\log \frac{\nu_T (\gamma_n)}{\mu_T (\gamma_n)} 
   - \log \frac{\nu_T (\gamma_{n+1})}{\mu_T (\gamma_{n+1})} 
   = - \log \frac{q(\gamma_{n+1})}{p(\gamma_{n+1})} \, .$$ Conditioning on $\G_n$, if $\gamma_{n+1} = u$, then the $\nu_T$-probability of $\gamma_n = v$ is $q(v)$, whence $$\E_{\nu_T} \left [ \log \frac{1}{M_{n+1}} - \log \frac{1}{M_n} 
   \,\bigg|\, \G_n \right ]
   = \sum_{v \text{ child of } u} - q(v) \log \frac{q(v)}{p(v)} = - X(u) 
   \, .$$ Taking the unconditional expectation, $$\E_{\nu_T} \left [ \log \frac{1}{M_{n+1}} - \log \frac{1}{M_n} \right ]
   = - \sum_{|v|=n} \nu_T (v) X(v)$$ and summing over $n$ shows that  implies that $\log (1 / M_n)$ has expectation bounded from below, establishing $\nu_T \ll \mu_T$. $\Cox$

\[cor:X\] Recall that $\gamma$ denotes the invasion path on $\TT$ and let $X_n$ denote $X (\gamma_n)$.

(i) If $\sum_{n=1}^\infty \E X_n < \infty$ then $\nu_\TT \ll \mu_\TT$ $\GW$-almost surely.

(ii) Define the filtration $\{ \G_n' \}$ on $(\Omega , \F)$ by letting $\G_n'$ be the $\sigma$-field generated by $\T$ together with $\gamma_1, 
    \ldots , \gamma_n$. Let $Y(v)$ be random variables such that $Y(v) \in \G_{|v|}'$ and $$\P [X(\gamma_n) \neq Y(\gamma_n) \mbox{ infinitely often} ] = 0 \, .$$ Then $\sum_{n=1}^\infty \E Y_n < \infty$ implies that $\GW$-almost surely, $\nu_\TT \ll \mu_\TT$.

  \
$(i)$ Writing $\E X_n = \E 