Convergence to a topology as what to a sigma algebra? Given the notion of some kind of convergence of sequences or nets on a set, one question is whether there exists a topology for that convergence and find it. The last section of Chapter 2 in Kelley's General Topology provides one characterization.
Since sigma algebra and topology are similar in many aspects, I was wondering if there is a concept for a sigma algebra, and/or from which people ask whether there exists a sigma algebra for a given collection of "convergent" objects and find it? Note that the concept needs not be a mimic of convergence for topology.
Although the concept may not be a mimic of convergence, I think to define something like convergence for a sigma algebra $\mathcal{F}$ on $\Omega$, one possibility is to 


*

*first define a "net" or "sequence" by considering a measurable mapping from a directed set $D$ or $\mathbb{N}$ with its discrete sigma algebra to the underlying set $\Omega$ of the sigma algebra $\mathcal{F}$, and 

*then define "convergence" of a "net" or "sequence" as $x \to \infty$ in $D$ or $\mathbb{N}$, and 

*then one can study what properties of the collection of all "convergent" "nets" or "sequences" has, and if they can in turn characterize the sigma algebra $\mathcal{F}$ from a given collection of  "nets" or "sequences" such that the "nets" or "sequences" "converge". What do you think?
Thanks and regards!
By the way, is this kind of questions okay at MO?
 A: It would be correct to say that continuous is to a topology as measurable is to a $\sigma$-algebra. However, this does not quite answer the question, which is driven by the fact that sometimes the notion of convergences comes before (or even without) a topology. A decent parallel for the latter is provided by alternative notions of an integral such as Henstock-Kurzweil integral or Daniell integral. With the Daniell integral, measurable functions are defined without an underlying measure space; however, one can always reconstruct a measure space from the integral. In the case of Henstock-Kurzweil, the process of integration does not correspond to Lebesgue integration with respect to any measure. 
Actually, we don't need to go far to see an integral that does not precisely correspond to any  $\sigma$-algebra. Our old friend the Riemann integral is such an example. Indeed, for any $a\in\mathbb R$ the function 
$$\chi_{\{a\}}(x) = \begin{cases} 1 \ \text{ if } x=a \\ 0 \ \text{ if } x\ne a\end{cases}$$
is Riemann integrable. Therefore, the $\sigma$-algebra generated by the set of Riemann integrable functions includes all countable sets. The function $\chi_{\mathbb Q}$ is measurable with respect to this $\sigma$-algebra, but is not Riemann integrable.
