Is the continuity of measures a topological continuity? I am wondering if there is an implicit topology in a $\sigma$-algebra that makes any finite measure defined on it a continuous function, since in this context continuity means limit-preserving (when the measure is finite) as occurs with first countable spaces.
 A: Quite surprisingly (to me at least) it turns out that the notion of convergence of sets given by $\liminf$ and $\limsup$ I mentioned earlier is the convergence induced by the weak topology of all finite measures. To see this I'll use the following characterizations:
$$x\in\liminf A_n \Leftrightarrow \exists\ N\in\mathbb{N} \text{ s.t. } x\in A_n \text{ for all }n\geq N $$
$$x\in\limsup A_n \Leftrightarrow x\in A_n \text{ for infinitely many } n$$
Now assume $A_n$ is a sequence of sets in the $\sigma$-algebra $\mathcal{A}$, $A\in\mathcal{A}$ such that $A_n\to A$ in the sense of the weak topology, i.e. such that $\mu(A_n)\to\mu(A)$ for all $\mu$. Let $x\in A$ and consider $\mu=\delta_x$, then we must have $\delta_x(A_n)\to1$; since $\delta_x(A_n)$ takes only values $0$ and $1$, this implies that definitely $\delta_x(A_n)=1$ and so $x\in\liminf A_n$. Conversely, let $y\in\limsup A_n$ and take $\mu=\delta_y$, then we have $\delta_y(A)=\lim \delta_y(A_n)=\limsup \delta_y(A_n)=1$ and so $y\in A$. So we have obtained
$$ A\subset \liminf A_n\subset \limsup A_n\subset A$$
which implies $A=\lim A_n$ in the set limit sense. The inverse fact that if $A=\lim A_n$, then $\mu(A_n)\to\mu(A)$ for all finite measures, is well known (for example quoted in the wikipedia link) and can be proved using monotone properties of the measure.
An example of a non trivial convergent sequence is the following: assume $x, x_n$ are all elements of the space such that $\{x\}, \{x_n\}\in\mathcal{A}$ and $x_n\neq x, x_n\neq x_m$ for all  $n\neq m$. Consider the sets $A_n=\{x,x_n\}$. Then we have $\liminf A_n=\limsup A_n=\{x\}$ and so $A_n\to \{x\}$ but $A_n\neq \{x\}$ for all $n$. 
The example above can be easily generalized whenever $\mathcal{A}$ contains a countable amount of pairwise disjoint sets (let $A_n$ be such sets, consider $B_n=A_0\cup A_n$ for $n\geq 1$, then $B_n\to A_0$). I think (but I don't have a proof of this) that such a sequence can aways be found if $\mathcal{A}$ is at least countable. Instead if $\mathcal{A}$ is finite then it will be generated by a finite partition of the space and in that case it's easy to see (take appropriate test measures $\mu$) that the weak topology induced on $\mathcal{A}$ is the discrete one.
