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Is it possible to give an explicit construction of a set function, defined on a $\sigma$-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?

Let me elaborate. By "explicit" I mean that the example should not appeal to non-constructive methods like the Hahn-Banach theorem or the existence of free ultrafilters. I'm aware that such examples exist, but I'm looking for something more concrete. If such constructions are not possible, I'm especially interested in understanding why that is so.

This question is similar, but, so far as I can tell, not identical to several other questions asked on this site and MO. For example, I've learned that proving the existence of the "integer lottery" on $P(\mathbb{N})$ requires the Axiom of Choice (https://mathoverflow.net/questions/95954/how-to-construct-a-continuous-finite-additive-measure-on-the-natural-numbers).

That's the sort of result I'm interested in, but it doesn't fully answer my question. My question doesn't require that the $\sigma$-algebra in question be $P(\Omega)$, and I'm interested in general $\Omega$, not just $\Omega = \mathbb{N}$.

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    $\begingroup$ You want to look at the proof that the Hahn-Banach theorem is equivalent to the existence of finitely additive measures like that. $\endgroup$ – Asaf Karagila Sep 14 '16 at 20:39
  • $\begingroup$ @AsafKaragila Fantastic! Thank you. Do you have a good reference for that result? $\endgroup$ – grndl Sep 14 '16 at 22:11
  • $\begingroup$ I don't remember off hand. I'll see what I can find later today. $\endgroup$ – Asaf Karagila Sep 15 '16 at 3:51
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The result that the Hahn-Banach is equivalent to the existence of a [nontrivial] finitely additive probability measure on an arbitrary Boolean algebra is due to Luxemburg. Note that we don't even require the Boolean algebra to be $\sigma$-complete. Going out on a limb, I'll guess that working with only $\sigma$-algebras you don't get the full strength of the HB theorem, but only something close enough.

You can find a reasonably detailed proof (along with many other equivalents of the Hahn-Banach theorem) in Eric Schechter's book "Handbook of Analysis and its Foundations" on page 620.

(When you look at the book, remember that Schechter calls a finitely additive measure a "charge".)

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The question is:

"Is it possible to give an explicit construction of a set function, defined on a σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?"

If you do not require the "measure" to be a "probability" (or to have only finite values), then the answer is YES. Here are two examples:

Example 1:

Consider $\mathbb{N}$ and $\Sigma=P(\mathbb{N})$. Clearly $\Sigma$ is a $\sigma$-algebra. Define $\mu : \Sigma \to [0,+\infty]$ by $\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.

It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure.

Note that $\mu$ is not a finitely additive probability. In fact, it is not even a finite finitely additive measure.

Example 2:.

Consider $[0,1]$ and $\Sigma$ be the set of countable or co-countable subsets of $[0,1]$. Clearly $\Sigma$ is a $\sigma$-algebra on $[0,1]$ Define $\mu : \Sigma \to [0,+\infty]$ by $\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.

It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure.

Note again that $\mu$ is not a finitely additive probability. In fact, it is not even a finite finitely additive measure.

Remark 1:

The examples above may be included in the following general case.

Let $\Omega$ be any infinite set and let $\Sigma$ be a $\sigma$-algebra on $\Omega$, such that there are infinitely many finite sets in $\Sigma$, in other words, the set $\{A \in \Sigma : A \textrm{ is finite }\}$ is infinite. Then, define $\mu : \Sigma \to [0,+\infty]$ by $\mu(A)=0$ if $A$ is finite and $\mu(A)=+\infty$ if $A$ is infinite.

It is easy to see that $\mu$ is a finitely additive measure, but not a (countable additive) measure.

Remark 2:

If the OP meant to ask:

Given ANY σ-algebra, is it possible to give an explicit construction of a set function, defined on that σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?

Then, the answer is NO, because any finitely additive measure on a finite σ-algebra is automatically a (countable additive) measure.

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  • $\begingroup$ Given an abstract $\sigma$-algebra, how do you do you construct a measure like that? $\endgroup$ – Asaf Karagila Sep 16 '16 at 19:09
  • $\begingroup$ @AsafKaragila What do you mean by an abstract $\sigma$-algebra? Are you asking if we can construct a "measure" like that in ANY $\sigma$-algebra? The answer is no. For instance, consider any finite $\sigma$-algebra. Any finitely additive measure on a finite $\sigma$-algebra is automatically a (countable additive) measure. $\endgroup$ – Ramiro Sep 16 '16 at 20:43
  • $\begingroup$ Sure, but any infinite one. Your answer is an example. Examples are not proofs. And this is not a question asking for a possible counterexample, but for a proof or lack thereof. $\endgroup$ – Asaf Karagila Sep 16 '16 at 20:45
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    $\begingroup$ @AsafKaragila The question is "Is it possible to give an explicit construction of a set function, defined on a σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?" The examples I gave shows the answer is YES. The question is NOT "Given ANY σ-algebra, is it possible to give an explicit construction of a set function, defined on that σ-algebra, with all the properties of a measure except that it is merely finitely additive and not countably additive?" $\endgroup$ – Ramiro Sep 16 '16 at 20:53
  • $\begingroup$ @Ramiro Thank you for your answer. I hadn't considered infinite measures, as I am ultimately interested in probability measures (though I didn't say this in the question, so your examples are good ones). Having read these comments, I now see that my question could have been put more carefully. Of course, I am not interested in the trivial case of finite $\sigma$-algebras. Although I would say that Ramiro has answered the question as stated, I'm inclined to leave it open for a bit as I find Asaf's line of response more interesting. $\endgroup$ – grndl Sep 17 '16 at 19:48
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The way I understand the question is this:

Can you construct

  • A measurable space $(\Omega,\Sigma)$ where $\Sigma$ is a $\sigma$-algebra
  • A finitely additive finite measure $\mu$ on $\Sigma$ that is not countably additive?

Previous answers showed that

  • The answer is yes if we allow $\mu$ to be infinite
  • The answer is no if $\mu$ is required to be finite and $\Omega$ is countable or finite.

But if I understand the owners intentions correctly from comments the exact question, where $\Omega$ may be uncountable and $\mu$ must be finite, has not been answered.

As I was interested in the question specified above I gave it some thought and found the following result, which means that the answer is essentially no:

Either it is not possible to construct such an example or if it is, then we cannot constructively show that the example is not countably additive.

Proof by contradiction:

Assume we have $(\Omega,\Sigma,\mu)$ as required. Assume also that we can construct a sequence $A_n$ of disjoint sets in $\Sigma$ which shows that $\mu$ is not countably additive, ie $\mu(\bigcup A_n)\ne\sum\mu(A_n)$.

Then we could construct another example where the underlying set is countable

  • $\Omega^\star=\{A_1,A_2,A_3,\dots\}$
  • $\Sigma^\star=2^{\Omega^\star}$
  • $\mu^\star(B)=\mu(\bigcup_{i:A_i \in B} A_i)$

Since $\mu^\star$ is essentially a restriction of $\mu$ it is also a finite, finitely additive measure and it is not countably additive by construction.

We already know from other answers that constructing an example with a countable $\Omega$ is not possible so we have a contradiction.

I would appreciate if anyone could tell which of the two possibilities is true.

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  • $\begingroup$ I agree with your assessment of the other answers -- they are relevant, but the main question hasn't been answered for finite-valued measures. Your result is very interesting and I agree it's true. I wonder if it is still possible to give a constructive example which clearly cannot be countably additive but we don't know at what point that countable-additivity breaks. $\endgroup$ – 6005 Jul 25 '17 at 15:56
  • $\begingroup$ @balazs.g You're right. Looking back at this thread, I think that I was too quick to accept Asaf's answer so I've unaccepted it. Thanks also for the interesting observation. $\endgroup$ – grndl Jul 25 '17 at 19:51
  • $\begingroup$ @balzas.g I do not understand why we shouldn't let $\Omega$ be countable. As a way of illustration let $\Omega=\mathbb{N}$, $\Sigma=2^\Omega$ and $\mathcal{F}\subset \Sigma$ be the family of co-finite sets, i.e. $F\in\mathcal{F}$ if and only if $F^c$ is finite. Then call $\mu\colon\Sigma\to [0,1]$ the function that assigns $1$ to elements in $\mathcal{F}$ and $0$ to those that do not belong to $\mathcal{F}$. Then $\mu$ is finitely additive but not $\sigma$-additive. $\endgroup$ – Niccolò Urbinati Apr 23 '18 at 16:14

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