# Does the following net define a finitely additive probability measure?

Let $$\mathcal X$$ be a set, and let $$\mathcal F$$ be the set of all finite subsets of $$\mathcal X$$ directed by subset inclusion.

For each finite set $$F \in \mathcal F$$, let $$\mu_F$$ be the probability measure defined on every subset $$X$$ of $$\mathcal X$$ by $$\mu_F(\{x\})=\begin{cases}1/|F|, \ x \in F,\\ 0, \ \text{otherwise.}\end{cases}$$

Does the net $$(\mu_F(X))_{F \in\mathcal F}$$ converge for all $$X \subset \mathcal X$$?

I am wondering because, if the net does converge, then it can be used to define a finitely additive probability measure $$\mu$$ on $$2^{\mathcal X}$$ by $$\mu(X) = \lim_{\mathcal F} \mu_F(X). \tag{1}$$

If $$X \in \mathcal F$$, then clearly the net converges. Indeed, for any $$Y \in \mathcal F$$ such that $$Y \supset X$$ we have $$\mu_Y(X) = |X|/|Y| \to 0$$.

So, if $$\mathcal A$$ is the finite/co-finite algebra, and $$\mu$$ is any extension to $$2^{\mathcal X}$$ of the probability on $$\mathcal A$$ that assigns finite sets measure $$0$$, then I can say that (1) holds for $$X \in \mathcal F$$, but this doesn't really answer my question.

• In the second sentence, you introduce a set $\mathcal X$, which hasn't been defined before. Aug 6, 2020 at 5:54
• @StephenMontgomery-Smith Thanks, corrected.
Aug 6, 2020 at 5:55
• Also, should $2^X$ be $2^{\mathcal X}$? Aug 6, 2020 at 5:56
• @StephenMontgomery-Smith Yes indeed
Aug 6, 2020 at 5:58

Consider $$\mathcal X = \mathbb N$$, and $$X$$ the set of even numbers. Then for any $$F \in \mathcal F$$, we can consider two sequences:

$$U_n = F \cup \{1,3,5,\dots,2n-1\}$$;

$$V_n = F \cup \{2,4,6,8,\dots 2n\}$$.

As $$n \to \infty$$, $$\mu_{U_n}(X) \to 0$$, and $$\mu_{V_n}(X) \to 1$$. So I don't see how the net can converge.

It seems to me that this kind of example will work when $$\mathcal X$$ is any infinite set, and $$X$$ is any set for which both $$X$$ and $$\mathcal X \setminus X$$ are infinite.

For $$\mathcal X$$ infinite, what holds is \begin{align}\limsup_{F\in\mathcal F}\mu_F(X)&=\begin{cases}1&\text{if }\lvert X\rvert\ge\aleph_0\\ 0&\text{if }\lvert X\rvert<\aleph_0\end{cases}\\ \liminf_{F\in\mathcal F}\mu_F(X)&=\begin{cases}0&\text{if }\lvert \mathcal X\setminus X\rvert\ge\aleph_0\\ 1&\text{if }\lvert\mathcal X\setminus X\rvert<\aleph_0\end{cases}\end{align}

The first proves the second because $$\liminf\limits_{F\in\mathcal F} \mu_F(X)=\liminf_{F\in\mathcal F} (1-\mu_F(\mathcal X\setminus X))=1-\limsup_{F\in\mathcal F} \mu_F(\mathcal X\setminus X)$$

and you've already proved that $$\limsup$$ is $$0$$ when $$X$$ is finite. If $$X$$ is infinite, consider any $$\varepsilon>0$$ and let $$F\in\mathcal F$$. We have $$\mu_F(X)=\frac{\lvert X\cap F\rvert}{\lvert F\rvert}$$. We can select a natural number $$n$$ such that $$\frac{\lvert X\cap F\rvert+n}{\lvert F\rvert+n}>1-\varepsilon$$, and then a finite subset $$V\subseteq X\setminus F$$ such that $$\lvert V\rvert=n$$. Then, $$\mu_{F\cup V}(X)=\frac{\lvert X\cap F\rvert+n}{\lvert F\rvert+n}$$ and $$F\cup V\supseteq F$$. The fact that this procedure can be done for all $$F$$ and $$\varepsilon$$ proves that $$\limsup\limits_{F\in\mathcal F}\mu_F(X)\ge 1$$.

Putting all together, if $$\mathcal X$$ is infinite:

1. if $$X$$ is finite, then $$\lim_{F\in\mathcal F}\mu_F(X)=0$$

2. if $$X$$ is co-finite, then $$\lim_{F\in\mathcal F}\mu_F(X)=1$$

3. if $$X$$ is neither finite nor co-finite, then $$\liminf\limits_{F\in\mathcal F}\mu_F(X)=0$$ and $$\limsup\limits_{F\in\mathcal F}\mu_F(X)=1$$.

Therefore, you don't have convergence in case (3).