The Cantor Lebesgue function is defined on the Cantor set $\mathcal {C}$ by $F(x)= \sum\limits _{k=1}^{\infty} \frac{b_k}{2^k}$ if $x= \sum\limits_{k=1}^{\infty}a_k 3^{-k}$,where $b_k=a_k/2$. in this definition, we choose the expansion of x in which $a_k=0$ or 2. show that $F$ is well defined and continuous on the Cantor set $\mathcal {C}$, and moreover $F(0)=0$ as well as $F(1)=1$.
First, to show it is well-defined, the only possible problem is some numbers have more than one ternary representation. However, such numbers can have only one representation that consists of all 0’s and 2’s. This is because the problem only arises when one representation terminates and another doesn't. Now if a representation terminates, it must end in a 2 if it contains all 0’s and 2’s. But then the other representation ends with 122222...and therefore contains a 1.
I don't get the gist of the overall argument. Normally to show well-definedness, we have to show that there is exactly one image for every one preimage. Still, I don’t get the overall argument, probably because I am still getting used to the expansion of the Cantor set. Can someone explain the overall logic of this proof? Thanks.
Your check for well-definedness is correct.
However, just for completeness, it might be better to also check that the infinite series that defines $F(x)$ converges for all $x \in \mathcal{C}$, in order to fully verify well-definedness.
But, this is also easy: $F(x)$ will converge if and only if $0.b_1 b_2 b_3 \dotsm$ is the binary expansion of a real number in $[0,1]$, simply from the definition of the infinite series defining $F(x)$. And, this is indeed the case since every infinite string $\{ b_1, b_2, b_3, \dotsc \}$ of $0$s and $1$s defines a real number in $[0,1]$ as $0.b_1 b_2 b_3 \dotsm$.