Proof that $\sum_{n=1}^{\infty}A_n/3^n$ has the Cantor distribution

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Cantor distribution is the probability distribution whose cumulative distribution function is the Cantor function, according to Wikipedia.

In this question, the answer says that the distribution of

$\displaystyle Z= \sum_{n=1}^{\infty}\frac{A_n}{3^n}\tag*{}$

where $A_n = 0,2$ with equal probability is the Cantor distribution.

How do I prove that? That is, how can I prove that the CDF of $Z$ is the Cantor function?

My research:

By using the Corollary 4.3(p.9) in this paper I can calculate the CDF of $Z-1/2(=S)$ by

$\displaystyle F(x) = \frac{1}{2} + \frac{1}{\pi} \int_{0}^{\infty} \frac{\sin{xt}}{t} \prod_{n=1}^{\infty} \cos{\frac{t}{3^n}}dt\tag*{}$

So I think I have to prove that this is a translation of the Cantor function, but I couldn't prove that.