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Jun 18, 2012 at 17:57 comment added François G. Dorais The second proof does not require choice; since $B$ is countable it is easy to construct ultrafilters on $B$...
Jun 18, 2012 at 16:57 comment added Marian 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?
Jun 17, 2012 at 0:29 vote accept provocateur
Jun 16, 2012 at 22:11 history edited François G. Dorais CC BY-SA 3.0
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Jun 16, 2012 at 22:00 history answered François G. Dorais CC BY-SA 3.0