Identifying random variables with their generated distribution function - Necessity of countable additivity?

Question

Let the state space $\Omega=[0,1]$ and $\lambda$ be the Lebesgue measure defined on the Borel $\sigma$-algebra on $[0,1]$. Consider measurable functions (random variables) $f:\Omega\to\mathbb{R}$ and let $\mathcal{D}$ be the Borel $\sigma$-algebra on $\mathbb{R}$.

Further, let a probability distribution P on $\mathbb{R}$ be a countably additive function from $\mathcal{D}$ to $[0,1]$ which satisfies $P(\emptyset)=0, P(\mathbb{R})=1$. Let $\mathcal{P}$ be a set of probability distributions.

In this setup, Wakker (1993) Proof of Corollary 4.5. argues that we can identify functions $f$ with their generated probability distributions. In another paper, Proof of Theorem 2, definition of the homomorphism they argue, that omitting the countable addivity of objects in $\mathcal{P}$ does not admit such a one-to-one relationship.

I've been thinking about this the last few days and I totally do not see where the countable additivity comes into play. Since I've been studying the mentioned papers deeply, let me know if you need further information about anything. I think that it is not necessary for you to read/work through both papers.

Cheers

Answer 1

I don't completely understand your question and have only skimmed the papers, but the relevant part in the second paper you mention seems to be

We, however, also allow for finite additivity, where distribution functions are more complex and do not fully characterize the probability measure.

If this is what you mean, then (probably) the following is meant:

If you consider a (probability) measure $P : \mathcal{B} \to [0,1]$, where $\mathcal{B}$ is the Borel-$\sigma$-algebra on $\Bbb{R}$, then this measure induces a so-called distribution function

$$ f_P (x) := P ((-\infty, x]). $$

This distribution function has the following properties:

1. it is continuous from the right,
2. it is monotonically increasing (non-decreasing),
3. $f_P (x) \to 0$ for $x \to -\infty$,
4. $f_P (x) \to 1$ for $x \to \infty$.

Conversely, for every function $f$ with these properties, there is a unique probability measure (countably additive) $P$ with $f = f_P$. (cf. http://en.wikipedia.org/wiki/Cumulative_distribution_function#Properties and/or http://en.wikipedia.org/wiki/Probability_distribution#Cumulative_distribution_function)

If we drop the assumption of countable additivity, terrible things can happen.

To see this, believe me that there is a finitely additive function

$$ \mu : {\rm Pot}(\Bbb{N}) \to \{0,1\} $$

which satisfies $\mu(E) = 1$ if $E^c$ is finite and $\mu(E) = 0$ if $E$ is finite.

(To see this, note that $\mathcal{F} := \{E \subset \Bbb{N} \mid E^c \text{ finite}\}$ is a filter (cf. http://en.wikipedia.org/wiki/Fr%C3%A9chet_filter) and hence extends to an ultrafilter $\mathcal{F}'$ which induces a function $\mu$ as above, cf. e.g. this answer (non)equivalence of definition of non-atomic measure for finitely additive measure)

Now define $P(M) := \mu(M \cap \Bbb{N})$ and observe that this is finitely additive, but the associated function $f_P$ only assumes the values $0,1$ (and no values in between). Hence, it can not be continuous from the right. One can probably also construct examples for which properties 3 and/or 4 fail.

Conversely, if one only requires finite additivity of $P$, then uniqueness of $P$ under the condition $f_P = f$ probably fails, although I do not have a counterexample at hand at the moment.