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Let $B$ be a Boolean Algebra.

A strictly positive measure on $B$ is a function $m$ from $B$ to $[0,1]$ such that (i) $m(b)=0$ iff $b=0$, (ii) $m(1)=1$, and (iii) $m(a+b)=m(a)+m(b)$ whenever $a$ and $b$ are disjoint.

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

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up vote 8 down vote accepted

Yes. Let $M$ be the space of all measures on $B$. This is a compact space when endowed with with the pointwise convergence topology since it is a closed subspace of $[0,1]^B$. If $b$ is a nonzero element of $B$, then the set $U_b = \lbrace m \in M : m(b) \gt 0 \rbrace$ is open and dense in $M$. By the Baire Category Theorem, the intersection of all these sets is nonempty.

For an explicit construction, let $b_1,b_2,\ldots$ enumerate $B\setminus\lbrace0\rbrace$ and for each $n$ let $m_n$ be a measure on $B$ such that $m_n(b_n) \gt 0$ (e.g. let $m_n$ be the characteristic function of an ultrafilter containing $b_n$). Then $m = \sum_{n=1}^\infty 2^{-n}m_n$ is as required.

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Thanks, this is very clear and decisive. –  provocateur Jun 17 '12 at 0:29
Both of these proofs require choice, but can't you produce a measure on $B$ by induction on $n$ (defining at stage $n$ the value of the measure on the subalgebra generated by $b_1,\ldots,b_n$) in a completely constructive, computable way? –  Marian Jun 18 '12 at 16:57
The second proof does not require choice; since $B$ is countable it is easy to construct ultrafilters on $B$... –  François G. Dorais Jun 18 '12 at 17:57

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