# Understanding a proof of Komlós's theorem

I'm reading a book about probability theory and they use a certain theorem, called Komlós's theorem, which states:

For a sequence $(\xi_n)$ of random variables on $(\Omega,\mathcal{F},P)$ with $\sup_n E|\xi_n| < \infty$. Then there is a random variable $\zeta \in L^1$ and a subsequence $(\zeta_k) = (\xi_{n_k})$ such that $$\frac{\zeta_1+\cdots+\zeta_k}{k} \to \zeta \text{ a.s. }\tag{1}$$ Moreover the subsequence $(\zeta_k)$ can be chosen in such a way that its further subsequence will also satisfy (1).

So I found a proof of this theorem in the book

"Two-Scale Stochastic Systems" of Yu. Kabanov and S. Pergamenshchikov.

The proof of the theorem can be found in the Appendix, on page 250. Unfortunately, it is not available online. However, I hope there's someone who owns this book and could help me.

The point, where I got stuck is on page 253.

It's clear that we are able to choose this increasing sequence $n_k$ such that for all $n \ge n_k$

$$E\eta^2_k \le E(\xi^{(k)}_n)^2 +2^{-k} \text{ and }|E(\xi^{(k)}_n-\eta_k | \gamma_{j_1},\dots,\gamma_{j_m})| \le 2^{-k}$$

for all $m\le k-1, j_1<j_2<\dots<j_m$, with $\gamma_j:= D_j(\xi^{(j)}_{n_j}-\eta_j)$.

Just for completeness, we set $\zeta_k:= \xi_{n_k}$.

Now I get confused, about the following 3 things:

1. Why is $|E(\gamma_k \mid \gamma_1,\dots,\gamma_{k-1})|\le 2^{-k+1}$ — the above inequalities hold for $\xi^{(k)}_n-\eta_k$ instead of $\gamma_k$?
2. What follows the first two inequalities is not clear:

$$\sum_{i=1}^\infty\frac{1}{k^2}E\gamma_k^2 \le 2\sum_{i=1}^\infty\frac{1}{k^2}E(\zeta_k^{(k)}-\eta_k)^2+ O(1) \le 4 \sum_{i=1}^\infty\frac{1}{k^2}E(\zeta_k^{(k)})^2 +O(1) < \infty.$$

In the last inequality, just calculating:

$$E(\zeta_k^{(k)}-\eta_k)^2 = E(\zeta_k^{(k)})^2 +2 E\,\zeta_k^{(k)}\eta_k + E\eta_k^2 \le 2 E(\zeta_k^{(k)})^2 +2 E\,\zeta_k^{(k)}\eta_k + 2^{-k}.$$

So the term $2^{-k}$ can be controlled, but I don't know how to bound the term $E\,\zeta_k^{(k)}\eta_k$.

I would appreciate it very much if someone could explain what's going on here.

thx & cheers

math

Since it seems to be difficult, I state the lemma's which the authors need for the proof. I cite:

$lemma$ 1 : Let $\eta _n$ be a sequence of random variables convergent weakly in $L^2$ to a random variable $\eta$. Then $$E|\eta| \le \lim\inf E|\eta_n| \tag{2}$$ $$E|\eta|^2 \le \lim\inf E|\eta_n|^2 \tag{3}$$

Now a definition:

$$\xi^{c}:=\xi 1\{|\xi|\le c\}$$ $$D_m(\xi):=\sum_{i=-\infty}^\infty i2^{-m} 1\{\xi\in (i2^{-m},(i+1)2^{-m}]\}$$

They call them truncation and discretization operators on $L^0$.

$lemma$ 2 : Assume $\sup_nE|\xi_n| < \infty$ and for every $k \in \mathbb{N}$ the sequence $(\xi_n^{(k)})$ converges weakly in $L^2$ to a random variable $\eta_k$. Then there exists $\eta \in L^1$ such that $\eta_k$ tends to $\eta$ a.s. and in $L^1$.

And the last lemma

$lemma$ 3 : Let $\mathcal{G}$ be a $\sigma$-algebra generated by a finite partition $A_1,\dots,A_N$ with $A_i \in \mathcal{F}$. Assume that a sequence of random variables $(\xi_n)$ converges weakly in $L^2$ to zero. Then for any $\epsilon >0$ there exists $n_0 =n_0(\epsilon)$ such that $$E(\xi_n|\mathcal{G})\le \epsilon$$ for all $n\ge n_0$

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Can you write the theorem? –  Davide Giraudo Dec 7 '11 at 15:59
I edited the question. But for the proof the authors need 5 lemmas, so I guess it's difficult just to have the main theorem. If it's helpful I can state the lemmas. –  math Dec 7 '11 at 16:20