NonLebesgue Measurable Sets
April 24, 2010  Posted by John Armstrong  Analysis, Measure Theory
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You say that all three sets (A, B, C) are subgroups. I see that for A (all n) and B (all even n). But how is C (all odd n) an additive subgroup — it’s not even an additive group since it doesn’t contain 0.
Or do I totally misunderstand?
Thanks!
Comment by marshall  April 25, 2010 
Sorry, I misspoke in my haste to get out the catchup post…
Comment by John Armstrong  April 25, 2010 
In case anyone is interested, the example you ended with is sometimes called a “maximally nonmeasurable set” or a “saturated nonmeasurable set”, and more examples and their properties can be found in the following posts:
Remarks on Bernstein sets
http://tinyurl.com/29sgs2b
http://tinyurl.com/27sqz2d [minor correction]
Now for something REALLY STRANGE. In 1917 Lusin and Sierpinski showed that the unit interval [0,1] can be partitioned into c = 2^(aleph_0) many pairwise disjoint sets each having Lebesgue outer measure 1. This shows a massive failure of additivity in the case of Lebesgue outer measure! And yes, they constructed c many such sets, not just uncountably many (which would give you “only” aleph_1 many such sets under CH).
Here’s their paper:
Nikolai N. Lusin and Waclaw Sierpinski, “Sur une décomposition d’un intervalle en une infinité non dénombrable d’ensembles non mesurables” [On a decomposition of an interval into a nondenumerably many nonmeasurable sets], Comptes Rendus Académie des Sciences (Paris) 165 (1917), 422424.
[JFM 46.0294.01] [available online]
http://www.emis.de/cgibin/JFMitem?46.0294.01
Comment by Dave L. Renfro  April 26, 2010 
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