Let $C$ denote the Cantor set and $f$ denote the Cantor-Lebesgue function on $[0, 1]$.
If $x \notin C$, then $x$ will be in exactly one of the intervals removed from $[0, 1]$, say, $(a, b)$. Then, $f(a) = f(b)$. Define $f(x) = f(a)$ for all $x \in (a, b)$. Verify that $f$ is monotone, continuous and satisfies the functional equation $f(x) = 2f(\frac x3)$ for all $x \in [0, 1]$.
I have proved that $f$ is monotone and continuous.
I require help to show that $f$ satisfies the functional equation $f(x) = 2f(\frac x3)$ for all $x \in [0, 1]$.
You only need to show this for $x=(0.(2d_1)(2d_2)...(2d_n))_3$ with a finite ternary representation with $d_k\in\{0,1\}$, i.e., digits $(2d_k)\in\{0,2\}$ only. Then $x/3=(0.0(2d_1)(2d_2)...(2d_n))_3$. Then $$ f(x)=(0.d_1d_2...d_n)_2\text{ and }\\ f(x/3)=(0.0d_1d_2...d_n)_2=f(x)/2. $$