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...(adsbygoogle = window.adsbygoogle || []).push({});

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?

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# Homework Help: Prove Cantor set is measure zero with style

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