# 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?

• Firstly, I don't think this question would be appropriate for MO (but of course, it is welcome here). If you get no good responses here then go ahead though. I don't have enough information to answer your question, but supposing you had a sequence $f_n$ of functions, you might want to ask if there is a measure space $(X,\Sigma,\mu)$ for which your functions are $\Sigma$-measurable, and that there exists $f$ such that $f_n \to f$ in $\mu$-measure. I don't really have any information on how you might do that or if it is an interesting question, but at least worthy of a comment. – nullUser Dec 31 '12 at 21:58
• so the way you wrote the question, I think you mean topology as the subject and not as the collection of open sets (or you would have included an article like 'a topology'). So in which case sigma algebra isn't a subject. This leads me to believe you rather ask "convergence is to a topological space as ______ is to a measure space" or the same thing as the questions with the articles included. I would say "convergence is to a topology as measurability is to a sigma algebra." This is because measure theory is a main employer of sigma algebras. – toypajme Dec 31 '12 at 23:06
• @toypajme: I should have said "convergence to a topology as what to a sigma algebra". Note that I think measurability to a sigma algebra as continuity to a topology, and to a topology, continuity and convergence are related but not the same concept. – Tim Dec 31 '12 at 23:10
• You can use the same definition of convergence, I think, but it is not clear to me whether the theorems carry over. – Zhen Lin Jan 1 '13 at 1:46
• @ZhenLin: Thanks! I think to define something like convergence for a sigma algebra, one possibility is to consider a measurable mapping from a directed set $D$ or $\mathbb{N}$ with its discrete sigma algebra to the underlying set of the sigma algebra, and then define "convergence" as $x \to \infty$ in $D$ or $\mathbb{N}$, and then one can study if what properties of such "convergence" has and if they can in turn characterize the sigma algebra. What do you think? – Tim Jan 1 '13 at 2:17

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.
• @Tim You can take any set $\mathcal{F}$ of real-valued functions on an abstract set $X$, and get a sigma-algebra on $X$ from it. Namely, the sigma-algebra generated by the sets $\{x\in X: f(x)<a\}$ where $f\in\mathcal{F}$ and $a\in\mathbb R$ are arbitrary. – user53153 Jan 1 '13 at 2:42
• Thanks! So it is the minimal sigma algebra which would make every $f \in \mathcal{F}$ to be measurable wrt the $B(\mathbb{R})$? Is it the same when an integral defines a sigma algebra? Then the integral defines a measure on it by integrating indicator function of every measurable subset in the sigma algebra? – Tim Jan 1 '13 at 2:46
• Thanks again! One thing I notice is that unlike the bijection between convergence classes of nets and topologies, there is no bijection between the families $\mathcal{F}$'s of functions and the sigma algebras. Maybe some additional requirements on the $\mathcal{F}$'s will make the correspondence bijective? – Tim Jan 1 '13 at 3:19