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I recently learned the basics of Riemann-Stieltjes integral, and based on the sources I found, we can define the expectation of random variables quite naturally with the R-S integrals: if $X$ is a random variable and $F_X$ is its distribution function defined as $P(X \leq x)$, then its expectation is $ E[X] = \int_{-\infty}^{\infty} x dF_X(x), $ or if $Y = g(X)$ such that $g : \mathbb{R} \rightarrow \mathbb{R}$ is measurable, then $ E[Y] = \int_{-\infty}^{\infty} g(x) dF_X(x), $ if the integral exists.

Now, consider the random variables $X_n(y)$ and $X(0)$ where $X_n(0) \rightarrow X(0)$ in the sense of weak convergence, $y \in [0,1]$, and their respective distribution functions $F_{X_n(y)}$ and $F_0$. My question is, for a fixed $x \in [0,1]$, how do I interpret the following R-S integral? Does the expectation interpretation above apply here? $$ \int_0^1 F_{X_n(y)}(x) dF_0 (y) $$

I made another post on MSE with more details on the context of the problem. The source paper claims that this integral is equal to $F_0(x)$, but I don't understand why.

This is my first post here, hopefully it's not too trivial...

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  • $\begingroup$ Hi gouhaha; this isn't my area of expertise, but I've upvoted to counteract the unexplained downvote in the hopes that someone more knowledgable in this area will come along to evaluate the question. $\endgroup$
    – Alec Rhea
    Commented Apr 19, 2021 at 4:06
  • $\begingroup$ Thanks very much @AlecRhea! $\endgroup$
    – gouhaha
    Commented Apr 19, 2021 at 4:11

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