Dudley’s exercise on finitely additive probabilities — why the measure exists

Note: This is a follow-up to this question.

For context, the problem in Dudley’s “Real Analysis and Probability” book is:

In a game, two players, Sam and Joe, each pick a nonnegative integer at random. For each, the probability that the number is in any set $$A$$ is $$\mu(A)$$, where $$\mu$$ is a finitely additive function defined on $$2^X$$ with $$\mu(A) = 0$$ for every finite set $$A$$ and $$\mu(X) = 1$$, and $$X = \mathbb{N}$$. The one who gets the larger number wins. A coin is tossed to determine whose number you find out first. It’s heads, so you find out Sam’s and still don’t know Joe’s. Now, who do you think will win?

It seems the answer is supposed to be “Joe will win,” which is supposed to show the absurdity of assuming only finite additivity. See the question linked above for a proof.

But how do we even know such a $$\mu$$ exists in the first place? If it doesn’t, any proof would be vacuous and therefore meaningless.

• Thanks for the comment — good point. The error is mine. I’ll delete my answer for now. – Theoretical Economist Mar 13 at 11:42
• See this question on MO. – Theoretical Economist Mar 13 at 11:50
• @TheoreticalEconomist Quick question: Are you convinced one way or the other that $\mu^*$ exists? Or are you unsure. – Yatharth Agarwal Mar 13 at 18:15
• What is $\mu^*$? Is it some $\mu$ that satisfies the conditions you outline in your question? If so, then the question I link to in my previous comment establishes existence, assuming you are willing to accept the axiom of choice. – Theoretical Economist Mar 13 at 18:18

The referenced problem in Dudley's "Real Analysis and Probability" provides such a construction (parts (a)-(c)). In a nutshell, consider the filter $$\mathcal{F}=\{A\subset \mathbb{N}:A^c\text{ is finite}\},$$ and let $$\mathcal{U}$$ be an ultrafilter containing $$\mathcal{F}$$. Define $$\mu$$ on $$\mathcal{P}(\mathbb{N})$$ as $$\mu(A)=1\{A\in \mathcal{U}\}$$. Then $$\mu(\mathbb{N})=1$$ ($$\because \mathbb{N}\in \mathcal{U}$$), $$\mu(A)=0$$ for any finite $$A$$ ($$\because A^c\in \mathcal{U}$$), and $$\mu$$ is finitely additive because for any finite partition $$\{A_1,\ldots, A_n\}$$ of $$\mathbb{N}$$ only one $$A_j$$ belongs to $$\mathcal{U}$$.

For a different construction you may look at this question.

Here's a Hahn-Banach construction.

Let $$C\subset \ell^{\infty}({\mathbb N})$$ be the subspace of convergent sequences.

Let $$T$$ be the linear functional on $$C$$ equal to the limit of a sequence. Then $$\|T\|=1$$.

By the Hahn-Banach, one can extend $$T$$ to a bounded functional on $$\ell^{\infty}({\mathbb N})$$ with norm $$1$$. This is important because of the following. If $${\bf a}\in \ell^\infty({\mathbb N})$$ is a sequence with nonnegative elements, then
$$|\|{\bf a}\| - T({\bf a}) | = |T(\|{\bf a}\|{\bf 1} - {\bf a}) | \le \|{\bf a}\| - \inf {\bf a},$$ which implies $$T({\bf a}) \ge \|{\bf a}\|-(\|{\bf a}\|-\inf {\bf a}) = \inf {\bf a} \ge 0.$$

Construction of $$\mu$$

For any set $$A \subseteq {\mathbb N}$$, let $$a^A= (a^A_1,a^A_2,\dots)\in \ell^\infty({\mathbb N})$$ be the sequence $$a^A_i = \delta_A(i)$$. That is $$a^A$$ is the indicator function of $$A$$. Define $$\mu(A)= T(a^A)$$.

Now:

1) $$\mu(A) \in [0,1]$$ for all $$A$$.

2) $$\mu({\mathbb N}) = T({\bf 1})=1$$.

3) If $$A\cap B=\emptyset$$, then $$\mu(A\cup B) = T(a^{A\cup B}) = T(a^A + a^B) = \mu(A) +\mu(B)$$ (this also implies $$\mu(\emptyset)=0$$).

4) If $$A$$ is finite, $$\mu(A)=T(a^A)=0$$ because $$a^A$$ converges to $$0$$.