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Thread: Measure Theory exercise...

  1. #1
    Member SENTINEL4's Avatar
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    1) Show that set {(x,y,z)$\displaystyle \in$ $\displaystyle \Re^3$ : z=0) is a zero subset of $\displaystyle \Re^3$

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

    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 -$\displaystyle \infty$ < x < +$\displaystyle \infty$
    Last edited by mr fantastic; May 12th 2009 at 03:40 AM. Reason: Merged posts
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  2. #2
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    Opalg's Avatar
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    Quote Originally Posted by SENTINEL4 View Post
    1) Show that set {(x,y,z)$\displaystyle \in$ $\displaystyle \Re^3$ : z=0) is a zero subset of $\displaystyle \Re^3$

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

    2) By the same argument as in 1), the set $\displaystyle \{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ has measure 0, for each rational number r. Now take the union over all such r.
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  3. #3
    Member SENTINEL4's Avatar
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    In 2), at $\displaystyle \{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ maybe it is t=r instead of z=r ??
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  4. #4
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    Quote Originally Posted by SENTINEL4 View Post
    In 2), at $\displaystyle \{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ maybe it is t=r instead of z=r ??
    Yes.
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  5. #5
    Newbie aleph1's Avatar
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    Quote Originally Posted by Opalg View Post
    1) Given m, n and k, the box $\displaystyle [m,m+1]\times[n,n+1]\times[-1/k,1/k]$ has volume 2/k. Taking the intersection over all natural numbers k, you see that the set $\displaystyle \{(x,y,z)\in\mathbb{R}^3:m\leqslant x\leqslant m+1,\ n\leqslant y\leqslant n+1,\ z=0\}$ has measure 0. Now take the (countable) union over all integers m and n.

    2) By the same argument as in 1), the set $\displaystyle \{(x,y,z,t)\in\mathbb{R}^4:z=r\}$ 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 $\displaystyle \{(x,y,z,t)\in\mathbb{R}^4:t=r\}$ has measure 0, for any rational number r. Now take the (countable) union over all integers m, n and p.
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