The complete statement is as follows:
Let $(C_n)$ the inductive sequence from $ C_0 = [0,1]$ to $C_n = C_{n-1} - J_n$ where $$ J_n = \bigcup_{k=1}^{3^n-2}\left(\frac{k}{3^n},\frac{k+1}{3^n}\right) $$
and define $(f_n)$ as $f_n = \chi_{C_n}$. Show that $$ f = \sum_{n=1}^{\infty}{f_n} $$ is measurable, not bounded, integrable and calculate its integral.
Here my questions about:
what means $\chi_{C_n} $? I think is the characteristic function of $C_n$ so $f_n$ is at most 1 and since $C_n$ is kind of cantor set, $f$ must be bounded because is the sum of characteristic functions on disjoint sets.
is that correct? some hint to prove this?
To answer your first question, $\chi_{C_n}$ does in fact mean the characteristic function of $f_n$. Observe that each $f_n$ is a simple function, as is each partial sum $\sum_{n=1}^N f_n$. Thus, the partial sums of $f$ are measurable, and hence it equals a limit of measurable functions and thus must be measurable.
To see that $f$ is unbounded, observe first that $f(x)$ is equal to the number of terms $f_n$ such that $f_n(x)=1$. Or in other words, it is equal to the number of sets $C_n$ such that $x\in C_n$. Since the sets form a decreasing sequence, it actually means that $f(x)$ is the largest value of $n$ such that $x\in C_n$, or infinity if $x\in \bigcap_n C_n$. In particular, this shows that $f$ is unbounded: certainly it is infinite on the Cantor set itself, $C:=\bigcap_n C_n$. But more than that, for any $k$ the set $\{x\colon f(x)>k\}$ has positive measure. Your mistake was in your belief that the sets $C_n$ were disjoint: they are not, and in fact the opposite is true: $C_n\subseteq C_m$ for all $n\geq m$, which means that they overlap "as much as possible"!
Since $f\geq 0$ and $f$ is measurable, it has a well-defined Lebesgue integral, although it may equal infinity. To find out if this is the case, we have to calculate the integral. By Tonelli's theorem, $$\int_{\mathbb R}f=\sum_{n=1}^{\infty}\int_{\mathbb R}f_n.$$ Now observe that the integral of $f_n$ is equal to the Lebesgue measure of $C_n$ (this is true in general for the integral of a characteristic function), so it suffices to compute the measure of $C_n$. But recall the definition of $C_n$: we obtain it from $C_{n-1}$ by removing the middle third of each of the intervals. So the measure of $C_n$ is two-thirds the measure of $C_{n-1}$. Since the measure of $C_0$ is $1$, it follows that $$ \int_{\mathbb R}f=\sum_{n=1}^{\infty}\Bigl(\frac{2}{3}\Bigr)^n=\frac{1}{1-\frac{2}{3}}-1=2. $$ Thus $f$ is integrable and its integral is $2$.
There is a neat interpretation of this fact. If you choose $X$ to be a uniformly random number in $[0,1]$, then the expected number of steps in the Cantor set construction before $X$ is removed is $2$.