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The Hahn-Banach theorem implies that Lebesgue measure can be extended give a "measure" on all subsets of [0,1], but this measure is only guaranteed to be finitely additive. It might magically turn out that this measure is countably additive, but this can only happen if the continuum is a real-valued measurable cardinal, a strong set-theoretic assumption. My question is: if it turns out that measure is countably additive on the measure zero sets, does this imply that the measure is countably additive everywhere?

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The answer is no. A proof can be found here.

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    $\begingroup$ Since this link seems to be outdated: Kumar + Kunen, "Induced ideals in Cohen and random extensions", Topology and its Applications 174, 2014 $\endgroup$
    – Mike Hall
    Commented Jun 24, 2015 at 0:27
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Note that even if the measure is only finitely additive, the ideal of null sets will still be countably saturated. The usual arguments involving Ulam matrices, such as can be found in Jech (Chapter 10) or other texts on set theory, show that if there is a countably additive, countably saturated ideal on some set then there is a weakly inaccessible cardinal. Hence, if no such cardinal exists then, as Emil has corrected me, the Hahn-Banach extension of Lebesgue measure will be countably additive on Lebesgue null sets but not on all nulls. This does not answer your question, but does show that the weaker hypothesis still requires large cardinals.

By the way, you may want to look at the paper by Joan Hart and Ken Kunen titled "Weak Measure Extension Axioms" in Topology and Applications 85 (1998) 219-246 to see what can be said about extensions of measures to measure certain sets.

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    $\begingroup$ I’m confused. Doesn’t your argument show that if no weakly inaccessible exists, then the extension will not be countably additive on null sets (= the null ideal is not countably additive)? $\endgroup$
    – Emil Jeřábek
    Commented Aug 24, 2011 at 13:06
  • $\begingroup$ Yes, you are right. The ideal of Lebesgue null sets remains countably additive, but there are, of course, new null sets. $\endgroup$
    – Juris Steprans
    Commented Aug 24, 2011 at 13:13
  • $\begingroup$ Thanks for the answer-- it points me in the right direction! $\endgroup$
    – Monroe Eskew
    Commented Aug 25, 2011 at 0:03

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