Why isn't this a $\sigma$-field? Suppose that $\Omega = Z$, where $Z$ is the set of integers, and define $F_n$ as the $\sigma$-field generated by the singletons {k} satisfying $k\leq n$. 
I've just read that $\cup_{n=1}^\infty F_n$ is necessarily not a $\sigma$-field, but I'm having a hard time seeing this. 
Interpreting the meaning of $\cup_{n=1}^\infty F_n$, I believe that it is the set of all possible unions of integers, along with the empty set. But if this were the case, it would then be a $\sigma$-field, so my interpretation must be wrong.
 A: Your union contains all the singletons, but it doesn't contain $2\mathbb Z$, the set of all even integers (which is a countable union of singletons). So it cannot be a $\sigma$-field.
(To see that it doesn't contain $2\mathbb Z$, consider that for each fixed $n$,
$$ G_n = \{ A\subseteq \mathbb N \mid (n+1\in A)\Leftrightarrow(n+2\in A) \}$$
is a $\sigma$-field, and that $F_n\subseteq G_n$ because $G_n$ contains all of the singletons $\{k\}$ with $k\le n$. But $2\mathbb Z$ certainly isn't in $G_n$, so it is not in $F_n$ either, and therefore not in the union of all the $F_n$s).
A: $F_n$ is the set of all subsets $S\subseteq\Bbb Z$ such that either $S\subseteq\{n,n-1,n-2,n-3,\cdots\}$ or $S\supseteq\{n+1,n+2,\cdots\}$.
Therefore, the singleton of every natural number is in $\bigcup_{n\ge 1} F_n$, but the set of even natural numbers isn't, because $2\Bbb N\notin F_n$ for any $n$.
A: Well, let's check what $F_n$ is. Clearly, it contains some copy of the power set of $\{1,...,n\}$ and hence, must also contain the complement of any such set in $\mathbb{Z}$. Let's argue that any set in $F_n$ can be written $A\cup C$ where $A\subseteq \{1,...,n\}$ and $\mathbb{Z}\setminus C\subseteq \{1,...,n\}$. Sets of this form are clearly closed under complements and contains $\mathbb{Z}$.
Now, let $(A_k\cup C_k)_{k\in \mathbb{N}}$ be a countable collection of such sets. Then, $\cup_{k=1} A_k\subseteq \{1,...,n\}$ naturally. By the same logic $\mathbb{Z}\setminus (\cup_{k=1} C_k)=\cap_{k=1} \mathbb{Z}\setminus C_k\subseteq \{1,..,n\}$. Hence, we have classified $F_n$ completely.
Now, we can see that $\cup_{n=1} F_n$ is, indeed, not the Power set. If $B\in \cup_{n=1} F_n$ then it has the form $A\cup C$ such that $A\subseteq \{1,..k\}$ for some $k$ and $\mathbb{Z}\setminus C\subseteq \{1,...,k\}$. Implying that $B$ must contain all integers greater than $k$. Such, for instance, the set of even numbers is not an element of $\cup_{n=1}^{\infty} F_n$.
Note that, you should probably force the singletons $\{-k\}$ for $k\in \mathbb{N}$ to be in $F_k$ from some point if you were hoping to create the power set.
