I'm studying finitely additive measures and have come across the following definition. A finitely additive measure $\mu$ on $(\Omega, \mathcal{F})$ is purely finitely additive if for all countably additive measures $\nu$ on $(\Omega, \mathcal{F})$ and all $A \in \mathcal{F}$, $0 \leq \nu(A) \leq \mu(A)$ implies $\nu = 0$.

Formally, I'm fine with this definition, but I feel I'm lacking intuition about it. I had expected a purely finitely additive measure to be a finitely additive measure that is not countably additive. That is, if $\mu$ is purely finitely additive, I expected that there exists a countably infinite disjoint sequence $\{ A_n: n \in \mathbb{N}\}$ in $\mathcal{F}$ such that $\sum_{n \in \mathbb{N}}\mu(A_n) < \mu(A)$, where $A = \cup_{n \in \mathbb{N}}A_n$.

I am unable to prove or disprove that this latter assertion is equivalent to the definition above. Could someone give me some hints or guidance?

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    $\begingroup$ Possible answer. Your source may follow the reasonably convention that a countably additive measure is finitely additive. You may think or want "purely finitely additive" to mean "finitely but not countably additive" when your source has the definition you quote in mind. $\endgroup$ – Ethan Bolker Feb 13 '17 at 16:45
  • $\begingroup$ @EthanBolker Yes, I think that is the intention. But I don't yet see how the two definitions are equivalent (if they are). $\endgroup$ – grndl Feb 13 '17 at 16:47
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    $\begingroup$ Maybe they are not. Someone who's looked at measure theory mich more recently than I may provide an answer. $\endgroup$ – Ethan Bolker Feb 13 '17 at 16:52
  • $\begingroup$ What is the reference you are reading? Is it the paper by Yosida and Hewitt? $\endgroup$ – Dominique R.F. Feb 13 '17 at 20:49
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    $\begingroup$ The definition does appear in that paper. I'm currently reading Rao and Rao's Theory of Charges. $\endgroup$ – grndl Feb 13 '17 at 21:17

I don't know much about purely finitely additive measures, but since no one has answered yet let me give my interpretation of the definition given by Yosida and Hewitt in their paper Finitely additive measures (1952).

As you expected, a (non-zero) purely finitely additive measure is clearly not countably additive. If it were, taking $\nu = \mu$ in the definition would imply $\mu=0$.

However, it seems that the right intuition is that purely finitely additive measures are those which are very far from being countably additive. Therefore, not being countably additive is not enough. As Yosida and Hewitt put it in their article, purely finitely additive measures are in fact "as unlike countably additive measures as possible".

Here is my (quite possibly wrong) interpretation of what they mean by this. After giving the definition of purely finitely additive measures, the authors prove the following theorem:

Theorem 1.15 (K. Yosida and E. Hewitt, Finitely additive measures, 1952):

Let $\phi \in \Phi$ have the property that for certain countably additive measures $\psi_1$, $\psi_2$, the inequality $\psi_1 \leq \phi \leq \psi_2$ obtains. Then $\phi$ is countably additive.

Here, we have fixed some set $X$ and an algebra of subsets of $X$, and $\Phi$ denotes the set of finite, signed, finitely additive measures defined on this algebra. For the following discussion, let me focus on positive measures. Then, if $\phi$ is a positive finitely additive measure, we have $\phi \geq 0$ where $0$ is the zero measure which is obviously countably additive. Now, if we had $\phi \leq \psi$ for some countably additive measure $\psi$, the theorem above would imply that $\phi$ is in fact countably additive. Therefore, if $\phi$ is a finitely additive measure that is not countably additive, there are no countably additive measures greater than it. However, we might still be able to approximate $\phi$ from below by countably additive measures. That $\phi$ is purely finitely additive means that you cannot approximate from below either: the only countably additive measure below $\phi$ is $0$!

Another way to say this is that, with respect to the natural partial order on measures, a (positive) purely finitely additive measure is one that is not comparable to any countably additive measure except the trivial one.

To recap, if $\phi$ is a positive purely finitely additive measure, there are no countably additive measures above it and the only one below it is the trivial measure. In this sense, we could say that $\phi$ is "far" from any countably additive measure. Let me say again that Yosida and Hewitt state very clearly that purely finitely additive measure should be thought of as very unlike countably additive ones, but that the explanation above is my own (nonexpert) interpretation and should not be taken at face value.

  • $\begingroup$ +1 Thanks for your thoughtful answer. I need to think on this for a bit and revisit the Yosida and Hewitt paper, but I'll reply with a more substantive comment soon. $\endgroup$ – grndl Feb 18 '17 at 3:46
  • $\begingroup$ This is helpful, thanks again. I think the question I asked is still open though. Namely, how does the Yoshida-Hewitt definition interact with countably infinite sequences of pairwise disjoint events? Given a purely finitely additive measure, can we construct such a sequence such that countable additivity fails? Does countable additivity always fail, for any such sequence? $\endgroup$ – grndl Feb 18 '17 at 18:27
  • $\begingroup$ Well, there exists at least one sequence where countable additivity fails, since, like I said, a purely finitely additive measure is not countably additive. Countable additivity cannot always fail, since you can take a sequence such that all sets are empty except finitely many and then additivity will work (this is a pretty trivial example, though). I guess in general it will be somewhere in between, but I'll have to think about this to make it precise... $\endgroup$ – Dominique R.F. Feb 18 '17 at 19:36
  • $\begingroup$ Anyway, you might want to look at theorems 1.18, 1.19 and 1.22 of their article, which give equivalent formulations of the definition. $\endgroup$ – Dominique R.F. Feb 18 '17 at 19:37
  • $\begingroup$ I think I did not fully appreciate your paragraph after the statement of Theorem 1.15 during my first reading. This is very helpful and gives me the intuition I was looking for. Thanks so much. $\endgroup$ – grndl Feb 18 '17 at 22:56

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