Given a random variable $X$ on $(\Omega, \mathscr{F}, \mathbb{P})$ and its law $\mathcal{L}_X$,
$$E[X] = \int_{\Omega} X d\mathbb{P} = \int_{\Omega} X(\omega) d\mathbb{P}(\omega)$$
Change of variable theorem allows us to compute as follows:
$$E[X] = \int_{\mathbb{R}} t d\mathcal{L}_X(t)$$
From Wiki:
What is $$\int_{\mathbb{R}} t d\mathcal{L}_X(t)$$ when, say, $X$ is Binomial or has the Cantor distribution?
Guess:
$$\int_{\mathbb{R}} t d\mathcal{L}_X(t)$$
$$= \int_{Range \ of \ X} t dF_X(t)$$
$$= \lim_{||P|| \to 0} \sum_{i=0}^{n-1} c_i [F(x_{i+1}) - F(x_{i})]$$ where $P = (\min(Range \ of \ X) = x_0 < x_1 < ... < x_n = \max(Range \ of \ X))$
And that holds for the F being the distribution function of either Cantor distribution or binomial distribution?
It is a standard result in measure theory that for nonnegative functions $f$
$$\int_A f d \mu = \lim_{n \to \infty} n \mu(f^{-1}([n,\infty)) + \sum_{k=1}^{n^2-1} \frac{k}{n} \mu(f^{-1}([k/n,(k+1)/n)).$$
(The details of the partitioning are not so important; the important matter is that the mesh size goes to zero and the upper bound goes to infinity.) For general measurable functions $f$, introduce $f^+=\max \{ 0,f \}$ and $f^-=-\min \{ 0,f \}$, then $\int_A f d \mu = \int_A f^+ d \mu - \int_A f^- d \mu$, provided at least one of these numbers is finite. Translated to the probability context, this is what you're trying to say.