Paths on $\mathbb{Z}^d$ Let's say a path must be non-self-intersecting, and that we have the usual lattice structure.  Then if $\sigma(n)$ is the number of paths of length $n$ then why do we have convergence of the sequence $\sigma(n)^{(1/n)}$ as is asserted in Grimmett's "Percolation?"
The same book also asserts that the property of having a path of length infinity is a tail-measurable event relative to the random variables that return value $1$ on open edges and $0$ on closed edges in the context of bond percolation on $\mathbb{Z}^d$.  I was hoping for a completely detailed explanation of why this is true.  It seems intuitively obvious because the event of having an infinite cluster should not depend on any finite list of the RVs, as the author notes.  But this event is only expressible as an uncountable union of elementary cylinders of the form $X_K=a_K$, $X_{K+1}=a_{K+1}$, etc.
I also want to make sure, as this is relevant to the second question, that when people talk about random graph theory, and thus graph-valued RVs, the implied sigma algebra on the space of subgraphs of a given graph ought to be the countable-cocountable sigma algebra.  is that true?
 A: Question 1.
This is an example of the omnipresent "Fekete's subadditive lemma", which states that if $f:\mathbb{N}\to\mathbb{R}$ is a function satisfying $f(m+n)\leq f(m)+f(n)$ then $f(n)/n$ converges to $\inf_m f(m)/m$ as $n\to \infty$. Proving this is a good Analysis I exercise.
In this case, let $\sigma(n)$ be the number of non-self-intersecting paths of length $n$ in $\mathbb{Z}^d$ starting at $0$. Observation: If we have a legal path of length $m+n$ then we can think of it as a legal path of length $m$ followed by a legal path of length $n$. Moreover the original path of length $m+n$ is completely determined by these two paths, so $\sigma(m+n)\leq\sigma(m)\sigma(n)$. (There is a further restriction that the latter path not intersect the former path, so equality need not hold.)
It follows that $f(n) = \log \sigma(n)$ is subadditive, so by the subadditive lemma $f(n)/n$ converges to $\inf_m f(m)/m$. Exponentiating, $\sigma(n)^{1/n}$ converges to $\inf_m \sigma(m)^{1/m}$.

Question 2.
Following our discussion in the comments, this may help. 
In detail, the reason the event "has an infinite cluster" is measurable at all (i.e., with respect to $\sigma(X_1,X_2,\dots)$) is that it is the union of the countably many measurable events "has arbitrarily large clusters containing $e$", where $e$ ranges over all possible edges. But note equally that it is the union of the countably many events "has arbitrarily large clusters containing $e$, even after erasing any of the first $499$ edges", where $e$ ranges over all possible edges except the first $499$. This latter description is manifestly in $\sigma(X_{500},X_{501},\dots)$.
