Prove Cantor set is measure zero with style

[benorin]

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 \frac{1}{3} of the unit interval and hence has measure \frac{2}{3} of the previous iteration. Thus, if C_{0} denotes [0,1], and C_{k} denotes the k^{\mbox{th}} iteration of removing middle thirds, then

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

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 A\subset\mathbb{R}^1 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?

 

[fourier jr]

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.

 

[AKG]

Do you have any good theorems on conditions for a subset of R to be Lebesgue Measurable?

 

[benorin]

I may use any theorems from Papa Rudin or Baby Rudin.

 

[AKG]

Could you share some?

 

[benorin]

Sure, gladly even

Could you share some?

Here's 5 pgs on it from Papa Rudin

 

[Tide]

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

D = \frac {\ln 2}{\ln 3} < 1

:)