\left[ \E_* X_n \right]$ we see that the hypothesis of $(i)$, namely $\sum_{n=1}^\infty \E X_n < \infty$, implies $\E \sum_{n=1}^\infty
\E_* X_n < \infty$. This implies $\sum_{n=1}^\infty \E_* X_n < \infty$ almost surely. A version of $\E_* X_n$ is $\sum_{|v| \in \TT_n} 
X(v) \nu_\TT (v)$, whence  holds for $\GW$-almost every $\TT$, implying almost sure absolute continuity of $\mu_\TT$ with respect to $\nu_\TT$.\
$(ii)$ The argument used to prove Lemma \[lem:X\] may be adapted as follows. Let $\disp M_n := \left. 
\frac{d \nu_\TT}{d\mu_\TT} \right |_{\G_n'}$, a version of which is the function taking the value $\disp \frac{\nu_\TT (v)}{\mu_\TT (v)}$ on $\{ \gamma_n = v \}$. Again $\{ M_n \}$ is a martingale and $\log (1/M_n)$ is a supermartingale which we need to show converges almost surely. The sequence $$S_M := \sum_{n=1}^M
   \left ( \log \frac{1}{M_{n+1}} - \log \frac{1}{M_n} \right )
   \, \one_{X(\gamma_n) = Y(\gamma_n)}$$ is a convergent supermartingale because its expected increments are either 0 or $-Y(\gamma_n)$; convergence of the unconditional expectations $\E Y (\gamma_n)$ implies almost sure convergence of the expected increments, implying almost sure convergence of the supermartingale $\{ S_M \}$. The hypotheses of $(ii)$ imply that the increments of $S_M$ differ from the increments of $\log (1 / 