This is a followup to:


[Strictly Positive Measures on Countable Boolean Algebras](https://mathoverflow.net/questions/99808/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?



Continuous Strictly Positive Measures on Countable Boolean Algebras
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 $\mu(C)=1$.


