$$\bQ {\left(  \eta^1 \in \cdot \mid \eps = g  \right)}$$ belongs to $ \sP_\mu $ for $ \bQ( \eps \in dg ) $-almost every $ g \in G $. By the assumption that $ \bP^1 $ is extremal, we obtain $$\bP^1(\cdot) = \bQ {\left(  \eta^1 \in \cdot \mid \eps = g  \right)} 
\label{decomp Q2}$$ for $ \bQ( \eps \in dg ) $-almost every $ g \in G $. We obtain the similar identity for $ \bP^2 $. Note that implies $$\bQ {\left(  \eta^1 = g \cdot \eta^2 \mid \eps = g  \right)} = 1 
\quad \mbox{for any $ g \in G $} . 
\label{decomp Q3}$$ Therefore we conclude that $$\bP^1 = T_g {\left( \bP^2 \right)} 
\quad \mbox{for $ \bQ( \eps \in dg ) $-a.e. $ g \in G $}.$$ This completes the proof.

In the above proof of Theorem \[Gal1\], take $ g_0 \in G $ such that $ \bP^1 = T_{g_0} {\left( \bP^2 \right)} $. Then it is obvious that the measure $ \bQ( g_0^{-1} \eps \in dg ) $ is the normalized Haar measure on $ H_\mu(\bP^2) $. This fact leads to the following paradox: [*The tail $ \sigma $-field $ \F^{\eta^1,\eta^2}_{-\infty } $ is non-trivial under $ 