Show the equality $\frac{d\mu_n}{d\nu_n}=\prod\limits_{m=1}^{n}q_m$ 
Show the equality $\frac{d\mu_n}{d\nu_n}=\prod\limits_{m=1}^{n}q_m$.

I don't understand why $\frac{d\mu_n}{d\nu_n}=\prod\limits_{m=1}^{n}q_m$ 
with the definitions:
$$F_n=\mu(\xi_n\le x),\quad G_n=\nu(\xi_n\le x),\quad q_n=\frac{dF_n}{dG_n},
$$ 
$$\mathcal F_n=\sigma(\xi_m:m\le n),$$ and $\mu_n,\nu_n$ are restrictions of $\mu$ and $\nu$ on $\mathcal F_n$.
 A: Calculate: For Borel subsets $B_1,\ldots,B_n$ of $\Bbb R$,
$$
\eqalign{
\mu_n(\xi_1\in B_1,\ldots,\xi_n\in B_n)
&=\mu(\cap_{m=1}^n\{\xi_m\in B_m\})\cr
&=\prod_{m=1}^n\mu(\xi_m\in B_m)=\prod_{m=1}^n\mu_m(\xi_k\in B_m)\cr
&=\prod_{m=1}^n\left(\int q_m(\xi_m) 1_{\{\xi_m\in B_m\}}\,d\nu_m\right).\cr
}
$$
On the other hand,
$$
\eqalign{
\int \left(\prod_{m=1}^n q_m(\xi_m)\right)1_{\{\xi_1\in B_1,\ldots,\xi_n\in B_n\}}\,d\nu_n
&=\int \left(\prod_{m=1}^n q_m(\xi_m)\right)\left(\prod_{m=1}^n 1_{\{\xi_m\in B_m\}}\right)\,d\nu\cr
&=\prod_{m=1}^n\left(\int   q_m(\xi_m)  1_{\{\xi_m\in B_m\}} \,d\nu\right)\cr
&=\prod_{m=1}^n\left(\int   q_m(\xi_m)  1_{\{\xi_m\in B_m\}} \,d\nu_m\right).\cr
}
$$
Comparing the two we see that
$$
\mu_n(\xi_1\in B_1,\ldots,\xi_n\in B_n)=\int \left(\prod_{m=1}^n q_m(\xi_m)\right)1_{\{\xi_1\in B_1,\ldots,\xi_n\in B_n\}}\,d\nu_n,
$$
which implies (by monotone class theorem) that
$$
\mu_n(F)=\int_F\left(\prod_{m=1}^n q_m(\xi_m)\right)\,d\nu_m
$$
for all $F\in\mathcal F_n$. This amounts to the statement that
$$
{d\mu_n\over d\nu_m}=\prod_{m=1}^n q_m(\xi_m).
$$
