\ll \mu_T$.

On the measure space $(\partial T , \B),$ define a filtration $\{ \G_n \}$ by letting $\G_n$\[G\_n-def\] denote the $\sigma$-field generated by the sets $\{ \gamma : \gamma_n = v \}$. The Borel $\sigma$-field $\B$ is the increasing limit $\sigma ( \bigcup_n \G_n )$. Let $$M_n := \left. \frac{d\nu}{d\mu} \right |_{\G_n} \, .\label{M_n-def}$$ In other words, $M_n (\gamma) = \nu (\gamma_n) / \mu (\gamma_n)$. Let $\Mbar := \limsup_{n \to \infty} M_n$. The Radon-Nikodym martingale theorem [@durrett4 Theorem 5.3.3] says that $\{ M_n \}$ is a martingale with respect to $(\partial T,
\B, \mu_T , \{ \G_n \})$ and that $\nu_T \ll \mu_T$ is equivalent to $\nu_T( \{ \Mbar = \infty \} )= 0$. This is equivalent to $\nu_T (\{ \ul{M}  = 0\}) = 0$ where $\ul{M} = 1 / \Mbar = \liminf_n 1/M_n$. The sequence $\{ 1/M_n \}$ is a $\nu_T$-martingale, therefore $\{ \log (1/M_n) \}$ is a $\nu_T$-supermartingale and to conclude that it $\nu_T$-a.s. does not go to negative infinity, it suffices to show that its expectation is bounded from below.

We compute the conditional expected increment of $\log (1/M_n)$. Letting $\gamma$ denote the ray $(\gamma_1 , \gamma_2, \ldots)$, $$\log \frac{1}{M_{n+1} (\gamma)}  - \log \frac{1}{M_n (\gamma)} = 
   