Criterion of two measures are absolute continuous or singular I was reading the Durrett's book: probability theory and example and stuck at some stages about the radon-nikodym derivatives related topic:
Here is the setting:

Let $\mu$ and $\nu$ be measures on sequence space $(\mathbb{R}^{\mathbb{N}}, \mathcal{R}^{\mathbb{N}})$ that make the coordinates $\xi_n(\omega) := \omega_n$ independent.
  Let  \begin{array}{l}
{F_n}(x): = \mu ({\xi _n} \le x)\\
{G_n}(x): = \nu ({\xi _n} \le x)
\end{array}
  Suppose $F_n \ll G_n$ and define  $q_n := \frac{d F_n}{d G_n}$ and let $\mathcal{F}_n:= \sigma(\xi_m: m \le n)$, let $\mu_n, \nu_n$ be the restrictions of $\mu$ and $\nu$ to $\mathcal{F}_n$, and let
  $${X_n} = \frac{{d{\mu _n}}}{{d{\nu _n}}} = \prod\limits_{m = 1}^n {{q_m}}$$
  Then the author states a theorem (Theorem 5.3.5) as a criterion for distinguishing the singular/absolute continuous measure: It says that
$\mu \ll \nu$ or $\mu \perp \nu$, according as $\prod_{m=1}^\infty \int \sqrt{q_m} dG_m>0$ or $=0$.

Here are some naive questions:
Q 1. I don't understand what is the purpose for defining the coordinate function $\xi_n$ and how to get the $X_n = \prod\limits_{m = 1}^n {{q_m}}$. 
Here is my thinking: consider $\;
{\mu _n} = \mathop  \otimes \limits_{m = 1}^n {F_n};\;
{\nu _n} = \mathop  \otimes \limits_{m = 1}^n {G_n}$. So by Fubini?.. (not sure, the notation getting confused..)
$${X_n} = \frac{{d{\mu _n}}}{{d{\nu _n}}} = \frac{{d\left( {\mathop  \otimes \limits_{m = 1}^n {F_n}} \right)}}{{d\left( {\mathop  \otimes \limits_{m = 1}^n {G_n}} \right)}} =?= \frac{{\prod\limits_{m = 1}^n {{F_n}} }}{{\prod\limits_{m = 1}^n {{G_n}} }} = \prod\limits_{m = 1}^n {{q_m}}$$
Q 2. Can I interpret the theorem above as follows? (not familiar with the word "... according as...")
$$\mu \ll \nu \text{ iff} \; \prod_{m=1}^\infty \int \sqrt{q_m} dG_m>0 \\ 
\mu \perp \nu \text{ iff}\;  \prod_{m=1}^\infty \int \sqrt{q_m} dG_m=0$$
Q 3. About the stated theorem, I was wondering is it possible to compute the integrand $\sqrt{q_m} = \sqrt{\frac{dF_m}{dG_m}}$ explicitly? can someone give me a simple example to show how this mechanism works? 
Thank you.
 A: The purpose of these notations is to define $X_n$ rigorously. The suggestion after Q1, while trying to achieve the same goal based on the same idea, is, as you note, more confused. 
The rephrasing in Q2 is correct. 
Re Q3, the Bernoulli case might be enlightening. Assume that $\mu$ and $\nu$ are products of Bernoulli measures on $\{0,1\}$, namely,$$\mu=\bigotimes_n\left(\tfrac12(1+a_n)\delta_1+\tfrac12(1-a_n)\delta_0\right),\qquad\nu=\bigotimes_n\left(\tfrac12\delta_1+\tfrac12\delta_0\right),$$ for some sequence $(a_n)$ such that $|a_n|\lt1$. Then 
$$q_n(0)=1-a_n,\quad q_n(1)=1+a_n,\quad G_n=\tfrac12\delta_1+\tfrac12\delta_0,$$ hence
$$\int\sqrt{q_n}\,\mathrm dG_n=\tfrac12\sqrt{1+a_n}+\tfrac12\sqrt{1-a_n}.$$
When $\alpha\to0$, $$\tfrac12\sqrt{1+\alpha}+\tfrac12\sqrt{1-\alpha}=1-\tfrac18\alpha^2+o(\alpha^2),$$ hence $\mu\ll\nu$ if and only if the series $\sum\limits_na_n^2$ converges.
An obvious case such that $\mu\ll\nu$ is when $a_n=0$ for every $n$. An obvious case such tht $\mu\perp\nu$ is when $a_n=a$ for every $n$, for some $a\ne0$, $|a|\lt1$. The theorem recalled in the question delineates the limit between these two regimes, quantifying how close to the null sequence the sequence $(a_n)$ should be, to ensure that $\mu\ll\nu$.
Another interesting conclusion of the theorem is that, in this independent setting, either $\mu\ll\nu$ or $\mu\perp\nu$, that is, no mixed case arises.
