# Total variation of complex measure is finite

Let $\mu$ be a complex measure on a measurable space $(X, \Sigma)$. Let $|\mu|$ be the total variation of $\mu$, defined by $|\mu|(E) = \sup \left\{ \sum_{j=1}^{\infty} |\mu(E_j)| : \{E_j\}_{j=1}^\infty\text{ is a pairwise disjoint,$\Sigma$-measurable partition of }E \right\}$.

I'm trying to show that $|\mu|(X) < \infty$. I know this is a very common result in most measure theory texts, but I don't have one on hand that deals with complex measures; and I'm having difficulty coming up with the proof. If anyone could give me a hint to start with, it would be greatly appreciated.

EDIT:

Note here I'm taking a complex measure to be a countably additive set map $\mu: \Sigma \to \mathbb{C}$ and hence $|\mu(E)| < \infty$, for every $E \in \Sigma$.

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You must be missing an additional assumption. Why should the variation of a measure be finite? Consider the Lebesgue measure on $\mathbb{R}$. – Christopher A. Wong Dec 25 '12 at 22:45
OK, I see the discrepancy. Some people (such as in Folland and other analysis textbooks) take the definition of "complex measure" to be slightly stronger than the definition I am used to. – Christopher A. Wong Dec 25 '12 at 22:55
I've edited the original question with a clarification on the definition of a complex measure. – anonymous Dec 26 '12 at 2:01

Hint: first write $\mu$ as $\mu_1+i\mu_2$ with $\mu_1,\mu_2$ real signed measures, and then apply Hahn decomposition theorem to $\mu_1,\mu_2$.