(i), we have $$\eta^i_k = \eta^i_{k-1} \cdot \xi_k , 
\quad k \in -\N , \ i=1,2.$$ Let $ k \in -\N $ and $ n \in \N $ be arbitrary numbers. Then $$\eta^i_k = \eta^i_{k-n} \cdot \xi_{k-n+1} \cdots \xi_k , 
\quad i=1,2.$$ Thus we have $${\left( \eta^1_k \right)} \cdot {\left( \eta^2_k \right)}^{-1} 
= {\left( \eta^1_{k-n} \right)} \cdot {\left( \eta^2_{k-n} \right)}^{-1} , 
\quad k \in -\N , \ n \in \N . 
\label{G action3}$$ Since the LHS is irrelevant to $ n \in \N $, there exists a random variable $ \eps $ which is $ \F^{\eta^1,\eta^2}_{-\infty } $-measurable such that $$\eps = {\left( \eta^1_k \right)} \cdot {\left( \eta^2_k \right)}^{-1} , \quad k \in -\N . 
\label{G action4}$$

If we denote by $ \bQ {\left(  \cdot \mid \eps = g  \right)} $ the regular conditional probability given $ \eps=g $, then we have the following disintegration: $$\bQ(\cdot) = \int_G \bQ {\left(  \cdot \mid \eps = g  \right)} \bQ( \eps \in dg ) . 
\label{decomp Q}$$ Hence we obtain an integral expression of $ \bP^1 $: $$\bP^1(\cdot) = \int_G \bQ {\left(  \eta^1 \in \cdot \mid \eps = g  \right)} \bQ( \eps \in dg ) .$$ By definition, we see that the law 