If $X$ is a compact Hausdorff space, we know that finite Baire measures uniquely extend to a Radon measure. Furthermore every Radon measure also restricts to a Baire measure. We also know that every tight finitely additive finite Borel measure (i.e. it satisfies inner regularity with respect to compacts, the Radon property) is also countably additive. Is it the case also that every finite finitely additive measure on the Baire $\sigma$-algebra on $X$ uniquely extends to a Radon measure?
2026-03-25 21:49:27.1774475367
Are finitely additive Baire measures the same as Radon measures
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The distinction between Baire and Borel measures is a red herring. We can consider a countable space $X$, where the Baire and Borel $\sigma$-algebras are both just the power set, i.e. every subset is both Baire and Borel. Then the question is just "must a finite, finitely additive measure on the power set also be countably additive"? And the answer is no, because a free ultrafilter provides a counterexample.