I've seen the equation below for several times, which seems obvious, but I can't prove it seriously.
Suppose $x_1, x_2, \cdots, x_n$ is a sample, $$F_n(x) = \frac{1}{n} \sum{I(x_i\le x)}$$ is EDF.
How to prove that $\bar{x} = \int x\,\mathrm dF_n(x)$?
I do not have enough expertise about measure-theoretic notation, so my approach might seem hand-wavy to some.
The differential $\mathrm{d}F_n$ of the ECDF is the distribution $\mathrm{d}F_n = f_n = \frac{1}{n}\sum_{i=1}^n\delta_{x_i}$. Here $\delta_{x_i}$ is the delta distribution, that is, for any function $g$, the dot product $\langle g, \delta_{x_i}\rangle$ is equal to
$$\langle g, \delta_{x_i}\rangle = \int g(x)\delta_{x_i}(x)\mathrm{d}x = g(x_i)$$
you can think of $\delta_{x_i}$ as having $0$ mass everywhere and all of its mass on the single point $x=x_i$. Then by definition, the expectation is
$$ \mathrm{E}_{F_n}[X] = \int x \mathrm{d}F_n(x) = \int x \dfrac{1}{n}\sum_{i=1}^n\delta_{x_i}(x)\mathrm{d}x = \dfrac{1}{n}\sum_{i=1}^n \int x \,\delta_{x_i}(x)\mathrm{d}x = \dfrac{1}{n}\sum_{i=1}^n x_i = \bar{X}_n$$
by the linearity of the integral and the fact that each integral $\int x \,\delta_{x_i}(x)\mathrm{d}x$ is equal to the dot product of the delta witht the identity function $\langle \mathrm{id}, \delta_{x_i} \rangle = \mathrm{id}(x_i) = x_i$.
