Cantor set: Lebesgue measure and uncountability I have to prove two things.
First is that the Cantor set has a lebesgue measure of 0. If we regard the supersets $C_n$, where $C_0 = [0,1]$, $C_1 = [0,\frac{1}{3}] \cup [\frac{2}{3},1]$ and so on. Each containig interals of length $3^{-n}$ and by construction there are $2^n$ such intervals. The lebesgue measure of each such interval is $\lambda ( [x, x + 3^{-n}]) = 3^{-n}$ therefore the measure of $C_n$ is $\frac{2^n}{3^n} = e^{(\ln(2)-\ln(3)) n }$ which goes to zero with $n \rightarrow \infty$. But does this prove it?
The other thing I have to prove is that the Cantor set is uncountable. I found that I should contruct a surjectiv function to $[0,1]$. But im totaly puzzeld how to do this.
Thanks for help

I decided I would answer the latter part of your question in a really pretty way. 
You can easily create a surjection from the Cantor set to $[0,1]$ by using binary numbers. Binary numbers are simply numbers represented in base $2$, so that the only digits that can be used are $0$ and $1$. If you take any number in the cantor set, which are numbers in base $3$ written exclusively with the digits $0$ and $2$, you can make the rule that if you see a $2$, turn it into a one. You have now mapped the Cantor set to the interval $[0,1]$ in binary (and therefore in any base), which proves uncountability.
