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This is a followup to:

Strictly Positive Measures on Countable Boolean Algebras

Suppose a countable Boolean algebra B is a subalgebra of the power set of the reals. (For example, let B be the Boolean algebra of definable sets of reals, in one of the various natural senses of definability.)

A strictly positive, continuous measure on B is a function m from B to [0,1] such that

(i) $m(b)=0$ iff $b=0$,

(ii) $m(1)=1$,

(iii) $m(a+b)=m(a)+m(b)$ whenever $a$ and $b$ are disjoint.

(iv) if $A_1$, $A_2$, $A_3$ ... is an increasing countable sequence of elements of B, and the union $A=\bigcup_{i=0}^{\infty}$ is also in $B$, then $m(A)=\lim_{i \rightarrow \infty} m(A_i)$

Is there a strictly positive, continuous measure on every such countable Boolean Algebra?

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    $\begingroup$ Doesn't the answer you linked to work? $\endgroup$
    – Jason Rute
    Commented Jun 26, 2012 at 23:16
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    $\begingroup$ Jason, it appears that Francois's answer provides only a finitely additive measure, rather than a countably additive one. For example, few of the functions $m_n$ in that answer are countably additive, and in the atomless case, none of them are. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27, 2012 at 0:18
  • $\begingroup$ Yes, that's the problem, I don't see how that construction guarantees countable additivity - in particular, the ultrafilters mentioned in the construction there need not be closed under countable limits. Hence my question. $\endgroup$
    – provocateur
    Commented Jun 27, 2012 at 0:52
  • $\begingroup$ Joel, I'm not sure none of them are. For example, consider the ultrafilter $\mathcal{U} = \lbrace a \in B : \sqrt2 \in a \rbrace,$ where $B$ is the Boolean algebra generated by the intervals $[p,q)$ where $p,q \in \mathbb{Q}$. It seems that the measure associated with $\mathcal{U}$ is actually countably additive, in the sense described in the question. $\endgroup$
    – François G. Dorais
    Commented Jun 27, 2012 at 1:46
  • $\begingroup$ François, you are right. I was thinking that one could get a measurable cardinal out of it, but your example shows that this version of countable additivity does not suffice for that. $\endgroup$
    – Joel David Hamkins
    Commented Jun 27, 2012 at 8:46

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We claim that there is no continuous such measure on the countable free algebras. Under Stone duality, one may consider the countable free algebra as the collection of all clopen sets in the cantor space. Let $C$ be the cantor space and let $B$ be the Boolean algebra of clopen sets in the cantor space. For the sake of contradiction, assume that $\mu$ is a continuous probability measure on $B$. If $x\in C$, then there is a decreasing sequence $(R_{n})_{n}$ of clopen subsets of $C$ with $\{x\}=\bigcap_{n}R_{n}$. In terms of the Boolean algebra $B$, we have $\bigwedge_{n}R_{n}=\emptyset$. Therefore by continuity, for all $\epsilon>0$ there is a clopen neighborhood $R_{n}$ of $x$ with $\mu(R_{n})<\epsilon$. Now let $(x_{n})_{n}$ be a dense subset of $C$. For each $n>0$, let $S_{n}$ be a clopen neighborhood of $x_{n}$ with $\mu(S_{n})<\frac{1}{2^{n+1}}$. Then we have $\bigcup S_{n}$ be a dense subset of $C$, so $\bigvee S_{n}=C$ in the Boolean algebra $B$. Therefore, by continuity and finite additivity, we have $\mu(C)\leq\sum_{n}\mu(S_{n})<\frac{1}{2}$. This contradicts the assumption that $\mu(C)=1$.

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  • $\begingroup$ In fact, this shows that a countable Boolean algebra carries a strictly positive continuous measure if and only if it is atomic. $\endgroup$
    – Emil Jeřábek
    Commented Jun 27, 2012 at 9:41

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