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

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

:)