I would to link the simple function and probabilistic approach for the calculation of the Fourier transform of the Cantor function.
Let $f:[0,1] \to [0,1]$ be the Cantor function.
In the simple function approach, at some point appears $ \hat f'(x) = \int e^{-ixt} df(t) $ where the integral is on $[0,1]$.
Write $t$ in base-$3$ with $ t=\sum_{n=1}^\infty t_n 3^{-n}, t_n \in T_n=\{0,1,2\}. $ Define a measure $g_n$ on $T_n$ by $dg_n(0)=dg_n(2)=1/2$ and $dg_n(1)=0$.
Then, viewing $f$ as a product measure on $ T=\otimes_n (T_n,\sigma(T_n),g_n) $ gives $ df(t)=\prod_n dg_n(t_n). $
This construction gives the correct result for $\hat f'(x)$.
But is the construction correct?
I find strange to work with $df$ and $dg_n$.
Usually measure are defined directly, not through the differential.
Also, wikipedia states that the cantor function has no discrete part, which seems to contradicts the construction (2nd paragraph, 2nd phrase in the Properties section).