The Unapologetic Mathematician

Mathematics for the interested outsider

Non-Lebesgue Measurable Sets

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April 24, 2010 - Posted by | Analysis, Measure Theory

7 Comments »

  1. 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 | Reply

  2. Sorry, I misspoke in my haste to get out the catch-up post…

    Comment by John Armstrong | April 25, 2010 | Reply

  3. In case anyone is interested, the example you ended with is sometimes called a “maximally non-measurable 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), 422-424.
    [JFM 46.0294.01] [available on-line]
    http://www.emis.de/cgi-bin/JFM-item?46.0294.01

    Comment by Dave L. Renfro | April 26, 2010 | Reply

  4. […] is), this condition reduces to asking that . If, further, , then we ask that . As an example, the maximally nonmeasurable set we constructed is […]

    Pingback by Measurable Subspaces II « The Unapologetic Mathematician | April 28, 2010 | Reply

  5. […] to add the requirement that . Also, the converse of this theorem is definitely not true; if is a non-measurable set, then the function is not measurable even though the absolute value is […]

    Pingback by Composing Real-Valued Measurable Functions I « The Unapologetic Mathematician | May 4, 2010 | Reply

  6. […] now we can take a thick, non-Lebesgue measurable set whose intersection with is itself a non-Lebesgue measurable set . However, , and has Lebesgue […]

    Pingback by Composing Real-Valued Measurable Functions II « The Unapologetic Mathematician | May 5, 2010 | Reply

  7. […] During this long period from 1904 to 1962, many bizzare results such as the Banach Tarsky paradox, existence of non-Lebesgue measurable sets etc were found. This put a suspicion in people’s minds as to whether this axiom was […]

    Pingback by The axiom of choice | Notes on Mathematics | September 19, 2012 | Reply


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