Results 1 to 2 of 2

Thread: Measure of countable co-countable set

  1. #1
    Super Member
    Joined
    Mar 2006
    Posts
    705
    Thanks
    2

    Measure of countable co-countable set

    Let X be an uncountable set, let $\displaystyle E \subset X $
    Consider the $\displaystyle \sigma $-algebra $\displaystyle \mathbb {M} = \{ E \ countable \ or \ E^c \ countable \} $

    Define $\displaystyle \mu : \mathbb {M} \rightarrow [0, \infty ] $ by $\displaystyle \mu (E) = 0 \ if \ E \ countable $ and $\displaystyle \mu (E) = 0 \ if \ E^c \ countable $

    Prove that $\displaystyle \mu $ is a measure.

    Proof so far.

    Let $\displaystyle E_1,E_2, ... \in \mathbb {M} $, claim: $\displaystyle \mu ( \bigcup _{n=1}^ \infty \mu (E_n) = \sum ^ \infty _{n=1} \mu (E_n)$

    Case 1) If $\displaystyle E_n $ is countable for all n, then $\displaystyle \bigcup _{n=1}^ \infty \mu (E_n) $ is countable and $\displaystyle \mu ( \bigcup _{n=1}^ \infty \mu (E_n) = 0 = \sum ^ \infty _{n=1} \mu (E_n)$ since $\displaystyle \mu (E_n) = 0 \ \ \ \forall n $

    Case 2) If $\displaystyle E_n ^c$ is countable. Since $\displaystyle ( \bigcup _{n=1}^ \infty (E_n) )^c \subset \bigcup ^ \infty _{n=1} E_n^c $ and the latter is countable, I then have $\displaystyle \mu ( \bigcup _{n=1}^ \infty (E_n) ) = 1 $

    But then I also have $\displaystyle \sum ^ \infty _{n=1} \mu (E_n) = \mu (E_1) + \mu (E_2) = . . . = 1 + 1 + 1 + . . . = \infty $

    So the two ain't equal... What did I do wrong?

    Case 3) Either $\displaystyle E_n$ is countable or $\displaystyle E_n^c$ is countable. How should I proceed with this one?

    Thank you!
    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 tttcomrader View Post
    Let X be an uncountable set
    (....)
    Let $\displaystyle E_1,E_2, ... \in \mathbb {M} $ be disjoint subsets, claim: $\displaystyle \mu ( \bigcup _{n=1}^ \infty E_n) = \sum ^ \infty _{n=1} \mu (E_n)$

    Case 2) If $\displaystyle E_n ^c$ is countable. Since $\displaystyle ( \bigcup _{n=1}^ \infty (E_n) )^c \subset \bigcup ^ \infty _{n=1} E_n^c $ and the latter is countable, I then have $\displaystyle \mu ( \bigcup _{n=1}^ \infty (E_n) ) = 1 $

    But then I also have $\displaystyle \sum ^ \infty _{n=1} \mu (E_n) = \mu (E_1) + \mu (E_2) = . . . = 1 + 1 + 1 + . . . = \infty $

    So the two ain't equal... What did I do wrong?

    Case 3) Either $\displaystyle E_n$ is countable or $\displaystyle E_n^c$ is countable. How should I proceed with this one?
    You can't expect equality if the subsets aren't disjoint.

    The key is given by the following question: is it possible that two sets $\displaystyle E_1,E_2$ are disjoint if they are such that $\displaystyle E_1^c$ and $\displaystyle E_2^c$ are countable?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: Jan 9th 2013, 06:14 AM
  2. Replies: 2
    Last Post: Nov 11th 2010, 04:56 AM
  3. [SOLVED] Countable union of closed sets/countable interesection of open sets
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: Oct 8th 2010, 01:59 PM
  4. Replies: 1
    Last Post: Feb 9th 2010, 01:51 PM
  5. Replies: 11
    Last Post: Oct 11th 2008, 06:49 PM

Search Tags


/mathhelpforum @mathhelpforum