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!



Lindeberg implies convergence of max of conditional variances in L1
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. 


