Fourier transform of the Cantor function Let $f:[0,1] \to [0,1]$ be the Cantor function.
Extend $f$ to all of $\mathbb R$ by setting $f(x)=0$ on $\mathbb R \setminus [0,1]$.
Calculate the Fourier transform of $f$
$$
\hat f(x)= \int f(t) e^{-ixt} dt
$$
where $dt$ is the Lebesgue measure on $\mathbb R$ divided by $2\pi$, and the integral is over $\mathbb R$.
I think this MO post says the result is
$$
\hat f (x)= \frac{1}{ix}-\frac{1}{ix}e^{ix/2}\prod_{k=1}^{\infty} \cos(x/3^k). \tag{1}
$$
To get this, I approximate $f$ by simple function
$$
f_n(x)= \sum_{i=1}^n \sum_{j=1}^{2^{i-1}} \frac{2j-1}{2^i}\chi_{E_{n,k}}
$$
where $E_{n,k}$ is the $k$th set removed during the $n$th stage of the Cantor process. Then
$$
\hat f_n(x) = \sum_{i=1}^n \sum_{j=1}^{2^{i-1}} \frac{2j-1}{2^i}\int_{E_{n,k}} e^{-ixt} dt  \tag{2}
$$
But I don't see how, in the limit, (2) simplifies to (1).  
 A: Since the excised intervals vary in their length while the remaining intervals are not, it seems easier to focus on the remaining intervals.
Let $I(n,k)$ for $n \geq 1$ and $0 \leq k \leq 2^n - 1$ be the remaining $2^n$ intervals after the $n$-th stage of the construction of the Cantor set. Then $|I(n, k)| = 3^{-n}$, and we can approximate $f$ by
$$f_n(x) = \int_{0}^{x} \left( (3/2)^n \sum_{k=0}^{2^n - 1} \chi_{I(n,k)}(t) \right) \, dt $$
To see this really approximates $f$, observe that $f_n$ increases only on $C(n) = \bigcup_{k=0}^{2^n-1}I(n,k)$ and on each subinterval $I(n,k)$, $f_n$ increases by exactly $2^{-n}$, as we can check:
$$ \int_{I(n,k)} (3/2)^n \chi_{I(n,k)}(t) \, dt = \frac{1}{2^n}.$$
Thus $f_n$ coincides exactly with the $n$-th intermediate function appearing in the construction of the Cantor-Lebesgue function $f$. Then $f_n \to f$ uniformly, and we have
$$ \begin{align*}
\int f(t) \, e^{-ixt} \, dt
&= \lim_{n\to\infty} \int f_n(t) \, e^{-ixt} \, dt \\
&= \lim_{n\to\infty} \left( \left[ -\frac{1}{ix} f_n(t) e^{-ixt} \right]_{0}^{1} + \frac{1}{ix} \int f_n'(t) \, e^{-ixt} \, dt \right) \\
&= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \lim_{n\to\infty} \int f_n'(t) \, e^{-ixt} \, dt \\
&= -\frac{e^{-ix}}{ix} + \frac{1}{ix} \left(\frac{3}{2}\right)^n \lim_{n\to\infty} \sum_{k=0}^{2^n - 1}  \int_{I(n,k)} e^{-ixt} \, dt
\end{align*}$$
Now, direct calculation shows that
$$\int_{a}^{a+\beta h} e^{-ixt} \, dt + \int_{a+(1-\beta)h}^{a+ h} e^{-ixt} \, dt
= 2 \cos\left(\frac{1-\beta}{2} hx\right)\frac{\sin(\frac{\beta}{2}hx)}{\sin(\frac{1}{2}hx)} \int_{a}^{a+h} e^{-ikt} \, dt. $$
Thus plugging $h = 3^{-n}$ and $\beta = \frac{1}{3}$, we have
$$\int_{I(n+1,2k)} e^{-ixt} \, dt + \int_{I(n+1,2k+1)} e^{-ixt} \, dt
= 2 \cos\left(\frac{x}{3^{n+1}}\right)\frac{\sin\left(\frac{x}{2\cdot 3^{n+1}}\right)}{\sin\left(\frac{x}{2\cdot 3^{n}}\right)} \int_{I(n,k)} e^{-ikt} \, dt. $$
Inductively applying this relation allows us to calculate the limit above, which I leave because I have to go out.
A: Let $\mu$ be the standard Cantor measure on the interval $(-1, 1)$. If we set $\mu(x)=\mu((-\infty, x))$, considering the self-similarity of $\mu$ on the first level, we easily obtain
$$
\mu(x)=\frac{1}{2}\Big(\mu(3x+2)+\mu(3x-2)\Big).
$$
Hence
$$(\mathcal F\mu)(3t)=\int \exp(3itx)\,d\mu(x)
=\frac{1}{2}\left(\int\exp(3itx)\,d\mu(3x+2)+\int\exp(3itx)\,d\mu(3x-2)\right) 
=\frac{1}{2}\left(\int\exp(it(y-2))\,d\mu(y)
+\int\exp(it(y+2))\,d\mu(y)\right) 
=\frac{1}{2}\Big(\exp(-2it)+\exp(2it)\Big)\times\int\exp(ity)\,d\mu(y) 
=\cos 2t\cdot(\mathcal F\mu)(t)$$
and
$$
(\mathcal F\mu)(t)
=\cos\frac{2t}{3}\times(\mathcal F\mu)\left(\frac{t}{3}\right)
=\cos\frac{2t}{3}\cdot\cos\frac{2t}{9}\times(\mathcal F\mu)\left(\frac{t}{9}\right)=\dots
=\prod_{n=1}^\infty \cos\frac{2t}{3^n},
$$
because the function $\mathcal F\mu$ is continuous at the origin and $(\mathcal F\mu)(0)=1$.
