Results 1 to 3 of 3

Thread: Measure Theory problem

  1. #1
    Junior Member
    Joined
    Oct 2008
    Posts
    41

    Measure Theory problem

    I know that the cardinality of the power set of R is strictly greater than c, and that the cardinality of the Borel sigma algebra on R is equal to c.

    I'm supposed to use this to show that there is a Lebesgue measurable subset of R that is not a Borel set.

    I'm a bit confused, because it seems like to do this I would have to show that there are Lebesgue measurable sets that are not Borel sets anyway apart from the cardinality argument, since not every subset of R is Lebesgue measurable.

    Any advice on this?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2008
    From
    Paris, France
    Posts
    1,174
    Quote Originally Posted by Diamondlance View Post
    I know that the cardinality of the power set of R is strictly greater than c, and that the cardinality of the Borel sigma algebra on R is equal to c.

    I'm supposed to use this to show that there is a Lebesgue measurable subset of R that is not a Borel set.

    I'm a bit confused, because it seems like to do this I would have to show that there are Lebesgue measurable sets that are not Borel sets anyway apart from the cardinality argument, since not every subset of R is Lebesgue measurable.

    Any advice on this?
    The idea is to show that the cardinality of the set of the Lebesgue measurable subsets of $\displaystyle \mathbb{R}$ equals the cardinality of the power set of $\displaystyle \mathbb{R}$. Because of what you wrote, this would be enough.

    This results from the following fact: there exists a negligible uncountable Borel subset $\displaystyle C$ of $\displaystyle \mathbb{R}$. More precisely, its cardinality is that of $\displaystyle \mathbb{R}$.

    Supposing this is true, then every subset of $\displaystyle C$ is a Lebesgue subset (because they're included in a negligible Borel subset). However, because $\displaystyle |C|=|\mathbb{R}|$, the set of the subsets (i.e. the power set) of $\displaystyle C$ is in bijection with the power set of $\displaystyle \mathbb{R}$. This proves what we want.

    Coming back to the announced fact, the usual example is a Cantor subset, and I'd bet you've already seen this. For instance, it can be the set of the numbers in $\displaystyle [0,1]$ such that all of their decimals (in base 10) are even. Or (the usual one) the set of the numbers in $\displaystyle [0,1]$ such that the only numbers appearing in their representation in base 3 are 0 and 2.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Junior Member
    Joined
    Oct 2008
    Posts
    41
    Thanks; that makes a lot of sense now!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Measure Theory Problem
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 6th 2010, 12:10 PM
  2. Measure Theory Problem
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: Sep 27th 2010, 08:16 PM
  3. Some measure Theory
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Apr 24th 2010, 08:39 AM
  4. More Measure Theory
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: Apr 9th 2010, 11:05 AM
  5. Measure Theory
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: Mar 25th 2010, 07:20 AM

Search Tags


/mathhelpforum @mathhelpforum