Lindeberg implies convergence of max of conditional variances in L1

Question

The following is taken from Dvoretzky, 1972, ASYMPTOTIC NORMALITY FOR SUMS OF DEPENDENT RANDOM VARIABLES, Equation 4.6.

$$\{X_{n,k}\}_{n=0,1,...;k=0,1...,k_n}$$ is a (triangular) array of r.v.'s /w

$$E[X_{n,k}|\mathcal{F}_{n,k-1}]=0$$ $$\sum_{k=1}^{k_n}E[X_{n,k}^2|\mathcal{F}_{n,k-1}]=1$$ $$\lim_n\sum_{k=1}^{k_n}E[X_{n,k}^21\{|X_{n,k}|>\epsilon\}]=0 \; \forall \epsilon>0 $$ Where the last condition implies that $\lim_n\sum_{k=1}^{k_n}P\{|X_{n,k}|>\epsilon\}=0 \; \forall \epsilon>0$

My goal is to show that then $$E[\max_{1\leq k\leq k_n}E[X_{n,k}^2|\mathcal{F}_{n,k-1}]] \rightarrow0 \textrm{ as } n\rightarrow \infty$$ which is equialent to showing that $$\sum_{k=0}^{k_n} P(E[X_{n,k}^2|\mathcal{F}_{n,k-1}]> \epsilon) \rightarrow 0$$ (right?). The unconditional Lindeberg doesn't imply this (or does it?).

I will just in a very short form state what my first attempt was and where my problem is. For $$E[\max_{1\leq k\leq k_n}E[X_{n,k}^2|\mathcal{F}_{n,k-1}]] \\ \leq \epsilon + \sum_{k=1}^{k_n} E[E[X_{n,k}^2|\mathcal{F}_{n,k-1}]1\{|X_{n,k}|>\epsilon\}]$$

my problem here is: How can I use the Lindeberg condition, as the additional set in not measurable w.r.t. $\mathcal{F}_{n,k-1}$.I tried playing around with the tower-property a bit but can't make it work. Maybe I need to use some Lemma? A way to show what I'm aiming for would be to show convergence in probability and uniform integrability of the $\max_{1\leq k\leq k_n}E[X_{n,k}^2|\mathcal{F}_{n,k-1}]$ for $n \in N$ , but that would be a lot harder (if it actually is implied) and I want to be sure I'm not missing something. I'm just confused because it sounds so trivial in the paper... Thanks in advance!

Answer 1

Let $u:=\epsilon$, $E_{k-1}Z:=E(Z|\mathcal F_{n,k-1})$, $$A_{n,k}:=E_{k-1}X_{n,k}^2,\quad B_{n,k}:=E_{k-1}X_{n,k}^2\,1\{|X_{n,k}|\le u\},\quad C_{n,k}:=E_{k-1}X_{n,k}^2\,1\{|X_{n,k}|> u\},$$ $$A_n:=\max_k A_{n,k},\quad B_n:=\max_k B_{n,k},\quad C_n:=\max_k C_{n,k}.$$
We need to check that $EA_n\to0$ (as $n\to\infty)$.

We have $A_{n,k}=B_{n,k}+C_{n,k}$, and so, $A_n\le B_n+C_n$. Obviously, $B_n\le u^2$ and hence $$EB_n\le u^2.$$ Next, $C_n\le\sum_k C_{n,k}$, and so, \begin{equation} EC_n\le \sum_k EC_{n,k}=\sum_k EX_{n,k}^2\,1\{|X_{n,k}|> u\}\to0, \end{equation} by the Lindeberg condition. Thus, $0\le\limsup_n EA_n\le \limsup_n EB_n+\limsup_n EC_n\le u^2$, for each real $u>0$. So, $EA_n\to0$, as desired.