Central limit theorem extends to absolutely continuous measures Let $(\Omega, \mathcal{F}, \mathbb{P})$ be a probability space. Let $\mathbb{Q}$ be a probability measure on $(\Omega, \mathcal{F})$ that is absolutely continuous w.r.t. $\mathbb{P}$. Let the Radon-Nikodym derivative $D= \frac{ d\mathbb{Q}}{d\mathbb{P}}$. Let
$(X_n)$ be a sequence of i.i.d random variables under $\mathbb{P}$. We define
\begin{equation}
M_n = \frac{1}{ \sigma \sqrt{n}} \sum_{i=1}^{n} (X_i -m),
\end{equation}
where $m = \mathbb{E}_{\mathbb{P}}[X_1]$ and $\sigma^2 = \mathbb{E}_{\mathbb{P}} [(X_1 -m )^2]$. We want to show that the central limit theorem holds under $\mathbb{Q}$, i.e. $\mathbb{E}_\mathbb{Q} [ f(M_n) ] \rightarrow \mathbb{E}_{\mathbb{Q}} [f(N)]$ as $n \rightarrow \infty$, where $N$ is under $\mathbb{Q}$ a $N(0,1)$ random variable and $f: \mathbb{R} \rightarrow \mathbb{R}$ is a bounded continuous function.
My approach:
Let $D_k = \mathbb{E} ( D | \sigma( X_1, X_2, \ldots, X_k))$ and write $\mathbb{E}_{\mathbb{Q}} [f(M_n)] = \mathbb{E}_{\mathbb{P}} [(D-D_k)f(M_n)] + \mathbb{E}_{\mathbb{Q}} [D_k f(M_n)]$. 
If $\sigma(X_1, X_2, \ldots, ) = \mathcal{F}$ (not sure if this is true), then the first term $\mathbb{E}_{\mathbb{P}} [(D-D_k)f(M_n)] $ converges to $0$, by Levy's upward theorem and the boundedness of $f$. However, I can't bound the second term. Any suggestions?  
 A: The question is non-trivial because the random variables are not independent under $ \mathbb Q$. We can use Theorem 16.2 of Billingsley's book Convergence of Probability Measures (1968). 

Theorem. Let $(\Omega,\mathcal B,\mathbb P)$ be a probability space and $E_n\in\mathcal B$. We assume that there exists a sub-field $\mathcal B_0$ of $\mathcal B$ and a constant $\alpha$ such that 
  
  
*
  
*$\mathbb P(E_n\cap E)\to\alpha \mathbb P(E)$ for each $E\in\mathcal B_0$;
  
*for each $n$, $E_n\in\sigma(\mathcal B_0)$.
If $\mathbb P_0\ll\mathbb P$, then $\mathbb P_0(E_n)\to\alpha$.

To see this, we first show that for each integrable function $g$,
$$\tag{*} \int_{E_n}g\mathrm d\mathbb P\to\alpha\int g\mathrm d\mathbb P.$$
It is true if $g$ is measurable with respect to $\sigma(\mathcal B_0)$ by approximation, and we apply this case to $\mathbb E[g\mid \sigma(\mathcal B_0)$ to get $(*)$ in full generality. Then we use this convergence with the Radon-Nikodym derivative of $\mathbb P_0$ with respect to $\mathbb P$.
To prove the central limit theorem under $\mathbb Q$, take $A$ a set such that the measure of its boundary is $0$ (for a standard normal distribution $N$, assuming that everything is centered and normalized). We choose a sequence $p_n$ such that $p_n\uparrow\infty$ and $p_n /\sqrt n\to 0$. We thus have that, 
$$\mathbb P\left(\frac 1{\sqrt n} \sum_{i=p_n+1}^nX_i  \in A\right) \to 
N(A).$$ 
We define  $\mathcal B_0 :=\bigcup_{n=1}^\infty \sigma(X_1,\dots,X_n)$. We thus have that for each $E\in\mathcal B_0$, 
$$\mathbb P P\left(\frac 1{\sqrt n} \sum_{i=p_n+1}^nX_i  \in A\cap E\right)\to N(A)\mathbb P(E).$$ 
By the theorem, we get the wanted conclusion.  
