Let $\chi_A$ be the characteristic function of the set $A$.
Then, for any $n\geq 1$,
- $\chi_{C_n}$ is a step function and its integral is equal to $\left(\dfrac{2}{3}\right)^n$ over $[0,1]$.
- $\chi_C\leq \chi_{C_n}$.
- the zero function is a step function lower than $\chi_C$ on $[0,1]$.
So, for any $n\geq 1$, $$0\leq \sup_{u\leq \chi_C} \int_0^1u \leq \inf_{u\geq \chi_C} \int_0^1u \leq \left(\dfrac{2}{3}\right)^n$$
where the $\sup$ and $\inf$ are taken over step functions on $[0,1]$.
Thus the upper integral $U(\chi_C)$ and the lower integral $L(\chi_C)$ of $\chi_C$ are equal and $\chi_C$ is integrable with $$\int_0^1\chi_C=0.$$
Suppose $X$ is Cantor distributed and we know that it is symmetric around $\frac{1}{2}$ and we have to rewrite as :
$$X - \frac{1}{2} = \sum_{k=1}^\infty \frac{X_k}{3^k}$$
where $\Pr[X_k = -1] = \Pr[X_k = 1] = 1/2$ and $X_k$ are iid for all $k$.
Remark: It is the same as what John Dawkins write, noted that $\frac{1}{2} = \sum_{k=1}^\infty \frac{1}{3^k}$.
For each $X_k$, $$\begin{align}\varphi_{X_k}(t) &= \sum\Pr[X_k=l]e^{itl}\\
&=\frac{1}{2}e^{-it} + \frac{1}{2}e^{it} = \cos(t)\\
\therefore \varphi_{X_k/3^k}(t)&=\varphi_{X_k}(t/3^k)\\
&=\cos\left(\frac{t}{3^k}\right)\end{align} $$
By independence, the sum of $X_k/3^k$ will be the product in the characteristic function,
$$\begin{align} \varphi_{X-1/2} (t) &= \prod_{k=1}^\infty \cos\left(\frac{t}{3^k}\right) \\
\varphi_{X} (t) &= e^{it/2} \prod_{k=1}^\infty \cos\left(\frac{t}{3^k}\right) \end{align}$$
For most mathematical beauty, a little bit of location-scale transformation is always needed.
References: Lukacs, Eugene (1970). Characteristic Functions, 2nd edition, Griffin, London.
Best Answer
I assume you already know what the cantor set is. In that case, the characteristic function is the function $f$ with $f(x)=1$ for $x \in C$ and $f(x)=0$ otherwise.