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Jun 1, 2020 at 18:28 history edited Robert Furber CC BY-SA 4.0
Removed wrong proof, as pointed out by Dmitri Pavlov. Added small comment about what happens without AC.
Jun 1, 2020 at 18:19 comment added Robert Furber @DmitriPavlov You are quite right - I went off half-cocked with that second "proof", I will edit the answer to remove it. Of course $\mathcal{BP}(\beta(\kappa)) = \mathcal{P}(\beta(\kappa))$. The measure $\nu$ has to be inner regular with respect to compact sets to define a normal state. (I will also fix the numbering of Fremlin's lemma, he added 451L at some point and bumped up all the later numbers.)
May 31, 2020 at 18:28 vote accept Dmitri Pavlov
May 31, 2020 at 18:26 comment added Dmitri Pavlov The union H' of all clopen subsets of H corresponding to finite subsets of κ is a dense open subset H'⊂H. The measure ν vanishes on all such clopen subsets. Thus ν cannot be a τ-smooth measure because v(H')=ν(H)≠0. In particular, ν is not a Radon measure and the integration functional cannot be normal, since the ν-integral of the characteristic function of an open subset coincides with its ν-measure.
May 31, 2020 at 18:24 comment added Dmitri Pavlov I have a question about your second proof, specifically, about the claim “Integration against a complex finite measure ν:BP(X)→C defines a normal linear functional on L∞(X) iff ν vanishes on meagre sets.” Consider the hyperstonean space H given by the Stone spectrum of the Boolean algebra 2^κ, where κ is a real-valued-measurable cardinal. Consider also a probability measure 2^κ→[0,1] that vanishes on all countable subsets of κ. Passing to H, we obtain a finite measure ν:BP(X)→C that vanishes on all meager subsets of H.
May 31, 2020 at 18:18 comment added Dmitri Pavlov Thanks a lot, Fremlin's Lemma 451Q indeed resolves the hyperstonean case! (The numbering appears to be different in the latest version.)
May 31, 2020 at 7:26 history answered Robert Furber CC BY-SA 4.0