Integration with respect to the empirical process?

Question

Let $X_1, X_2, \dots, X_n$ be i.i.d from distribution $F(x)$. Let $v_n (x)$ be the corresponding empirical process

$$v_n (x) = \sqrt{n} \left[ F_n (x) - F(x) \right]$$

where $F_n (x)$ is the empirical distribution function.

Then I would like to understand the integral of a deterministic function $g(x)$ with respect to $v_n (x)$. That is, the integral

$$ \int_\mathbb{R} g(y) dv_n (y)$$

Intuitively, when I first saw this integral I interpreted it as

$$ \int_\mathbb{R} g(y) dv_n (y) = \sqrt{n} \left( \int_\mathbb{R} g(y) dF_n (y) - \int_\mathbb{R} g(y) dF(y) \right)$$

Which in turn gives

$$ \frac{1}{\sqrt{n}} \sum_{i=1}^n g(X_i) - \sqrt{n} \int_\mathbb{R} g(y) dF(y)$$

Is this the correct interpretation?

Answer 1

First, the empirical process, in general, is not what you mean. The empirical process is a process index by a function space. Check this classical book chapter 1 for an understanding.

Return to your question though, you are integrating w.r.t to a cdf function. Therefore, if you are lucky it can be well defined, it is just a Lesbesgue integration. Therefore, everything you know about Lesbesque measure applies.