Prove inequality bound for truncated random variables $X_{nk\delta}, X_{nk1}$ I need to prove a theorem from "The General Central Limit Theorem for Real and Banach valued random variables" (auth. E. Giné, A. Araujo) that gives necessary and sufficient conditions for weak convergence.
More specifically I need help proving the inequality below.
The inequality:

Notation etc.

*

*$0 < \delta < 1 $

*$ X_{nk1} = X_{nk} \ 1_{\{ X_{nk} \leq 1 \}} $

*$X_{nk\delta} =    X_{nk} \ 1_{\{ X_{nk} \leq \delta \}}$

*$\mu_{nk} = \mathcal{L}(X_{nk})$ $ \ \ \ $ (the distribution of $X_{nk}$)
My initial computation:
\begin{align*}
& \ \ \ \ \ \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} (x-E[X_{nk1}])^2 d\mathcal{L}(X_{nk1})(x)-\int _{\{ \vert x \vert \leq \delta \}} (x-E[X_{nk\delta}])^2 d\mathcal{L}(X_{nk\delta})(x)\bigg\vert \\[1em]
&= \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk1})(x) + \int _{\{ \vert x \vert \leq \delta \}} (E[X_{nk1}])^2 d\mathcal{L}(X_{nk1})(x) \\[1em]
& \quad \quad \quad - 2\int _{\{ \vert x \vert \leq \delta \}} xE[X_{nk1}] d\mathcal{L}(X_{nk1})(x)- \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x) \\[1em]
& \quad \quad \quad - \int _{\{ \vert x \vert \leq \delta \}} (E[X_{nk\delta}])^2 d\mathcal{L}(X_{nk\delta})(x) + 2\int _{\{ \vert x \vert \leq \delta \}} xE[X_{nk\delta}] d\mathcal{L}(X_{nk\delta})(x) \bigg\vert\\[1em]
&= \sum_{k=1}^{k_n} \bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-2E[X_{nk1}]\int _{\{ \vert x \vert \leq \delta \}} x d\mathcal{L}(X_{nk})(x) + \mu_{nk}([-\delta,\delta])(E[X_{nk1}])^2 - \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-\mu_{nk}([-\delta,\delta])(E[X_{nk\delta}])^2 + 2E[X_{nk\delta}]\int _{\{ \vert x \vert \leq \delta \}} x d\mathcal{L}(X_{nk})(x) \bigg\vert \\[1em]
\end{align*}
Thoughts: Now I can't quite see how to go from here. It also puzzles me quite a bit that I need the complements $[-1,1]^c,[-\delta,\delta]^c $ somehow. At some point we will probably use the triangle inequality and pull out max from the sums. Am I on the right track and if so can anyone help me finish the computation?
 A: I don't have a complete answer, but at least I have a comment that I think goes most of the way. A simple version of this problem is when $X_{nk}$ is symmetric and supported on $[-1,1]$, so that $E(X_{nk})= E(X_{nk1}) = E(X_{nk\delta})=0$, and $X_{nk} = X_{nk1}$ in which case the problem reduces to (when $k_n=1$)
$$\bigg\vert \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)-\int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x)\bigg\vert \le 2 \delta \mu_{nk}[-\delta, \delta]^c. $$
Intuitively $\int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk\delta})(x)>  \int _{\{ \vert x \vert \leq \delta \}} x^2 d\mathcal{L}(X_{nk})(x)$, since the law of $X_{nk\delta}$ puts more mass on subsets of $[-\delta, \delta]$. Also, for any measurable subset $A$ of $[-\delta, \delta]$, $\mu_{nk\delta}(A) - \mu_{nk}(A) \le \mu_{nk}([-\delta,\delta]^c)$ (expand $\mu_{nk\delta}(A)$ as a sum of two terms depending on whether $X_{nk} \in [-\delta,\delta]$). Use this fact and the definition of the integral to get the bound in this simple case.
