MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.