Is there a bijection from $[0,1]$ to the Cantor set $C$? Does there exist a bijective function $\phi$ from the unit interval $[0,1]$ to the Cantor set $C$? If so, how can it be constructed? I could then proceed to build a measure space ($C$, $\mathcal{M}_\phi$, $m_\phi$) where $m_\phi(E)$ = $m(\phi^{-1}(E))$ for $E \subset C$. How would $m_\phi(C) = 1$? 
 A: We look only at the less interesting part of the question, that of finding an explicit bijection from $[0,1]$ to the Cantor set $C$.
Certainly Cantor-Bernstein-Schroeder is perfectly constructive, and can be used to give an explicit, albeit messy, bijection. 
But the "near-bijection" described by @mixedmath can be tweaked to produce an explicit bijection.
For any real number $x$ in $[0,1]$, let the canonical representation of $x$ be defined as follows.  If $x$ is not a dyadic rational (that is, if $x$ does not have an eventually constant binary representation), the canonical representation of $x$ is the ordinary binary representation.  If $x\ne 1$, and $x$ is a dyadic rational, the canonical representation of $x$ is the one that is eventually $0$'s.  Finally, the canonical representation of $1$ is the representation with all $1$'s.
Now we define the bijection.  If $x$ is not a dyadic rational, map $x$ to the number $y$ whose ternary representation replaces the $1$'s in the binary representation of $x$ by $2$'s, and leaves the $0$'s unchanged.
Suppose now that $x$ is a dyadic rational.  Write down its canonical representation.  
Let $x$ be a dyadic rational whose canonical representation begins with $0$.  Remove the $0$, move the remaining bits one to the left. Then replace the $1$'s by $2$'s. Call the number which has this ternary representation $y$, and map $x$ to $y$.
Let $x$ be a dyadic rational whose canonical representation begins with $1$. Represent it in the "non-terminating form" that is, after a certain point all $1$'s.  Remove the initial $1$, move all bits one to the left. Then replace the $1$'s by $2$'s.  Call the number which has this ternary representation $y$, and map $x$ to $y$.
A: There happens to be a very simple near-bijection between the two. In short, consider the binary decimal representation of all the numbers in $[0,1]$ and consider the ternary representation of all the elements of the Cantor Set. Since the Cantor Set can be attained by removing the middle third repeatedly, the elements of the Cantor Set are exactly those whose ternary representations have only 0s and 2s. 
Then a number such as $0.202202...$ in the Cantor set could be associated with the number $0.101101...$ in $[0,1]$. So the function is simply to send $c \to (\frac{c}{2})$, taking account of the obvious change of base of course.
There is a problem with double-representation, as pointed out by Q. Yuan. Namely, both $0.0222...$ and $0.2$ (in ternary) get sent to $0.1$ (in binary). But perhaps there is a quick work-around here.
However, this does not complete the second half of your question. And, in addition, this is not in the least continuous.
