Hint:
- There is a theorem that if $C_n$ is a descending sequence of measurable sets,
$C = \bigcap C_n$ then $\lim_{n \rightarrow \infty}m(C_n)= m(C)$.
Here, you know $C_n$ is a descending sequence, $m(C_n) $= (2/3)^n, and you want to know m(C)
- to prove $C$ is uncountable:
Express a number $x$ between $[0,1]$ in base $3$:
$x =0.x_1x_2x_3...$
In the 1st step, we remove the middle third from $[0,1]$
We express 0=.0, $\frac{1}{3} = .1, \frac{2}{3} = .2$, 1= .222222...
We have 3 intervals: [.0 , .1] , (.1, .2), [.2 , .22222...] and we remove the middle interval. The removed interval (.1 , .2) consists of all numbers with $x_1$ = 1, except the endpoint $.1$ of [.0 , .1 ], however, we can express .1 as .02222.....
So we can use the rule: whenever we have a number of form 0.x1(x is a sequence consisting of 0,1,2) as the end point of an interval, we express as 0.x0222....So in this step, we remove all numbers with $x_1$ = 1. The remaining intervals are [.0, .0222...] and [.2, .222...]
Similarly, we can prove that in the n-th step, we keep only those numbers with $x_n$ = 0 or 2:
So the Cantor set contains of all numbers of the form $.x_1x_2..$ with $x_i =0 $ or 2.
There exists a bijection between $E$ and $[0,1]$:
If you consider the new set E, with each member of E is a member of the Cantor set with every digit is divided by 2. E consists of all sequence with $x_i$ is 0 or 1. $|E| = |C|$
There exists an injective map from $[0,1]$ to this new set E. So you can prove it's uncountable.
Yes, it is.
Note that the Cantor set is in fact a compact group, and the product measure $\mu$ is the Haar measure, that is, the invariant finite, regular Borel measure, which is by Haar's theorem unique up to a scaling factor.
That Lebesgue measure restricted to the fat Cantor set is invariant is an easy exercise (it follows easily from the fact that the Lebesgue measure is translation invariant).
It is also regular, because it is a restriction of a regular measure, so by Haar's theorem it is a scalar multiple of the product measure.
Best Answer
Look at the unit interval after the $n$th chopping. Let $C_n$ be the length of the set resulting from $n$ removal of the middle third. Then $$|C_n| = \left({2\over 3}\right)^n.$$ Now arrive at your conclusion. (I am using $|\cdot |$ for Lebesgue measure).