# Prove Cantor set is measure zero with style

1. Nov 11, 2005

### 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?

Last edited: Nov 11, 2005
2. Nov 11, 2005

### 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.

Last edited: Nov 11, 2005
3. Nov 11, 2005

### AKG

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

4. Nov 11, 2005

### benorin

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

5. Nov 11, 2005

### AKG

Could you share some?

6. Nov 11, 2005

### benorin

Here's 5 pgs on it from Papa Rudin

#### Attached Files:

• ###### Papa Rudin (pg. 60-64, Lebesgue Measure).pdf
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7. Nov 11, 2005

### Tide

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

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

:)