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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).

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In the usual (probabilistic) construction of the uniform distribution $\mu$ on the standard Cantor set $C$, one introduces a sequence $(X_n)_{n\geqslant1}$ of i.i.d. random variables with uniform distribution on $\{0,2\}$, and one considers $ X=\sum\limits_{n\geqslant1}3^{-n}X_n. $ Then $X$ is uniformly distributed on $C$, hence the Fourier transform $\widehat\mu$ of the measure $\mu$ is the function defined by $ \widehat\mu(t)=\int_\mathbb R\mathrm e^{\mathrm itx}\mathrm d\mu(x)=\mathbb E(\mathrm e^{\mathrm itX}). $ (Different normalizations are used but the steps below can be easily adapted.)

Now, by definition of stochastic independence, $ \mathbb E(\mathrm e^{\mathrm itX})=\prod_{n\geqslant1}\mathbb E(\mathrm e^{\mathrm it3^{-n}X_n})=\prod_{n\geqslant1}\varphi(t/3^{n}), $ where $\varphi(t)=\mathbb E(\mathrm e^{\mathrm itX_1})$, that is, $\varphi(t)=\frac12(1+\mathrm e^{2\mathrm it})$. Finally, $ \widehat\mu(t)=\prod_{n\geqslant1}\frac{1+\mathrm e^{2\mathrm it/3^n}}2=\mathrm e^{\mathrm it/2}\cdot\prod_{n\geqslant1}\cos(t/3^n). $ Nota: Indeed, $\mu(\{x\})=0$ for every $x$, as can be seen by including $\{x\}$ in smaller and smaller triadic intervals of length $3^{-n}$, whose measure is, by definition, either $2^{-n}$ or zero, for every $n\geqslant1$. Likewise, recall that $C$ has Lebesgue measure zero hence $\mu$ and the Lebesgue measure are mutually singular.