Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A probability measure defined on a sample space $\Omega$ has the following properties:

  1. For each $E \subset \Omega$, $0 \le P(E) \le 1$
  2. $P(\Omega) = 1$
  3. If $E_1$ and $E_2$ are disjoint subsets $P(E_1 \cup E_2) = P(E_1) + P(E_2)$

The above definition defines a measure that is finitely additive (by induction) but not necessarily countably additive.

What is a probability measure that would be finitely additive but not countably additive (for a countable sample space $\Omega$)?

The example that I have seen most commonly on forums (this and elsewhere) is to set $P(E) = 0$ if $E$ is finite and $P(E) = 1$ if $E$ is co-finite. But that is not a probability measure as defined above since it is not defined on every subset of $\Omega$.

So an example of such a probability measure, or what is the reasoning that a finitely additive probability measure is not always countably additive?

share|cite|improve this question
    
I doubt that a general (probability-) measure has to be defined on all subsets of $\Omega$. If it is defined for a set $A$, then $A$ is said to be measurable. – Berci Sep 30 '12 at 13:41
    
@Berci I believe that would make sense for uncountable sets and the restriction exists because a general measure is countably additivity. But for a countable set, you should be able to define a probability measure for each subset. I edited the question to make it clarify countable sample space. – Kostub Deshmukh Oct 1 '12 at 6:31
up vote 8 down vote accepted

Let $\mathcal{U}$ be a free ultrafilter on $\mathbb{N}$. Let $P(A)=1$ if $A\in\mathcal{U}$ and $P(A)=0$ if $A\notin\mathcal{U}$. I think it is impossible to give an explicit example of a finitely additive measure on a $\sigma$-algebra that is not countably additive, but our resident set theorists might be able to tell you more about that.

share|cite|improve this answer
    
Thanks very helpful. I can see why a free ultrafilter will have this property, while a principle ultrafilter will not. Moreover, the Ultrafilter Lemma asserts the existence of a free ultrafilter (which contains the co-finite filter). However I cannot quite visualize what such a free ultrafilter on $\N$ will contain. And why can't such an ultrafilter be explicitly described? – Kostub Deshmukh Oct 1 '12 at 6:04
1  
@kostub: There are models without the axiom of choice in which there are no free ultrafilters (at all!), so it is impossible to give an explicit description of a free ultrafilter. Using the axiom of choice we can prove such filter exists, but not much more. – Asaf Karagila Oct 1 '12 at 7:33

In the following note the author shows that a finitely additive diffused measure on $\mathcal{P}(\omega)$ can be used to define a non Ramsey family. Combining this with a result of Mathias, it follows that it is consistent with $ZFC$ that there is no (ordinal) definable finitely additive diffused total measure on $\mathcal{P}(\omega)$.

share|cite|improve this answer

In this interesting note, the author proves the following: It is consistent to have a finitely additive total extension of Lebesgue measure on $[0, 1]$ such that, although, the measure zero sets form a sigma ideal, there is no real valued measurable cardinal below continnum.

share|cite|improve this answer

This question was from years ago, but I was just about to ask a similar question (I found this page from the stackexchange list of similar questions). My own question is whether it is possible to have an explicit example. The answers above are all non-explicit. Here is another (non-explicit) answer in a different form that I found to be helpful. It uses a Banach limit from functional analysis.

Define the natural numbers $\mathbb{N} = \{1, 2, 3, \ldots\}$ and define $2^{\mathbb{N}}$ as the set of all subsets of $\mathbb{N}$. Define $P:2^{\mathbb{N}}\rightarrow\mathbb{R}$ as follows: For each set $A \subseteq \mathbb{N}$, define $P(A)$ as a Banach limit of the sequence $\left\{\frac{|A \cap \{1, 2, ..., k\}|}{k}\right\}_{k=1}^{\infty}$.

Banach limit properties:

A Banach limit can be proven to exist and to have the following properties:

1) It is defined for all bounded real-valued sequences $\{x_k\}_{k=1}^{\infty}$, regardless of whether or not $x_k$ has a limit. In fact, the Banach limit is always a real number between $\liminf_{k\rightarrow\infty} x_k$ and $\limsup_{k\rightarrow\infty} x_k$.

2) The Banach limit is the same as the regular limit whenever the regular limit exists.

3) The Banach limit of a sum of two bounded sequences is the sum of the Banach limits of the individual sequences.

4) The Banach limit is nonnegative whenever $x_k \geq 0$ for all $k$.

There are many ways to define a real-valued function on bounded sequences that satisfies the above properties, so it is implicitly assumed that we consistently use one such function. The value of that function on a given bounded sequence is what we shall call the "Banach limit" of that sequence. Proofs of existence of such functions use nonexplicit things like axiom of choice or ultrafilters.

Using these properties:

Now if $A$ is a finite subset of $\mathbb{N}$ then $\frac{|A \cap \{1, ..., k\}|}{k} \leq \frac{|A|}{k}\rightarrow 0$, so the limit exists and $P(A)=0$. In particular, $P(\{n\})=0$ for all $n \in \mathbb{N}$. So:
$$ 1=P[\mathbb{N}] = P[\cup_{n=1}^{\infty} \{n\}] \neq \sum_{n=1}^{\infty}P[\{n\}]=0$$

Furthermore, $P(A)$ is nonnegative for all $A\subseteq \mathbb{N}$ (by the 4th property of Banach limits above). It also satisfies $P(A \cup B)=P(A)+P(B)$ whenever $A$ and $B$ are disjoint (which can be shown by the 3rd property above). So this $P(A)$ function is indeed a finitely-additive measure on all subsets of $\mathbb{N}$, but not a countably-additive one.

Remaining question:

The above pushes a bit more towards an explicit answer, but still uses Banach limits and hence is not explicit. Can a more explicit answer can be given? Now I'm not sure if I should formally ask this question on stackexchange or not, I suspect I would just get pointers back to your question.

share|cite|improve this answer
    
Read more about the necessity of the axiom of choice, or rather weaker versions of it, for proving the existence of Banach limits, as well for the Hahn-Banach theorem. – Asaf Karagila Jan 18 at 23:01
    
@AsafKaragila : I'm not sure if your comment is intended as a general pointer to more info, or if it is intended as a (full? partial?) answer to my "Remaining question" above. In the latter case, note that it is not clear to me that Banach limits are needed to produce such an example. Perhaps there is some easier construction of an example? – Michael Jan 18 at 23:03
    
Michael, as I am currently writing something on that topic (nothing particularly interesting though), I can direct you to Schechter's HAF and to let you search for an answer there, where it is certainly likely to exist. – Asaf Karagila Jan 18 at 23:05

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.