Results 1 to 4 of 4

Math Help - How can the measure of a measurable set be bounded?

  1. #1
    Newbie
    Joined
    Oct 2013
    From
    MA
    Posts
    4
    Thanks
    1

    How can the measure of a measurable set be bounded?

    Helle Everyone,

    In one book, the Lebesgue measure is said to possess the following properties (among others):


    1) the measure of any measurable set can be approximated from above by open sets; that is, for any measurable M


    \mu (M) = inf \{ \mu(O): M \subset O, O is open \}

    2) the measure of any measurable set can be approximated from below by compact sets; that is, for any measurable M


    \mu(M) = sup \{ \mu( C ): C \subset M, C is compact \}


    Can any one provide a very simple example of this property? Like in \mathbb{R}^{1}?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2013
    From
    MA
    Posts
    4
    Thanks
    1

    Re: How can the measure of a measurable set be bounded?

    Any one?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Jan 2014
    From
    New Zealand
    Posts
    2

    Re: How can the measure of a measurable set be bounded?

    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,450
    Thanks
    1864

    Re: How can the measure of a measurable set be bounded?

    Quote Originally Posted by john1844 View Post
    Helle Everyone,

    In one book, the Lebesgue measure is said to possess the following properties (among others):


    1) the measure of any measurable set can be approximated from above by open sets; that is, for any measurable M


    \mu (M) = inf \{ \mu(O): M \subset O, O is open \}
    If all you want is an example, take [0, 1]. It can be "approximated from above" by the open sets (-1/n, (n+1)/n). Each of those has measure (n+1)/n+ 1/n= 1+ 2/n which goes to 1, the measure of [0, 1], as n goes to infinity.

    2) the measure of any measurable set can be approximated from below by compact sets; that is, for any measurable M


    \mu(M) = sup \{ \mu( C ): C \subset M, C is compact \}
    Again, use [0, 1] which can be "approximated from below" by the compact sets [1/n, (n-1)/n]


    Can any one provide a very simple example of this property? Like in \mathbb{R}^{1}?
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. outer measure of disjoint measurable sets question
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: February 20th 2013, 08:11 PM
  2. Let f be a bounded measurable function on [a,b]...
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: May 6th 2010, 10:10 AM
  3. Bounded, measurable?
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: April 25th 2010, 08:52 AM
  4. Replies: 3
    Last Post: March 14th 2010, 08:58 AM
  5. Outer measure and measurable sets.
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 3rd 2009, 07:57 AM

Search Tags


/mathhelpforum @mathhelpforum