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I'm looking for a space $(\Omega ,\mathcal F)$ and a function $\mathbb P:\mathcal F\to [0,1]$ s.t. $\mathbb P(\Omega )=1$ and $\mathbb P$ is addtitive but not $\sigma -$additive.

Of course I got inspired but several post on mathstack exchange by $\Omega =\mathbb R$, $$\mathcal F=\{A\subset \mathbb R\mid A\text{ or $A^c$ is countable}\},$$ and $\mathbb P(A)=0$ if $A$ is countable an $\mathbb P(A)=1$ if $A$ is not countable. But unfortunately this is a probability measure. Any example ?

My question is not a duplicate of this post (as several thought) because the answers don't provide probability spaces.

Rhys Steele
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John
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  • @Mark: If you read careful both answer, they don't provide a counter example. For example, in saz answer, $\mu$ is not defined on $2\mathbb N\in \mathcal P(\mathbb N)$. For the other answer, it's not a probability space. – John Jul 24 '19 at 13:26
  • @saulspatz: Same remark than for mark (it's the same link) – John Jul 24 '19 at 13:27
  • @John The site puts up this message when one votes to close a duplicate. Mark made the comment, but didn't vote to close. – saulspatz Jul 24 '19 at 13:31
  • What is your definition of a premeasure? The definition on the wikipedia page here would trivially imply that any premeasure defined on a $\sigma$-algebra is a measure so you must mean something other than this. – Rhys Steele Jul 24 '19 at 15:22
  • @RhysSteele: I don't know why I wrote premeasure... What I wanted to say by premeasure is that $\mathbb P$ is additive but not $\sigma -$finite (I really don't know that premeasure was a real term, sorry for the mistakes). – John Jul 24 '19 at 15:26
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    To people voting to close: This is still not a duplicate of the linked post since that doesn't ask for an example with total measure $1$ and doesn't get one. All $3$ answers rely very much on having sets of infinite measure. – Rhys Steele Jul 24 '19 at 19:13

1 Answers1

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This answer will show only that such a set function must exist. The reason that it is difficult to write one down explicitly is that the existence of an example on $\mathbb{N}$ requires something more than ZF. I suspect that the existence of any example will require more than just ZF but I am not sure of this. I will use the ultrafilter lemma$^*$.

Let $\Omega = \mathbb{N}$ and let $\mathcal{F} = \mathcal{P}(\Omega)$. Let $\mathcal{G}$ be a non-principal ultrafilter on $\mathbb{N}$ (recall that an ultrafilter on $\mathbb{N}$ is called principal if it is of the form $\mathcal{G} = \{A \subseteq \mathbb{N} : a \in A\}$ for some fixed $a$).

Define $$\mathbb{P}(A) = \begin{cases} 1 \qquad A \in \mathcal{G} \\ 0 \qquad \text{otherwise} \end{cases}$$ Then $\mathbb{P}$ is a finitely additive set function (see this question). However since $\mathcal{G}$ is non-principal, it cannot contain any finite set. In particular, $\mathbb{P}(\{i\}) = 0$ for every $i \in \mathbb{N}$. Hence $\mathbb{P}$ is not countably additive since $$1 = \mathbb{P}(\mathbb{N}) \neq \sum_{i=0}^\infty \mathbb{P}(\{i\}) = 0$$

$^*$ which is equivalent to the Boolean Prime ideal theorem and hence strictly weaker than choice. I need the ultrafilter lemma to see that a non-principal ultrafilter exists on $\mathbb{N}$. One way to get this is to use the ultrafilter lemma to get an ultrafilter extending the cofinite filter. This ultrafilter then cannot be principal.

As pointed out in the answer here and the references therein, it is consistent with ZF+DC that there are no "non-principal measures" on $\mathbb{N}$. A non-principal measure on the powerset of a set $X$ is a set function $\mu: \mathcal{P}(X) \to [0,1]$ such that $\mu$ is finitely additive, $\mu(X) = 1$ and $\mu(\{x\}) = 0$ for all $x$. The $\mathbb{P}$ in this answer is exactly an example of a non-principal measure on $\mathbb{N}$. The answer to my follow up question here then shows that it is consistent with ZF+DC that you cannot construct an example on $\mathbb{N}$.

Rhys Steele
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  • I was under the impression that the asymptotic density $m: \mathcal P(\mathbb N) \to\mathbb [0,1]$ defined by $$m(A) = \lim_{n\to\infty} \frac{1}{n} | A \cap { 1,\dots, n}|$$ was an example of a non-principal finitely additive measure with 'mass' 1. Am I wrong, or where is the DC here?(I assume DC is some kind of Choice? not a set theorist, sorry) – Calvin Khor Jul 25 '19 at 10:15
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    @CalvinKhor I think the problem with that is that the limit doesn't exist for all subsets of $\mathbb{N}$ so $m$ isn't actually defined on $\mathcal{P}(\mathbb{N})$. In fact, the subset of $\mathcal{P}(\mathbb{N})$ for which the limit exists isn't a $\sigma$-algebra. DC is dependent choice. I'm not a set theorist either, so I don't want to pretend I have a detailed idea of what's going on. The last part of my answer is just put together from reference hunting. – Rhys Steele Jul 25 '19 at 10:21
  • oh, of course! thanks – Calvin Khor Jul 25 '19 at 10:25