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Homework Help: Prove Cantor set is measure zero with style

  1. Nov 11, 2005 #1

    benorin

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    One of my HW questions asks me to prove that the usual "middle thirds" Cantor set has Lebesgue measure 0. I know two ways, but they lack style...

    They are (that you may post): #1) The recursive definition of the Cantor set (call it C) removes successively [tex]\frac{1}{3}[/tex] of the unit interval and hence has measure [tex]\frac{2}{3}[/tex] of the previous iteration. Thus, if [tex]C_{0}[/tex] denotes [0,1], and [tex]C_{k}[/tex] denotes the [tex]k^{\mbox{th}}[/tex] iteration of removing middle thirds, then

    [tex]m(C_{k})=\left( \frac{2}{3}\right)^{k}m([0,1]) \rightarrow 0 \mbox{ as } n\rightarrow \infty [/tex]

    thus m(C)=0.

    #2) same jazz only summing measures of the removed portions (the middles thirds) as a geometric series that converges to 1, and hence m(C)=0.

    Blah, blah, blah.... no style.

    I'm looking for interesting, in the context, using this theorem to prove it would qualify:

    Thm. If [tex]A\subset\mathbb{R}^1[/tex] and every subset of A is Lebesgue measurable then m(A)=0.

    Any suggestions as to how I might pull that off?

    Or are there any proofs the PF-math community would like to share?
     
    Last edited: Nov 11, 2005
  2. jcsd
  3. Nov 11, 2005 #2
    i kind of like the first one ^
    the connected components of the cantor set are points, which have measure zero. i haven't thought about that real hard but it might go somewhere.

    edit: actually both those proofs look fine.
     
    Last edited: Nov 11, 2005
  4. Nov 11, 2005 #3

    AKG

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    Do you have any good theorems on conditions for a subset of R to be Lebesgue Measurable?
     
  5. Nov 11, 2005 #4

    benorin

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    I may use any theorems from Papa Rudin or Baby Rudin.
     
  6. Nov 11, 2005 #5

    AKG

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    Could you share some?
     
  7. Nov 11, 2005 #6

    benorin

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    Sure, gladly even

    Here's 5 pgs on it from Papa Rudin
     

    Attached Files:

  8. Nov 11, 2005 #7

    Tide

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    How about using the fact that the fractal dimension of the set is

    [tex]D = \frac {\ln 2}{\ln 3} < 1[/tex]

    :)
     
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