$K$ : Cantor set
$f$ : $[0,1] \to K$
We can write $\displaystyle x=\sum_{k=1}^\infty \frac{\varepsilon_k}{2^k} \ (\varepsilon_k = 0,1)$ for all $x \in [0,1]$ by binary expansion.
Define
$$ f(x)=f \left( \sum_{k=1}^\infty \frac{\varepsilon_k}{2^k} \right) = \sum_{k=1}^\infty \dfrac{2\varepsilon_k}{3^k}.$$
Then, prove that $\{f >a \}=\{x\in [0,1] \mid f(x) >a \}$ is Lebesgue measurable for all $a\in \mathbb{R}$.
If $a< 0,$ $\{ f>a \}=[0,1]$ and it is Lebesgue measurable.
But what about for $0\leqq a$?
I don't know how $\{f >a \}$ is expressed for $0\leqq a$.
Your function $f$ is increasing.
Therefore $\{x\in[0,1]\mid f(x)>a\}$ is either empty or has the form $(b,1]$ or $[b,1]$ for some constant $b$ that depends on $a$.
Both of $(b,1]$ and $[b,1]$ are obviously Lebesgue measurable.