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.

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

2 Answers 2


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).


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