In the book 'An Introductory Course in Functional Analysis' by Bowers and Kalton, they give the definition of positive measure at page $209:$

Let $(X,\mathcal{A})$ be a measurable space, A set function $\mu:\mathcal{A} \rightarrow [0,\infty]$ is said to be countably additive if $$\mu\left( \bigcup_{j=1}^\infty A_j \right) = \sum_{j=1}^\infty \mu(A_j)$$ whenever $(A_j)_{j=1}^\infty$ is a sequence of pairwise disjoint measurable sets. A countably additive set function $\mu:\mathcal{A} \to [0,\infty]$ such that $\mu({\emptyset})=0$ is called a positive measure.

The authos give definition of complex measure at page $213:$

Let $(X,\mathcal{A})$ be a measurable space, A countably additive set function $\mu:\mathcal{A} \rightarrow \mathbb{C}$ is called a complex measure. When we say $\mu$ is countably additive, we mean
$$\mu\left( \bigcup_{j=1}^\infty A_j \right) = \sum_{j=1}^\infty \mu(A_j)$$ whenever $(A_j)_{j=1}^\infty$ is a sequence of pairwise disjoint measurable sets in $\mathcal{A},$ where the series is absolutely convergent.

At the same page, the authors mentioned the following:

There is a significant difference between positive measures and complex measures. In the definition of complex measure, we require that a complex measure $\mu$ be finite; that is, $|\mu(A)|<\infty$ for all $a\in\mathcal{A}.$ This was not a requirement for a positive measure.

Question: Why do we need to assume that $|\mu(A)| < \infty$ for complex measure while we do not need to assume it in positive measure? I strongly believe that purpose of inequality is not to ensure the series converges absolutely only, but it has other purpose.

  • 14
    $\begingroup$ To avoid having things like $\infty-\infty$. $\endgroup$ Commented Sep 9, 2017 at 17:38
  • $\begingroup$ Can you elaborate more on your statement? Can you give an example where we encounter the mentioned situation? $\endgroup$
    – Idonknow
    Commented Sep 10, 2017 at 0:28

1 Answer 1


Having a complex measure without imposing the condition of it being finite gives us an extremely weak object, because it doesn't need to be monotone.

In the case of positive measures we had that $\mu(A) \leq \mu(B)$ whenever $A\subseteq B$, and we would like to recover a similar if yet weaker property for complex measures.

Such property is the Hahn descomposition theorem, that allows us to break a complex measure into two measures that are again monotone. Moreover, every proof that I have seen of this descomposition theorem uses that the complex measure is finite. I can't think of a counter example right now, but I think that the result is false if one removes the hypothesis of it being finite.


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