I'm currently reading through the paper Asymptotic normality of nearest neighbor regression function estimates and am struggling to understand the asymptotic equalities that were shown in the proof of Lemma 4 (p. 923) where the last one was proven by using integration by parts:
$\begin{align*}&\qquad \frac{m(x_0)}{\sqrt{a_n^3}}\int \left[\alpha_n(x_0)-\alpha_n(x)\right]K'\left(\frac{F(x_0)-F(x)}{a_n}\right)F(d x)\\ &= -\frac{m(x_0)}{\sqrt{a_n^3}}\int \alpha_n(x)K'\left(\frac{F(x_0)-F(x)}{a_n}\right)F(d x)\\ &= -\frac{m(x_0)}{\sqrt{a_n}}\int K\left(\frac{F(x_0)-F(x)}{a_n}\right)\alpha_n(dx). \end{align*}$
Here $m(x) = E[Y|X=x]$, $\alpha_n(x) = \sqrt{n} [F_n(x)-F(x)]$ denotes the empirical process of the random variable $X$ (having continuous distribution function $F$), $K$ being a twice continuously differentiable probability kernel with bounded support and $na_n^3 \to \infty$, $a_n \to 0$ for $n \to \infty$.
I know that I could expand the second to last term like so: $\begin{align} &\quad-\frac{m(x_0)}{\sqrt{a_n^3}}\int \alpha_n(x)K'\left(\frac{F(x_0)-F(x)}{a_n}\right)F(d x)\\ &= -\frac{m(x_0)\sqrt{n}}{\sqrt{a_n^3}}\int F_n(x)K'\left(\frac{F(x_0)-F(x)}{a_n}\right)F(d x) + \frac{m(x_0)\sqrt{n}}{\sqrt{a_n^3}}\int F(x)K'\left(\frac{F(x_0)-F(x)}{a_n}\right)F(d x)\\ \end{align}$
where the second integral contains transformations with $F$, so a transformation of variables and the integral a la $u = F(x)$ is possible and then application of the "classic" integration by parts yields the first of the following terms and a "constant" term.
$\begin{align} &= \frac{m(x_0)\sqrt{n}}{\sqrt{a_n}}\int K\left(\frac{F(x_0)-F(x)}{a_n}\right) F(d x) -\frac{m(x_0)\sqrt{n}}{\sqrt{a_n}}\int K\left(\frac{F(x_0)-F(x)}{a_n}\right) F_n(d x)\\ &= -\frac{1}{\sqrt{a_n}}m(x_0)\int K\left(\frac{F(x_0)-F(x)}{a_n}\right) \alpha_n(d x)\end{align}$
It would be nice if someone could provide me an explanation on the upper equalities since I'm not even sure if I'm on the right track. I only consider the classic integration by parts, maybe the Lebesgue-Stieltjes variant is more helpful.