: First it is a complex function defined on a measurable space(do I need to explain what is measurable space or what is a complex number?). Second it satisfies the countable additivity(do I need to explain what is countable additivity or what is countable?). Third, the value of empty set is zero(do I need to explain what is empty set or why such a set exists?). ...(to be continued) –  zzzhhh Mar 21 '11 at 7:06
    
I think this definition is a natural extension of traditional measure to complex field, so I think there is no need to explain what a complex measure is and no need to develop other versions of the definition. That's why I didn't explain it. As for your second question "What are your thoughts about this problem?", I just have no idea about this question, so I ask. ...(to be continued) –  zzzhhh Mar 21 '11 at 7:08
    
Finally, even if there are many definitions, it is enough to point out this problem, and I will appreciate you very much if you can show me the drastical consequence resulting from different definitions, but it is an over-reaction to vote to close my question. I don't mean to offend anybody, but this kind of over-reaction is arrogant and rude and does offend me. I can not understand this behavior because this behavior does not accord with what a people having high reputation should do. –  zzzhhh Mar 21 '11 at 7:08

1 Answer 1

up vote 1 down vote accepted

Let $(E_i)_{i\in\mathbb N} \subset E$ be a sequence of disjoint measurable sets covering $E$. Let $\chi_{E_i}$ be the characteristic function of $E_i$, that is $\chi_{E_i}(x) = 1$ for $x\in E_i$ and $\chi_{E_i}(x) = 0$ otherwise, and $\lambda_i$ the unique complex number which satisfies $|\lambda_i|=1$ and $\lambda_i \nu(E_i) = |\nu(E_i)|$.

We will now consider $f^k = f^k_{(E_i)_{i\in\mathbb N}} := \sum_{i=1}^k \lambda_i * \chi_{E_i}$.

We have:

1) $|f^k| \leq 1$

2) $\int_E f^k \;d\nu = \sum_{i=1}^k |\nu(E_i)|$

q.e.d.

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I get it! The point is to find the $\lambda_i$ that has unit amplitude but can turn any complex number into its absolute value, wonderful! so the left hand set is a subset of the right hand set, therefore has a smaller sup. Thank you, Alexander Thumm, it's very nice and ingenious of you! –  zzzhhh Mar 21 '11 at 8:18

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