measure theory - Do limits of sequences of sets come from a topology?

In measure theory we frequently see the following definitions:

$$\limsup_{n\to\infty} A_n = \bigcap_{n=1}^{\infty}\left(\bigcup_{j=n}^{\infty} A_j\right)$$

$$\liminf_{n\to\infty} A_n = \bigcup_{n=1}^{\infty}\left(\bigcap_{j=n}^{\infty} A_j\right)$$

where $(A_n)_n$ is a sequence of measureable sets i.e. $\forall n: A_n\in\mathcal{M}$, where $\mathcal{M}$ is a $\sigma$-algebra on $X$, for example $\mathcal{M} = 2^X$. Therefore it makes sense to also define:

$$\lim_{n\to\infty}A_n = \limsup_{n\to\infty} A_n = \liminf_{n\to\infty} A_n$$

when the last two agree. If $\mu$ is a finite (positive, to keep things simple) measure, it is easy to see that under such definition we have $\mu(\lim_{n\to\infty}A_n) = \lim_{n\to\infty}\mu(A_n)$, whenever $\lim_{n\to\infty}A_n$ exists, which looks like some kind of continuity.

Does this kind of convergence of sequences of measurable sets arise from a (preferably Hausdorff, so that limits are unique) topology on $\mathcal{M}$? If such a topology exists, is $\mu:\mathcal{M}\to[0,\infty)$ in fact a continuous function?

(A related question that may be of interest would be: what happens if we allow arbitrary sets? Can we make the Von Neumann universe $V$ into a topological space in such a way?)

Answer

There is such a topology. Simply give $\mathcal{M}$ the subspace topology induced by the product topology on $2^X$.

It may help to think of $2^X$ as the set of functions from $X$ to $\{0,1\}$, by identifying a set with its indicator function. Then we have $1_{\limsup A_n} = \limsup 1_{A_n}$ and so on. Since the product topology is just the topology of pointwise convergence, this behaves as desired.

However, the map $A \mapsto \mu(A)$ is not in general continuous with respect to this topology. For instance, the finite sets are dense in $\mathcal{M}$ with this topology, and so any nontrivial non-atomic measure gives a discontinuous map.