Liverani's CLT (a question) Let $(\Omega,\mathcal{F},P)$ be a probability space where $\Omega$ is a complete separable metric space, let $T:\Omega\to \Omega$ ` be an ergodic transformation,  let $\hat{T}:L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the (linear, isometric) operator $\hat{T}X:=X\circ T$, and let $\hat{T}^{*}: L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the adjoint of $\hat{T}$. 
Supose $\mathcal{F}_{0}\subset \mathcal{F}$ is a given sigma field and that, upon definning 
$$\mathcal{F}_{k}:= T^{-k}\mathcal{F}_{0}:=\{T^{-k}A: A\in \mathcal{F}_{0}\}$$
 one has $\mathcal{F}_{k}\subset \mathcal{F}_{k-1}\subset \mathcal{F}$ for all $k\in \mathbb{Z}$, and let $E_{n}: L^{2}_{_P}(\Omega)\to L^{2}_{_P}(\Omega)$ be the projection 
$$E_{n}(X):=E[X|\mathcal{F}_{n}]$$
onto the subspace of $\mathcal{F}_{n}-$measurable functions. 
In his paper "Central Limit Theorem for Deterministic Systems", C.Liverani presents the following theorem (be aware! I made significant changes in notation)
Theorem 1 Let $X_{0}\in L^{\infty}_{_P}(\Omega)$ be an $\mathcal{F}_{0}-$measurable function such that $EX_0=0$, and define $X_{n}:=\widehat{T}^n X_{0}$. Then the Central Limit Theorem for $\{X_{n}\}_{n\geq 0}$ holds under the following assumptions:
(0) $E_{1}\hat{T}\hat{T}^{*}=E_{1}$
(1) $\sum_{n=0}^{\infty} |E[X_{0}X_{n}]|<\infty$
(2) $P[\sum_{n=0}^{\infty} |E_{0}[\hat{T}^{*n}X_{0}]|=\infty]=0$
I.e, the assumtions $(0)-(2)$ imply that
 $$\frac{1}{\sqrt{n}}\sum_{k=0}^{n}X_{k}\Rightarrow N(0,\sigma^{2}) $$
where
(A) $\sigma^{2}\leq EX_{0}^2+2\sum_{k=1}^{\infty}EX_{0}X_{k}$, and 
(A') equality (for $\sigma^2$) occurs if, for the series in $(2)$,  we strengthen to convergence in  $L^{1}_P(\Omega)$. 
(B) $\sigma=0$ if and only if $\hat{T}X_0= (\hat{T}-Id)Y_0$
for some $\mathcal{F}_{0}-$measurable function $Y_0$.
I have been working in this result and I have found myself stuck in the following part of his argument (I just summarize here, assuming that he who can help me will go to the original paper if she doesn't know the details. I will give more explanations upon request): let be $\lambda>1$ be given and let be $Y_{0}(\lambda)$, $D_{1}(\lambda)$, $\mathcal{F}_{0}$ measurable such that $E_{1}D_{1}(\lambda)=0$ and, if  $Y_{k}(\lambda):=\hat{T}^{k}Y_{0}(\lambda)$, $D_{k}(\lambda)=\hat{T}^{k-1}D_{1}(\lambda)$ ($k\geq 1$)`, then
$$X_{k}=D_{k}(\lambda)+Y_{k}(\lambda)-\lambda^{-1}Y_{k-1}(\lambda)$$
Indeed, this gives $Y_{0}(\lambda)=\sum_{n\geq 0}\lambda^{-n}E_{0}\hat{T}^{*n}X_{0}$, $D_{k}(\lambda)$ happens to be a reverse martingale difference with respect to $\{\mathcal{F}_{k}\}_{k}$ (this follows from $E_{0}D_{1}(\lambda)=D_{1}(\lambda)$ and $E_{1}D_{1}(\lambda)=0$), and $D_{1}:=\lim_{\lambda\to 1} D_{1}(\lambda)$ is a (well defined) measurable function which he proves has second moment.
Here is what I do not understand, and is essential to his argument: how does one know, under the theorem's hypothesis, that $E_{1}D_{1}=0$? An exchange of limits and conditional expectation seems to be involved here, and I can't justify it without passing to the hypothesis (A'). Can you give some light on this? 
Unfortunately, my imagination is too short and can't see an example where (2) holds but (A') does not, so I don't even know how to start looking for possible counterexamples to the theorem as stated here.
PS: The reason of my concerns is that, with those things that I've understood, this theorem (or better: its restricted version, assuming (A')) can be extended to the case $X_{0}\in L^2$ by means of just a couple of very simple observations to Liverani's argument. As I find this fact worth communicating I decided to try writing this extension down, but I realized I was missing that step, which I can't fill neither see in the paper's proof.
Regards.
 A: (Important editions were made following the discussion in the comments. My acknowledgments to Ian Morris for his enlightment on this question.)
Indeed, Liverani proves that 
$$E[(D_{1}(\lambda))^2]\leq E[X_{0}^2]+2\sum_{n\geq 1}E[X_{0}X_{n}]$$
In particular, $\{D_{1}(\lambda)\}_{\lambda>1}$ is a bounded set in $L^{2}$ and there exists, by Alaoglu Theorem, a sequence $\lambda_{n}\to 1$ and a weak-$L^2$ limit for $D_{1}(\lambda_{n})$, say $D'$.
For such a limit one has in particular that, if $I_{A}$ denotes the characteristic function of a measurable set $A$ then
$$E(D_{1}(\lambda_{n})I_{A})\to E(D'I_{A})$$
and in particular one has that, for all $A\in \mathcal{F}_{1}$, $E(D'I_{A})=0$. 
This proves that $E_{1}D'=0$, and the result follows from the fact (see the discussion below in the comments) that a.e limits and weak limits coincide when both exist. 
A: Which instance of $\mathbb{E}(Y_1|\mathcal{F}_1)-0$ (or $E_1D_1=0$ in your notation) is problematic? I can find one instance shortly after (1.6) but I think that this is a typo, and the text there should read
`$\ldots$since $\mathbb{E}(\hat{T}f - \hat{T}g(\lambda) + \lambda^{-1}g(\lambda)|\mathcal{F}_1) = \mathbb{E}(Y_1(\lambda)|\mathcal{F}_1)=0$.''
This statement can be justified by some straightforward manipulations since the sums involved now converge in $L^1$. Conversely, at this stage in the argument $g$, and hence $Y_1$, is only measurable, and I can see no reason for $\mathbb{E}(Y_1|\mathcal{F}_1)$ to even be a well-defined expression.
Once one has established $\mathbb{E}(Y_1^2)<\infty$, it is then implicitly necessary to go back and show that indeed $\mathbb{E}(Y_1|\mathcal{F}_1)=0$. Is this the problematic step? (I am certainly not quite sure at this stage how to do it.)
