= Solution
For each fixed $m\ge0$ and $T\ge m$, remove the first $m$ weight factors but retain the same walk law:
$$
Z_T^{(m)}=\left\langle\prod_{j=m+1}^T(1+\varepsilon h(j,\xi_j))\right\rangle.
$$
This quantity is measurable with respect to $\mathcal G_{m+1}=\sigma(h(j,x):j\ge m+1,x\in\mathbb Z^d)$, since the averaging over all walk positions is deterministic and involves no earlier environment variables. The bounds on each deleted factor give, pointwise for every environment,
$$
(1-\varepsilon)^mZ_T^{(m)}\le Z_T\le(1+\varepsilon)^mZ_T^{(m)}.
$$
Both constants are finite and strictly positive. Taking lower limits gives the exact identity of events
$$
\{\liminf_TZ_T=0\}=\{\liminf_TZ_T^{(m)}=0\}\in\mathcal G_{m+1}.
$$
There is no exceptional-set issue in this comparison. Choose the version of $\zeta$ to equal $\liminf_TZ_T$ when this is finite and to equal $1$ otherwise; it agrees with the almost-sure <martingale> limit, and its zero event is precisely the left-hand event above. Since the identity holds for every $m$, $\{\zeta=0\}$ belongs to the printed <tail sigma-algebra> $\bigcap_{n\ge1}\mathcal G_n$. This proves the <tail event for a vanishing positive path-weight limit> without pretending that the whole value of $\zeta$ is tail-measurable.
The <independent> time layers of the environment are <independent> random vectors, so the <Kolmogorov zero-one law> applies to their <tail sigma-algebra>. It yields $\mathbb P(\zeta=0)\in\{0,1\}$. Under the small-noise condition from part (b), $\mathbb E\zeta=1$ and $\zeta\ge0$, so that probability cannot be $1$. Hence
$$
\boxed{\mathbb P(\zeta=0)=0,\qquad \zeta>0\text{ almost surely}.}
$$
The tail-event argument itself works for every $0<\varepsilon<1$; the deduction of positivity here uses the mean-one conclusion supplied by part (b).
Back to article page
