Results 1 to 5 of 5

Math Help - Measure Theory exercise...

  1. #1
    Member SENTINEL4's Avatar
    Joined
    Mar 2009
    From
    Kastoria, Greece
    Posts
    92
    1) Show that set {(x,y,z) \in \Re^3 : z=0) is a zero subset of \Re^3

    2) Show that set {(x,y,z,t) \in \Re^4 : - \infty<x<+ \infty, - \infty<y<+ \infty, - \infty<z<+ \infty, t \inQ) is a zero subset of \Re^4

    I have these two problems in my Measure Theory lesson... If anyone can help i would appreciate it...

    I would be happy if someone tell me something to begin with these exercises if not the full solution...
    Especially what i am doing with - \infty < x < + \infty
    Last edited by mr fantastic; May 12th 2009 at 04:40 AM. Reason: Merged posts
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by SENTINEL4 View Post
    1) Show that set {(x,y,z) \in \Re^3 : z=0) is a zero subset of \Re^3

    2) Show that set {(x,y,z,t) \in \Re^4 : - \infty<x<+ \infty, - \infty<y<+ \infty, - \infty<z<+ \infty, t \inQ) is a zero subset of \Re^4
    1) Given m, n and k, the box [m,m+1]\times[n,n+1]\times[-1/k,1/k] has volume 2/k. Taking the intersection over all natural numbers k, you see that the set \{(x,y,z)\in\mathbb{R}^3:m\leqslant x\leqslant m+1,\ n\leqslant y\leqslant n+1,\ z=0\} has measure 0. Now take the (countable) union over all integers m and n.

    2) By the same argument as in 1), the set \{(x,y,z,t)\in\mathbb{R}^4:z=r\} has measure 0, for each rational number r. Now take the union over all such r.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member SENTINEL4's Avatar
    Joined
    Mar 2009
    From
    Kastoria, Greece
    Posts
    92
    In 2), at \{(x,y,z,t)\in\mathbb{R}^4:z=r\} maybe it is t=r instead of z=r ??
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by SENTINEL4 View Post
    In 2), at \{(x,y,z,t)\in\mathbb{R}^4:z=r\} maybe it is t=r instead of z=r ??
    Yes.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie aleph1's Avatar
    Joined
    Feb 2009
    From
    Virginia Beach VA
    Posts
    11
    Quote Originally Posted by Opalg View Post
    1) Given m, n and k, the box [m,m+1]\times[n,n+1]\times[-1/k,1/k] has volume 2/k. Taking the intersection over all natural numbers k, you see that the set \{(x,y,z)\in\mathbb{R}^3:m\leqslant x\leqslant m+1,\ n\leqslant y\leqslant n+1,\ z=0\} has measure 0. Now take the (countable) union over all integers m and n.

    2) By the same argument as in 1), the set \{(x,y,z,t)\in\mathbb{R}^4:z=r\} has measure 0, for each rational number r. Now take the union over all such r.
    Opalg's eplanation is clear. If I understand Opalg's 1), then by parrallel argument, should 2) be unioned over three indexes?:
    2) By the same argument as in 1), the set \{(x,y,z,t)\in\mathbb{R}^4:t=r\} has measure 0, for any rational number r. Now take the (countable) union over all integers m, n and p.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. measure theory
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: July 5th 2011, 09:37 AM
  2. Measure Theory
    Posted in the Differential Geometry Forum
    Replies: 5
    Last Post: May 1st 2010, 11:51 AM
  3. Measure Theory
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 25th 2010, 08:20 AM
  4. exercise in Number Theory
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: May 22nd 2008, 01:44 AM
  5. measure theory
    Posted in the Advanced Math Topics Forum
    Replies: 2
    Last Post: June 14th 2007, 02:47 AM

Search Tags


/mathhelpforum @mathhelpforum