The task is thus:
We let $P$ be the probability measure defined by the uniform distribution on [1/3, 2/3]. We must then compute the integral: $\int_R X_C dP$, with $X_C$ being the characteristic function of the Cantor set.
I'm weary of this problem because either it's too simple, or I don't really understand it. Since the measure puts all it's "mass" in the interval [1/3, 2/3], and the Cantor set has no points there, wouldn't the integral be simply zero?
Yes, that's correct: the function $X_C$ is $0$ almost everywhere with respect to $P$, so its integral is $0$.
(What you've said isn't quite literally true, though: the Cantor set actually does have two points in $[1/3,2/3]$, namely $1/3$ and $2/3$. But the measure of two points is still $0$.)